Select the option that represents the number of parallelograms in the given figure.

 

 

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.

First ten odd numbers form an arithmetic progression of the form
1,3,5,7,9......
here a = first term = 1
d = common difference = 2
Average of first n numbers = (2a + (n - 1)d)/2
n th term of the AP = a + (n - 1)d
Substituting a = 1, d = 2 and n = 10 in the above formulas
average of first 10 numbers = 10
10 th term of the AP = 19
Therefore average of first 10 terms is 19 - 10 = 9 greater than
the last term. Hence the average is greater than the sum by
9/19 X 100 = 900/19 %