sum of first n natural numbers = n(n+1) / 2

Average = n(n+1) / 2n = (n + 1)/2

2 * Average = (n + 1) ~ n

So this should be in [1, n] ]]>

For example how many times we write the digit 6 when we write from 1 to 10000

We have 4 zeros, so the answer would be 4 x 10^3 = 4000

In other cases (like the one you mentioned), we have to go in the traditional way.

Will try something more interesting..

How many times we write 0 if we write from 1 to 10000 ?

]]>Ladder Theorem states 1/A + 1/B = 1/h

]]>If possible please post more question solving techniques like it .

Secondly, sir which books did you refer for your cat prep.

Thanks ]]>

Odd is of the form (2k+1)

(2a+1) + (2b+1) ... + (2e+1) = 35

2a + 2b +....+ 2e = 30

a + b + ... + e = 15

Whole number Solution = 19c4

**If N = 2^3 * 3^2 then how many sets of 2 distinct factors of N are co-prime to each other**

N => 2^3 * 3^2

No. Of co primes => [(2 * 3 + 1) ( 2 * 2 + 1) - 1]/2

(7 * 5 -1)/2

=> 17

If N is of the form a^x * b^y, Number of co-primes = [ (2x +1) (2y+1) - 1]/2

**116 people participated in a singles tennis tournament of knock out format. The players are paired up in the first round, the winners of the first round are paired up in second round, and so on till the final is played between two players. If after any round, there is odd number of players, one player is given a bye, i.e. he skips that round and plays the next round with the winners. Find the total number of matches played in the tournament.**

116 ppl.

1 winner 115 loser.

In 1 match 1 loser is decided. So 115 matches to decide 115 losers and 1 winner.

**Find the one hundredth smallest positive integer that can be written using the digits 1,3 and 5 in base 7?**

1,3,5

1 digit : 3

2 digit: _ _ =>9

3 digit : _ _ _ =>27

Total 39 so far

1 _ _ _ => 27

3 _ _ _ => 27

Total 93 so far

5 1 1 _ => 3

5 1 3 _ => 3

So 99 done

100th will be 5 1 5 1 Ans

**If ‘x’ and ‘y’ are real numbers such that 3x + 4y = 48. Then, what is the maximum possible value of xy?**

x + x + x + 4y/3 + 4y/3 + 4y/3 /6 >= (x^3y^3 * 64/27)^1/6

8^6 * 27/64 >= (xy)^3

2^12 * 3^3 >= (xy)^3

xy = 2^4 * 3 = 48

**N! is ending with m zeroes.(N+2)! is ending with (m+2) zeroes. Also 90 < = N < = 190.Find possible values of N?**

Increment of 2 zeroes as result of multiple of 25

Means if you check for 24! =>

24/5 => 4

25/5 => 5

5/5 => 1

So 5 +1 => 6

Same for 26!

So

98! And 100!

99! And 101!

Similarly

148! --> 150!

149! --> 151!

And

173! --> 175!

174! --> 176!

Remember you don't have to consider 125! (As it results into increment of 3 zeroes )

So 6 possibilities

**Which of the following statements is false?a) (23!)^2 > 23^23b) (20!∗19!∗18!) < 57!c) (33!)^4 < 33^60d) None of these**

B is clearly right since 20 + 19 + 18 = 57

A is right coz

Lhs=> 23!^2 = (23 * 1)(22 * 2)(21 * 3)...(2 * 22)(3 * 23)

However in rhs each terms is =>23 * 23 * 23...

Certainly lhs > rhs

Now in 3rd

33!^2 > 33^33

So 33!^2 * 33!^2 > 33^33 * 33^33

33!^4 > 33^66 hence 3rd is false

OA:C

**In a plane there are 37 straight lines of which 13 passes through Point A and 11 passes through Point B. Besides, no line passes through 3 points and no line passes through both A and B and no two are parallel. Find the total number of point of intersections of the straight lines**

Total number of points of intersection of 37 lines is 37c2

But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting (13c2) points, we get only one point A.

Similarly 11 straight lines out of the given 37 straight lines intersect at point B.

Therefore instead of getting (11c2) points, we get only one point B.

Hence the number of intersection points of the lines is = 37c2 - 13c2 -11c2 + 2 = 535

**Find the number of ways of selection 4 letters from the word EXAMINATION**

EXAMINATION has 11 letters in total

And out of which

E --> 1

X --> 1

M -->1

T ---> 1

O ---> 1

(A) --> 2

I --> 2

N --> 2

8 distinct letters.

- 4 letters selected, which are all distinct: 8C4 = 70
- 2 letters alike, and 2 distinct

(eg: AAEM)

= 3c1 x 7c2 = 63 - 2 letters alike, and 2 letters alike

(eg: NNII)

= 3c2 = 3

So answer is,

70 + 63 + 3 = 136

**Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.**

Method:1 (Not recommended)

_ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers

342 numbers added till 999

1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315

3 0 _ _ : 49 added

3 1 _ _ : 49 added

3 3 _ _ : 49 added

3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147

Total 294 more. Remaining 21

3 9 0 _ : 7

3 9 1 _ : 7

3 9 3 _ : 7

3939

Method:2

Base method approach

0 -> 0

1 -> 1

2 -> 3

3 -> 4

4 -> 7

5 -> 8

6 -> 9

1000th term : 1000 in base 7 = 2626 < - > 3939

a. 0.1591

b. 0.2727

c. 0.375

d. None of these

We can simply consider the original sample set to be 10000 (as the values are in the form of 0.00x and so, we get nice values to work with). Standard are 5000 with a probability of 0.01 of dying so, it is probable that 50 among these will die. Similarly, among preferred, 24 will die and among ultra preferred, 14 will die. So, there are 88 deaths happening out of which, 24 are those of the preferred holders. So, 24/88 would be 0.2727

**A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?a. 1.5b. 2.5c. 3d. 5e. 5**

Use alligations. When mixing 40% chocolate and 100% chocolate to get 50% chocolate, the mix should be in the ratio 5:1. This means 5 parts 40% chocolate + 1 part 100% chocolate gives 6 parts 50% chocolate. So to get 15 parts 50% chocolate, we need 15/6 = 2.5 cups of pure chocolate.

**Twenty-four men can complete a work in sixteen days. Thirty-two women can complete the same work in twenty-four days. Sixteen men and sixteen women started working for twelve days. How many more men are to be added to complete the work remaining work in 2 days?a. 16b. 24c. 36d. 48e. 54**

Let work done by 1 man = m units

Let work done by 1 woman = w units

24 men can do 24m units of work in 1 day

In 16 days they can do 24m * 16 units which is the total work

Similarly, total work will be 32w * 24

Since work done is the same, we can equate the two

24m * 16 = 32w * 24

m=2w

16 m and 16 w working for 12 days should do (16m+16w) * 12 = (16m * 8m) * 12 = 24m * 12.

x men join and they work for 2 more days.

So 24m * 12 + 24m * 2 +xm * 2 = 24m * 16

Solving this, we get x = 24

**There are 140 students in a b-school. The number of students who specialize in Marketing, Finance and HR is 50, 80 and 70 respectively. The ratio of the number of students who have taken more than one of the specializations to the number of students who have taken up all three specializations is 3:2. If each student of the school specializes in at least one subject, then how many students specialize in exactly two subjects?**

It's better to do these using the a+2b+3c and a+b+c logic (people vs. combinations). In this case, a+b+c=140 and a+2b+3c=200. Also, (b+c)/c=3/2 and so, c=2b. On solving, you will get b=12. (a is the number of students who take exactly one specialization, b is the number of student who take exactly two specializations and c is the number of students who take all three)

**The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,…. has one odd number followed by the next two even numbers, then the next three odd numbers followed by the next four even numbers and so on. What is the 2017th term of the sequence?**

Just need to observe that the first series ends at 1, second series at 4, third series at 9 and so on. So, the 63rd series will end at 3969. The total number of terms till the end of the 63rd series will be 2016. So, the 2017th term will be the next even number i.e. 3970.

**How many points of intersection are there for the two functions defined as f(x) = 1/(x^2 - 1) and g(x) = f(x - 1)?**

Equate the two functions. You will get

1/(x^2-1)=1/((x-1)^2-1)

x^2-1=x^2-2x

X=1/2

So only one point

**To create an army comprising of a giant, 5 barbarians and 7 archers, it takes 84 gems. To create another army comprising of a giant, 7 barbarians and 10 archers, it takes 114 gems. If a player wants to create an army of 9 giants, 23 barbarians and 30 archers, how many gems would that player need to spend?**

Let the multiplier of the first one be x and that of the second one be y. Equate the coefficients. So we will get x+y=9, 5x+7y=23 and 7x+10y=30. If you get a common solution as is true in this case, you will get the answer. The other way is of course taking values for archers in terms of multiples of 10. 2 or 3 iterations should do it i suppose. (Answer is 426)

**A is a set having k elements. X and Y are subsets of A such that X is a subset of Y. In how many ways can we choose X and Y? Your answer will be in terms of n**

Take a set of some 3 elements and {1, 2, 3}.

If there are no elements in Y, there will be only one combination possible.

If there is 1 element in Y, there will be 3c1(1c0+1c1) combinations possible.

If there are 2 elements in Y, there will be 3c2(2c0+2c1+2c2) combinations possible.

If there are 3 elements in Y, there will be 3c3(3c0+3c1+3c2+3c3) combinations possible.

So, 1+6+12+8=27.

The basic point is, you will get nc0+nc1(2)+nc2(4)+nc3(8).... which will translate to 3^n.

**x^4 - 12x^3 + ax^2 - bx + c = 0. if it is known that its roots are +ve real numbers then what is maximum value of a + b + c ?**

p + q + r + s = 12... so, the sum of products of roots taken 2 at a time, 3 at a time and 4 at a time would be maximum when they are all equal to 3.

pq+pr+ps+qr+qs+rs+pqr+pqs+prs+qrs+pqrs will be 243.

**Chord PQ of a circle is the perpendicular bisector of chord XY of the same circle such that they intersect at point M inside the circle. XY = PM = 14. What is the area of the circle?**

PQ will be the diameter of the circle. Let QM = x.

7 * 7 = x * 14.

So we get x = 7/2. Diameter will be 14 + 7/2. So, area of the circle will be pi*(35/4)^2

100! / 60! * 40!

12 = 2^2 * 3

max power of 3 in 100! = 100/3+100/9+100/27+100/81 = 33+11+3+1 = 48

max power of 3 in 60! = 60/3 + 60/9 + 60/27 = 20 + 6 + 2 = 28

max power of 3 in 40! = 40/3 + 40/9 + 40/27 = 13 + 4 + 1 = 18

3^48 / 3^28 * 3^18 = ( 3^2 )

max power of 4 in 100! = 100/4 + 100/16 + 10/64 = 32

max power of 4 in 60! = 60/4 + 60/16 = 18

max power of 4! in 40! = 40/4 + 40/16 =12

2^32 / 2^18 * 2^12 = 2^2

so highest power is 1

**A jet is flying 2400 miles from Hawaii to San Francisco. In still air, it flies at 600 mph. There is 40 mph tailwind in the same direction. Exactly how many hours after takeoff would it becomes neutral for the plane to either go to San Francisco or to return to Hawaii in the case of an emergency?a. 1.25 hoursb. 1.5 hoursc. 1.75 hoursd. 2 hours**

d / 640 = 2400 - d / 560

d = 1280 miles

1280 = 640 * t

t = 2 hours

**Find the sum, S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/2^32**

Integrate it!

Integration 1/x = [ ln x ] , x goes from 0 to 2^32 : so answer simply is 32 * ln2 = 32 * 0.693 = 22

**Two motorists set out at the same time to go from A to B, a distance of 100 miles. They both followed the same route and travelled at different, though uniform speeds of an integral number of miles per hour. The difference in their speeds was a prime number of miles per hour and after they had been driving for 2 hours, the distance of the slower car from A was 5 times that of the faster car from B. At what speed did the two motorists drive?**

A-------100miles------------B

Sp and Sq and | Sp - Sq | = prime number

Sp * 2 = b

Sq *2 = 500 - 5b

10Sp + 2 Sq = 500

(42, 40) satisfies

**If there are "n" students in a circular arrangement facing toward the center. It is known that their roll numbers are Integral value from 1 to "n" and they are sitting in the order of increasingroll numbers. Then what would be value of "n" if Roll number 13 is just opposite to Roll Number 29 ?**

14 to 28 : 15 students

30 , 31 ,32 and 1 to 12 : 15 students

so total 32

**If x/(2a+b) = y/(2b+c) = z/(2c+a) = 8 and x/2a = y/2b = z/2c = k where a+b+c not equal to 0 then k=?**

x = 16a + 8b

y = 16b + 8c

z = 16c + 8a

x/2a = 8 + 4b/a

y/2b = 8 + 4c/b

z/2c = 8 + 4a/c

all are equal :

b/a = c/b = a/c : numbers are equal

8 + 4 = 12

**How many liters of a 12 liter mixture containing milk and water in the ratio 2:3 be replaced with pure milk so that the resultant mixture contains milk and water in equal proportion?**

milk = 4.8 + 3x/5

water = 7.2 - 3x/5

4.8 + 3x/5 = 7.2 - 3x/5

x=2

**There are two vessels A and B, both containing vinegar solution of 40% concentration. I add some pure vinegar to A to bring the concentration to 50%. In the vessel B, I take out some quantity of the solution and replace it with an equal quantity of pure vinegar, to bring the concentration to 50%. What is the ratio of the amount of vinegar added in A and B, if the quantity of initial solutions in A and B are in the ratio of 1 : 2?**

For A

vinegar : 40

water : 60

20 ml vinegar added

conc of vinegar = 50%

For B

vinegar =80 + 3x/5

water = 120-3x/5

80+3x/5 = 120 - 3x/5

x = 100/3 ml was added

20 : 100/3 = 3:5

**In a polygon, the sum of all the exterior angles is equal to sum of all the interior angles. Each interior angle is equal to X degrees or its supplementary, where X is not equal to 90. How many of the angles must be X degrees?**

Sum of all interior angles= Sum of all exterior angles = 360

x + x + 180 - x + 180 - x

total 2 angles have x degree

**Two tangents pa and pb are drawn from an external point p length Length of pa=156 and length of ab =120 . Find radius of circle**

Let O be the center and Angle APB = θ

Let AB cuts PA at X

AX =XB = 120/2 = 60

Sin θ = 60/156

Sin(90-θ) = 60/ r

Solve above : r =65

Total ways is 6!

For every arrangement, we will have an internal order of abd, adb, bad, bda, dab, and dba.

Out of these, adb and bda are what we need.

So, 1/3rd of the total arrangements.

**Fifty inabitants of the shire was surveyed to note down their posessions of Arkenstones, Elfstones and rings of power. Of then, 22 own an Arkenstone, 15 own an Elfstone and 14 own a ring of power. Nine of these inhabitants own exactly two items out of an Arkenstone, an Elfstone and a ring of power; and, one inhabitant owns all three. How many of the fifty inhabitants own none of three: Arkenstone, Elfstone or a ring of power?**

Total number of objects will be given by a+2b+3c where a is the number of people who possess only one object, b is the number of people who possess exactly two objects and c is the number of people who possess exactly three objects. So, in this case, total number of objects is 22+15+14=51. Number of people who have exactly 2 objects is 9 and so, 2b=18. Similarly, number of people who have exactly 3 objects is 1 and so, 3c=3. So, a+18+3=51 and so, a=30. So, a+b+c=30+9+1=40 and so, 10 do not have any objects

**A, B, C, D, E are five students who took CAT. Following are the sums of their overall scores, taken three at a time: 119, 121, 124, 125, 123, 126, 127, 128, 129 and 132. What is the highest and lowest score among the scores of A, B, C, D, E?**

Let a > b > c > d > e

a + b + c = 132......(1)

c + d + e = 119......(2)

a + b + d = 129......(3)

b + d + e = 121......(4)

We know the values of all possible triplets. So, 5c3=10 cases in total. So, each element occurs 6 times.

6a + 6b + 6c + 6d + 6e = 1254

a + b + c + d + e = 209.....(5)

From (1), (2) and (5)

c = 42

From (2), (3) and (5)

d = 39... putting these values in (2)

e = 38

b = 44

a = 46

**An institute conducts 32 tests. The number of tests attempted by three students studying in the class is as follows:Neil - 16Nitin - 18Mukesh - 20The number of tests written by more than one student is at least:a) 8b) 11c) 13d) 16e) 15**

Let a be the number of students who have taken exactly 1 test, b be the number of students who have taken exactly 2 tests, c be the number of students who have taken exactly 3 tests. Total number tests will be equal to a + 2b + 3c = 16 + 18 + 20 = 54

Also, 32 tests have been conducted in total. So a+b+c=32

From the two equations

b + 2c = 22

To minimize overlap, we need to maximize c.. so c=11.

