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.