Pyramid e&c Interview Questions, Process, and Tips

Friends,
The Answer given by ‘Gaurav Sharma’ is correct and
the approach (bottom to top) suggested by ‘Shailesh’ is
good. But with minor correction we can arrive the solution
using this approach:
After 5th loot, No. of breads left = 3
after 4th loot, no. of breads left = (3+0.5)x2 = 7
after 3rd loot, no. of breads left = (7+0.5)x2 = 15
after 2nd loot, no. of breads left = (15+0.5)x2 = 31
after 1st loot, no. of breads left = (31+0.5)x2 = 63
So, before 1st loot, no. of breads left = (63+0.5)x2 = 127

3000

20

b

FYI

Error is at 48. Correct once is 58

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

pages # figures
1000 1000 4
from 999 to 100 (999-99)*3 2700
from 99 to 10 (99-9)*2 180
from 9 to 1 (9-0)*1 9
total sum= 2893

sunday

8400

1!=1
2!=2×1=2
3!=3x2x1=6
4!=4x3x2x1=24
5!=5x4x3x2x1=120
6!=6x5x4x3x2x1=720
The answer is 120.

9654

Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.

108