**There are 49 zeros, 51 ones and 53 twos written on the board randomly. A student is blindfolded and then asked by his teacher to touch any two numbers on the board arbitrarily. The teacher deleted those two numbers and replaced them by a single number in the following manner:If the pair is Replaced by(0, 0) → 0(1, 1) → 0(2, 2) → 2(1, 2) → 1(0, 1) → 1(0, 2) → 0If they continued this process what was the number left on the board in the end?(a) 0(b) 1(c) 2(d) Cannot be determined**

Essentially, when we are replacing 2 zeros, we do not make any changes to the existing sum of all the elements on the board, when we replace 2 ones, we bring down the total by 2, when we replace 2 twos, we bring down the total by 2, when we replace a one and a two, we bring down the total by 2, when we replace a zero and a one, we do not change the total and when we replace a zero and a two, we bring down the total by 2. So, we bring down the total by a 0 or a 2 with each successive replacement. As the total at the beginning is 157 and we lose zeros or twos, the remaining number will be 1.

Alternatively, you can actually cancel out pairs of zeros, ones, and twos and get the final answer.

**On a race track a maximum of 5 horses can race together at a time. There are a total of 25 horses. There is no way of timing the races but you can see the horses as they cross the finish line. What is the minimum number of races we need to conduct to get the top 3 fastest horses?A. 6B. 7C. 8D. 10**

5 winners from the first five races. Another race to determine who is the fastest. Now, positions 2 and 3 need to be figured out. The important thing to not here is that the horses who came 2nd and 3rd need not be the 2nd and 3rd fastest as they have competed with only the winner from the particular group and not the other horses. So, to be sure that a 'deserving' horse has got through to the final, we select those horses who have a chance of being the 2nd and 3rd fastest.

Let the horses be represented by

p1, p2.... p5

p6, p7.... p10

.

.

p21, p22.. p25

Let p1, p6, p11, p16 and p21 win the races. As p16 and p21 have come 4th and 5th in this race, they cannot be among the three fastest horses. So, the entire branch of p16 and p21 is out. Now, p1 is the fastest horse in the first race and so, there is a chance that p2 and p3 are the second and the third fastest overall. Similarly, p6 is the fastest horse in the second race and there is a chance that p7 could be the third fastest (understand that p1 is the fastest and so, the best case scenario for p6 would be if he comes in the 2nd place and the best case scenario for p7 is when he comes in the 3rd place. As the 3 positions would have been taken, p8 cannot feature in the contenders' list). Finally, p11 would be a part of the final race having won his round and being in the 3rd position in the previous race. So, the final race will have p2, p3, p6, p7, p11 in it the top two of whom would win.

Total of 5+1+1=7 races.

**There are one thousand students at the George Washington High School. Each student is assigned a locker, numbered 1 through 1000. On the first day of school each year, the students participate in an unusual ritual: All the lockers are closed in the beginning. The students then enter the school through one door, parade past the all the lockers, and then exit through another door. While in the school, the first student reverses the door position of each locker - if the door is open, he closes it, and if it is closed, he opens it. The second student reverses the door position of every other locker, starting with locker number 2. The third student reverses every third locker, starting with locker number 3, etc. After all 1000 students have completed this ritual, how many lockers will be left open?**

All the lockers are opened and closed the same number of times as they have factors. As they are all closed at the start, an odd number of operations have to be performed on the ones that are open at the end. So, if N = a^x * b^y * c^z... the number of factors will be (x+1)(y+1)(z+1)... now if this has to be odd, all of x+1, y+1, z+1 will be odd. So, x, y, z will be even. So, N has to be a perfect square. As there are 31 perfect squares less than 1000 (1-961), we understand that 31 lockers will be open after this exercise.

**There is a staircase of 10 steps. In how many ways can Amit climb the staircase if he can take a maximum of 3 steps at a time?a. 274b. 275c. 276d. 277**

If there were just 1 step, there would have been just 1 way to go up

For 2 steps, either (1,1) or (2)

For 3 steps, either (1,1,1)(2,1)(1,2)(3)

For 4 steps, (1,1,1,1)(2,1,1)(1,2,1)(1,1,2)(2,2)(3,1)(1,3)

If you look at it, to reach step 5, we need to be at either step 2, step 3 or step 4 before we make the last move. We could be at step 2 in 2 ways, at step 3 in 4 ways and at step 4 in 7 ways, So, s(5) = 2+4+7=13

So basically, it is a Fibonacci sequence with each term being equal to sum of the three preceding terms

1, 2, 4, 7, 13, 24, 44, 81, 149, 274

**If a clock gains 5 minutes every hour, how many times do the hands meet during the day ?**

The hands meet when the minute hand gains 360 degrees over the hour hand. 360/(11/2)= every 720/11 minutes . So we have to figure out how many such packets we have in a day. The clock goes ahead by 65 minutes for every 60 minutes that a normal clock would take. So, if the normal clock moves by 1440 minutes, the faulty clock would have moved 1440 * 65/60 = 1560 minutes.

So 1560/(720/11)

**Amul and Cadbury run a 12km cycling race on a circular track of circumference 750metre.Amul can beat Cadbury in a race of 2,000metre by 500metre.If Amul gives a headstart of 500 metre to cadbury,then how many times will amul overtake cadbury in the race?**

Speed of Amul : Speed of Cadbury = 4/3

For them to meet, Amul has to travel 750 m more

(x + 750)/x = 4/3

x = 2250. So, every time Amul covers 3000 m, he overtakes Cadbury. So, for 12 km, he will meet 4 times (that 500 m bit is required as the fourth time they meet, Amul will overtake Cadbury)

If you calculate and check with Sa = 40 m/s and Sc=30 m/s you will get the first meeting at 50 seconds and every subsequent meeting after 75 seconds.

Bus .......... A....... Vikram ...B

AB is a tunnel

And suppose person is at point C

AC : CB = 1:3

Now the time at which bus reaches A

Vikram would reach from point C to point A

And the time at which bus reaches to B

Vikram reaches from C to B

Let assume

Length of tunnel is 4km

So if Vikram moves 1 km forward towards B

Bus would reach at A

So remaining distance covered by Vikram is 2km

And bus covered the whole distance 4km(length of tunnel) in the same time with a speed of 80km/hr

So speed of Vikram = 80/2 => 40km/hr

**Find the ten's place digit of 625^246 - 441^128**

(a5)^even ends with 25

Last two digits

(41)^28

=> (81)^14

=> (61)^7

=> 21

Or (41)^28

=> (4 * 8)1

=> 21

So 25-21 = 04

digit at tenth place = 0

**25 males and 12 females can complete a piece of work in 12 days. They worked together for 8 days and the women left the work.the remaining men completed the work in 6 days. Find out in how many days 15women can complete the entire work?**

They worked for 8 days

I.e completed 2/3 work

So Remaining (1/3) work is completed by men in 6 days

I.e males complete this work in 18 days

So 18 * X /(18+X) = 12

=> 3x /(18+X) = 2

=> 3x = 36 + 2x

=> X = 36

I.e 12 women complete this work in 36 days

So 15 women in 12 * 36/15

I.e 144/5 days

**X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day one man whose capacity to do the work is twice that of X, joined. On the third day another man whose capacity is thrice that of X, joined and the process continues till the work is completed. In how many days will the work be completed, if everyone works for four hours a day?**

Total work = 20 * 7 = 140 units

Efficiency of X => 1unit /hour

First day

1 * 4

Second day

Another man joined

Whose efficiency is 2

So total work on second day

(1+2) * 4

So basically

4 [ 1 + 1+2 + 1+2+3 + 1+2+3+4 + 1+2+3+4+5 ] = 140

So 5 days

**How many multiples of 18 are there which are less than 3500 and also 2 more than the square of a natural number**

18k = t^2 + 2 ---(1)

Now t^2 mod 18

Min Possibility 0

and max possibility - 17

so for t^2 +2 to be divisible by 18

t^2 must give remainder as 16 .

so t must be of type 18k+ 4 or 18k - 4 ..

√3600 = 60

√3500 = 59 (approx)

So maximum value of t can be 59

so t = 4 (18 * 0 +4)

14 (18 * 1 - 4)

22 ,

32,

40,

50 ,

58(18 * 3 + 4 )

So 7

Area of triangle using medians

M = (16+20+24)/2 => 30

Area of triangle

=> 4/3 √ [30 * (30-16)(30-20)(30-24) ]

=> 4/3 √ [ 30 * 14 * 10 * 6]

=> 80 √7

Now we know medians divide

Triangle into 6 equal parts

So area of one smaller triangle who height is BG

Is 1/6 (80√7)

And base = 1/3 of (16)

So

(1/2) * (1/3) * 16 * h = 1/6 (80√7)

=> H = 5√7

**If a > 0, b > 0 then the number of real roots of the equation 2x^7 - ax^5 - 3x^4 - bx^2 + 7 = 0 can bea) 2b) 3c) 4d) 0**

Use Descartes rule

And see the sign changes in F(X)

And F(-x)

If Number of sign changes in F(X) =k

number of Positive roots can be k , k-2 ,k-4 and so on

Similarly

Number of sign Changes in F(-x) = k

number of negative roots = k , k-2 , k-4 and so on

**Find the largest 5 digit number which when divided by 8 leaves a remainder of 5 and when divided by 7 leaves a remainder of 2.**

7k +2 => 8p +5

=> 7k = 8p + 3

For p = 4

It satisfies

I.e least number which satisfies the condition is

8 * 4 +5 = 37

And number is of the form

56k + 37

Now to find largest 5 digit number

99999/56

=> 1785.xx

So k = 1785

56 * 1785 + 37

=> 99997

**From the first 25 natural numbers, how many arithmetic progressions of 6 terms can be formed such that common difference of the AP is a factor of the 6th term.**

Let the first term of the AP be a and the common difference be d.

The sixth term of the series will be a+5d

Given that d should be a factor of a+5d

=> a+5d is divisible by d

=> a should be divisible by d

So the required cases are

d = 1, a = 1,2,3.......20

d= 2 , a = 2,4,6.......14

d = 3, a = 3, 6,9

d= 4, a = 4

So the required number of AP’s are 20+7+3+1 = 31

**101010101010..... (94 digits) / 375. Find the remainder.**

375 => 125 * 3

Use CRT

With 3

Sum of digits

47 mod 3 = 2

Now ,

With 125

You have to check only last three digits

As 10^3 mod 125 = 0

So last three digits would be 010

So 125k+10

3b +2 = 125k +10

=> For k = 2

It satisfies

So 260

35 * 36 * ...67 , cancel one 17 with 51.

so remaining: (35 * 36....50) * 3 * (52 * 53....67) mod 17

(16!) * 16! * 3 mod 17 = 1 * 1 * 3 = 3 mod 17 (using Wilson)

since we cancelled 17, so 17 * 3= 51.

**Q2) Find the number of integer solutions for |x| + 2|y| + |z| = 4**

|x|+2|y|+|z|=4

for |y|=0, |x|+|z|=4 so 4n=4 * 4=16 cases

for |y|=1, y=+-1 and |x|+|z|=2 so 2 * 4 * 2=16 cases

for |y|=2 y=+-2 |x|+|z|=0 so 2 * 1 = 2 cases

so 16+16+2 = 34 integer solutions.

**Q3) If N = 2^3 * 3^4 * 5^9, then find the number of trailing zeroes at the end of product of all factors of N which are not divisible by 12.**

number formed from 2 * 3^3 * 5^9 = 80 factors.

so product of these = 12^80 * (2 * 3^3 * 5^9)^40 = 2^200 * 3^83 * 5^360.

total product = (2^3 * 3^4 * 5^9)^(100) = 2^300 * 3^400 * 5^900

so not divisible ke liye divide the above two.

2^100 * 3^x * 5^540 so 10^100

100 trailing zeros.

**Q4) Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + acb = 2536, what is c+b+a ?**

221a + 212b + 122c = 2536

122(a+b+c) + (99a + 90b) = 2536

2536 mod 9 = 7

99a + 90b mod 9 = 0

122(a+b+c) mod 9 = 7

5(a+b+c) mod 9 = 7

so : a+b+c = 14 as 70 mod 9 = 7

**Q5) Find the number of ordered pairs of integers for : x^2 + y^2 - xy = 727**

x^2 + y^2 - xy = 727

x^2 - yx + (y^2 - 727 ) = 0

D = y^2 - 4(y^2 - 727)

D = 2908 - 3y^2

2907-3y^2+1 should be a perfect square

3(969-y^2)+1 should be a perfect square

clearly y=31 , which gives 2 possible x = 13 and x= 18

values can be interchanged here :

y =13 will give x=31 , x=18

y=18 will also give x=31 , x=13

same will be possible when all these values are of negative sign

so total 3 * 2 * 2 = 12 values

**Q6) Find the maximum power of 12 in C(100,60) ?**

100! / 60! * 40!

12 = 2^2 * 3

max power of 3 in 100! = 100/3+100/9+100/27+100/81 = 33+11+3+1 = 48

max power of 3 in 60! = 60/3 + 60/9 + 60/27 = 20 + 6 + 2 = 28

max power of 3 in 40! = 40/3 + 40/9 + 40/27 = 13 + 4 + 1 = 18

3^48 / 3^28 * 3^18 = ( 3^2 )

max power of 4 in 100! = 100/4 + 100/16 + 10/64 = 32

max power of 4 in 60! = 60/4 + 60/16 = 18

max power of 4! in 40! = 40/4 + 40/16 =12

2^32 / 2^18 * 2^12 = 2^2

so highest power is 1

**Q7) Two motorists set out at the same time to go from A to B, a distance of 100 miles. They both followed the same route and travelled at different, though uniform speeds of an integral number of miles per hour. The difference in their speeds was a prime number of miles per hour and after they had been driving for 2 hours, the distance of the slower car from A was 5 times that of the faster car from B. At what speed did the two motorists drive?**

Sp and Sq and | Sp - Sq | = prime number

Sp * 2 = b

Sq *2 = 500 - 5b

10Sp + 2 Sq = 500

(42, 40) satisfies

**Q8) If Ap is the sum to the first p terms of the series A = 12^144 + 12^143 + 12^142 + ………, then find Bp, which is the sum to the first p terms of the series A1 + A2 + A3 ...?**

sum of P terms :

12^144 + 12^143 +....12^145-P

12^144 [ 1 - (1/12)^P ] / [ 11/12 ]

12^145 [ 1 - 1/12^P ] / 11 = Ap

Bp = A1 + A2 + .... Ap

12^145 [ 1 - 1/12^P ] / 11

12^145 / 11 [ 1 - 1/12 + 1-1/12^2+....1/12^P ]

12^145 / 11 [ 11P/11 - 1/11 + (1/12)^P/11 ]

12^145/121 [ 11P -1 + (1/12)^P ]

**Q9) Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?**

a0 , b0 , c0 , d0 , e0

a0 : a1 to a24 + sara

b0 : b1 to b24+ sara

c0 : c1 to c24 + sara

d0 : d1 to d24 + sara

e0 : e1 to e24+ sara

2 friends of sara are connected to each other : say a-b

24 * 5 + 3 = 123

all five friends are mutual friends

each has 20 distinct friends and give people a0,b0,c0,d0,e0 : 20 * 5 + 5 = 105

**Q10) There are two vessels A and B, both containing vinegar solution of 40% concentration. I add some pure vinegar to A to bring the concentration to 50%. In the vessel B, I take out some quantity of the solution and replace it with an equal quantity of pure vinegar, to bring the concentration to 50%. What is the ratio of the amount of vinegar added in A and B, if the quantity of initial solutions in A and B are in the ratio of 1 : 2?**

For A

vinegar : 40

water : 60

20 ml vinegar added

conc of vinegar = 50%

For B

vinegar =80 + 3x/5

water = 120-3x/5

80+3x/5 = 120 - 3x/5

x = 100/3 ml was added

20 : 100/3 = 3:5

profit on sale = profit on SP = Profit/SP *100%

CP=50, MP=80.

