Topic : Quant Mixed Bag

Solved ? : Yes

Source : MathOratory ]]>

Topic : Quant Mixed Bag

Solved ? : Yes

Source : MathOratory ]]>

So there's a net reduction in total of 5

This net reduction will be carried forward throughout all the years

If nobody would've retired or joined the average of the 25 people would've been 45+10=55, as we are moving 10 years ahead

But remember there's a net reduction of 5 years in total

As there is no change in the number of teachers, the net reduction in average would be 5/25 = 1/5

So the current average should be 55 - 1/5 = 54 4/5

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

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

]]>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 ]]>

i.e., n!! = 1! * 2! * 3! * ... n!

Find the highest power of 2014 that divides 2014!!, without leaving a remainder. ]]>

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

]]>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% ]]>

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 ]]>

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% ]]>

But the shopkeeper returns a sachet worth rs. 1.5 to the customer, so the net SP = 24

CP = 24/1.2 = 20 ]]>

Thus, P/SP = 75/100

If P = 75, SP = 100, CP = 25

Thus, P/CP = 75/25, or 300% ]]>

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) ]]>

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% ]]>

So, effectively = (2/3) * (21/20) = 7/10, which implies a reduction of 3/10 = 30% decrease ]]>

a) Decrease by 13.64%

b) Decrease by 15%

c) Decrease by 18.75%

d) Decrease by 30% ]]>