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Math Expert
Joined: 02 Sep 2009
Posts: 50572
A man buys some shirts and some ties. The shirts cost $7 each and the [#permalink] ### Show Tags 08 Feb 2018, 05:35 1 00:00 Difficulty: 5% (low) Question Stats: 88% (01:57) correct 12% (03:18) wrong based on 110 sessions ### HideShow timer Statistics 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] ### Show Tags 08 Feb 2018, 08:07 1 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.
Intern
Joined: 03 Jul 2015
Posts: 8
Re: A man buys some shirts and some ties. The shirts cost $7 each and the [#permalink] ### Show Tags 08 Feb 2018, 11:23 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

E
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Affiliations: Target Test Prep
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Posts: 2830
Re: A man buys some shirts and some ties. The shirts cost $7 each and the [#permalink] ### Show Tags 12 Feb 2018, 16:39 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.

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