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The product of 2 positive integers is 720. If one of the numbers is in

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The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post Updated on: 04 Dec 2014, 04:18
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The product of 2 positive integers is 720. If one of the numbers is increased by 10 and the other reduced by 1, the product remains unaltered. What is the value of the smaller of the two integers?

A) 6
B) 7
C) 8
D) 9
E) 10

Originally posted by desaichinmay22 on 04 Dec 2014, 04:05.
Last edited by Bunuel on 04 Dec 2014, 04:18, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post 04 Dec 2014, 04:27
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desaichinmay22 wrote:
The product of 2 positive integers is 720. If one of the numbers is increased by 10 and the other reduced by 1, the product remains unaltered. What is the value of the smaller of the two integers?

A) 6
B) 7
C) 8
D) 9
E) 10


\(ab = 720\);
\((a + 10)(b - 1) = 720\) --> \(ab +10b - a - 10 = 720\) --> \(720 + 10b - a - 10 = 720\) --> \(10b - a - 10 = 0\).

\(10b - a - 10 = 0\) --> \(10b - \frac{720}{b} - 10 = 0\) --> multiply by b: \(10b^2 - 10b - 720 = 0\) --> \(b^2 - b - 72 = 0\) --> \((b + 8)(b - 9) = 0\) --> \(b = 9\) (discard negative root since we know that a and b must be positive) --> \(a=80\) --> smaller number is 9.

Answer: D.
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Re: The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post 04 Dec 2014, 05:04
desaichinmay22 wrote:
The product of 2 positive integers is 720. If one of the numbers is increased by 10 and the other reduced by 1, the product remains unaltered. What is the value of the smaller of the two integers?

A) 6
B) 7
C) 8
D) 9
E) 10


let two numbers be x an y
thus, xy=720.------------1)
also, (x+10)(y-1)=720
from 1, we have (x+10)(y-1)=xy
xy-x+10y-10=xy
10y=10+x
or y=1+x/10

now for x=80 y becomes 9. and product becomes 720. thus value of smaller integer =9
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Re: The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post 04 Dec 2014, 19:06
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Answer = D = 9

Say one of the number = x, other would be \(\frac{720}{x}\)

\(x * \frac{720}{x} = 720\)

x is increased by 10, \(\frac{720}{x}\) is decreased by 1

Setting up the equation

\((x+10)(\frac{720}{x} - 1) = 720\)

\(x^2 + 10x - 7200 = 0\)

x = 80

\(Smaller = \frac{720}{80} = 9\)
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Re: The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post 25 Jan 2016, 10:35
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Hi All,

This question can be solved by TESTing THE ANSWERS.

We're told that the product of 2 positive integers is 720. Next, we're told that If one of the numbers is increased by 10 and the other reduced by 1, the product remains the SAME. We're asked for the value of the smaller of the two integers.

Since the number 7 does NOT divide evenly into 720, Answer B CANNOT be correct. So...

Let's TEST Answer D: 9

IF....
Smaller number = 9
Larger number = 720/9 = 80

Decreasing the smaller number = 9-1 = 8
Increasing the larger number = 80+10 = 90
(8)(90) = 720
This is an exact MATCH for what we were told, so this MUST be the answer.

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Re: The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post 18 Mar 2019, 21:38
PareshGmat wrote:
Answer = D = 9

Say one of the number = x, other would be \(\frac{720}{x}\)

\(x * \frac{720}{x} = 720\)

x is increased by 10, \(\frac{720}{x}\) is decreased by 1

Setting up the equation

\((x+10)(\frac{720}{x} - 1) = 720\)

\(x^2 + 10x - 7200 = 0\)

x = 80

\(Smaller = \frac{720}{80} = 9\)


Hello!

Does anyone know how to factorize the following?

\(x^2 + 10x - 7200 = 0\)


Thank you so much!
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Re: The product of 2 positive integers is 720. If one of the numbers is in  [#permalink]

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New post 19 Mar 2019, 20:23
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Hi jfranciscocuencag,

That type of Algebra isn't actually necessary for this question, but you can factor this equation without too much trouble. Here's how:

X^2 + 10X - 7200 = 0

The "-7200" means that we'll have one positive factor and one negative factor. Since the middle term is "+10X", we need two numbers that only differ by 10 on the 'absolute' scale. Thus -72 and 100 would NOT be 'close enough' (since they differ by 28 in absolute terms), so we have to look for two values that are closer. Consider how "72" = (8)(9)... and you should be able to see how 7200 = (80)(90).... and +90 and -80 as you're factors...

X^2 + 10X - 7200 = 0
(X + 90)(X - 80) = 0

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Re: The product of 2 positive integers is 720. If one of the numbers is in   [#permalink] 19 Mar 2019, 20:23
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