# Quant Boosters - Quant Mixed Bag - Set 1

• [Credits : Saquib Hasnain]
Last and first place has to be 5.
5 _ _ _ _ _ _ _ 5
10 x 10 x 10 x 1 x 1 x 1 = 1000

• Q6) A and B working separately can do a piece of work in 9 and 12 days respectively. If they work for a day alternately,A beginning in how many days,the work will be completed ?

• [Credits : Saquib Hasnain]
let the total work be 36 units.
A can do 36/9 = 4 units/day
B can do 36/12 = 3 units/day
In two days, the can complete 7 units
to complete 35 units, they will take 10 days..
left units is 1 which A will complete on the 11th day, taking 1/4th day.
so 10 1/4 days

• Q7) When a 2-digit number is reversed, the new number is reduced by 63 and the sum of these individual numbers is 9. Now if these same numbers are converted into base x the larger number becomes 5 times that of the smaller one.The value of x?

• [Credits : Saquib Hasnain]
9(a-b) = 63
a - b = 7
a + b = 9
a =8 , b = 1
8x+1 = 5(x+8 )
3x = 39
x = 13

• Q8) 817 ^ 673 mod 100 = ?

• [Credits : Saquib Hasnain]
817^673 mod 4 = 1
817^673 mod 25 = 12
4k+1 = 25q +12
k = 9..
so remainder = 37

• Q9) If the equation x^4 - 4x^3 + ax^2 + bx + 1 = 0 has 4 positive roots, then a + b = ?

• [Credits : Saquib Hasnain]
Product is 1..sum is 4..all roots are positive..so all of them have to be 1..
a = 4C2 = 24/4 = 6
b = -4C3 = -4
so a+b = 6-4 = 2

• Q10) A train after covering 1200 km meets with an accident and due to this it proceeds at 2/3 of its original speed and arrives at the station in 5 hours. Had the accident taken place after covering a distance of 300 km further, it would have arrived 2.5 hours earlier. What is the speed of the train ?

• [Credits : Saquib Hasnain]
So say the speed is v and 2/3v.. the guy saves 2.5hours..if the accident happens 300 further..so he saving that 2.5 hours in that 300m..
Time taken to cover 300m at speed v is 300/v
Time taken to cover 300m at speed 2/3v is 300/(2v/3)
Difference is of 2.5hours..
So 300 (3/2v - 1/v) = 2.5
300 (1/2v) = 2.5
v = 60

• Q11) After three successive raises, Aftab’s salary became equal to 378/125 of his initial salary. By what percentage was the salary raised the first time if the third rise was twice as high (in percentage) as the second rise and the second raise was twice as high (in percentage) as the first rise?
(a) 10%
(b) 15%
(c) 20%
(d) 25%

• [Credits : Saquib Hasnain]
378 = 6 x 63 = 6 x 7 x 9
125 = 5 x 5 x 5
so, (1 + 1/5) (1 + 2/5) (1 + 4/5)
1/5 = 20%, 2/5 = 40% and 4/5 = 80%

• Q12) Find the sum of all the three digit numbers that contain only 1, 2, 3 or 4. It can contain the digits more than once in the same number too.

• [Credits : Sabyasachi Mishra]

Say a 3 digit number is abc.

Now there are 4 ways to choose a, 4 ways for b and 4 ways for c = > There are 4 *4 * 4 = 64 three-digit numbers that can be made using 1,2,3 and / or 4.

Let’s write all these numbers one below another to add them up. Now each digit would appear 16 times in ones place.

Their sum = 16 *(1 + 2+ 3 + 4) = 160

Similarly the tens place would add up to 160 and the hundreds place too.

So the actual sum = 160 (1 + 10 + 100) = 160 * 111 = 17760

• Q13) What is the Probability of selecting 3 numbers from 1-50 such that no two are consecutive?

• [Credits : Sabyasachi Mishra]

Imagine you have three 0s and 45 1s.

Every sequence you make would look somewhat like this.

10111111101111 … 1011 or 10011111 … 1011

Let’s say after every sequence, I put one 1 after the first zero and another 1 after the second zero.

Now I have a sequence of 50 digits, all in the form of 1 or 0. And no two zeroes are consecutive.

Now 1 represents not-chosen and 0 represents chosen. So, I have picked three numbers between 1 and 50 such that no two of them are consecutive. (each 1 or 0 representing the status of numbers from 1 to 50 in order)

The question simplifies into finding the number of ways three 0s and 45 1s can be arranged = > (48! / 45! 3!) or (48 C 3)

Now ways of selecting 3 numbers of the 50 = 50 C 3

Probability = 48C3 / 50C3 = 48 * 47 * 46 / 50 * 49 * 48 = 47 * 46 / 50 * 49 = 0.88245

• Q14) how many numbers are there from 1 to 1000 which are not divisible by the digits 2, 3 and 5?
a) 166
b) 266
c) 375
d) 366

• [Credits : Sabyasachi Mishra]

Yes, this can be solved using a Venn diagram and intersections. But, if you are going to solve it in an exam, I would suggest a better and faster approach.

Here, in our case LCM of 2,3 and 5 is 30.

So, we can say that, if we write from 1 to 30 and find out the pattern, the same would be repeated across sets of 30 numbers thereafter.

Step 1

Write all numbers from 1 to 30

Step 2

Start with 2, the smallest divisor and cross out the numbers that are divisible. Then move to the next smallest one, in this case 3, circle out the numbers divisible by 3. And then do it for 5, use squares. The sheet would look like this.

Step 3

Now that you need the numbers that are not divisible by either of these factors, let’s list them out.

Step 4

1000 = 30 * 33 + 10

So, the numbers not divisible by 2, 3 or 5 would be 8 * 33 + 2 = 264 + 2 = 266

Note:

1. This method can also be used to find the number of numbers between ranges that are divisible by specific factors. We need to start the count from the smallest number in the range and then list down a series of consecutive numbers (total count = LCM of factors)
2. This method can also be used to find the number of numbers divisible by 2,3 and 5. In that case, instead of listing down the left ones, list down the crossed / highlighted ones.
3. This method works faster when you have to do it on a small subset and extend the results. If the LCM is more than 100 or the factors are large numbers, Venn diagram method would be the way to go ahead.

• Q15) How many AP can be formed with common difference 1 and sum is 840?

65

58

61

61

61

64

61