Friday, 28 February 2020

The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?

Aptitude Solved Problems


1. The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?

Answer: 30 years

Explanation:
Let present age of the elder person = x
Present age of the younger person = ( x - 16 )
By given condition ,
( x - 6 ) = 3 * ( x - 16 - 6 )
( x - 6 ) = 3 * ( x - 22 )
( x - 6 ) = ( 3x - 66 )
2x = 60
x = ( 60 / 2 )
x = 30
∴ Present age of the elder person is 30 years.

2. A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

Answer: 88 feet

Explanation:
Area of the field = 680 sq.feet
length = 20 feet.
( length * breadth ) = 680 sq.feet
( 20 * breadth ) = 680
Breadth = ( 680 / 20 )
Breadth = 34 feet
Length of fencing = ( length + ( 2 * breadth ) )
= ( 20 + ( 2 * 34 ) )
= ( 20 + 68 )
= 88 feet
∴ Area to be fenced is 88 feet.

3. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division B is 35 kg. What is the average weight of the whole class?

Answer: 37.25 kg

Explanation:
Given that,
Average weight of divisions A is 40 kg.
Average weight of divisions B is 35 kg.
Total weight of students in division A = ( 36 * 40 )
Total weight of students in division B = ( 44 * 35 )
Total students = ( 36 + 44 )
= 80
Average = ( Total weight of students in division A + Total weight of students in division B ) / ( Total students )
= ( 36 * 40 ) + ( 44 * 35 ) / 80
= ( 9 * 40 ) + ( 11 * 35 ) / 20
= ( 9 * 8 ) + ( 11 * 7 ) / 4
= ( 72 + 77 ) / 4
= ( 149 / 4 )
= 37.25
∴ Average weight of the whole class is 37.25 kg.

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