Gmg Interview Questions, Process, and Tips

9

Let x be fathers present age and y be son present age.
5 yrs ago, the age of father and son be x-5 & y-5.
Then,
x-5+y-5=40
x+y-10=40
x+y=50
y=50-x ———–> (1)
ratio between father and son in present age
x:y=4:1
x/y=4/1
x=4y
Apply eq (1) ,
x=4(50-x)
x=200-4x
x+4x=200
5x=200
x=200/5
x=40,,
=> The present age of father is 40.

9 and 16

( C ) Crawl

B, 8mm x 5 mm = 40 mm^2
1000mm x 1000mm = 1,000,000mm^2
= 1,000,000 mm^2/40mm^2
= 25,000

1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.

Suppose that total 100 employees are in company….. out of that 35 are man and remaining 65 are women.
20% of man 35 = 20*35/100
40% of women = 40*65/100
total employees = 7+26 = 33 out of 100
so, Ans = 33 %

Gain=(profit/c.p)*100

Gain =selling price-cost price =900-675
=225.
Gain=(225/675)*100.
=33.3%

put 1 red marble in one jar and all the rest (99) in the
other.
This gives you 50% from the first jar (if they pick that
jar they will get red 100% of the time). For the other jar
the chances are 49/99 or 49.494949%. Divide that by 2 and
its 24.7474%. Total odds are 50% plus 24.7474% = 74.7474%

10 x 27=270

Explanation:

The number of ways of selecting three men, two women and three children is:

= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 ways.

D=5, G=1

19

to find the root of f(x) = 0;

20&130