QQuestion
What is the rate of simple interest per annum, when the interest of some money in 10 years will be \frac{2}{5} part of its amount?
Exam-ready answer • 02 Mark
✓Answer
Given: In 10 years, interest = \frac{2}{5} of the amount.
Let principal = P, rate = r% p.a., interest in 10 years = I.
Relation with amount:
I = \frac{2}{5}(P + I)
⇒ 5I = 2P + 2I
⇒ 3I = 2P
⇒ I = \frac{2}{3}P
Simple interest formula for 10 years:
I = P × r × 10 / 100
So, P × r × 10 / 100 = \frac{2}{3}P
⇒ r × 10 / 100 = \frac{2}{3}
⇒ r = \frac{2}{3}\times \frac{100}{10}=\frac{20}{3}
Related Questions
More WBBSE Moderate level questions
At the same rate of simple interest in percent per annum, if a principal becomes the amount of Rs. 7,100 in 7 years and Rs. 6,200 in 4 years, determine the principal and rate of simple interest in percent per annum.
The annual interest is part of its principal, then determine the interest of ₹ 690. The annual interest is for 8 months.
After retirement Gobinda Babu got ₹ 5,00,000. He deposited a part of it in post office at 7.2% simple interest p.a. and the other part in a bank at 6% simple interest p.a. Every year he got ₹ 33,600 in total as interest from bank and post office. Find the amounts he deposited in bank and post office separately.
If the annual rate of simple interest decreases from 5.5% to 4.5%, then the total interest is decreased by ₹ 250. Find the capital.
Find the rate of simple interest per annum when the interest of some money in 5 years will be part of its principal.
Simple interest of ₹ Y for Z month at the rate of X% per annum is (a) ₹ XYZ/1200 (b) ₹ XYZ/100 (c) ₹ 200 (d) ₹ XYZ/120
In how many years a sum of money at 6 % simple interest per annum would be 4 double?
If the amount of ₹ 180 after one year will be ₹ 198, then the rate of simple interest is _____
Find x, if sin x = cos (x – 20°)
The Median of first (2n + 1) natural numbers is . Find the value of n.
The ratio of the length of radius of two solid right circular cylinders is 2:3 and the ratio of their heights is 5:3. Find the ratio of the area of curved surfaces.
If cos⁴θ – sin⁴θ = , find the value of 1 – 2 sin²θ.
Small steps build strong concepts.