CMAT 2015 Solved Question Paper

(d) Let CP of the cow be ₹ x and that of buffalo be ₹ y
∴ SP of cow at loss of 10% = (x - 10% of x)
$$ = $\left(\mathit{x}-\frac{10}{100}\times \mathit{x}\right)$ = ₹ $\frac{9\mathit{x}}{10}$
Also, SP of buffalo at profit of 20% = (y +20% of y)
$$ = $\left(\mathit{y}+\frac{20}{100}\times \mathit{y}\right)$ = ₹ $\frac{6\mathit{y}}{5}$
According to the question,
$$$\frac{9\mathit{x}}{10}+\frac{6\mathit{y}}{5}$ = (x + y)
⇒$$$\frac{9\mathit{x}+12\mathit{y}}{10}$ = x + y
⇒$$9x + 12y = 10x + 10y
⇒$$ x = 2y $$ ......(i)
Now, SP of cow at 5% profit = (x + 5% of x)
$$ = $\left(\mathit{x}+\frac{5}{100}\times \mathit{x}\right)$
$$ = $\left(\mathit{x}+\frac{\mathit{x}}{20}\right)=\mathrm{₹}\frac{21\mathit{x}}{20}$
Also, SP of buffalo at 15% loss = (y - 5% of x)
$$ = $\left(\mathit{y}-\frac{15}{100}\times \mathit{y}\right)$ = $\frac{17\mathit{y}}{20}$
According to the question,
$$(x+ y ) - $\left(\frac{21\mathit{x}}{20}+\frac{17\mathit{y}}{20}\right)$ = 800
⇒20x + 20y - 21x - 17y = 800 x 20
⇒$$ -x + 3y = 16000 $$.........(ii)
⇒ -2 y + 3 y = 16000 [From Eq. (i)]
$$y = 16000
So, the required selling price of buffalo is ₹ 16000.