The following question is based on the given data for an examination.

A. Candidates appeared                 10500

B. Passed in all the five subjects     5685

C. Passed in three subjects only     1498

D. Passed in two subjects only       1250

E. Passed in one subject only           835

F. Failed in English only                       78

G. Failed in Maths only                      275

H. Failed in Physics only                    149

I. Failed in Chemistry only                147

J. Failed in Biology only                     221

 

How many candidates passed in at least four subjects ?

Let x be number of houses and y be colors.

on the basis of question
x = 4y+1
x = 5(y-1)
on solving these 2 equations x = 25

Let’s break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 – sum(2, 2, 18) = 22
2, 4, 9 – sum(2, 4, 9) = 15
2, 6, 6 – sum(2, 6, 6) = 14
2, 3, 12 – sum(2, 3, 12) = 17
3, 4, 6 – sum(3, 4, 6) = 13
3, 3, 8 – sum(3, 3, 8 ) = 14
1, 8, 9 – sum(1,8,9) = 18
1, 3, 24 – sum(1, 3, 24) = 28
1, 4, 18 – sum(1, 4, 18) = 23
1, 2, 36 – sum(1, 2, 36) = 39
1, 6, 12 – sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 – sum(2, 6, 6) = 14
3, 3, 8 – sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

3T+12C = 48,250----(1)

21T+36C = ?

multiply eq (1) with 3
i.e,, 3(7T+12C) = 48,250*3
21T+36C = 1,44,750/-

If there are x girls, then
x + 15x = 1200
16x = 1200
x= 1200/16 = 75

Here given 120 men = 200 children

therefore 1 child = 3/5 of Man------ step I

Now given that the food was consumed by 150 children

From step I we get 150 children =90 Men ( 1 child = 3/5 Man ie , 150 X 3/5 = 90)
So the food is for (120-90) men = 30 Men

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then, .

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108 6R = 108 R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

It is given that the first digit is 1/3 of the second. There are 3 such possibilities

1)1 and 3
2)2 and 6
3)3 and 9
Now, the third digit is the sum of the first and second digits.
1)1 + 3 = 4
2)2 + 6 = 8
3)3 + 9 = 12
It is clear that option 3 is not possible. So we are left with only two options. Also, the last digit is three times the second, which rules out the second option.

Hence, the answer is 1349.