Amf Interview Questions, Process, and Tips

1261

let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer

Profit one with 3 & 2 nd 2.4

Answer: 10
Reason:
The best case happens ONLY when each rat dies just as they
taste the FIRST bottle given to them (you can imagine it a
miracle 😀 )
In this case, the very first 10 attempts reveal the
poisonous bottles. So the answer is 10.

Explanation:

72 — 9
? —- 1

=> Down Stream = 8

45 —- 9
? —- 1

=> Up Stream = 5

Speed od current S = ?

S = (8 – 5)/2 = 1.5 kmph.

The first 10 odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31.

Sum of the odd prime numbers = (3+5+7+11+13+17+19+23+29+31)

= 158

Number of odd prime numbers = 10

We know, Average = (sum of the 10 odd prime numbers ÷ Number of odd

prime numbers)

Average =

= 15.8

∴ The Average of first 10 prime numbers which are odd is 15.8

Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.

b

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

b

A whole number

23 Teams

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