3 * (SP-50) * 100/SP = 2 * D (D=discount%)

Also, SP = (1-D/100) * 80 = 80 * (100-D)/100

so 3 * {80(100-D)/100 -50 } * 100 = 2D * {80 * (100-D)}/100

150 { 75-2D} = D {200-2D}

D^2 - 250D + 5625 = 0

D= (250-200)/2 = 25%

**Q) Find the product of the irrational roots of the equation (2x – 1) (2x -3) (2x – 5) (2x – 7) = 9.**

(2x-1)(2x-7)(2x-3)(2x-5) = 9

(4x^2-16x+7) (4x^2-16x+15)=9

Let 4x^2-16x=t

(t+7)(t+15)=9

t^2+22t+96=0

t=-16, t=-6

so 4x^2-16x+16=0 and 4x^2-16x+6=0

x^2-4x+4=0 2x^2-8x+3=0

1st gives x=2 and second eqn gives both irrational roots. so c/a= 3/2

**Q) For how many ordered triplet (x,y,z) of positive integers less than 10 is the product xyz divisible by 20?**

20=5 * 2^2 so one 5 should be there and rest we have to distribute

(5,2,2)....(5,2,8) so 3+6 * 3 = 21

(5,4,4)...(5,4,8) so 3+2 * 6 = 15

(5,6,6,), (5,6,8) = 3+6=9

(5,8,8)= 3 so till here= 48.

ab (5,4,1/3/5/7/9) = 3+6 * 4 = 27

and (5,8,1/3/5/7/9) = 3+6 * 4 = 27

so 54 extra.

hence 48+54=102

**Q) The points (3,7), (6,2) and (2,k) are the vertices of a triangle. For how many values of k is the triangle a right triangle ?**

A=(3,7), B(6,2), C(2,k)

AB^2 = 9+25 = 34

AC^2 = 1+(7-k)^2 = k^2-14k+50

BC^2 = 16+(2-k)^2 = k^2-4k+20

if AB is hyp. 2k^2-18k+70 = 34

k^2--9k+18= 0

2 real values here.

if AC is hyp: k^2-4k+54 = k^2-14k+50

10k=-4 so k=-0.4 so one value

if BC is hyp, k^2-14k+84=k^2-4k+20 so k=6.4 so one value.

so (2+1+1)= 4 values

**Q) Choti has 11 different toys and Bade has 8 different toys.Find the number of ways in which they can exchange their toys so that each keeps their initial number of toys?**

Goods have to be "exchanged". 19C8 or 19C11 is the total possibilities, but since they have to be "exchanged" so it is 19C8-1.

(original configuration not to be counted)

**Q) In how many ways can seven balls of different colours be put into 4 identical boxes such that none of the box is empty?**

(4,1,1,1): 7C4 * 3 * 2 * 4 = 840

(3,1,1,2): 7C3 * 4C2 * 2C1 * 4!/2! = 5040

(2,2,2,1): 7C2 * 5C2 * 3C2 * 4 = 2520

so(840+5040+2520)/4! = 350

**Q) A team is planning to participate in a Dahi-handi competition. On average, they succeed in breaking the handi on 2 out of every 11 attempts. How many attempts will they have to make to have a fair chance of succeeding in breaking the handi?**

2/11 + 9/11 * 2/11 + 9/11 * 9/11 * 2/11..n terms.

r=9/11

{ 1-(9/11)^n} > = 0.5

(9/11)^n < = 0.5

so n=4

PS: Fair chance means >=50% probability

**Q) A person has just sufficient money to buy either 30 guavas, 50 plums or 70 peaches. He spends 20% of the money on travelling, and buys 14 peaches, 'x' guavas and 'y' plums using rest of the money. If x, y > 0, what is the minimum value of the sum of x and y?**

30a =50b = 70c =5T

4T = 14c + xa + yb

4T = 14 * 5T/70 + x * 5T/30 + y * 5T/50

4 = 1+ x/6 + y/10

x/6+y/10 = 3

5x+3y = 90

so min. sum when x=15, y=5 so 20

**Q) How many ordered pairs of integers (a,b), are there such that their product is a positive integer less than 100?In the above question, if (a,b) is not different from (b,a), then how many such pairs are possible?**

a) when both are +ve:

1 * 1 ... 1 * 99 and reverse = 99 * 2 - 1 = 197

2 * 2...2 * 49 and reverse = 48 * 2 - 1 = 95

3 * 3....3 * 33 = 31 * 2 - 1 = 61

4 * 4...4 * 24 = 21 * 2 - 1 = 41

5 * 5...5 * 19 = 15 * 2 - 1 = 29

6 * 6....6 * 16 = 11 * 2 - 1 = 21

7 * 7...7 * 14 = 8 * 2 - 1 = 15

8 * 8 .... 8 * 12 = 5 * 2 - 1 = 9

9 * 9...9 * 11 = 3 * 2 - 1 = 5

so total = 473

same when both are -ve so 473 * 2 = 946

b) Unordered = 99+48+31+21+15+11+8+5+3= 241 so 241*2= 482

**Q) Find the number of positive solutions for 2x + 3y more than 60 and 2y + 3x less than 60**

Draw 3x+2y = 60 and 2x+3y = 60

find the common region as per question

a triangle will be formed with cordinates (12,12 ) ( 0,20 ) ( 0,30 )

area of this triangle = 60

use picks theorem : A = i + b/2 - 1

61 = i + b/2

calculate the number of boundary pts of triangle boundary , total 20 boundary pts will be there

9 on the vertical line between 0,20 and 0,30

3 on the line between 0,20 and 12,12

5 on the line between 0,30 and 12,12

3 vertices

total 20 boundary points

61 - 20/2 = 51 integral points

Yes When r = 1, s can take 37 values [371]

When r = 2, s can take 17 values [352]

When r = 3, s can take 11 values [333]

When r = 4, s can take 7 values [314]

When r = 5, s can take 5 values [295]

When r = 6, s can take 4 values [276]

When r = 7, s can take 3 values [257]

When r = 8, s can take 2 values [238]

When r = 9, s can take 2 values [219]

When r = 10, s can take 1 values [1910]

When r = 11, 12, 13 s can take one value each.

Totally, there are 92 values possible.

**A boss decides to distribute Rs. 2000 between 2 employees. He knows X deserves more that Y, but does not know how much more. So he decides to arbitrarily break Rs. 2000 into two parts and give X the bigger part. What is the chance that X gets twice as much as Y or more?**

If he divides randomly 2000 into 2 parts then the larger part ranges uniformly from (1000,2000) we want the probability that the larger part lie in ( 1000*(4/3) , 2000 ) for X to get at least twice as much as Y. Therefore the required probability would be: (2000 - (4/3)*1000) / 1000 = 2 - (4/3) = 2/3 that is 66.67%

**A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?**

If you assume that any digit is in fixed position, then the remaining four digits can be arranged in 4! = 24 ways.

So, each of the 5 digit will appear in each of the five places 24 times. So, the sum of the digits in each position is 24(1+3+5+7+9) = 600

The sum of all such numbers will be 600(1 + 10 + 100 + 1000 + 10000) = 6666600

**Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year.Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90 of the time. When it doesn't rain, he incorrectly forecasts rain 10 of the time. What is the probability that it will rain on the day of Marie's wedding?**

We want to know P(A1|B), the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below.

P(A1|B)=P(A1)P(B|A1) / P(A1) P(B|A1)+P(A2) P(B|A2)

P(A1|B)=(0.014×0.9 / ( 0.014×0.9+ 0.986×0.1) )

P(A1|B)=0.111

**A swimmer jumps from abridge over a canal and swims 1 km upstream. after that first km he passes a floating cork. He continues swimming for half an hour and then, turns around and swims back to the bridge. The swimmer and the cork arrive at the same time. What is the speed of the water if the speed of the swimmer has been constant?**

If the swimmer is swimming away from the cork for half an hour (up stream), it will take him another half hour to swim back to the cork again. Because the swimmer is swimming with constant speed (constant relatively to the speed of the water!) you can look at it as if the water in the river doesn't move, the cork doesn't move, and the swimmer swims a certain time away from the cork and then back. So in that one hour time, the cork has floated from 1 kilometer up stream to the bridge.

Conclusion: The water in the canal flows at a speed of 1 km/h.

**What is the digit at the ten's place of the number 6^11^7 ?a) 3b) 1c) 5d) 9**

The digit at the ten’s place of 6^2, 6^3, 6^4, 6^5 and 6^6 are 3, 1, 9, 7 and 5 respectively.

Also, the digit at the ten’s place of 6^7 is 3.

⇒ The cyclicity of the ten’s digit of 6^N is 5. (N ≥ 2) The remainder when 11^7 is divided by 5 is 1.

⇒ The digit at the ten’s place of the given number N will be

same as the digit at ten’s place of 6^6 which is 5.

**xy + zy = 37 and xz + zy = 72.Find the number of ordered triplets (x,y and z) such that x, y and z are positive integers.a) 0b) 1c) 2d) 3**

y(x+z) = 37=1×37

It is obvious that x + z > 1, therefore ‘y’ has to be equal to 1. ⇒x+z = 37 and y=1

⇒xz + z =72

⇒ z(37 – z) + z = 72

⇒38z – z2 =72

⇒ z2 – 38z + 72=0 ⇒ (z – 2)(z – 36) = 0

⇒z= 2 or 36 So,there are 2 solutions: (x=35,y=1,z=2) and (x=1, y=1,z=36).

**A function f(x) is defined as f(x) = f(x – 2) + x(x – 2) for all the integer values of ‘x’. Given that f(1) + f(6) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?**

Given that f(x)=f(x–2)+x(x–2)So,

At x = 3, f(3) = f(1) + 3(1) = f(1) + 3

and f(5) = f(3) +5(3) = f(3) +15 = f(1) + 18 Similarly,

f(6)=f(4)+6(4)=f(4)+24 Hence, f(4) = f(6) – 24

Also, f(4) = f(2) + 4(2) = f(2) + 8 Hence, f(2) = f(4) – 8 = f(6) – 32 Therefore,

f(1) + f(2) + f(3) + f(4) + f(5) + f(6)

= f(1) + {f(6) – 32} + {f(1) + 3} + {f(6) – 24} + {f(1) + 18} + f(6) =3{f(1)+f(6)}+ {–32+3–24+18}

= 3(0) – 35

= –35.

**How many divisors of 25200 can be expressed in the form 4n + 3, where n is a whole number?**

25200 = 2^4 ×3^2 ×5^2 ×7^1

As the required divisors when divided by 4 leave remainder 3, the power of 2 in the divisors has to be 0. Therefore, any such divisor is of the form 3^a × 5^b ×7^c, which when divided by 4 leaves the remainder (–1)^a × 1^b × (–1)^c.

For the remainder to be 3 i.e. –1, one of ‘a’ or ‘c’ must be even/0 and the other should be odd. Also, ‘b’ can take all the three possible values without making a difference to the remainder.

The nine possibilities are listed below:

a=0,b=0,c=1

a=2,b=0,c=1

a = 1, b = 0, c = 0

a = 0, b = 1, c = 1

a=2,b=1,c=1

a = 1, b = 1, c = 0

a=0,b=2,c=1

a=2,b=2,c=1

a = 1, b = 2, c = 0

**What is the total number of ways of selecting twenty balls from an infinite number of blue, green and yellow balls?**

Let the number of blue, green and yellow balls picked be x, y and z respectively.

∴ x + y + z = 20

So the number of ways = (20+3–1)C(3–1)= 22C2 = 231

Make one positive max so 49, so other two should add to -19. hence -9 and -10

49 * (-9) * (-10) = 4410

**Q2) Find the range of x where ||x - 3| - 4| > 3.**

case 1: |x-3| - 4 > 3

so |x - 3| > 7

so x > 10 or x < -4

so -(inf, -4) U (10,inf)

case 2: |x - 3| - 4 < -3|x - 3| < 1

so -1 < x - 3 < 1

2 < x < 4

take union of the three regions.

**Q3) a and b are roots of the equation x2 - px + 12 = 0. If the difference between the roots is at least 12, what is the range of values p can take?**

D= p^2 - 48

roots = p + - rt (p^2-48) /2

so given: rt (p^2-48) >= 12

p^2 - 48 >= 144

(p-8rt3) (p+8rt3) >= 0

so (-inf, -8 rt 3] U [8 rt3, inf)

**Q4) (|x| - 3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?**

(a-3)(b+4) = 1 * 12, -1 * -12, 2 * 6, -2 * -6, 3 * 4, -3 * -4

a >= 0, b >= 0

so a = 4, b = 8 so 4 cases here (+- for mod)

a=5, b=2 so 4 cases here.

a=6, b=0 so 2 cases here.

Total = 10

**Q5) [x] = greatest integer less than or equal to x. If x lies between 3 and 5, what is the approximate probability that [x^2] = [x]^2?**

between 3 and 4, [x]^2=9

works till root 10, so from 3 to 3.16

between 4 and 5, [x]^2=16

works till root 17 = 4.12 so from 4 to 4.12

so (0.16+0.12)/2 = 0.14

**Q6) f(x + y) = f(x)f(y) for all x, y.f(4) = 3 , what is f(–8)?**

f(0 + 4) = f(0)f(4) so f(0)=1

f(4 - 4) = f(4)f(-4) so f(-4) = 1/3

so f(-8)= f(-4) * f(-4)= 1/3 * 1/3 = 1/9

**Q7) An escalator is moving downwards at a speed of 4 steps/minute. Neerja takes 6 minutes less to reach the bottom from the top of the escalator, if he comes down on the moving escalator, as compared to when he does so on the stationary escalator. Gia takes 6 more min to reach the top from the bottom of the escalator if he goes up on the escalator moving downward as compared to when he does so on the stationary escalator. They start simultaneously from the top and the bottom of the escalator, moving downward, respectively and meet after 4 minutes. How many steps are there in the escalator?a. 60b. 56c. 48d. Cannot be determined**

total steps be T

T/N -T/(N+4) =6

3N^2+12N-2T=0 N=-12+rt (144+24T) /6 --(1)

and T/(G-4)-T/G = 6

3G^2 -12G-2T=0

G= 12+rt (144+24T) /6 ---(2)

relative speed = N+G

so T= 4N+4G --(3)

using 1 and 2,

T = 4 * rt (144+24T)/3

9T^2 = 16 (144+24T)

9T^2 - 384T-2304 = 0

T = (384 + 480)/18 = 48 steps

**Q8) Two cars P and Q are moving at uniform speeds, 50 km/hr and 25 km/hr respectively, on two straight roads intersecting at right angle to each other. P passes the intersecting point of the roads when Q has still to travel 50 km to reach it. What is the shortest distance between the cars during the journey?a. 20√5 kmb. 50 kmc. 25 kmd. 25√2 km**

P's distance be Y, Q's be X

Y/50 = (X-50)/25

Y=2X-100

say Y=100, X=100

distance at any time=d

d^2 = (100-50t)^2 + (100-25t)^2

diff wrt t for minima

(100-50t)*2 +(100-25t) = 0

t=12/5

d^2 = 20^2 + 40^2 = 2000 so d= 20 rt 5

**Q9) Two men are walking towards each other alongside a railway track. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this man is 18 seconds. Assume the velocities are constant throughout. How long after the train has passed the second man will the two men meet?A. 89.7 minutesB. 90 minutesC. 90.3 secondsD. 91 seconds**

Train speed= T, men speed: v1, v2. length L.

so L/(T-v1)=20 and L/(T+v2) = 18

so 20T-20v1=18T+18v2

T= 9v2+10v1 --(1)

Ditsane between the men = 600 (T+v2)

Distance covered in these 10 mins = 600(v1+v2)

so reqd time = {600(T+v2)-600(v1+v2)}/(v1+v2)

= (6000-600)= 5400 sec= 90 mins ( using 1)

**Q10) A and B started running from the same point and in opposite directions around a circular track of radius 24.5 m. A’s speed was twice that of B’s speed. They met each other aиer 14 seconds. What was A’s speed?**

A=2x, B=x

length= 2pi*r = 49pi = 154 m.

so given, 154/3x = 14 x=11/3 so 2x=22/3

a) 1111 * 23 * 54

b) 1111 * 23 * 48

c) 1111 * 23 * 42

d) 1111 * 23 * 36

Let us consider that the 4 digit number is abcd

Sum of all its arrangements = 1111 * 6 * (a + b + c + d) = 1111 * 6 * 23

Now, if there are 'n' ways of selecting the values for abcd, your answer would be 1111 * 23 * 6 * n

So, let's try and figure that part out.

9851, 9842, 9761, 9752, 9743, 9653, 8762,8753,8654

are the only ones which are there. So, n = 9

So, our answer is 1111 * 23 * 54

**If the sum of the first 'n' terms of an Arithmetic Progression is 100 and the sum of the next 'n' terms of the Arithmetic Progression is 300, then what is the ratio of the first term and the common difference?**

Let's say 'n' is 5

Sum of first 5 terms is 100, so 3rd term is 20

Sum of next 5 terms is 300, so the middle of those 5 terms, which will be the 8th term overall is 60

=> 5d = 60 - 20

=> d = 8

=> a = 20 - 2d = 4

Required ratio is 1:2

**A is 80% more efficient than B who is 60 % more efficient than C. A takes 40 days less than B to complete a work. A starts the work and works for 25 days and then B takes over. B then works for next 30 days and then stops. In how much more time can C complete the remaining work?**

Let us say efficiency of C is 1

=> Efficiency of B = 1.6 = 8/5

=> Efficiency of A = 1.8 * 1.6 = 9/5 of B = 72/25 of A

A takes 40 days less than B because A takes 5/9th time of B

=> A takes 4/9 days less than B

=> B takes 90 days and A takes 50 days

I guess you can work out the rest from here.

