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akay: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Sun May 06, 2007 5:03 am

GMAT past paper - Algebra

by akay » Sat Aug 11, 2007 8:15 am

I am close but for some reason not being able to get this.

If the sum of two positive integers is 24 and the difference of their squares is 48, what is the product of the two integers

(A) 108
(B) 119
(C) 128
(D) 135
(E) 143

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Source: Beat The GMAT — Problem Solving | Sat Aug 11, 2007 8:15 am

RM: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Sat Jun 16, 2007 10:58 pm

by RM » Sat Aug 11, 2007 9:52 am

Let the two positive integers be x and y.

Sum of two positive integers is 24, i.e. x + y = 24 ... (1)
Difference of their squares is 48, i.e. x2- y2 = 48
Or, x2-y2 = (x+y)(x-y) = 48 ... (2)

Solving equation (1) & (2):
(x+y)(x-y) = 24(x-y) = 48
(x-y) = 2 ... (3)

Solving equation (1) & (3):
x = 13
y = 11

Product of the two integers = x.y = 13x11 = 143, Choice E

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akay: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Sun May 06, 2007 5:03 am

by akay » Sat Aug 11, 2007 9:33 pm

thanks RM appreciate your taking the time.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Fri Dec 22, 2017 10:01 am

If the sum of two positive integers is 24 and the difference of their squares is 48, what is the product of the two integers

(A) 108
(B) 119
(C) 128
(D) 135
(E) 143

We can let x = one integer and y = the other integer and create the following equations:

x + y = 24

and

x^2 - y^2 = 48

Notice that x^2 - y^2 = (x + y)(x - y), and so:

(x + y)(x - y) = 48

Since x + y = 24:

24(x - y) = 48

x - y = 2

So, we have x + y = 24 and x - y = 2. If we add these two equations together, we have 2x = 26 or x = 13. Once we know x = 13, y must be 11 since x + y = 24. Thus, xy = (13)(11) = 143.

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]