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LUCIFER1703
Joined: 06 Jan 2018
Last visit: 13 Dec 2022
Posts: 28
Given Kudos: 51
Location: India
Schools: AIM '22 (A)
16 Jan 2020, 09:46
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C
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A school wants to spend equal amounts of money on the purchase of two types of copiers. The cost of one is $320 and the cost of the other is $560. What is the minimum number of total copiers the school can purchase?
(A) 13
(B) 12
(C) 11
(D) 7
(E) 4
(A) 13
(B) 12
(C) 11
(D) 7
(E) 4
16 Jan 2020, 20:24
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If they buy x of the cheaper type and y of the more expensive type, and spend the same amount in total on each type, then we have:
320x = 560y
and cancelling down, we arrive at
4x = 7y
So 4x and 7y are the same whole number, and 4x and 7y must therefore have the same divisors. Since 4 is clearly a divisor of 4x, it must be a divisor of 7y, and thus of y, so the smallest possible value of y is 4, which makes x equal to 7, and the answer is 4+7 = 11.
320x = 560y
and cancelling down, we arrive at
4x = 7y
So 4x and 7y are the same whole number, and 4x and 7y must therefore have the same divisors. Since 4 is clearly a divisor of 4x, it must be a divisor of 7y, and thus of y, so the smallest possible value of y is 4, which makes x equal to 7, and the answer is 4+7 = 11.
mykrasovski
Joined: 17 Aug 2018
Last visit: 17 Apr 2022
Posts: 340
Given Kudos: 253
Location: United States
Schools: Tuck '22 (M) Kellogg '22 (D)
WE:General Management (Other)
16 Jan 2020, 20:39
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We can solve this question using factoring. The school wants to spend equal amounts of money to buy X copiers of model 1 and Y copiers of model 2.
Mathematically, this means that the total cost of model 1 should be equal to the total cost of model 2, i.e. $320*X = $560*Y. When would the costs be the same?
320*X = 560*Y
32*X = 56*Y
\(2^5*X = 7*2^3*Y\)
It is clear that the left hand side needs to be multiplied by 7 and the right hand side needs to be multiplied by \(2^2\) for the equation to be right. So, 7+\(2^2\) = 11. Answer (C).
Mathematically, this means that the total cost of model 1 should be equal to the total cost of model 2, i.e. $320*X = $560*Y. When would the costs be the same?
320*X = 560*Y
32*X = 56*Y
\(2^5*X = 7*2^3*Y\)
It is clear that the left hand side needs to be multiplied by 7 and the right hand side needs to be multiplied by \(2^2\) for the equation to be right. So, 7+\(2^2\) = 11. Answer (C).