**In how many ways can 10 boys be selected from 20 if Rajesh and Ramesh are not selected together?**

If only Rajesh is selected, 18C9

If only Ramesh, is selected, 18C9

If none of them is selected, 18C10

Total ways = 18C9 + 18C9 + 18C10

**A alone would take 8 hours more to complete a work than when A and B would work together. If B worked alone it would take him 9/2 hours more to complete the work than when A and B would work together. What time would they take to finish the work if they worked together?**

You can use this concept for such questions,

If A takes 'a' hours more than A & B combined and B takes 'b' hours more than A & B combined, then A & B together will take sqrt(ab) hours to do the job.

You will get the answer directly and quickly.

If you wish to understand how this formula came, assume that they together take 'x' hours.

So, A will take 'x + a' hours and B will take 'x + b' hours.

The equation will be 1/(x+a) + 1/(x+b) = 1/x

Solving this equation, you would get x = sqrt(ab)

Applying the same in this question, x = sqrt(8 * 9/2) = sqrt(36) = 6

**A man bought an article and sold it at a gain of 5%. If he had bought it for 5% less and sold it for Rs 1 less, he would have made a profit of 10%. Find CP of the article**

Let us say that the cost price is 100x, he sold it for 105x

If he had purchased it at 5% less or at 95x and sold it at 1 rs less or at 105x - 1, he would have gained 10%

=> 1.1 * 95x = 105x - 1

=> 105x - 104.5x = 1

=> 0.5x = 1

=> x = 2

So, the cost price of the article = 100x = 200 Rs.

**If a^3 + b^3 + c^3 = 36 then the maximum value of 3abc will be?**

When the sum of numbers is constant, the product is maximum when they are equal.

=> When a^3 + b^3 + c^3 = 36

=> Maximum value of a^3 * b^3 * c^3 = 12 * 12 * 12

=> Maximum value of abc = 12

=> Maximum value of 3abc = 36

**A person was appointed for a 50 days job on a condition that he will be paid Rs. 12 for every working day but he will be fined Rs. 6 for every day he remains absent. After the completion of the work, he got Rs. 510. For how many days, he worked?(a) 15 days(b) 45 days(c) 30 days(d) 20 days(e) None of the above**

12 * present - 6 * absent = 510

=> 2 * present - absent = 85

Also, present + absent = 50

Adding these two equations 3 * present = 85 + 50 = 135

=> present = 135/3 = 45

**There was 120 Litres of pure milk in a vessel. Some quantity of milk was taken out and replaced with 23 Litres of water in such a way that the resultant ratio of the quantity of milk to that of water in the mixture was 4:1. Again 23 Litres of the mixture was taken out & replaced with 28 Litres of water. What is the ratio of milk to water in the resultant mixture ?a. 58 : 37b. 116 : 69c. 46 : 29d. 101 : 37e. 53 : 23**

Initially, it had 120 liters of pure milk.

Suppose 'x' liters was taken out and 23 liters of water was added. The ratio of milk : water became 4:1

=> (120 - x):23 = 4:1

=> 120 - x = 92

=> x = 28 liters.

Now, 23 liters of mixture is taken out. This 23 liters has milk and water in the ratio of 4 : 1.

=> 92/5 liters of milk was taken out and 23/5 liters of water was taken out.

Now, 28 liters of water was added. The total quantity is back to 120 now

New quantity of milk = 92 - 92/5 = 368/5

New quantity of water = 120 - 368/5 = 232/5

Ratio of milk: water = 368 : 232 = 46 : 29

**What is the remainder when the infinite sum (1!)^2 + (2!)^2+ (3!)^2 + ··· is divided by 1152?**

We have to find out the remainder when (1!)^2 + (2!)^2+ (3!)^2 + ··· is divided by 1152

1152 = 2^7 * 3^2

= (6!)^2 is divisible by 1152

= All (n!)^2 are divisible by 1152 as long as n > 5

So, our problem is now reduced to

Rem [((1!)^2 + (2!)^2 + (3!)^2 + (4!)^2 + (5!)^2)/1152]

= Rem[(1 + 4 + 36 +576 + 14400) / 1152]

= Rem [15017/1152]

= 41

If you are looking for a solution to the given question, just jump to the last paragraph in the answer. If you want to understand the concept, read from the top.

This is a simple case of conditional probability.

Let's look at all the items that we have at hand.

We have 100 items from A, 90 are fine and 10 are defective.

We have 200 items from B, 160 are fine and 40 are defective.

Now, let's try and answer some questions

Q. What is the probability of the superviser picking an item which is defective?

Ans. Total cases = total items available = 100 + 200 = 300

Favorable cases = defective items available = 10 + 40 = 50

Required probability = 50/300 = 1/6

Q. What is the probability of the superviser picking an item which is fine?

Ans. Total cases = total items available = 100 + 200 = 300

Favorable cases = fine items available = 90 + 160 = 250

Required probability = 250/300 = 5/6

Q. What is the probability of the superviser picking an item which is from B?

Ans. Total cases = total items available = 100 + 200 = 300

Favorable cases = B items available = 200

Required probability = 200/300 = 2/3

While these questions dealt with the case of normal probability, let us look at some cases of conditional probability.

Q. What is the probability of the superviser picking an item which is defective given it is a machine B item?

Ans. Total cases = total B items available = 200

Favorable cases = defective B items available = 40

Required probability = 40/200 = 1/5

Notice the difference here. The total cases have changed. Let's try another example.

Q. What is the probability of the superviser picking an item which is fine given it is a machine A item?

Ans. Total cases = total A items available = 100

Favorable cases = fine A items available = 90

Required probability = 90/100 = 9/10

So, the point that I am trying to make is - in case of conditional probability, the total cases might change. They will be the total cases given the condition is applied.

Keeping that in mind, let us try to look at the question given to us.

Q. Find the probability that the defective item came from Machine A

Ans. Now we know that the item is defective.

Total cases = total defective items available = 50

Favorable cases = defective A items available = 10

Required probability = 10/50 = 1/5

**If x + 1/x = 3 then x^5 + 1/x^5 = ?**

We are given x + 1/x = 3

The trick to solving these questions is to find out higher powers one by one.

Basically, we should try and find out the values of x^n + 1/x^n

When n = 2

To get the value of n = 2, we need to square the original equation

(x + 1/x) = 3

=> (x + 1/x)^2 = 3^2

=> x^2 + 1/x^2 + 2 = 9

=> x^2 + 1/x^2 = 7

When n = 3

To find the value, we need to multiply the equations where n = 1 and n = 2

(x + 1/x) * (x^2 + 1/x^2) = 3*7

=> x^3 + 1/x^3 + x + 1/x = 21

=> x^3 + 1/x^3 + 3 = 21

=> x^3 + 1/x^3 = 18

When n = 4

To find the value, we can square the equation we got at n = 2

(x^2 + 1/x^2)^2 = 7^2

=> x^4 + 1/x^4 + 2 = 49

=> x^4 + 1/x^4 = 47

When n = 5

To find the value, we can multiply the equations we got at n = 1 and n = 4

(x + 1/x) * (x^4 + 1/x^4) = 3*47

=> x^5 + 1/x^5 + x^3 + 1/x^3 = 141

=> x^5 + 1/x^5 + 18 = 141

=> x^5 + 1/x^5 = 123

So, the value we were trying to find out was 123

Alternatively, you can just multiply the values that you have got at n = 2 and 3 to get the answer.

(x^2 + 1/x^2) * (x^3 + 1/x^3) = 7*18

=> x^5 + 1/x^5 + x + 1/x = 126

=> x^5 + 1/x^5 +3 = 126

=> x^5 + 1/x^5 = 123

**If 5 letters are posted for 5 different addresses, how many ways are there for each of the letters to reach wrong addresses?**

Number of ways in which 'n' objects can be placed on 'n' positions in such a manner that none of them is correct is given by the Dearrangement formula.

Dearr(n) = n!(1/0! - 1/1! + 1/2! - 1/3!.... 1/n!)

In this question, we need to place 5 objects (letters) in 5 positions (addresses) such that none of them is correct. This can be done in

Dearr(5) = 5! (1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)

=> Dearr(5) = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)

=> Dearr(5) = 60 - 20 + 5 - 1 = 44

So, we can send the 5 letters, such that all are delivered at wrong adresses, in 44 different ways.

**An alloy contains copper, zinc and nickel in the ratio of 5 : 3 : 2. The quantity of nickel in kg that must be added to 100 kg of this alloy to have the new ratio 5 : 3 : 3 is?**

The total quantity of alloy before the addition of Nickel is 100 kg.

This 100 kg is composed of 50 kg Copper, 30 kg Zinc, and 20 kg Nickel

Now we want the new ratio to be 5:3:3 by adding some quantity of Nickel.

You can just look at it and say that the quantity of Nickel that should be added is 10 kg.

However, let us try and understand the method for complicated cases.

Let us say that 'x' kg of Nickel is added so that Nickel becomes 20 + x and the total quantity of the alloy becomes 100 + x

Now, Nickel is 3/(5 + 3 + 3) = 3/11 th of the total mixture.

=> (20 + x)/(100 + x) = 3/11

=> 220 + 11x = 300 + 3x

=> 8x = 80

=> x = 10

=> Quantity of Nickel that should be added is 10 kg.

**A container has 100 liters (mixture of milk and water) in the ratio of 3:2. When 40 liters of mixture is taken out and replaced with the same amount of water, what is the ratio of milk and water left in the container?**

Total solution in the mixture = 100 liters

Quantity of milk = (3/5) * 100 = 60 liters

Quantity of water = (2/5) * 100 = 40 liters

When 40 liters of mixture is taken out, 40% of everything is taken out

Quantity of milk = 60% of original = 0.6 * 60 = 36 liters

Quantity of water = 60% of original = 0.6 * 40 = 24 liters

Same quantity (40 liters) of water is added

Quantity of water = 24 + 40 = 64 liters

New Ratio of milk : water = 36 : 64 = 9 : 16

**Walking at 5 kmph I missed my train by 7 min . Walking at 6 kmph I reached the station 5 min early . How far is the station from the house?**

Let us say that the actual distance between the house and station is 'd' km and the ideal time to cover the distance is 't' hours.

When I walk at 5 kmph, I reach 7 minutes (or 7/60 hours) late

=> d/5 = t + 7/60

=> t = d/5 - 7/60

When I walks at 6 kmph, he reaches 5 minutes (or 1/12 hours) early

=> d/6 = t - 1/12

=> t = d/6 + 1/12

Using the above equations, we can say

d/5 - 7/60 = d/6 + 1/12

=> d/5 - d/6 = 7/60 + 1/12

=> d/30 = (7 + 5)/60

=> d = 30*(12/60) = 6 km

**The price of a gold nugget is directly proportional to the square of its weight. If a person breaks down the gold nugget in the ratio of 3:2:1 and gets a loss of Rs. 4620, what is the initial price of the gold nugget?**

We are given that the price of gold is directly proportional to square of its weight.

=> Price = k*weight^2

Let us say initially the weight of the nugget was 6 units

=> P(original) = k*6^2 = 36k

After breaking it down into three pieces, we get the weights as 3, 2, and 1

=> P(3) = k * 3^2 = 9k

=> P(2) = k * 2^2 = 4k

=> P(1) = k * 1^2 = k

=> P(new) = 9k + 4k + k = 14k

Loss = 36k - 14k = 4620 Rs.

=> 22k = 4620 Rs.

=> k = 210

=> Price (original) = 210 * 36 = 7560 Rs.

**How can I divide 25000 coins of Rs. 1 each into 15 buckets such that I can obtain any amount just by picking up buckets and without moving the coins?**

In questions like these the idea is to move in powers of 2 as long as required.

If the total number of coins was 15, we will first find out the power of 2 it is lesser than

=> 15 = 2^4 - 1

=> We will need 4 buckets with the distribution as 1, 2, 4, 8

If the total number of coins was 100, we will first find out the power of 2 it is lesser than

=> 100 = 2^7 - 28

=> We will need 7 buckets with the distribution as 1, 2, 4, 8, 16, 32, 37

Here the total number of coins is 25000

We know that the number of buckets is 15

The distribution will be

1, 2, 4, 8, .... 2^13, and 25000 - (2^14 - 1) = 8617

**How many ways are there for 6 men and 7 women to stand in line so that none of the women are next to each other?**

First, let us arrange the men. We can do that in 6! ways.

Let us say they are A, B, C, D, E, and F

The arrangement that we have made is:

BCFEAD

Now, we need to place women along with the men so that no two women are together. They can go on the positions indicated by blanks

_ B _ C _ F _ E _ A _ D _

7 women can be arranged on 7 positions in 7! ways.

Total ways = 6!*7!

**Which is greater, 70^71 or 71^70?**

70^71 (the one with the higher power)

As a rule, if you are given two natural numbers a > b > 1

then the one with the higher power will be bigger

=> a^b < b^a

The exceptions to this are

a) 3^2 > 2^3

b) 2^4 = 4^2

4 digit let abcd

base 6 so 216a+36b+6c+d

where each digit is less than 6

now in base 10 , 1000a+100b+10c+d

1000a+100b+10c+d=4*(216a+36b+6c+d)

so 136a=44b+14c+3d

when a=1 then b=2,c=3,d=2

when b=2

272=44b+14c+3d

but max value of b=5 so 14c+3d=52

which is not possible for b and c 10 which is not in option so no need to check further!

(x+1)(x+9)+8=0

x^2+10x+17=0

so a+b=-10

and ab=17

now roots of x^2+a+b)x+ab-8=0

x^2-10x+9=0

so x=1,9

positive roots

so D>=0

q^2-4p^2>=0

(q-2p)(q+2p)>=0

now sum of roots and product of roots wil be positive

so -q/p so p > 0 , q < 0

so q-2p < 0 is true

total people =42

so number of handshake =42c2

now subtract those cases when handshakes between 4 members of same family

so 10 * 4c2=60

now one handshake between Chris and his wife should be subtracted

so OA=42c2-60-1=800

let 1 men in start so total work=M

now in first 8 hours

8(1+1/2+1/4+1/8+1/16)=M

8+4+2+1+1/2=15+(1/2)

let log_5a=k

so 6^k-16^(1/k)=32

6^k * -2^(4/k)=32

so k=2

so so log_5a=2

so a=25

Quick Approach- 72=2^3*3^2

so 3a and 2b will be present in our ans

only 4th choice have that

so OA=4

Quick Approach = Product of roots =@+1/@ minimum value of which is 2

now one roots is 2 so other root will be greater than 1

so only possible option 1

n * (n+1)/2 -x=756

approx value of n= 39

so 39 * 40/2=780

so missed value =780-756=24

f(x)=ax^2+bx+c

g(x)=dx^2+ex+m

now f(x)-g(x)=x^2(a-d)+x(b-e)+c-m

let a-d=M

b-e=N

c-m=T

so f(x)-g(x)= Mx^2+Nx+T

so M+N+T=1

4M+2N+T=2

9M+3N+T=5

now find M,N,T= M=1 N=-2 and T=2

so F(x)-g(x)=x^2-2x+2

so f(4)-g(4)=10

7429 = 17 x 19 x 23

12673 = 19 x 23 x 29

∴ four prime numbers are 17 x 19 x 23 x 29.

∴ The sum is = 17 + 29 = 46.

**Hypoteneus of a Right angle triangle is 26. What could be its perimeter?(a) 51(b) 56(c) 45(d) 60**

We know 5 - 12 - 13 is a Pythagorean triplet

∴ 10 – 24 – 26.

Will also be a Pythagorean triplet. ∴ Perimeter is 60. Hence (d).

**P, 2P, q + P, 2P – 2q – 6 is an A.P. then what is its 99th term?(a)–99(b) –98(c) 99(d) 100**

A.P. So differences will be equal

2P – P = q + P – 2P = 2P – 2q – 6 – q – P

P = q – P = P – 3q – 6

∴ q = 2P and 2P – 6 = 4q

On solving, we get P = – 1 and q = – 2

∴ A.P. is –1, –2, –3………….

