There are B number of boys and G number of girls in a class. What is

Question

There are B number of boys and G number of girls in a class. So total number of students is B+G. What is the value of G?

(1) B is 85% of G.

(2) Value of G is more than 36 but less than 54.

chetan2u · Accepted Answer

There are B number of boys and G number of girls in a class. So total number of students is B+G. What is the value of G?

(1) B is 85% of G. 
If G is 100, B=85
If G is 200, B=170
Insufficient

(2) Value of G is more than 36 but less than 54.
Can be any value 40,50,etc
Insufficient

Combined
 B=\frac{85G}{100} ,
  B=\frac{17G}{20} 
Therefore for B to be an integer, G should be a multiple of 20 as 17 and 20 are co-prime.
There is ONLY one multiple of 20 between 36 and 54..
Answer is 2*20=40
Sufficient

C

MayankSingh · Answer

From (i)- B=0.85G or B=(17/20)*G ---- 
From(ii)- 36<G<54 ------

Combining (i)&(ii)

B=(17/20)*G ; For B to be integer, G= 20/40/60.... multiple of 20

From (ii) permissible value for G=40  
Therefore, G=40

Ans: Both statement together are sufficent to ans the ques.

imaityaroy · Answer

Hi,
Can we solve this way ?
S1 alone is insufficient because it only gives % but we want a value
S2 alone is insufficient because it only gives us a range of G
S1&S2 together 
B:G ratio is 85x:100x
In reduced form the ratio is 17x/20x
If x = 1 then ratio simplifies to 17:20. This does not come in the range provided in S2. The range is 36 to 54 exclusive.
If x = 2 then the ratio simplifies 34:40. This does come in the range provided in S2.
Any x >= 3 will be above the range provided 
hence G = 40 
so both together is sufficient. Hence C

chetan2u · Answer

Yes, that would be a correct approach.

egmat · Answer

Hi imaityaroy,

Yes, your approach is completely valid, and it lands on the exact same place as chetan2u's divisibility argument. You just arrived there from a slightly different angle.

Here's why your method works. Since B and G are counts of students, both must be whole numbers, and the ratio B:G reduces to  17:20 . Because  17  and  20  share no common factor, the only way to keep both B and G integers is to scale by a whole number x:

- B =  17 x
- G =  20 x

So G can only be  20, 40, 60, ...  - that's the same conclusion as "G must be a multiple of  20 ." Your two methods are really one method.

Then you apply the range  36  < G <  54 :

- x =  1  → G =  20  (too small)
- x =  2  → G =  40  (fits)
- x =  3  → G =  60  (too big)

Only G =  40  survives, so together the statements pin down a single value -  sufficient . Perfect.

One tiny label fix:  where you wrote "the range provided S1," you meant  S2  - the range ( 36  to  54 ) comes from Statement 2, and the ratio comes from Statement 1. Your logic is right; just the statement labels got swapped.

The one thing to always keep in mind with this move: it only works because both quantities are forced to be integers. If B and G could be any real values, the ratio alone would allow infinitely many G's even inside the range, and you'd be back to  insufficient . Here the integer requirement is what makes x jump in whole steps and isolate a single answer.

Nicely reasoned overall.

Answer: C

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