Short way for solving problem

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larrybird: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Sat May 24, 2008 7:44 am

Short way for solving problem

by larrybird » Sat Jul 04, 2009 4:38 pm

Experts: Can you expect to get a problem like this on the real exam? It seems like it`s a little long. Maybe I am missing a shortcut... Could anybody show an intuitive way of "Guesstimating" is case you take too long to solve the problem ?
Thanks!!!

The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If the number of boys in Class B is one less than the number of boys in Class A, and if the number of girls in Class B is two less than the number of girls in Class A, how many girls are in Class A?

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Source: Beat The GMAT — Problem Solving | Sat Jul 04, 2009 4:38 pm

atulkumar79: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Tue Mar 17, 2009 3:43 pm; Thanked: 1 times

Re: Short way for solving problem

by atulkumar79 » Sat Jul 04, 2009 5:09 pm

larrybird wrote:Experts: Can you expect to get a problem like this on the real exam? It seems like it`s a little long. Maybe I am missing a shortcut... Could anybody show an intuitive way of "Guesstimating" is case you take too long to solve the problem ?
Thanks!!!

The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If the number of boys in Class B is one less than the number of boys in Class A, and if the number of girls in Class B is two less than the number of girls in Class A, how many girls are in Class A?

I would assume you can expect this question on the exam, and solve it in under 2 mins. The trick in this question is that you are given too much information to solve the question. The shortcut to the solving is that you can skip the middle part where it mentions the combined ratio of the 2 classes, this can save you some unneeded calculation. You can setup 2 X and Y equations based on the rest of the information provided and solve for X and Y.

Here's how I would solve it:
Class A: No. of boys = 3x
No. of girls = 4x

Class B: No. of boys = 4y
No. of girls = 5y

now B has 1 boy less than A, so the first equation would be:
3x - 4y = 1

and also B has 2 girls less than A, so the seconds equation would be:
4x - 5y = 2

Now it is pretty straight forward to solve for x which would be = 3
and hence the answer would be : 12. No. of girls in A.

This is the simplest I can think of. Comments Welcome!!!!

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ghacker: Master | Next Rank: 500 Posts; Posts: 148; Joined: Wed Jun 03, 2009 8:04 pm; Thanked: 18 times; Followed by:1 members

by ghacker » Sat Jul 04, 2009 5:32 pm

The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If the number of boys in Class B is one less than the number of boys in Class A, and if the number of girls in Class B is two less than the number of girls in Class A, how many girls are in Class A?

Give the answer choices , so that I can give a very short method

Since the answer choices are not available we can solve the question with out using equations

Logic says that in class A the total number of children ( G And B ) must be a multiple of 7

Similarly in Class B its a multiple of 9

When A and B are combined the combined class must be a multiple of 39

Ok

So total number of children is a multiple of 39 hence we check the multiples of 39 which satisfies the other two conditions

We take 39 then A will be 21 and B will be 18

If A is 21 then Boys = 9 Girls = 12 , Since B is 18 , Boys = 8 and Girls will Be 10

Also it satisfies the next two conditions given in the squib

So the answer is 12

***If you observe carefully you can see that one can disregard the higher multiples of 39 ( example 39*2,39*3 ect.... Y ? the differences will be multiplied by 2,3 ect ....)

Last edited by ghacker on Sat Jul 04, 2009 5:55 pm, edited 5 times in total.

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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Sat Jul 04, 2009 5:41 pm

The biggest difficulty in word problems is to figure out the equation. Then the rest is pretty simple.

I don't know how you can solve this with pure intuition, maybe someone can show us how to do it.

I worked out the problem algebraically and IMO it takes <=2 min.

lets call class A = class 1 and class B = class 2

G1::B1 = 3x::4x
G2::B2 = 4y::5y

B2 = B1 - 1
G2 = G1 - 2

so we have
4y = 3x - 1
5y = 4x - 2

[(3x-1) + 3x]/[(4x-2) +4x] = 17/22

solve for x.

132x - 22 = 136x - 34

4x = 12

x = 3

G1 = 4x = 4*3 = 12