Tips And Tricks And Shortcuts on Perimeter Area Volume Questions

Question 1:If length of the rectangle is increased by 50% and breadth is decreased by 20%. Then what is the percentage change in the area?

Options: 

A. 70% decrease

B. 30 % increase

C. 20% increase

D. 20% decrease

Solution:     Original area = l * b

New length = 50% increase = \frac{150}{100}l

= \frac{3}{2}l
New breadth = 20% decrease = \frac{80}{100}b = \frac{4}{5}b

Therefore, new area = \frac{3}{2}l  * \frac{4}{5}b

New area =\frac{6}{5} l b

Change in Area = New Area – Original Area

Change in Area = \frac{6}{5}lb – lb

Change in Area = \frac{1}{5}lb

Percentage change in area = \frac{\frac{1}{5}lb}{lb} * 100

Percentage change in area = \frac{1}{5} * 100 = \frac{100}{5} = 20% 

Since, the answer is positive, it means there is increase in the area.

Correct option: C

Type 4: How to solve cost related problems

Question 1 A wall of trapezium shape has height 8 m. The parallel sides of trapezium are 4 m and 6 m. If the rate of painting per square meter is Rs.50 than find the cost painting the complete wall?

Options: 

A. Rs. 400

B. Rs. 420

C. Rs. 540

D. Rs. 450

Solution:    Area of trapezium =\frac{1}{2} * (sum of parallel sides) * distance between them

Area of trapezium = \frac{1}{2} * (4 + 6) + 8

Area of trapezium = \frac{1}{2} * (18)

Area of trapezium = 9 square meter

Rate of painting per square meter is Rs.50

Therefore, to paint 9 square meter, total cost of painting = 9 * 50 = Rs. 450