Question

A container contains 120 lit of Diesel. From this container, 12 lit of Diesel was taken out and replaced by kerosene. This process was further repeated for two times. How much diesel is now there in the container ?
A) 88.01 lit
B) 87.48 lit
C) 87.51 lit
D) 87.62 lit

Answer 1

Answer:   B) 87.48 lit 

Explanation:

For these type of problems,

Quantity of Diesel remained = (q(1-pq)n)   

 

Here p = 12 , q = 120 

 

=> (120(1-12120)3) = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.

Answer 2

A container contains 120 lit of Diesel. From this container, 12 lit of Diesel was taken out and replaced by kerosene. This process was further repeated for two times. How much diesel is now there in the container ?
A) 88.01 lit
B) 87.48 lit
C) 87.51 lit
D) 87.62 lit

=> Option B

Description:

Quantity of Diesel remained = (q(1−p/q)^n)
 
Here p = 12 , q = 120 
 
=> (120(1−12/120)^3)

=>120 x 0.9 x 0.9 x 0.9 

=> 87.48 lit.