Sw6 associates Interview Questions, Process, and Tips

A. 4

Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? 28 m

say the work is w and let no of days taken by b is ‘x’ which
we have to calculate
so work done by a in one day is w/6
work done by b in one day is w/x
a and b together can do work in 4 days ie=(w/6)+(w/x)=(w/4)
solving equation x=12
so no of days taken by b=12

E. Copper

13

4 1 1 4
5 1 1 4
+ 0 1 1 4
———-
9 3 4 2

the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;

4

Doug won the race
Explanation-
The ratio of steps covered by me, Doug, and Anne is = 6:7:8

Since Distance covered in Me 21 steps = Distance covered by Doug 24 steps = Distance covered by Anne 28 steps

LCM (21, 24, 28) = 168

For a distance of 168 units, the ratio of the distance covered in each step is;
= 8 : 7 : 6
Therefore the ratio of speeds is;
= 8×6: 7×7: 6×8
= 48 : 49 : 48

Here, the speed of the Doug is more than the speed of me and Anne
Hence, Doug won the race

26

1997

Initially (2*33 + 1*24)/(2+1) = 30p a gm
So, cost for 100g = 100*30 = 3000p

Later, (1*33 + 2*24)/(1+2) = 71/3 p a gm.
So, cost for 100g = 100*(71/3) = 2366.67p

He saves 3000 – 2366.67 = 633.33p

Concept of weighted average.