Management Aptitude Test Sep 2013 Paper

(b) Number of questions for team A = 300
Also, number of questions for team B = 90% of 300
= $\frac{90\times 300}{100}$ = 270
Now, let the questions attempted per hour by A = x
Then, questions attempted per hour by B = x + 7
By using the conditionof given in question,
$\frac{270}{\mathit{x}}+3=\frac{300}{\mathit{x}+7}$ + 6
$\frac{270}{\mathit{x}}=\frac{300}{\mathit{x}+7}$ + 3
$\frac{270}{\mathit{x}}=\frac{300+3\left(\mathit{x}+7\right)}{\mathit{x}+7}$
270 (x + 7) = (300 + 3x+ 21)x
270x + 1890 = 321x + 3${\mathit{x}}^2$
${\mathit{x}}^2$ + 51x - 1890 = 0
${\mathit{x}}^2$ + 17x - 630 = 0
${\mathit{x}}^2$ + 35x - 18x - 630 = 0
x(x + 35) - 18 (x + 35)= 0
(x - 18) (x + 35) = 0
⇒ x = 18, -35
So, team A attempted 18 questions per h.
(∵ Negative value of x is not permitted)