CoCubes Aptitude Questions

In the final state, the number of girls should be 1.5 * the number of boys.
When 50% of the boys are taken as girls, let the number of boys = x
Number of girls = 1.5x
Total number of students = 2.5x
Original number of boys = 2x (50% of boys = x)
Original number of girls = 0.5x
Girls form 20% of the overall class

Initially, net flow in the tank = 40 litres/ 8 minutes = 5 litres/minute
Now, the increase in outflow should be greater than 5 litres/min for the tank to fill up.
Increase in outflow = 1.5x – x = 0.5x
Only in option C, 0.5 * 12 = 6 litres/min > 5 litres/min

Let HCF of (x, y) be h. Then we can write x = h * a and y = h * b. Furthermore, note that HCF (a, b) = 1. This is a very important property. One that seems obvious when it is mentioned but a property a number of people overlook.

So, we can write x = 35a; y = 35b

x + y = 1085 => 35(a + b) = 1085. => (a + b) = 31. We need to find pairs of co-prime integers that add up to 31. (Another way of looking at it is to find out integers less than 31 those are co-prime with it or phi(31) as had mentioned.

Since 31 is prime. All pairs of integers that add up to 31 will be co-prime to each other. Or, there are totally 15 pairs that satisfy this condition.

Let us assume Car A travels at a speed of a and Car B travels at a speed of b.
Further, let us assume that they meet after t minutes.
Distance traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.
Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.
Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).
=> at = 54b ---------- (1)
and bt = 24a -------- (2)
Multiplying equations 1 and 2

we have ab * t² = 54 * 24 * ab
=> t² = 54 * 24
=> t = 36

Sum of first 12 terms is equal to sum of first 14 terms.
Sum of first 14 terms = Sum of first 12 terms + 13^th term + 14^th term
=> 13^th term + 14^th term = 0

Let us assume 13^th term = k, common ratio = r. 14^th term will be kr.
k + kr = 0
k (1 + r) = 0
=> r = -1 as k cannot be zero

Common ratio = -1.
Now, if the first term of this GP is a, second term would be -a, third would be a and so on
The GP would be a, -a, a, -a, a, -a,...

Sum to even number of terms = 0
Sum to odd number of terms = a

Sum to 17 terms is 92 => a = 92
Third term = a = 92