E. m. zalamea actuarial services, inc. Interview Questions, Process, and Tips

Fathers before age=3*before sons age—-(1)
After 2.5 years..
Fathers before age+2.5 years =2.5*sons age—-(2)
Substitute (1) in (2),
3*sons before age+2.5 years=2.5*sons age
0.5*sons age=2.5 years
Sons before age=5years
So current age after 2.5 years,
Sons before age+2.5 =5+2.5=7.5 years

Suppose the lengthier arm of weighing pan is of x cm and other arm is y cm .Also let weight if each melon be m kg.
so applying equilibrium of torque principles ,we get
case 1:-
1x-8my=0
case 2:-
2mx-1y=0
using case 1 equation , we substitute value of x into case 2 equation;
16mmy-1y=0
(16mm-1)y=0
y is length of weighing arm and cannot be 0,
therefore ,
16mm-1 =0
16mm=1
mm=1/16
m= square root (1/16)
m=+-1/4
m is weight of melon and cant be negative.
Hence m, weight of one melon is 1/4 kg

e. 56 is 7*8 which is the product of 7 with even no.

11

( b ) 216.90

(34/17)*3+34

A certain sum amounts to Rs. 1725 in 3 years

and amounts to Rs.1875 in 5 years

so interest of 2 years = 1875 -1725

= 150

so interest of 1 year = 75

so interest of 3 years = 75 × 3 =225 rs

so , Principal = Amount – SI

= 1725 – 225

= 1500 rs

now ,

S.I. = P × N × R /100

75 = 1500 × 1 × R /100

R = 75 / 15

R = 5%

A

24

A

16.8 sec

if rs 12 and rs 15 per meter then 750*12+750*15

One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.

Robin =72
Dravid = 98
Azhar = 22
Sachine = 98
Total score = 290

48,400

94

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