∴ T99 = a + 98d = – 1 + 98 x –1 = – 99

**What is the range of x satisfying x^2 + 2x – 15 > 0 and x^2-3x-4 < 0(a)-5 < x < 4(b) -1 < x < 4(c) -5 < x < 3(d) 3 < x < 4**

x^2 + 2x – 15 > 0 or (x + 5)(x - 3) > 0

∴ x < - 5 or x > 3

x^ 2 – 3x – 4 < 0

(x – 4) (x + 1) < 0

⇒ -1 < x < 4

∴ common solution is 3 < x < 4

**There are 100 people in a room. Some or all of them shake hands. Probability that there will be two people in the room who have shaken hands with the same number of people?**

No of people = 100

Possible no of handshakes = 0 to 99 => 100

So it seems all 100 ppl can have different no of handshakes, but notice 0 and 99 both won't b possible. If out of 100 ppl, there is one person who shakes hand with 0 ppl(none) then can there be any person among them who will shake hand with 99 ppl(all except himself? Not possible since he won't b shaking hand with himself and that guy who didn't shake hand)

Hence it means 0 and 99 both won't coexist, instead any one of those two will come

Hence now we got 100 people and 99 different no of handshakes. So even if 99 people got those 99 different number, 100th person will get no of handshakes same as one of those 99, and hence there will be atlst two people with same no of handshakes no matter what. Hence it's a certain event and probability is 1.

**If x + (1/x) = 1, find x^127 + (1/x)^127**

(x + 1/x) = 1

x^2 + 1 = x

x^2 + 1 - x = 0

Now try to observe

Where have you seen such terms

a^3 + b^3 = (a+b) (a^2 + b^2 - ab)

So here we have to multiply by

(x+1) on both sides

(x+1) (x^2 -x +1 ) = 0

=> (x^3 +1) = 0

x^3 = -1

Now x^127 => (x^3)^42 * x

=> (-1)^42 * x

=> x

So x + (1/x) which is equal to 1

**A natural number equals 75 times the average of its digits. How many three digit numbers satisfy this condition**

If average is 1

Then number is 75 (which can't be possible )

If average is 2

Then number is 150 ( as yes it's average is 2 also )

If average is 3

Number is 75*3 => 225 ( yes average of the digits is also 3)

Average is 4

Number is 300 ( but average of digits is not 4)

Average is 5

Number is 375 ( average of digits is 5 ,so possibile )

Average is 6

75 * 6 => 450 ( average of digits is not 6)

Average is 7

75 * 7 => 575 ( average is not 7)

Average is 8

75 × 8 => 600 ( again not possible )

Average is 9

75 * 9 => 675 ( but average can't be 9)

So.here Basically two concepts

- Table of 75
- Average of digits can't be more than 9

As it is a three digit number

If you take average as 10 then sum.of digits would be more than 27 which can't be possible

So three numbers are possible - 150, 225 , 375

**The regional passport office has a waiting list of 6665 applicants for passport. The list shows that there at least 6 males between any two females. The largest possible number of females in the waiting list is**

First person is female

Then eight person is female

So one combination of

1F MMMMMM

Repeats

6665 /7 => 952(1/7)

So 952 combinations will have 952 female

And one more female in the last

So 952 + 1 = 953

**In an arithmetic progression, the sum of the first N terms is T and the sum of the first 2N terms is 6T. If the sum of the first 3N terms is kT, then find k**

Take N = 1

Sum of first term is T

Sum of first two terms is 6T

I.e first term = T

Second term = 5T

So third term has to be 9T

Sum of first three terms = T + 5T + 9T

=> 15T

So k = 15

**A clock was correct at 2 p.m, but then it began to lose 30 minutes each hour. It now shows 6 pm, but it stopped 3 hours ago. What is the correct time now?**

The clock loses 30 minutes per hour.

i.e 30 minutes of this faulty clock = 60 minutes of the correct clock

From 2 p.m to 6 p.m ,

total number of hours = 4 hours

4 hours of this faulty clock => 4 x 60/30= 8 hours of original clock

So, The correct time when the clock show 6 p.m = 6 p.m + 4 = 10 p.m

But The clock stopped 3 hours ago ,

So present time is 10 p.m + 3 hours = 1 a.m

Let us say that A and B started from point P and point Q respectively and met at point R in the middle after time 't'.

Speed of A is given to us as 40 kmph.

Let us assume the speed of B to be 's'.

If we consider the distance PR, it was covered by A in 't' hours and B in 25 hours.

=> PR = 40 * t = s * 25

If we consider the distance RQ, it was covered by B in 't' hours and A in '16' hours.

=> RQ = s * t = 40 * 16

Let us divide the two equations. We will get

40t/st = 25s/640

=> 40/s = 5s/128

=> s^2 = 40 * 128/5 = 8 * 128 = 1024

=> Speed of B = s = 32 kmph

**What degree occurs at a time 3:30 in the clock?**

At 3:30, the hour hand would be exactly between 3 and 4

=> 15 deg from both 3 and 4

At 3:30, the minute hand would be exactly at 6

The gap between the hour hand and the minute hand will be

30 deg (between 5 and 6) + 30 deg (between 4 and 5) + 15 deg (between hour hand and 4) = 75 degrees

**You were a quarter of my age when I was twice your present age" said A to B. In five years from now, our ages will add up to 45. How old was I when you were born?**

Let us say my current age is a and your current age is b

When I was 2b when you were 2b/4 or 0.5b

=> a - 2b = b - 0.5b

=> a = 2.5b

In 5 years, I will be a + 5 and you be b + 5

Sum of our ages will be 45

=> a + 5 + b + 5 = 45

=> a + b = 35

Substitute the value of b

=> 2.5b + b = 35

=> 3.5b = 35

=> b = 10

=> a = 25

So, when you were born i.e. 10 years ago; I was 25 - 10 = 15 years old

**What is the largest three-digit number that when divided by 6 leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?**

Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of

LCM(x,y,z)*n + constant

The key in these questions is finding out the value of 'constant'. If all of them leave the same remainder 'r', constant = r. It can also be looked at as the smallest number satisfying the given property.

In this question, we are given

Remainder from 6 is 5

Remainder from 5 is 3

So, the number N = LCM(6,5)*n + constant = 30n + constant

To figure out the constant, look at the numbers which give a remainder of 5 from 6.

They are 5, 11, 17, 23, 29....

Among these, find the one which leaves a remainder of 3 from 5. It is 23.

So, our number N should be of the format of 30n + 23

Biggest three digit number will occur when n is 32 = 30*32 + 23 = 983

**How many prime numbers less than 75 will leave the odd reminder when divided by 5?**

The number should leave a remainder of 1 or 3 from 5

=> It is of the form of 5k + 1 or 5k + 3

=> The number ends in 1 or 3

I will just count out such numbers under 75, which are 16

1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71 and 73

From these, I would remove the non prime ones which are 1, 21, 33, 51 and 63

So, I will be left with 11 numbers which fit the bill.

3, 11, 13, 23, 31, 41, 43, 53, 61, 71 and 73

**What is the unit digit of LCM of (13^501 – 1) and (13^501 + 1)?**

13^501 – 1 and 13^501 + 1 are two consecutive even numbers.

One of them will be of the format 4k and the other one will be of the format 4k + 2.

They will only have 2 as a common factor.

=> HCF (13^501 – 1, 13^501 + 1) = 2

We also know that HCF * LCM = Product of two numbers

=> LCM = (13^501 – 1)(13^501 + 1)/2 = (13^1002 – 1)/2

Now, let’s try and find out the last digit of the LCM

Last digit of 13^1002

= Last digit of 3^1002

= Last digit of 3^(4k + 2)

= 9

Last digit of 13^1002 – 1 = 9 – 1 = 8

Last digit of (13^1002 – 1)/2 = 8/2 = 4

**What is the sum of all the integers less than 100 which leave a remainder 1 when divided by 3 and a remainder of 2 when divided by 4?**

Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of LCM(x,y,z) * n + constant

The key in these questions is finding out the value of ‘constant’. If all of them leave the same remainder ‘r’, constant = r. It can also be looked at as the smallest number satisfying the given property.

In this question, we are given

Remainder from 3 is 1

Remainder from 4 is 2

If you look at the negative remainders

Remainder from 3 is -2

Remainder from 4 is -2

So, the number N = LCM(3,4) * n – 2 = 12n – 2

So, any number which is of the format of 12n – 2 will satisfy the given conditions.

Positive Integers less than 100 which satisfy the above condition are

10, 22, 34… 94

Sum of these integers

= (No. of terms/2) * (First term + Last Term)

= (8/2) * (10 + 94) = 416

**Find the number of positive integral solutions of |x| + |y| = 10.**

Let |x| = a and |y| = b. First find the positive integral solution of a + b = 10.

Number of positive integral solutions= (10-1)C(2-1) = 9.

Now for each solution (a1, b1), the values of (x,y)= (a1, b1), (-a1, b1), (a1, -b1) and (-a1, -b1).

So total number of positive integral solutions= 4 × 9 = 36.

**Find the total number of integral solutions of IxI + IyI + IzI = 15.**

First, let a = |x|, b = |y|, c = |z|.

Now, we need to find the number of positive integral solutions of a + b + c = 15. The number of solutions are 14C2 = 91. Now for each value of a,b and c we will have two values of x, y and z each. Therefore, the total number of solutions = 91 x 2 x 2 x 2= 728.

Now let one of the variables be equal to 0. For example, let x = 0 and |y| and |z| be at least equal to 1. Therefore, we need the positive integral solution of b + c = 15, where b = |y| and c = |z|. The number of solutions is 14C1 = 14. Each of these solutions will give two values of y and z and there are 3 ways in which we can keep one of the variables equal to 0. Therefore, total number of ways are 14 x 2 x 2 x 3 = 168.

Now let two of the variables be equal to 0. In this case, the total number of solutions is equal to 6.

Therefore, the total number of integral solutions = 728 + 168 + 6 = 902.

**Let g(x) = max (5 − x, x + 2). The smallest possible value of g(x) isa) 4.0b) 4.5c) 1.5d) None of these.**

To solve such kind of questions, in most cases, all you need to do is to equate the two values inside the function

=> 5 – x = x + 2

=> x = 3/2 = 1.5

Please note that 1.5 is not the answer to the question.

To find out the answer to the question, we need to find out the value of g(1.5)

When we put x = 1.5, we get g(x) = max (5 – 1.5, 1.5 + 2) = max (3.5, 3.5) = 3.5

So, our answer is 3.5 Option D – None of these

I would strongly recommend that you watch the below video for better understanding of the solution of the question

We need to find out the units digits of

2! + 4! + 6! + .... 98!

Let us look at the units digit for all of them

2! = 1 * 2 = 2; unit's digit is 2

4! = 1 * 2 * 3 * 4 = 24; unit's digit is 4

6! = 1 * 2 * 3 * 4 * 5 * 6 = 720; unit's digit is 0

8! = multiple of 6!, divisible by 10; unit's digit is 0

10! = multiple of 6!, divisible by 10; unit's digit is 0

12! = multiple of 6!, divisible by 10; unit's digit is 0

.

.

.

98! = multiple of 6!, divisible by 10; unit's digit is 0

Unit's digit of 2! + 4! + 6! + .... 98!

= Sum of unit's digits of individual terms

= 2 + 4 + 0 + 0 + 0 ..... 0

= 6

**What is the remainder when 97! is divided by 101?**

Wilson's Theorem says For a prime number 'p'

Rem [ (p-1)! / p] = p-1

This can be extended to say,

Rem [ (p-2)! / p] = 1

Let us use that here. We need to find out Rem [97! / 101] = r

We know from the above theorem,

Rem [99! / 101] = 1

=> Rem [99 * 98 * 97! / 101] = 1

=> Rem [ (-2)*(-3)*r / 101] = 1

=> Rem [6r / 101] = 1

=> 6r = 101k + 1

We need to think of a value of k, such that 101k + 1 is divisible by 6.

If we put, k = 1, we get 101 + 1 = 102, which is divisible by 6.

=> 6r = 102

=> r = 17

**If a person starts writing all 4 digit numbers, how many times has he written the digit 2?**

Let us call the 4 digit number as abcd

When a = 2, b / c / d can be filled in 10 ways each.

So, 2 will appear in place of 'a' 10 * 10 * 10 = 1000 times

When b = 2, a can be filled in 9 ways whereas c / d can be filled in 10 ways each.

So, 2 will appear in place of 'b' 9 * 10 * 10 = 900 times

Similarly, 2 will appear in place of 'c' 900 times and in place of 'd' 900 times.

Total number of times digit '2' will appear = 1000 + 900 + 900 + 900 = 3700 times

**What is the number of digits in 2^2009?**

Using Logarithms is probably the fastest and the best way to get to the answer.

If Log(x) = a.bcde, number of digits in x is a + 1

Log (2^2009) = 2009 Log(2) = 2009*0.3010 = 604.709

=> 2^2009 has 604 + 1 = 605 digits

**How many digits are there in (2ABC)^4 where 2ABC is a 4 digit number?**

We need to find out the number of digits in (2ABC)^4

The smallest such number will be 2000^4 = 16*10^12

This will have 14 digits.

The biggest such number will be 2999^4. While that is hard to find out, we can find out 3000^4 easily.

3000^4 = 81*10^12. It has 14 digits

=> 2999^4 will also have 14 digits.

As the smallest and the biggest number of the format (2ABC)^4 has 14 digits, we can say that there are 14 digits in (2ABC)^4

**The product of two numbers is 7168 and their highest common factor is 16. How many pairs of numbers are possible such that the above condition is satisfied?**

Let us say that the numbers are x and y.

Both of them are multiples of 16 as their HCF is 16

So, we can say that they are 16a and 16b.

Also, HCF (a,b) = 1. They have no factor in common otherwise the HCF of 16a and 16b won't be 16. In other words, a and b are coprime to each other.

x * y = 7168

=> 16a * 16b = 7168

=> a * b = 28

Now, we need to split 28 into coprime factors

28 = 1 * 28 or 4 * 7

=> (x, y) = (16, 448) or (64, 112)

So, there are two pairs that satisfy the given conditions.

Note: I have considered only positive numbers for the calculation shown above. If you wish to consider negative numbers as well, the answer would double

**31 consecutive leaves were torn off. which of following could be the sum of 62 page numbers on these leaves?a) 1955b) 2201c) 2079d) none**

From the leaves that are torn off, the first page number should be odd.

Let us assume it is 2a + 1.

Since there are total of 62 pages which are torn off, the last page number will be 2a + 62

Sum of 'n' terms in an Arithmetic Progression = n/2[First term + Last Term]

=> Sum of page numbers = 62/2 [2a + 1 + 2a + 62]

=> Sum of page numbers = 31(4a + 63)

Now, let's look at the options

Option (a) 1955 is ruled out because it is not divisible by 31

Option (b) 2201

=> 31(4a + 63) = 2201

=> 4a + 63 = 2201/31 = 71

=> 4a = 71 - 63 = 8

=> a = 2

=> 2201 is a valid value

Option (c) 2079 is ruled out because it is not divisible by 31

**How many times is a key of a typewriter pressed in order to type the first 299 natural numbers by pressing the space bar once between any two successive natural numbers?**

Number of 1 digit natural numbers {1, 2, 3 ... 9}= 9

Keystrokes required to type them = 9*1 = 9

Number of 2 digit natural numbers {10, 11, 12 ... 99} = 90

Keystrokes required to type them = 90*2 = 180

Number of 3 digit natural numbers {100, 101, 102 ... 299} = 200

Keystrokes required to type them = 200*3 = 600

Number of spaces between the numbers = 298

Keystrokes required to type them = 298*1 = 298

Total keystrokes required = 9 + 180 + 600 + 298 = 1087

**Which number is greater, 80^50 or 60^54?**

To solve this, it would help if you know Log(2) = 0.3010 and Log(3) = 0.4771

If we have to compare a and b, we can compare Log(a) and Log(b).

Log (60^54) = 54 Log (60) = 54 [Log (2) + Log(3) + Log(10)] = 54[0.3010 + 0.4771 + 1] = 54 * 1.771 = 95.634

Log (80^50) = 50 Log(80) = 50 [Log(8) + Log(10)] = 50[3 Log(2) + 1] = 50[3*0.3010 + 1] = 50 * 1.9030 = 95.15

So, Log (60^54) > Log (80^50) => 60^54 > 80^50

**How many 6 digit numbers can be made using digits 1, 2, 3, 4, 5, and 6 without repetition such that the hundred digit is greater than the ten digit, and the ten is greater than the one digit?**

Let us assume that the number that we have is abcdef

The condition given is d > e > f

Let us first select three digits for a, b, and c.

