Question

A farmer spends pounds $752$ in buying horses and cows; if each horse costs pounds $37$ and each cow pounds $23$, how many of each does he buy?

• A. Horses $=11$, cows $=15$
• B. Horses $=12$, cows $=12$
• C. Horses $=10,cows$=9$$
• D. Horses $=7,cows$=19$

Answer 1

The correct option is A Horses $=11$, cows $=15$

Let the number of horses be $x$ and and cows be $y$

According to question, we have

$37x+23y=752$ ....(i)

$\frac{37x}{23}+y=\frac{752}{23}x+\frac{14x}{23}+y=9+\frac{16}{23}x+y+\frac{14x-16}{23}=9$

$x$ and $y$ are integers, because number of animals can only be an integer value

$\Rightarrow \frac{14x-16}{23}=$ integer

$\Rightarrow \frac{70x-80}{23}=$ integer

$3x-3+\frac{x-11}{23}=$ integer

$\Rightarrow \frac{x-11}{23}=$ integer

Let the integer be $p$

$\frac{x-11}{23}=px=23p+11$ ........(ii)

Substituting value of $x$ in (i)

$37(23p+11)+23y=752\Rightarrow 23y=345-851p\Rightarrow y=15-23p$ .......(ii)

We can see from (ii) that $x<0$ for integer $p<0$ and from (iii) $y<0$ for integer $p>0$

So, $p$ can only be equal to $0$

Substituting $p=0$ in (ii) and (iii)

$\Rightarrow x=11,y=15$

So, the number of horses are $11$ and cows are $15$.