Quant Boosters  Sibanand Pattnaik

Dudul n Gudul make round trips between Cuttack n Bhubaneswar respectively ..They first met at 10kms from Bhubaneswar, they reached their respective destinations n returned back n on their way , they meet for the 2nd time at 300 metres from Cuttack ..Find the distance between Cuttack n Bhubaneswar ?
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) 1225
b) 192
c) 200
d) 872x + y + z = 48
total whole number solutions = n+r1 C r1 = 48+31 C 31 = 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 0so 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 likeX < 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
=> 1056C22*5C2 = 1051520 = 70
=> 70/900 = 7/90The 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: 37How 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 52How 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 coprimes "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^(n1)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 = 9How 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 = 10^9920 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 = 2k1 k is positive
=> 4x + 5 (2k1) + 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) = 146The 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: 27HCF of 3 numbers is 6 and their sum is 120. How many such triplets exist ?
6x + 6y + 6z = 120
=> x + y + z = 20
19c2 casesHCF > 1
only possible case for HCF = Either 2 and 5 ..
case 1 (when HCF =2)
2x + 2y + 2z = 20
=> x + y + z = 10
9c2Case 2 , (When HCF = 5)
5x + 5y + 5z = 20
=> x + y + z = 4
so 3c2Therefore , required cases = 19c2  9c2  3c2 = 132 (ORDERED )
Now remove the 2 same 1 cases where HCF is 1
cases , 1 , 1 ,18 ; 3, 3 , 14 ; 7,7,6; 9,9,2 ... each of these arranged in 3!/2! ways so , 12 ways ..
so (132  12)/6 + 4 = 24
(for un ordered) ..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 = 14If 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 = 2Find 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 = 890How 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 = 52X 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 coprime 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) = (BC) * (B+C)
Now Let (BC) = X/Y * (CA) , (1)
Then , (B+C) = Y/X * (C+A) (2)
GCD (X,Y) = 1
this way we get to 169E = x  1 + 2x  1 + 3x  1 + 4x  1 + ... + 19x  1 + 20x  1 attain a minimum value
a) 1/10
b) 1/12
c) 1/14
d) 1/16 (Credits  Test Funda)E = 1x  1 + 2x  1/2 + 3x  1/3 + ... + 20x  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 = (t1)^2 = 0
So t = 1, x1/x = 1
x^2  1 = x, x^2  x  1 = 0
x = ( 1 + root(5))/2what is the coefficient of a^204 in (a1)(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 coefficient 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^11) * (x^55)" product of constant term = 5
 (x^22) * (x^44)" product of constant term = 8
 (x^11)" (x^22)" (x^33)" 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 atleast 1
(x+1) + b + c + d + e = 12
=> x + b + c + d + e = 11 so 15c4 casesnow remove the cases when x > = 9
so, (x + 9) + b + c + d + e = 11
=> x + b + c + d + e = 2 i.e 6c4 casesand 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 casesRequired numbers = 15c4  4 * 5c4  6c4
= 1365  20  15
= 1365  35
= 1330There 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 digitsTotal 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 = 10Alternate method : Coefficient of x^15 in (x + x^2 + x^3 + ... + x^6)^3
= x^12 in (1  x^6)^3 * (1x)^3
= 14c2  3c1 * 8c2 + 3c2
= 10Find 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 = 10Method 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 = 10In how many ways can an amount of rs. 100 be paid exactly using coins of denomination 1,2 and 5 such that atleast 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 (3^53)/3! + 1 = 41How 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) * (n48) = 12 * 48 = 2^6 * 3^2
now for "n" to be odd , (n48) 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 oddFind 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
= 375For more detailed explanation of the concept, refer the video.