# Quant Boosters - Soumya Chakraborty - Set 3

• B,C,D is 460% of A, so A,B,C,D must be 560% of A
As A,B,C,D together is 56 lakhs, and that corresponds to 560% of A
100% of A corresponds to 10 lakhs (that's A's share)

A,C,D is 366.66%, which is 3 + 2/3 = 11/3 of B
So, A,B,C,D must be 1+11/3 = 14/3 of B

14 corresponds to 56 lakhs, so 3 must correspond to 12 lakhs (B's share)

C is 2/5 of A,B,D
so, A,B,C,D must be 1 + 2/5 = 7/5 of A,B,D
7 corresponds to 56 lakhs
5 corresponds to 40 lakhs (that's sum of A,B, D share)

out of which A = 10
B = 12
So, D = 40 - (10+12) = 18

• Q3) Weekly income of two persons are in the ration 7:2 and their expenses are in the ratio 5:2. If each of them save 300 rupees per week then the weekly income of the first person is ?

• Observe few things first: If we have two numbers. Both are increased/decreased by the same value. The difference of the two values should remain constant.

Now, let us apply it here. But before we do that, understand that we can apply it here as both person saves equal amount. And Income - Savings = Expenditure

So, the numerator and the denominator of the first ratio MUST reduce by a constant value to get the second ratio. If that happens the difference between the numerator and denominator must remain constant.

Now, we have the first ratio 7/3 and the second one 5/2
In the first ratio, the difference between numerator and denominator is 4 and the in the second one is 3

So multiply both parts in the first ratio by 3, and both parts in the second ratio by 4, to make the difference equal.

we now have 21/9 and 20/8

Now, if we observe the numerator reduce by 1 and so does the denominator. but it should actually reduce by 300. So, we need to multiply by 300, effectively everything here by 300

As, we are looking for the income of the first person, that is 21 parts: it would be 21 * 300 = 6300

• Q4) The ratio between present age of A and B is 3:4. The present age of B is 10 years more than the age of A 5 years ago. Find out the present age of B

• If B's age is 10 years more than A's age 5 years back
Then B's age should be 5 years more than A's present age
Now, A's and B's present ages are in the ratio 3:4
and if the gap is 5 years, which corresponds to 1 part in the ratio
B's age must be 20 years, as it is 4 parts in the ratio

• Q5) n!! is defined as the product of the first n!
i.e., n!! = 1! * 2! * 3! * ... n!
Find the highest power of 2014 that divides 2014!!, without leaving a remainder.

• 2014 = 2 * 19 * 53
check for the highest power of largest prime, here 53
2014! contains 38 53's
so 2014!! = 1! 2! 3! ... 2014! will contain 53 (1 + 2 + 3 ... + 37) + 38 = 37297

First thing that you need to understand is getting the highest power of a certain prime that a given factorial will be divisible by

Let's say what is the maximum power of 2 that will divide 50!

In order to solve that, we will successively divide 50 by 2, and keep collecting the quotients. The answer will be sum of all the quotients

50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1

Thus answer is 25 + 12 + 6 + 3 + 1 = 47

Now, if we need to figure out the max power of a certain composite, we need to check for each prime in the composite ...

In a certain factorial, if we need to check the max power of 2015 = 5 * 13 * 31

we need to check for each primes. But, in any factorial the number of 31s cannot exceed the number of 13s and the number of 5s. Thus, the number 2015s will be determined by 31s. Thus, checking by 31 alone is sufficient

If the question says 2015!!

31! will have one 31, so will 32!, so will 33! ... till 61!
So each of these are giving 1 power each, in total 31*1

62! to 92!, we will get 2 powers of 31 each... in total 31*2

This will carry on till 31 * 30

Now, let us consider the numbers: 960! and (961)!
960/31 = 30

961/31 = 31
31/31 = 1

Thus, although 960! gives 30 powers of 31,
961! gives 31 + 1 = 32 powers

This happens because the first quotient will leave a quotient when divided by 31

Again we will have this series: 31 * 32, 31 * 33, ... till the first quotient leaves a quotient of 1
62/31 = 2
But 61/31= 1

thus 31 * 32, 31 * 33, ... 31 * 62

Extending this logic ... we get the answer

• Q6) The MRP of a chocolate is Rs 65. A customer purchased the chocolate for 56 Rupees 16 Paise. He got two successive discounts of which one is 10%. Find the other discount of the discount scheme that was given by the seller.

• Observe that both 65 and 56.16 are multiples of 13
dividing both by 13, we have 5 and 4.32
reduce 10% of 5 => you get 4.5
4.5 to 4.32 is a reduction of .18
which is 4%

• Q7) If all the sides of a cuboid increases by 8%, its volume increases by x%
If all the sides of a different cuboid reduces by 8%, its volume reduces by y%
Which of the following is true?
a) x = y
b) x > y
c) x < y
d) We need the initial side ratio of the two cuboids that we begin with

• Understand this LOGICALLY
when we have successive change over +a and +a, the net change will be more than 2a
when we have successive change over -a and -a, the net change will be less than 2a
Think over the successive change formula, you'd understand
So, the x% must be greater than y%

• Q8) A grocer sells a soap markerd Rs 30 at 15% discount and gives a shampoo sachet costin Rs 1.5 free with each soap. He then makes 20% profit. His cost price per soap is

• After the 15% discount over 30, the new SP would have been 25.5.
But the shopkeeper returns a sachet worth rs. 1.5 to the customer, so the net SP = 24
CP = 24/1.2 = 20

• Q9) A trader makes a profit equal to the selling price of 75 articles when he sold 100 of the articles. What % profit did he make in the transaction?

• Profit of 100 = SP of 75
Thus, P/SP = 75/100
If P = 75, SP = 100, CP = 25
Thus, P/CP = 75/25, or 300%

• Q10) The fees of a school increased 9.09%, then by 8.33%, and then by 7.69% over 3 years. If the current school fees is Rs. 3,360, then find the approximate school fees 3 years back.

• The multiplication factors are:
9.09% increase = 12/11
8.33% increase = 13/12
7.69% increase = 14/13
So, the value becomes
(12/11) * (13/12) * (14/13) = 14/11
14 is the new part so that corresponds to 3360 = 14 * 240
11 should then correspond to 2640 (=11 * 240)

• Q11) A student scored 43% in an exam and failed by 12 marks. Another student scored 48% marks in the same exam and passed by 18 marks. Find the passing percentage in the exam.

• The percentage points difference is 5%, which corresponds to 30 marks
So each percentage point should be 6 marks. Now, someone who failed by 12 marks, must have failed 2 percentage points. As, he gets 43%, the passing percentage must be 45%

• Q12) In September 2009, the sales of a product were 2/3rd of that in July 2009. In November 2009, the sales of the product were higher by 5% as compared to September 2009. How much is the percentage of increase in sales in November 2009 with respect to the base figure in July 2009.

65

64

61

1

61

61

58

71