Questions & Answers: "Unitary Method and Simple Interest"

Complete guide to "Unitary Method and Simple Interest" for Math students. Below you will find important questions and model answers to help you prepare.

5 Questions

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Question 1

1 Mark

Rs. 9700 are invested in the bank with simple interest at the rate of 5 p.c.p.a. To get the same amount that will be received in 4 years in 2.5 years only, by how much should the rate of interest needs to be increased?

Options

Option A

8%

Option B

5%

Option C is correct

3%

Option D

2.5%

Question 2

1 Mark

Instead of investing Rs. 3000 with simple interest for two and half years, if Rs. 4000 are invested with simple interest for two years, interest of Rs. 25 is earned more. What is the rate of interest?

Options

Option A

10 %

Option B

6 %

Option C

4 %

Option D is correct

5 %

Question 3

1 Mark

A vehicle covers a distance of 42 km in one hour. Then how much distance will it cover at the same speed in 20 minutes?

Options

Option A

21 km

Option B

840 m.

Option C is correct

14 km

Option D

20 km

Question 4

1 Mark

Santoshrao deposited Rs. 2,10,000 in bank for 5 years at simple interest. After the end of period, he received Rs. 3,36,000 from the bank. Then what was the rate of interest?

Options

Option A

11 %

Option B

10 %

Option C is correct

12 %

Option D

14 %

Question 5

1 Mark

The amount of certain principal at certain rate of interest after 5 years is Rs. 10800 with simple interest and the amount for the same principal and at the same rate of interest after 3 years is Rs. 9680 with simple interest. Find the principal and the rate of interest.

Options

Option A

P = ₹ 7000 R = 8%

Option B

P = ₹ 8000 R = 9%

Option C

P = ₹ 9000 R = 7%

Option D is correct

P = ₹ 8000 R = 7%

Explanation

Let P be the principal and R be the rate of interest per annum.The amount (A) with simple interest is given by A=P+P×R×T100A = P + \frac{P \times R \times T}{100}.Given:Amount after 5 years (A5A_5) = Rs. 10800P+P×R×5100=10800(1)P + \frac{P \times R \times 5}{100} = 10800 \quad (1)Amount after 3 years (A3A_3) = Rs. 9680P+P×R×3100=9680(2)P + \frac{P \times R \times 3}{100} = 9680 \quad (2)Subtract equation (2) from equation (1):(P+5PR100)(P+3PR100)=108009680\left(P + \frac{5PR}{100}\right) - \left(P + \frac{3PR}{100}\right) = 10800 - 9680 2PR100=1120\frac{2PR}{100} = 1120 This means the simple interest for 2 years is Rs. 1120.So, the simple interest for 1 year (SI1SI_1) is 11202=560\frac{1120}{2} = 560 Rs. Now, we can find the simple interest for 3 years (SI3SI_3):SI3=3×SI1=3×560=1680 Rs.SI_3 = 3 \times SI_1 = 3 \times 560 = 1680\text{ Rs.}Using equation (2), A3=P+SI3A_3 = P + SI_3:9680=P+16809680 = P + 1680 P=96801680P = 9680 - 1680 P=8000 Rs.P = 8000\text{ Rs.} Now, we can find the rate of interest (R) using SI1=P×R×1100SI_1 = \frac{P \times R \times 1}{100}:560=8000×R×1100560 = \frac{8000 \times R \times 1}{100} 560=80R560 = 80R R=56080R = \frac{560}{80} R=7%R = 7\% Thus, the principal is Rs. 8000 and the rate of interest is 7%.