Numbers of students in section A = x
? Numbers of students in section B and C = (100 ? x)
? x 70 + (100 ? x) 87.5 = 84 100
=> 70x + 87.5 100 ? 87.5x = 8400
=> 8750 ? 17.5x = 8400
=> 17.5x = 8750 ? 8400 => x = 20.

The third proportional of two numbers p and q is defined to be that number r such that

p : q = q : r.

Here, required third proportional of 10 & 20, and let it be 'a'

=> 10 : 20 = 20 : a

10a = 20 x 20

=> a = 40

Hence, third proportional of 10 & 20 is 40.

Excluded number = (27 x 5)  -  (25 x 4) = 135 - 100 = 35.

Assume Hebah has Rs. M

Since 25% more money at Poonam

=> Money at Poonam = M + (25% of M)

=> Money at Poonam = M + 0.25M = Rs. 1.25M

Money with Navaneet is thrice the money with Poonam,

=> Money at Navaneet = 3(1.25M) = Rs. 3.75M

Now, sum of all Money = (M + 1.25M + 3.75M) = Rs. 6M

But given the average of the money is Rs. 350

=> 6M/3 = 350

=> 2M = 350

=> M = 350/2 = Rs. 175

=> Amount of money Navaneet has = Rs. 3.75M = 3.75 x 175 = Rs. 656.25.

Marks in English  = 85×7 ? 83×6

= 595 ? 498

= 97

We Know that sum of the n natural numbers = n ( n + 1 ) 2

Then sum of 50 natural numbers= 50 ( 51 ) 2= 1250

Average of the 50 natural numbers = 1250 50 = 25.5

Age of the teacher = (37 * 15 - 36 * 14) years = 51 years.

Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers.

Given that their average is 52

Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6

6x + 30 = 312

x = 47

So Product = 47 × 57 = 2679

Let the number of boys = x

From the given data,

=> [21x + 16(18)]/(x+16) = 19

=> 21x - 19x = 19(16) - 16(18)

=> 2x = 16

=> x = 8

Therefore, the number of boys in the class = 8.

You could solve this by drawing a Venn diagram. A simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking French to find the number taking only French. Likewise find those taking only German. Then we have:Total = only French + only German + both + neither

78 = (41-9) + (22-9) + 9 + neither.

Not enrolled students = 24