Sum of 3 integers

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money9111: Legendary Member; Posts: 2109; Joined: Sun Apr 19, 2009 10:25 pmFollowed by: 79 members; Location: New Jersey; Thanked: 109 times; GMAT Score:640

by money9111 » Sun Jan 31, 2010 10:01 pm

The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

a. 1104
b. 972
c. 672
d. 294
e. 192

again if someone can tell me how to hide the OA I'll post it within the post..

thanks!

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Source: Beat The GMAT — Problem Solving | Sun Jan 31, 2010 10:01 pm

papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Sun Jan 31, 2010 10:12 pm

Let x be the middle number

Largest = 3*x

Smallest = 3x - 23

Sum = 3x-23+x+3x = 40

7x = 63

x = 9

Largest = 27, Smallest = 4, Middle = 9

So, product of all three integers = 27*4*9 = 972 (B)

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money9111: Legendary Member; Posts: 2109; Joined: Sun Apr 19, 2009 10:25 pmFollowed by: 79 members; Location: New Jersey; Thanked: 109 times; GMAT Score:640

by money9111 » Sun Jan 31, 2010 10:14 pm

OA = B

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Sun Jan 31, 2010 10:20 pm

let the no be x,y,z

z=3y
x=z-23=3y-23

x+y+z=40
--->3y-23+y+3y=40

y=9
z=27
x=4
x*y*z=4*9*27=972
B

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Mon Feb 01, 2010 1:26 am

I don't know why you all are making fuzz for just a simple problem...
@sars72 you were going correct with the algebric solution but you missed the thing out in last step... Its very simple follow the steps below..

For solving |x-4| = 4-x,
Just break it in three steps..
1st step : Let x-4 be a positive no. so x>=4 and |x-4| = x-4
thus |x-4|=4-x --> x-4=4-x --> x=4.
but x>=4 hence there is only one solution for this condition i.e. x=4 ------- (i)

2nd step : Let x-4 be a negative no. so x<4 and |x-4| = -(x-4)
thus |x-4| = 4-x --> -(x-4) = 4-x --> 4=4 which is always true under specified condition...
thus x<4 is always true.. ------------------------------------------(ii)

so from (i) & (ii) we get that the overall solution of the equation |x-4| = (4-x) is x<=4 .

so u can easily splve it algebrically ... u can also use plugging in method as above..

[/quote]

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Mon Feb 01, 2010 1:33 am

Hey, you have posted in a wrong topic by mistake.

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by Scott@TargetTestPrep » Wed Dec 20, 2017 11:37 am

money9111 wrote:The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

a. 1104
b. 972
c. 672
d. 294
e. 192

We can create the following 3 equations in which a = the smallest integer, b = the middle integer, and c = the largest integer:

a + b + c = 40

and

c = 3b

c/3 = b

and

c - 23 = a

Thus:

c - 23 + c/3 + c = 40

2c + c/3 = 63

Multiplying by 3, we have:

6c + c = 189

7c = 189

c = 27, so a = 27 - 23 = 4 and b = 27/3 = 9.

Thus, the product of the 3 integers is 4 x 9 x 27 = 972.

Answer: B

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Dec 20, 2017 12:16 pm

money9111 wrote:The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

a. 1104
b. 972
c. 672
d. 294
e. 192

An alternate approach is to GUESS AND CHECK.

Let S = the smallest integer, M = the middle integer, and L = the largest integer.

Case 1: S = 1, with the result that L = 1+23 = 24 and that M = 40-1-24 = 15
Since the value of L (24) is not 3 times the value of M (15), Case 1 is not viable.

Since the value of L must be 3 times the value of M, L must be a MULTIPLE OF 3.
Thus, the next greatest option for L is 27.

Case 2: L=27, with the result that S = 27-23 = 4 and that M = 40-27-4 = 9
Success!
The value of L (27) is 3 times the value of M (9).

Resulting product:
27*4*9 = 972.

The correct answer is B.

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