From the given 6 digits (1,2,3,4,5,6), 3 digits for a,b, and c can be selected in 6C3 = 6!/3!3! = 20 ways.

These three selected digits can be arranged on three positions of a, b, and c in 3! = 6 ways

The remaining 3 digits are automatically selected for d, e, and f.

So, we have 1 way of selecting them.

The biggest digit will be allocated to 'd'.

The second biggest digit will be allocated to 'e'.

The third biggest digit will be allocated to 'f'

So, we have 1 way of arranging them on the three positions of d, e, and f.

Total ways = 20 * 6 * 1 * 1 = 120

]]>(1) 25 : 64

(2) 25 : 84

(3) 45 : 112

(4) 35 : 92

Let the number of students in Batch A in January be 2x . Let number of students in Batch A in February be 5y.

Total numbers of students in Batch A and Batch B combined in January and February are 5x and 13y respectively. Therefore 13y : 5x = 26:5 or y : x = 2:1 or y = 2x.

From here, we can compile the following table:

**If 5x + 2y + z = 81, where x, y and z are distinct positive integers, then find the difference between the maximum and the minimum possible value of x + y + z.(1) 44(2) 48(3) 55(4) 60(5) 64**

Minimum possible value of x + y + z will be when we maximize the value of ‘x’. Maximum possible value of x will be 15 and since x, y and z are distinct positive integers, y = 1 and z = 4. So,minimum possible value of x+y+z = 15+1+ 4 = 20.

Maximum possible value of x + y + z will be when the value of z is maximized. Maximum possible value of z will be when y = 2 and x = 1, i.e. z = 72.

Required difference is 75 – 20 = 55.

**How many two-digit numbers have exactly four factors?(1) 30(2) 31(3) 32(4) 28**

Since, the two-digit number has exactly four factors, therefore the number has to be a product of two prime numbers or a perfect cube.

Case I:

When one of the numbers in the product is 2.

The other number in the product can be 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

There are 13 such possible products.

Case II:

When one of the numbers in the product is 3.

The other number in the product can be 5, 7, 11, 13, 17, 19, 23, 29 and 31.

There are 9 such possible products.

Case III:

When one of the numbers in the product is 5.

The other number in the product can be 7, 11, 13, 17 and 19. There are 5 such possible products.

Case IV:

When one of the numbers in the product is 7. The other number in the product can be 11 and 13. There are 2 such possible products.

Case V:

When the two-digit number is a perfect cube. There is only one such number, i.e. 27.

Therefore, in all there are 30 two-digit numbers that have exactly 4 factors.

**There were twelve teams participating in a football tournament where each team played exactly one match with each of the other eleven teams. Each match was played between two teams. The winner of each match was awarded 3 points while no point was awarded to the team that lost the match. In case of a tie, both teams got 1 point each. If the aggregate number of points awarded to all the teams at the end of the tournament was 189, then how many matches ended in a tie?(1) 7(2) 8(3) 9(4) 6(5) 4**

Total number of matches played in the football tournament = 11 + 10 ...+ 2 + 1 = 66.

Maximum possible number of points awarded to all the teams is 66 × 3 = 198. This is possible if no match ended in a tie. But the total points awarded to all the teams is 189.

So the number of matches that ended in a tie = 198 – 189 = 9. As a tied match generates only 2 points, which is one less than number of points generated by a win/loss match.

**There are two numbers A and B. A can be expressed as a product of 13 and a two-digit prime number and B can be expressed as a product of 17 and a two-digit prime number. If the unit’s digit of the product of A and B is 7, then how many distinct products of A and B are possible?(1) 48(2) 55(3) 80(4) 110(5) 120**

Assume the numbers to be 13×N and 17×M, where N and M are two-digit prime numbers.

There are two possible ways in which 7 can be the unit’s digit of the product i.e. 1 × 7 or 3 × 9.

Case I: If units digit of N is 3, then units digit of M will be 9. Then, N = 13, 23, 43, 53, 73 or 83 and M = 19, 29, 59, 79, 89

So, number of distinct products = 6 × 5 = 30

Case II: If units digit of N is 1, then units digit of M will be 7 Then N = 11, 31, 41, 61 or 71 and M = 17, 37, 47, 67 or 97

So, number of distinct products = 5 × 5 = 25

Case III: If unit digits of N is 7, then units digit of M will be 1. Then N = 17, 37, 47, 67 or 97 and M = 11, 31, 41, 61 or 71

So number of distinct products = 5 × 5 = 25

Case IV: If unit digit of N is 9, then unit digit of M will be 3. Then N=19,29,59,79 or 89 and M =13,23,43,53,73 or 83

So number of distinct products = 5 × 6 = 30

Here case III and case IV will give the same products as case II and case I respectively.

∴ Total number of distinct products = 55.

**A and B play a dice game using two dice viz. ‘X’ and ‘Y’. Die ‘X’ has 1,2,3,4,5 and 7 printed on its six faces whereas die ‘Y’ has 2, 3, 4, 5, 6 and 8 printed on its six faces. There is only one number printed on every face of the two dice. In turns each of A and B rolls both the dice simultaneously and records the product of the two numbers appearing on the top of the two dice, as their respective scores. If the sum of the scores of players A and B is an even number in a round then how many distinct scores A could have in that round?(1) 32(2) 18(3) 22(4) 24**

Let X1 and Y1 be the sets having all the numbers printed on

die X and die Y respectively. X1 ={1,2,3,4,5,7}

Y1 = {2, 3, 4, 5, 6, 8}

For the sum of scores of players A and B in a particular round to be even, the individual scores of both A and B should either be odd or even.

Case I: A and B both have scores that are odd numbers. Each of A and B could have any of the following 7 scores: 3, 5, 9, 15, 21, 25 and 35.

Case II: A and B both have scores that are even numbers. Each of A and B could have any of the following 17 scores 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 28, 30, 32, 40, 42 and 56.

Hence, each of A and B could have 17 + 7 = 24 distinct score in that round.

**A person receives payment P(n) on day ‘n’. If P(n) = 2 × P(n – 1) and the person receives Rs.12 on day two, then what is the total payment received by that person for 10 days starting from the day one?**

This is nothing but a geometric progression whose common ratio is 2 and P(2) = 12.

∴P(1) = 6

⇒ P(1) + P(2) + ... + P(10) = 6 + 12 + ... till 10 terms

Ans - 6138

**There are three natural numbers X, Y, and Z such that the LCM of (X, 120) is 1320, LCM of (Y, 120) is 1680 and LCM of (Z, 120) is 1800. Which of the following statements is true?(1) X, Y, and Z all three can be perfect squares.(2) Only Y and Z can be perfect squares.(3) Only Y can be both a perfect square and a perfect cube.(4) Z is definitely a perfect square.(5) Z can be a perfect square.**

Factorizing the numbers:

120=23 ×31×51 1320=120×11= 2^3×3^1× 5^1×11

⇒ X definitely has to be a multiple of 11 but it cannot be a perfect square as X cannot contain (112)

1680 = 120 × 2 × 7 = 2^4 × 3^1 × 5^1 × 7^1

⇒ Y should be multiple of 2^4 × 7^1 but it cannot be a perfect square as Y cannot contain 72.

1800=120×3×5= 2^3×3^2×5^2.

Since the LCM of 120 and Z is 1800, therefore Z can be a perfect square. One of the possible values of Z can be 3^2 × 5^2. But it is not necessary that Z has to be perfect square as it can be 2×3^2 ×5^2 also.

Hence, option (5) is the correct choice.

**A trader used to make 5% profit on an item selling at usual marked price. One day, he trebled the marked price of the item and finally offered a discount of 30%. Find the percentage profit he made on the item that day.(1) 120.5%(2) 100%(3) 99.5%(4) 94.5%(5) None of these**

Let the earlier cost price of the item = Rs.100 ⇒ Earlier marked price = Rs.105.

On that day, 30% discount is offered on Rs. 3 × 105 = Rs.315 Thus, new selling price = Rs.220.50

New Profit percentage = 120.50.

**If ‘a’ is an odd number and ‘b’ is an even number, then what is the total number of solutions of the equation ab = 2a + b + 598?a) 16b) 12c) 5d) 6e) 8**

The given equation is ab + 2 = 2a + b + 600.

⇒ ab – 2a – b + 2 = 600 ⇒ (a – 1)(b – 2) = 600

Now, it is given that ‘a’ is an odd number and ‘b’ is an even number which implies that both (a – 1) and (b – 2) are even numbers.

Therefore, the possible pairs of values of (a – 1) and (b – 2) that satisfy the given equation are (2, 300); (4, 150); (6, 100); (10, 60) (12, 50). (20, 30); (30, 20); (50, 12); (60, 10); (100, 6); (150, 4); and (300, 2).

Therefore, there are 12 solutions for the given equation. Hence, option (2) is the correct choice.

**The sum of the coefficients of the polynomial (x – 1)^7 (x – 2)^2 (x – 4) is(a) 0(b) 16(c) –16(d) None of these**

Putting x = 1 in the expansion (1+x)^n = nC0 + nC1 x + nC2 x^2 + ... + nCx x^n, we get,

2^n = nC0 + nC1 x + nC2 + ... + nCn.

We kept x = 1, and got the desired result i.e. ∑nr = 0 Cr = 2^n

5 letters can go in 5 envelopes in 5! or 120 ways.

Only one of these 120 ways is correct and in the other 119 ways, at least two letters are in the wrong envelopes.

=> a = 119

All letters in wrong envelopes is Dearrangement of 5 letters in 5 envelopes.

=> b = Dearr (5) = 5! (1/2 - 1/6 + 1/24 - 1/120) = 60 - 20 + 5 - 1 = 44

Value of a - b = 119 - 44 = 75

**What is the minimum value of |x - 1| + |x - 2| + |x - 3| + ... + |x - 9| + |x - 10|?**

Put x as the middle value to get the answer.

Here you could put x as 5 or 6 or any value in between, you will get the answer.

If I put x = 6, I get 5 + 4 + 3 + 2 + 1 + 0 + 1 + 2 + 3 + 4 = 25

**If p(x) = ax^2 + bx + c is a quadratic eqation such that p(1) = p(2) and p(4) = 0 then find:a) m is not = 4 that p(m) = 0b) p(5)c) p(0)/p(3)**

p(1) = p(2)

=> a + b + c = 4a + 2b + c

=> b = -3a

p(4) = 0

=> 16a + 4b + c = 0

=> 16a - 12a + c = 0

=> c = -4a

p(x) = ax^2 + bx + c

=> p(x) = a(x^2 -3x - 4) = a(x-4)(x+1)

a) The other root is m, which is -1

b) p(5) = 6a. Since it depends on 'a', a unique value can't be determined.

c) p(0)/p(3) = a(-4)(1)/a(-1)(4) = 1

**If a and b are positive numbers such that (a^2 + b^2) is divisible by 7, then the largest number by which (a^2 + b^2) is divisible isa) 7b) 49c) 21d) 14**

Remainder of n from 7 can be 0, 1, 2, 3, 4, 5, or 6

Remainder of n^2 from 7 will be 0, 1, 4, 2, 2, 4, 1 (These correspond to the above values)

Remainder of a^2 + b^2 from 7 can be 0 if and only if both a and b are divisible by 7 (Try adding two numbers from the previous list to get a 0 or 7. The only possibility is 0 + 0)

So, a^2 will be divisible by 49 and b^2 will be divisible by 49.

So, a^2 + b^2 will be divisible by 49.

**Set S is formed by selecting some of the numbers from the first 110 natural numbers such that the HCF of any two numbers in the set is the same. If every pair of numbers of set S has to be relatively prime and set S has the maximum number of elements possible, then in how many ways can the set S be selected ?**

There are 25 primes till 100 and 29 primes till 110. Set S will have 30 elements = 1 and 29 primes or their powers. From 2, 4, 8, 16, 32, 64; any one can be included in set S. We have 6 choices. From 3, 9, 27, 81; any one can be included in set S. We have 4 choices. From 5, 25; any one can be included in set S. We have 2 choices. From 7, 49; any one can be included in set S. We have 2 choices. Total number of choices = 6 * 4 * 2 * 2 = 96.

**A/3 is an integer but A/6 is not. B/5 is an integer but B/10 is not. Which of the following may not be an integer?a) (5A-3B)/15b) (5A-3B)/30c) (5A-B)/10d) (5A-B)/20**

A/3 is an integer but A/6 is not

=> A is an odd multiple of 3

=> A = 3(2n + 1) = 6n + 3 {Here 'n' is an integer}

B/5 is an integer but B/10 is not

=> B is an odd multiple of 5

=> B = 5(2m + 1) = 10m + 5 {Here 'm' is an integer}

Now, using the options, we need to look at 5A - 3B and at 5A - B

Case 1:

5A - 3B = 30n + 15 - 30m - 15 = 30(n - m)

=> 5A - 3B is a multiple of 30. Option (a) and (b) are ruled out.

Case 2:

5A - B = 30n + 15 - 10m - 5 = 30n - 10m + 10 = 10 (3n - m + 1)

=> 5A - B is a multiple of 10. Option (c) is ruled out

Since, options (a), (b), and (c) are ruled out, our answer is Option D

**The sum of lengths of hypotenuse and one of the perpendicular sides of a right angled triangle is L. The area of triangle s maximum when angle between those two sides isa) 45b) 22.5c) 60d) None**

Consider the side as a and hypotenuse as (L - a)

Calculate third side and find out area.

You will find out that it depends on your ability to maximize a * a * (L-2a)

Its sum is constant at L.

=> Product will be maximum when all the three terms are equal.

=> Hypotenuse = Twice of side

=> Angle between them is 60 degrees

**The positive integer N and N^2 both end in the same sequence of four digits 'abcd' when written in base 10 where digit 'a' is not zero. find three digit number abc.a) 984b) 937,c) 987d) 321**

N^2 - N will end in 000

=> N(N-1) will be divisible by 625

=> abc5 or abc0 is divisible by 625

=> From the options only 9375 is divisible by 625

=> So, 937 is the answer

**If p, q and r are in A.P & x, y and z are in G.P, Then x^(q-r) * y^(r-p) * z^(p-q) = ?**

Using options and assuming values works best in such cases.

Take, p = 1, q = 2, r = 3 and x = 1, y = 2, z = 4

=> x^(q-r) * y^(r-p) * z^(p-q) = 1^(-1) * 2^2 * 4^(-1) = 1

To solve it properly, take the AP values as (a-d), a, (a+d) and the GP values as b/r, b, br

=> x^(q-r) * y^(r-p) * z^(p-q)

= (b/r)^(-d) * b^(2d) * (br)^(-d)

= b^(-d + 2d - d) * r^(d + 0 - d)

= b^0 * r^0

= 1

**LCM of 2 numbers is 315. Which of the following cannot be the sum of two numbers, given that their HCF is prime number > 3a) 322b) 98c) 320d) 126**

First of all, find out the factors of 315

315 = 3^2 * 5 * 7

=> HCF can be 5 or 7

Case 1: HCF is 5

=> The two numbers are 5a and 5b, where a & b are coprime to each other

=> 5 * a * b = 315

=> a * b = 63

=> a & b are (1,63) or (7,9)

=> Sum of the numbers = 5a + 5b = 5 * 64 or 5 * 16 = 320 or 80.

=> This eliminates option C.

Case 2: HCF is 7

=> The two numbers are 7a and 7b, where a & b are coprime to each other

=> 7 * a * b = 315

=> a * b = 45

=> a & b are (1,45) or (5,9)

=> Sum of the numbers = 7a + 7b = 7 * 46 or 7 * 14 = 322 or 98

=> This eliminates options A & B

=> Answer is Option D

First time when Dudul n gudul meet , together they cover then entire distance between Cuttack n Bhubaneswar.. 2nd time when they meet they covered thrice the distance. So I can say 3 * 10000 - 300 = 29.7 kms is the distance between 2 cities

Else s1/s2 = 10000/L - 10000

And in second meet

S1/S2 = (L + 300)/2L - 300

So equating 2 equations we can find L.

**How many whole number solutions exist for the equation x + y + z = 48 such that x< y < z?a) 1225b) 192c) 200d) 872**

x + y + z = 48

total whole number solutions = n+r-1 C r-1 = 48+3-1 C 3-1 = 50C2

But 50C2 includes "ALL POSSIBLE WAYS IN WHICH 48 CAN BE DIVIDED INTO 3 GROUPS "

So broadly there are 3 type of cases in which 48 can be divided into 3 groups ..

