Smith school of business at queen's university Interview Questions, Process, and Tips

56.25 % is the right answer.
area of square = side ^ 2
at first, it was 100% , means 1
then, a side was enlarged to 25%, means now, 125% means new side is 1.25
taking square will give 1.5625
1.5625 – 1 = 0.5625
so, 56.25% increment in area.

12

20

(b) 16.66%

Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar

(As there was an increase in price so, x + 20%x = 1.2x)

At current rates for ₹x you can purchase y/1.2 gm of sugar

So the reduced consumption is y-(y/1.2)

Percentage change = (reduced consumption/ original consumption ) *100

That is (0.2/1.2) *100 = 16.66% (approx)

In 3500 g rod contains 74% of silver = 3500 * 74/100 = 2590 g
Then 3500 g + 500 g of rod contains 84% of silver
Let x be the silver contained in 500 g of silver
(2590/3500 * 100) + (x/500 * 100) = 84
74 + x/5 = 84
(370 + x) /5 = 84
370 + x = 420
x = 50
Then the percentage of silver contained in 500 g of rod = 50/500 *100 =10%

2/3 * 15 miles = 10 miles then

time for 10 miles travel is t1 = 10/40 = 1/4

then remaning distance is 5 miles

time for 5 miles travel is t2 = 5/60 = 1/12

t1+ t2= 1/3 in seconds * 60 = 20 min

5*(root2)

3

Assume Q can do the work alone in Q days:

work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.

324

the sum of all = 10
Accept
324 = 3+2+4 =/=10 (NOT EQUAL TO 10)

we need to take half tabulate among the 4 tablets. den it ll be like 1 tabulate is of fever and one tabulate is of cough

Conclusion 1 follows

A parabola is a conic section.
So cut a cone obliquely.

let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240