Ipse Interview Questions, Process, and Tips

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

39/50

Ans =18
Explanation:
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X – 4)
bullets each.
But it is given that,after they shot 4 bullets each, total
number of bullets remaining is equal to the bullets each
had after division i.e. X
Therefore, the equation is
3 * (X – 4) = X
3 * X – 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18

First write equations from info:
(A) (Mon + Tue + Wed)/3 = 111 Rearrange as ——–> Tue + Wed = 111 – Mon
(B) (Tue + Wed + Thu)/3 =102 Rearrange as ——–> Tue + Wed = 102 – Thu
(C) Thu = 0.8(Mon)

Substitute equation C into B:
(B) Tue + Wed = 102 – 0.8(Mon)

At this point I changed the values for clearer algebra:
Mon = x
Tue + Wed = y

Re-write equations A & B with new values:
(A) y = 111 – x
(B) y = 102 – 0.8x

Solve simultaneous equations:

111 – x = 102 – 0.8x
111 – 102 = x – 0.8x (Re-arraged)
9 = 0.2x
x = 45

Thus, Mon = 45C

Thu = 0.8(45)
Thu = 36C

So the answer is it was 36C on Thursday

X-Y=-1

8 is the answer
b’coz no two have the same age
8+6+5+4+3=26

D. Rs. 187500

15/46=0.32

40
40
30
60
– 10 cows
+ 20 days
if 40 cows can eat for forty days
then 20 cows can eat for 80 days

9

2

No of men employed = N

No of days to finish the work = 9 days

No of men after increase = (N + 10)

No of days to finish the work = 6 days

Equating mandays

9N = (N+10)*6

9N — 6N = 60

3N = 60

N = 20

No of men employed = 20

6.35

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