CASE - 1 --> WHEN X , Y , Z are all equal i.e X = Y = Z = 16 .. so it can be done in only 1 way

CASE - 2 ---> WHEN 2 GROUPS ARE SAME AND 1 GROUP is different .. so cases like

0 0 48

1 1 46

2 2 44

.......and so on till

24 24 0

so total 0 to 24 = (24 - 0 ) + 1 = 25 cases .. .. Just for your easy understanding i ve counted else you can directly do 2a + b = 48 so "a" can take values from 0 till 24 so 25 cases .

And now each of these 25 cases would have been arranged in 3!/2! ways ..why ?? because " 2 are same and 1 is different" .. so total 25 * 3 = 75 cases

BUT THESE 75 SOLUTIONS INCLUDE THAT "16 , 16 , 16" CASE ASWELL..so exclude that .. So finally 75 - 3 = 72 CASES ..

CASE - 3 ---> WHEN ALL THE GROUPS ARE DIFFERENT

This includes cases like 3 , 24 , 21 ; i.e where X , Y ,Z are different ..

Now each these "ALL DIFFERENT ELEMENT " sets like ( 3 , 24 , 21 ) can be arranged among themselves in 3! i.e 6 ways ..But we need only ONE TYPE OUT OF THOSE 6 WAYs ..In simple words those 6 ways will include cases like

X < Y > Z ; X < Y < Z ; X > Y < Z ; X < Z < Y ; X > Z > Y ;X > Y > Z

BUT WE NEED ONLY "X > Y > Z" type cases so 1 out of 6 i.e 1/6 OF (ALL DIFFERENT CASES )

So X > Y > Z CASES = (TOTAL - CASE 2 - CASE 1 )* 1/6

= (50c2 - 72- 1)/6 = [192]

**What is the probability of selecting a number is selected [100,999] such that sum of digits of number is 14.**

a+b+c = 14

1st digit cannot be 0.

(a'+1)+b+c = 14

a'+b+c= 13 total = 15C2=105

a' cannot be 9 or more since (a'+1) would be 10 or more then

-> (9+a')+b+c = 13

a'+b+c= 4 -> 6C2

b and c cannot be 10 or more.

a'+(b+10)+c = 13

a'+b+c= 3 -> 5C2 and same for c

=> 105-6C2-2*5C2 = 105-15-20 = 70

=> 70/900 = 7/90

**The volume of a mixture of milk and water was increased by 68.75% by adding pure milk to it. If the ratio of milk n water in the resultant solution is 3:1, find the ratio of milk n water in the original solution ?**

68.75 = 11/ 16

let Vol 1= 16 then vol 2 = 27

We know water proportion initially * Volume 1 = Water proportion finally *Volume 2

=> water proportion initially * 16 = (1/4) * 27

=>water proportion initially = 27/64

so water : milk = 27: 37

**How many pairs of factors of N=360 will be coprime to each other?**

360 = 36 * 10 = 2^3 * 3^2 * 5

ORDERED PAIRS = (2 * 3 + 1)(2 * 2 + 1) (2 * 1 + 1) = 105

UNORDERED PAIR = [(2 * 3 + 1)(2 * 2 + 1) (2 * 1 + 1) -1 ]/2 = 52

As in this case variables are implicit so Its UNORDERED PAIRS .. so 52

**How many ways can 360 be expressed as Product of 2 coprimes?**

Let us suppose you have to express the number 360 as product of 2 co primes i,e a and b

So 360 = 2^3 * 3^2 *5

SO TOTAL 3 PRIMES Are there in 360 ..

We have to express as product of 2 co-primes "a" and "b"

So each of the primes i.e 2 , 3 and 5 has just 2 choices i.e either they can go "a" or to "b" .. so 2 * 2 * 2 = 8

there these 8 include cases like (1. 360 ) and (360,1) i.e "repeat cases

so 8/2 = 4

the formula is (2^n)/2 = 2^(n-1)

**Real madrid and Manchester united play a football match in which REAL beats MANCHESTER 4 - 2 ..In how many Ways can the goals be scored provided MANCHESTER never had a lead over REAL during the match?**

if Manchester never had a lead means 1st goal was scored by REAL .. So score card at this juncture would ve been 1 - 0 ..The only way MANCHESTER could take lead is by scoring next 2 goals , i.e a pattern like RMMRRR.. Except this one case , MANCHESTER Can no way take lead as such ..

So (5!)/3! *2! = 10

NOW excluding this one case of RMMRRR ... We have 10 - 1 = 9

**How many integers P are there between 0 and 10^100 such that units digit of P^3 is 1 ?**

So , each number from 1 - 10 has a distinct unit digit for its cube ..

occurrence of each digit in cube is 1/10 times

So 0 - 10^100 it will be 10^100 * 1/10 = 10^99

**20 groups of five shooters each compete in an shooting championship. A shooter finishing in Nth position contributes N points to his team, and there are no ties. The team that wins will be the one that has the least score . Given that , the 1st position team's score is not the same as any other team, the number of winning scores that are possible is**

The group’ scores must sum to 1 + 2 + . . . + ..+99+100 =5050. The winning group can be ATMOST 1/20 *5050 = 252.5 and is at least 1 + 2 + 3 + 4 + 5 = 15. However, not all scores between 15 and 252 inclusively are possible because all teams must have integer scores and no team can tie the winning team.

number of scores possible = 251 - 15 + 1 ...= 237

**X varies inversely as y and x varies directly as the square of z. If y decreases by 43.75% and z decreases by 75%, by what percent does x change?a. 88.88%b. 50%c. 45.45%d. 54.54%e. 44.44%**

Here my approach would be

X varies inversely as Y so XY is Constant

X varies directly with z^2 so x/z^2 is a constant

so (XY)/Z^2 = constant

X = (constant * Z^2 )/Y

So multiplying factor of X = (1 * multiplying factor of Z^2 )multiplying factor of Y

=> multiplying factor of X = [1 * (1/4)^2] / (9/16)

=> multiplying factor of X = 1/9

As mulpliying factor of x is 1/9 so change is "x" is 8/9 .. i.e 11.11 * 8 = 88.88 %

**Total positive integral solutions of 4x + 5y + 2z = 111?**

Answer by Bruce Wayyne

y needs to be odd, else LHS would become even.

say y = 2k-1 k is positive

=> 4x + 5 (2k-1) + 2z = 111

2x + 5k +z = 58

Let k = 1,

2x+z= 53 so (1, 51)...(26, 1) so 26

k=2

2x+z=48 so (23, 2).....(1, 46) so 23

k=3

2x+z=43 so (1,41)....(21, 1) so 21

k=4

2x+z= 38 so (1, 36)....(18,2) so 18

k=5

2x+z= 33 so (1,31)....(16,1) so 16

Hence the pattern is +3, +2, +3, +2....so on.

so (26+21+16+11+6+1) + (23+18+ 13 + 8 + 3) = 146

**The volume of a mixture of milk and water was increased by 68.75% by adding pure milk to it..if the ratio of milk n water in the resultant solution is 3:1.. find the ratio of milk n water in the original solution ?**

68.75 = 11/ 16

let Vol 1= 16 then vol 2 = 27

We know water proportion initially * Volume 1 = Water proportion finally *Volume 2

=> water proportion initially * 16 = (1/4) * 27

=>water proportion initially = 27/64

so water : milk = 27: 37

or

milk : water = 37: 27

**Fresh grapes contain 84% water while raisins contain 20% water.How many kg of raisins can be prepared from 80 kg of fresh grapes?**

16% of Grapes = 80% of raisins

=> Raisins = 1/5 * Grapes

Grapes = 80

=> Raisins = 1/5 * 80 = 16

Ratios method:

84% = 21 / 25

if weight = 25 ; then water = 21 and rest = 4

20% = 1/5

if weight = 5 : then water = 1 and rest = 4

So ratio of weight of Grapes : raisins = 5:1

so i need 80 kg of Grapes that mean 5 becomes 16 times so 1 should also become 16 times ... so 16 kgs

**Ram bought a few mangoes and apples spending an amount of at most 2000.If each mango cost 4 and Apple 6 and Ram bought at least 1. Find the different possible amounts could have spent in purchasing fruits?**

Sum of 3 number is even so either all three are even or 2 odd and 1 even ..

7^4 > 2002 .. so it must be less than this ... 5^4 is 3 digit number so we MUST TAKE 6^4 else how can you form a 4 digit number .. ??

that way 6^4 is included , now only 2 cases are possible

either the other 2 number ll be even or they have to be odd ... By common sense , we ignore 4 and 2 so it must be 3^4 and 5^4

so 6+ 5+ 3 = 14

**If 1/a + 1/b + 1/c + 1/d = 2, where a , b , c , d are distinct natural numbers, what is the value of a + b + c + d ?**

If sum of Factors of N excluding N is equal to N then N is called a perfect number ..

for Eg 28 = 1 + 2 + 4 + 7 + 14

496 = 1 +2 +4 +8 + 16 + 31 + 62 + 124 + 248

PERFECT NUMBERS show another property

The sum of the reciprocal of the factors of a perfect number INCLUDING THE NUMBER ITSELF = 2

Again i repeat "INCLUDING THE NUMBER ITSELF"

for eg : 28 is a perfect number

1+ 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 2

**Find the largest value of a for which 100 + a^3 becomes perfectly divisible by 10 + a**

by "Factor theorem" we know ,if 100 + a^3 is divisible by 10 + a then

substitute , a = -10 in 100 + a^3 = 100 -1000 = 900

Now 10 + a MUST be a FACTOR is 900

so let K * (10 + a) = 900

Now to obtain the GREATEST value of "a" , we HAVE TO put K as 1 so

10 + a = 900

=> a = 890

**How many ways can 22 identical balls be distributed in 3 identical boxes ?**

It is unordered distribution of a+b+c = 22

so total cases = 24c2

2 same n 1 different cases .. will be .. 0 0 22 ; 1 1 20 ....till 11 11 0 .

so total 0 - 11 that is 12 cases and each of them ll be arranged in 3!/2! i.e 3 ways

why ?? because they are of type A A B ..

so( 24c2 - 36 ) / 6 + 12 = 52

**X and Y have some chocolates with them which they wish to sell . The cost of each chocolate is equal to the number of total chocolates with both of them together initially. Together they sell all the chocolates and after that they start distributing the money collected in this particular fashion. First X takes a 10 rupee note, then Y takes a 10 rupee note and so on. In end it's turn of Y who didn't get any more 10 rupees. How much rupees Y get in his last turn?**

As X started the distribution part and took a 10 rupee note first and in the end also, he is able to take a 10 rupee note. That means Total amount, which needs to be a perfect square for n mangoes @ n rupees per mango, is odd multiple of 10 plus some more which is less than 10. That means ten's place digit of the perfect square is ODD. So certainly unit digit of perfect square is 6.

**25 is smallest square number that is average of two other co-prime square numbers i.e. 1 & 49. What's the next such square number?** (Credits - Kamal Lohia sir)

2C^2 = A^2 +B^2

=> C^2 - A^2 = B^2 - C^2

=>(C - A) * (C + A) = (B-C) * (B+C)

Now Let (B-C) = X/Y * (C-A) , --------(1)

Then , (B+C) = Y/X * (C+A) --------(2)

GCD (X,Y) = 1

this way we get to 169

**E = |x - 1| + |2x - 1| + |3x - 1| + |4x - 1| + ... + |19x - 1| + |20x - 1| attain a minimum valuea) 1/10b) 1/12c) 1/14d) 1/16** (Credits - Test Funda)

E = 1|x - 1| + 2|x - 1/2| + 3|x - 1/3| + ... + 20|x - 1/20|

All zeros occur at 1/n (n is a natural number less than 20)

Let say minimum of E occurs at x = 1/t

This is analogous to considering that there are 1 + 2 + 3 ... + 20 = 210 points on the number line. We need to select a point such that the sum of the distances of this point from all the 210 point will be a minimum. Thus, we need to take the median of these 210 points. i.e, any point between 105th point and 106th point. If we move to the left on the 105th point or to the right of the 106th point, the sum of the distances will increase.

Now the 105th point will be 1 + 2 + 3 + ... + 14 = 105

Thus the 105th point is 1/14 and the 106th point is 1/15. The expression, E will be minimum if we take x as any of these two values. i.e, at x = 1/14 or x = 1/15 (or any value in between)

Here, answer is 1/14

**Solve for x if (x - 1/x)^1/2 + (1 - 1/x)^1/2 = x**

Domain x - 1/x > = 0,

1 - 1/x > = 0, x > 0

So (x - 1)/x > = 0 , (x^2 - 1)/x > 0

x > = 1 is Domain.

now squaring twice n simplifying

Replace x - 1/x = t

t^2 + 2 = x^2 + 1/x^2 so

t^2 + 2 - 2t - 1 = t^2 - 2t + 1 = (t-1)^2 = 0

So t = 1, x-1/x = 1

x^2 - 1 = x, x^2 - x - 1 = 0

x = ( 1 + root(5))/2

**what is the coefficient of a^204 in (a-1)(a^2 - 2) (a^3 -3)....(a^20 -20)**

Max power of "a" in the expression = 1+2+3+4+.....+20 = (20 * 21 )/2 = 210

To get co-efficient of a^204 we have to find the constant term of expression having power a^6 or remove factors of a^6" , which are

- (x^1-1) * (x^5-5)" product of constant term = 5
- (x^2-2) * (x^4-4)" product of constant term = 8
- (x^1-1)" (x^2-2)" (x^3-3)" product of constant term = -6

4 (x^6 - 6) * 1 .. product of constant terms = -6

so your required coefficient = 5 + 8 - 6 - 6 = 1

**How many 5 digit numbers are there whose sum of digits is 12 ?**

a + b + c + d + e = 12

now a should be at-least 1

(x+1) + b + c + d + e = 12

=> x + b + c + d + e = 11 so 15c4 cases

now remove the cases when x > = 9

so, (x + 9) + b + c + d + e = 11

=> x + b + c + d + e = 2 i.e 6c4 cases

and when b, c , d , e > = 10 , we have to remove those cases as well .. isn't it ?

so , x + (b+10) + c + d + e = 11

=> x + b + c + d + e = 1 i.e 5c4 cases

Required numbers = 15c4 - 4 * 5c4 - 6c4

= 1365 - 20 - 15

= 1365 - 35

= 1330

**There are exactly 2005 ordered pairs (x, y) of positive integers satisfying 1/x + 1/y = 1/N. Find the number of digits in smallest possible value of N if N is a natural number.**

2005 = 5 * 401

N=a^p * b^q

N^2 = a^(2p) * b^(2q)

So (2p + 1)(2q + 1)= 5 * 401

p = 200, q = 2

N = 2^200 * 3^2

[LogN] + 1 = [200log2 + 2log3] + 1 = [61.16] + 1 = 62 digits

**Total no of integral solution for x + y + z = 15 where 1≤ x, y, z ≤ 6**

(6 - a ) + (6 - b) + (6 - c) = 15

=> a + b + c = 3

5c2

Total no . of integral solutions = 10

Alternate method : Coefficient of x^15 in (x + x^2 + x^3 + ... + x^6)^3

= x^12 in (1 - x^6)^3 * (1-x)^-3

= 14c2 - 3c1 * 8c2 + 3c2

= 10

**Find the coefficient of x^14 in (x^3 + x^6 + x^9 +...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)**

Method - 1 :

If you are good in binomial then do it directly. Else you can also think this in these lines. Is this not same as 3x + 2y + z = 14 and is this not same as finding the positive distinct integral solutions of the equation a + b + c = 14? so (13c2 - 3 * 6)/6 = 10

Method 2 :

If you want to solve it by binomial then check below

coefficient of x^14 in (x^3 + x^6 + x^9 + ...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)

= coefficient of x^14 - 6 (1 + x^3 + x^6 + ...) * (1 + x^2 + x^4 + ...) * (1 + x + x^2 + x^3 + ...)

= coefficient of x^8 in (1 + x^2 + x^3 + x^4 + ... + x^8 +...) * (1 + x^2 + x^3 + x^4 + ... + x^8 + ....)

= 1 + 1 +1 + 1 + 1 + 2 + 1 + 2 = 10

**In how many ways can an amount of rs. 100 be paid exactly using coins of denomination 1,2 and 5 such that at-least one coin of each denomination is used?**

Basically the problem is to find +ve integer solutions to A + 2B + 5C = 100

So A + 2B can take values 5 , 15, 25 ,......95

The number of solutions corresponding to this will be 2 + 7 + 12 ... + 47 = 245

Also, A + 2B can be 10 , 20 , 30 ... 90

The number of solutions corresponding to this will be 4 + 9 + 14 ... + 44 = 216

so total = 245 + 216 = 461.

**In how many ways can 2310 be expressed as a product of 3 factors?**

2310 = 2 * 3 * 5 * 7 * 11

These 5 primes to be distributed into 3 places.

