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A man buys some shirts and some ties. The shirts cost $7 each and the

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A man buys some shirts and some ties. The shirts cost $7 each and the  [#permalink]

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New post 08 Feb 2018, 06:35
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A man buys some shirts and some ties. The shirts cost $7 each and the ties cost $3 each. If the man spends exactly $81 and buys the maximum number of shirts possible under these conditions, what is the ratio of shirts to ties?

(A) 5:3

(B) 4:3

(C) 5:2

(D) 4:l

(E) 3:2

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A man buys some shirts and some ties. The shirts cost $7 each and the  [#permalink]

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New post 08 Feb 2018, 09:07
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Bunuel wrote:
A man buys some shirts and some ties. The shirts cost $7 each and the ties cost $3 each. If the man spends exactly $81 and buys the maximum number of shirts possible under these conditions, what is the ratio of shirts to ties?

(A) 5:3

(B) 4:3

(C) 5:2

(D) 4:l

(E) 3:2


Method 1: algebraic

let number of shirts be \(x\) & number of ties be \(y\) and we need \(\frac{x}{y}\) i.e. \(x:y\)

so we have \(7x+3y=81\)---------(1)

\(3y=81-7x =>y=\frac{(81-7x)}{3} => y= 27-\frac{7x}{3}\)

For \(y\) to be integer \(x\) has to be a multiple of \(3\). so pairs of possible \(x\) & \(y\) are

(3,20), (6,13), (9,6).

Now we need maximum shirt so \(x=9\) & \(y=6\)

Hence ratio\(=\frac{9}{6}=3:2\)

Option E

----------------------------------------------
Method 2: use substitution. Work through options.

For option E: let shirts be \(3x\) & tie be \(2x\)

so we have \(3x*7+2x*3=81 =>x=3\)

For other options you will not get the common factor as integer, hence can be rejected.
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Re: A man buys some shirts and some ties. The shirts cost $7 each and the  [#permalink]

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New post 08 Feb 2018, 12:23
Bunuel wrote:
A man buys some shirts and some ties. The shirts cost $7 each and the ties cost $3 each. If the man spends exactly $81 and buys the maximum number of shirts possible under these conditions, what is the ratio of shirts to ties?

(A) 5:3

(B) 4:3

(C) 5:2

(D) 4:l

(E) 3:2



A basic equation can be formed as below...no of shirts be S, no of ties be T
7S+3T=81
=> 7S=3(27-T)
=> 27-T has t be multiple of 7. To maximize S, minimise T ,

Minimum T cab be 6 so that 27-T ie, 21 is multiple of 7.

=> S=9 & T=6
=> S:T = 3:2

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Re: A man buys some shirts and some ties. The shirts cost $7 each and the  [#permalink]

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New post 12 Feb 2018, 17:39
Bunuel wrote:
A man buys some shirts and some ties. The shirts cost $7 each and the ties cost $3 each. If the man spends exactly $81 and buys the maximum number of shirts possible under these conditions, what is the ratio of shirts to ties?

(A) 5:3

(B) 4:3

(C) 5:2

(D) 4:l

(E) 3:2


We can create the equation:

7S + 3T = 81

7S = 81 - 3T

7S = 3(27 - T)

S = 3(27 - T)/7

The greatest value of S will be obtained when T is as small as possible. The smallest value of T which produces a number divisible by 7 after substitution is 6, so when T is 6 we have:

S = 3(21)/7 = 9

So, the ratio of S to T is 9 : 6 = 3 : 2.

Answer: E
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Re: A man buys some shirts and some ties. The shirts cost $7 each and the &nbs [#permalink] 12 Feb 2018, 17:39
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