Total : 3^5 ways , but this is ordered ( arranged) and we need unordered.

We know that any triplet with distinct elements (a,b,c) is arranged in 3! ways, so all the distinct triplets in 3^5 = 243 are arranged in 3! Ways.

Except for one triplet (1, 1, 2310) which is arranged in 3!/2= 3 ways

So remove this from total and we are left with only distinct, is divide by 3! ,

So (3^5 - 3)/3! , and just add that one case of 1, 1, 2310

So (3^5-3)/3! + 1 = 41

**How many pairs of positive integers (m, n) satisfy 1/m + 4/n = 1/12. Where n is an odd integer less than 60.**

(m -12) * (n-48) = 12 * 48 = 2^6 * 3^2

now for "n" to be odd , (n-48) has to be odd

so n - 48 can be either 1 or 3 or 3^2 (because we need ODD)

(m -12) * (n - 48) = 2^6 * 3^2 * 1

OR (m -12) * (n - 48) = 2^6 * 3 * 3

or (m -12) * (n - 48) = 2^6 * 3^2

when n - 48 = 1, n = 49 (acceptable)

when n - 48 = 3, n = 51 (acceptable)

when n - 48 = 3^2, n = 57 (acceptable)

only these 3 values are possible for n < 60 and odd

**Find number of whole number solutions of a + b + c + d = 20 where a, b, c and d ≤ 8**

Cofficient of x^20 in (1 + x + x^2 + x^3 + ... + x^8)^4

= coff of x^20 in (1 - x^9)^4 * (1 - x)^-4

= 23c3 - 4 * 14c3 + 4c2 * 5c3

= 1771 - 1456 + 60

= 375

For more detailed explanation of the concept, refer the video.

]]>let total work=512 unit

32 men per day work=512/16=32 unit

so 1 men per day=1 unit

in same way per day work of women=8/9 units

now 16 M+ 36 W = 8 day work=16 * 1 * 8 + 36 * 8 * 8/9 = 384 units

so remaining work=128 units

let x more men needed

so (16+x) * 2 + (36 * 8/9 * 2) = 128

so x = 16

**The roots of x^3 - ax^2 + bx - c = 0 are p, q and r while the roots of x^3 + dx^2 + ex - 90 = 0 are p+3, q+3 and r+3. what is the value of 9a + 3b + c ?**

x^3 - ax^2 + bx - c = 0

Sum of roots p + q + r = a - - - -- - -(i)

pq + qr + pr = b - - - - - - - -(ii)

pqr = c-- - - - - - -(iii)

x^3 + dx^2 + ex - 90 = 0

(p+3)(q+3)(r+3) = 90

pqr + 3(pq + pr + qr) + 9(p + q + r) + 27 = 90

Put (i), (ii) and (iii) in the above equation, It will be

9a + 3b + c = 63

**A test has ten questions. Points are awarded as follows:• Each correct answer is worth 3 points.• Each unanswered question is worth 1 point.• Each incorrect answer is worth 0 points.How many different scores are possible ?a) 31b) 30c) 29d) 28e) 27**

a= correct

b= incorrect

c= unanswered

a+b+c=10

now 3a+c= k where k is different score

when a=10 then max score = 30

when a=9 then you can give rest 1 to b or c when b=1 then score =27

when c=1 then score 27+1=28

now a=8 then rest 2 can be distributed as b=2, c=0 or b=0 , c=2 or b=1,c=1

so score 26,25,24

so now you are getting consecutive values so stop here

so missed out values = 29

so total 0 to 30 except 29 so 30 values

**For how many positive integer values of ‘x’ is ||||x – 1| – 2| – 3| – 4| < 5**

Approach-1

Manual counting will work here

Approach - 2

We can clearly see that x = 1 is satisfying, so minimum value of x will be 1.Now for the highest possible value of integer ‘x’ which satisfies the given inequality, say x - maximum , all the modulus brackets will open with a positive sign

|x – 1| > 0, ||x – 1| – 2| > 0, |||x – 1| – 2| – 3| > 0

When all the brackets open with a positive sign, x – 1 – 2 – 3 – 4 < 5 or x < 15 or x's maximum value will be 14

Total values will be 1-14 so 14 values

**It is given that a straight line L intersects the curve y = 3x^3 - 15x^2 + 7x - 8 at three distinct points (x1, y1), (x2, y2), (x3, y3). Find the value of x1 + x2 + x3**

Straight line is y = mx + c

so 3x^3 - 15x^2 + 7x - 8 = mx + c

so 3x^3 - 15x^2 + (7 - m)x - (8 + c) = 0

so sum of roots = 15/3=5

**Find all real numbers x such that x[x[x[x]]]= 88 ([.]greatest integer less than equal to x)**

(x-1)^4 < x[x[x[x]]] =< x^4

so (x-1)^4 < 88 < = x^4

so (x-1) < 88^(1/4) < = x

so x-1 > 88^(1/4) so x>1+88^(1/4)

so 88^(1/4)< = x < (88)^1/4 +1

88/x= [x[x[x]]

so 88/x should be integral value

and x will be in 88^1/4 < = x < 88^(1/4)+1

approx value of 88^1/4=3.066 and approx value of 88^1/4 +1 =4.066

and 88/x should be integer so look for those values of x which will give 88/x as integers

because we are looking for x in form of m/n where m is factor of 88

x could be 4 or 44/13 or 44/12 or 44/11 or 88/23 or 88/25 or 88/27 , 88/28 ..

now out of these only x=88/28 is satisfying our condition so OA=88/28

**Three are n – 1 red balls, n green balls, and n + 1 blue balls in a bag. The number of ways of choosing two balls from the bag that have different colours is 299. What is the value of n?**

(n-1)n+n(n+1)+(n+1)(n-1)=299

so 3n^2-1 = 299

so 3n^2 = 300

so n = 10

**If x^2 + x = 19, Find (x+5)^2+1/(x+5)^2**

x^2 + x = 19

(x+5)^2=x^2+10x+25

x^2=19-x

so x^2+10x+25=19-x+10x+25=9x+44

(9x+44)+1/(9x+44)

(9x+44)^2 +1)/(9x+44)

(81x^2+44^2+88 * 9 * x+1)/(9x+44)

((81((19-x)+44^2+88 * 9 * x+1))/9x+44

= 79

**If x and y are real numbers such that x^2 - 10x + y^2 + 16 = 0. Determine the maximum value of the ratio y/x**

Let y/x =k so y=kx

So x^2 -10x+(k^2x^2)+16=0

x^(1+k^2)-10x+16=0

x is real so D>=0

100 - 4 * 16 * (1+k^2)>=0

So 100/64 >=1+k^2

So 36/64 >=k^2

So -6/8 < = k < = 6/8

So max k = 3/4

**For positive integers x and y, 3x + 8y = 68. Find the maximum value of x^2 y^3**

3x+8y=68

we want to maximize x^2 * y^3

3x/2 =8y/3 =68/5

so x=68 * 2/(15)

y= (68 * 3)/40

so x integral will be 12 and y integral will be 4

max (x^2 y^3) would be 9216

This is like distributing 20 identical chocolates between 4 kids A, B, C, and D. You arrange these 20 chocolates in a line

C C C C … C

Now, you put in partitions in between them. Let me denote the partitions with P

C C C P C P C C C C C P C C C C C C C C C C C

A gets the chocolates before the first partition, B gets the chocolates between the first two partitions, C gets the chocolates between the second and the third partition and D gets the chocolate after the third partition. In the arrangement shown above A gets 3 chocolates, B gets 1, C gets 5, and D gets 11. Any rearrangement of the above, will lead to a new distribution of chocolates.

The above can be rearranged in 23! / 20! 3! We got this because there are a total of 23 entities out of which 20 Cs are identical and 3 Ps are identical.

Now, to extend this concept, what if we have to distribute ‘n’ chocolates in ‘r’ kids. After putting the n chocolates in a line, we would need r - 1 partitions. This would mean that there will be a total of n+r-1 entities out of which n would be identical (of one type) and the other r - 1 would be identical as well (of another type). So, the number of ways in which that would be possible = (n + r - 1)! / n! (r-1)!

This, in other words, is (n + r - 1) C (r - 1)

And now, we can use this formula to solve any similar questions

a + b + c + d = 20

Case 1: a, b, c, d are non-negative integers.

Number of solutions = (20 + 4 - 1) C (4 - 1) = 23 C 3 = 1771

Case 2: a, b, c, d are positive integers

We allocate at least a value of 1 to a, b, c, d.

So, we can say a = a’ + 1, b = b’ + 1, c = c’ + 1, d = d’ + 1 where a’, b’ c’, d’ are non-negative integers

=> a’ + 1 + b’ + 1 + c’ + 1 + d’ + 1 = 20

=> a’ + b’ + c’ + d’ = 16

=> Number of solutions = (16 + 4 - 1) C (4 - 1) = 19 C 3

Case 3: a, b, c, d are non-negative integers such that a > 5 and b > 2

We allocate at least a value of 5 to a and 2 to b

So, a = a’ + 5 and b = b’ + 2

=> a’ + 5 + b’ + 2 + c + d = 20

=> a’ + b’ + c + d = 13

=> Number of solutions = (13 + 4 - 1) C ( 4 - 1) = 16 C 3

Case 4: a, b, c, d are non-negative integer such that a > b

Let us first consider the situation where a = b

If a = b = 0, c + d = 20. This has 21 solutions

If a = b = 1, c + d = 18. This has 19 solutions

If a = b = 2, c + d = 16. This has 17 solutions

.

.

If a = b = 10, c + d = 0. This has 1 solution

So, the total number of a solutions when a = b is 21 + 19 + 17 … + 1 = 11/2*(21 + 1) = 121

We know that the number of solutions when a, b, c, and d are non-negative integers is 1771. Out of these 1771 cases, in 121 cases a = b.

So, in 1771 - 121 = 1650 cases a is not equal to b.

In half of the above cases a will be greater than b whereas in the other half of the cases a will be less than b.

So, number of solutions where a > b is 1650/2 = 825

**There is a escalator and 2 persons move down it.A takes 50 steps and B takes 75 steps while the escalator is moving down. Given that the time taken by A to take 1 step is equal to time taken by B to take 3 steps.Find the no. of steps in the escalator while it is stationary?**

Let us say that the escalator moves at the rate of n steps per second. Let us also say A takes 1 step per second and B takes 3 steps per second.

Case 1: When A is coming down

A will take 50 seconds to complete 50 steps.

In 50 seconds, escalator would have moved 50n steps.

Total number of steps on the stationary escalator = 50 + 50n

Case 2: When B is coming down

B will take 25 seconds to complete 75 steps

In 25 seconds, escalator would have moved 25n steps.

Total number of steps on the stationary escalator = 75 + 25n

Total number of steps on the stationary escalator is a constant

=> 50 + 50n = 75 + 25n

=> 25n = 25

=> n = 1

Total number of steps = 50 + 50 = 75 + 25 = 100

**Of 100 people, 86 ate eggs, 75 had bacon, 62 had toast and 82 had coffee. Minimum number who had all 4?**

We are given that there are 100 people. Let us call them P1, P2, P3 ... P100

Number of people who ate Egg = 86

=> Those who didn't = 100 - 86 = 14

=> 14 people did not eat eggs.

Let us number those P1, P2, P3 ... P14

Similarly, 25 people did not eat bacon

Let us number those P15, P16, P17 ... P39

Similarly, 38 did not have toast

Let us number those people P40, P41, P42 ... P77

Similarly, 18 people did not have coffee

Let us number those people P78, P79, P80 ... P95

Now, we are left with 5 people (P96, P97, P98, P99, P100) who ate / drank all of the four dishes / drink involved. This is the minimum possible configuration.

**Find the last 2 digits of (123)^123!?**

For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.

After that, we can use the property

Last two digits of (...a1)^(...b) will be [Last digit of a * b]1

Let us try and apply this concept in the given question

Last two digits of 123^123!

= Last two digits of 23^123!

= Last two digits of (23^4)^(123!/4)

= Last two digits of (529^2)^(123!/4)

= Last two digits of (...41)^(a large number ending in a lot of zeroes)

= Last two digits of (...01) {Here I have used the concept mentioned above}

= 01

**How many 6 digit numbers can be made using digits 1, 2, 3, 4, 5, and 6 without repetition such that the hundred digit is greater than the ten digit, and the ten is greater than the one digit?**

Let us assume that the number that we have is abcdef

The condition given is d > e > f

Let us first select three digits for a, b, and c.

From the given 6 digits (1,2,3,4,5,6), 3 digits for a,b, and c can be selected in 6C3 = 6!/3!3! = 20 ways.

These three selected digits can be arranged on three positions of a, b, and c in 3! = 6 ways

The remaining 3 digits are automatically selected for d, e, and f.

So, we have 1 way of selecting them.

The biggest digit will be allocated to 'd'.

The second biggest digit will be allocated to 'e'.

The third biggest digit will be allocated to 'f'

So, we have 1 way of arranging them on the three positions of d, e, and f.

Total ways = 20 * 6 * 1 * 1 = 120

**What is the units digit of 2! +4! +6! ...+98!?**

We need to find out the units digits of

2! + 4! + 6! + .... 98!

Let us look at the units digit for all of them

2! = 1 * 2 = 2; unit's digit is 2

4! = 1 * 2 * 3 * 4 = 24; unit's digit is 4

6! = 1 * 2 * 3 * 4 * 5 * 6 = 720; unit's digit is 0

8! = multiple of 6!, divisible by 10; unit's digit is 0

10! = multiple of 6!, divisible by 10; unit's digit is 0

12! = multiple of 6!, divisible by 10; unit's digit is 0

.

.

.

98! = multiple of 6!, divisible by 10; unit's digit is 0

Unit's digit of 2! + 4! + 6! + .... 98!

= Sum of unit's digits of individual terms

= 2 + 4 + 0 + 0 + 0 ..... 0

= 6

**A team has a food stock for N days. After 20 days 1/4th of the team quits and the food stock lasts for another N days. What is the value of N?**

Let us say that there were '4x' members in the team.

Total amount of food = N * 4x = 4Nx

Food consumed in 20 days = 20 * 4x = 80x

Food left after 20 days = 4Nx - 80x

The remaining food was consumed by 3x men in N days

=> Food left was = N * 3x = 3Nx

=> 4Nx - 80x = 3Nx

=> Nx = 80x

=> N = 80 days.

**What is the sum of the series, 1.(2) ^1 + 2.(2) ^2+3.(2) ^3+...+100.(2) ^100?**

S(1) = 1.(2)^1 = 2

S(2) = 1.(2)^1 + 2.(2)^2 = 2 + 8 = 10

S(3) = 1.(2)^1 + 2.(2)^2 + 3.(2)^3 = 2 + 8 + 24 = 34 = 2 + 2.(2)^4

S(4) = 1.(2)^1 + 2.(2)^2 + 3.(2)^3 + 4.(2)^4 = 2 + 8 + 24 + 64 = 98 = 2 + 3.(2)^5

and so on.

So, S(n) = 2 + (n-1).2^(n+1)

So, S(100) = 2 + 99.2^101

**Suppose you like a book. It is available on the Flipkart website at a 30% discount and on the Amazon website at a 20% discount. If you buy it via the mobile app, Flipkart offers an additional 20% discount on the reduced price whereas the Amazon app offers an additional 30% discount. So, which app you should use to buy?**

Let us say the book costs 100 Rs.

Via Flipkart: Website offers a 30% discount bringing down the price to 70 Rs. The flipkart app offers another 20% discount ( on 70 Rs.) i.e. a discount of 14 Rs. This reduces the price to 56 Rs.

Via Amazon: Website offers a 20% discount bringing down the price to 80 Rs. The Amazon app offers another 30% discount (on 80 Rs.) i.e. a discount of 24 Rs. This reduces the price to 56 Rs.

As you can see, both of them are offering it to you at the same price. So go ahead, buy it from Amazon - because it is better.

**What is the common ratio in a geometric progression in which: the first term is 7; the last term is 448; and the sum of terms is 889?**

Let us consider the 'n' terms in the Geometric Progression as a, ar, ar^2 ... ar^(n-1)

In this question, we know that the first term is 7

=> a = 7

The last term in the series is 448

=> ar^(n-1) = 448

=> r^(n-1) = 64

Now, this could be 2^6 or 4^3 or 8^2

To figure this part out, we also have one more information that the sum of the series is 889

=> a(r^n - 1)/(r - 1) = 889

Now, we can solve the above equations to get the answer. Or we can just assume some values and solve this out

Let us assume that the common ratio, r = 2

=> Sum of the GP = 889

=> 7 + 14 + 28 + 56 + 112 + 224 + 448 = 889

This fits. So, the common ratio is 2

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