WBBSE Class 10 Maths Arithmetic Chapter 2 Compound Interest And Uniform Rate Of Increase Or Decrease Multiple Choice Questions

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WBBSE Class 10 Maths Arithmetic Chapter 2 Compound Interest And Uniform Rate Of Increase Or Decrease Multiple Choice Questions

Example 1. The interest on Rs. 700 for 2 years at 10% compound interest per annum is

  1. ₹ 147
  2. ₹ 126
  3. ₹ 126
  4. ₹ 105

Solution: Answer is 1. ₹ 147

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Example 2. The time in which the simple interest and compound interest will be same on a certain sum of money in an equal rate of interest is

  1. 3 years
  2. 2 years
  3. 1 year
  4. \(\frac{1}{2}\) year

Solution: Answer is 3. 1 year

WBBSE Class 10 Maths Arithmetic Chapter 2 Compound Interest And Uniform Rate Of Increase Or Decrease Multiple Choice Questions

Example 3. The depreciation rate of a Machine that cost Rs. 200 in 1st year is 15% and 2nd year is 10%. The cost of the machine after 2 years will be

  1. Rs. 150
  2. Rs. 147
  3. Rs. 157
  4. Rs. 153

Solution: Answer is 4. Rs. 153

Geometric Progression Mcqs With Solutions

Example 4. The present price of a machine is Rs. x and the rate of depreciation is 4y% per annum. The price of the machine will be after Z years is

  1. \(x\left(1+\frac{y}{25}\right)^z\)
  2. \(x\left(1-\frac{y}{25}\right)^z\)
  3. \(2 x\left(1-\frac{y}{100}\right)^z\)
  4. \(2 x\left(1-\frac{y}{25}\right)^{2 z}\)

Solution: Answer is 2. \(x\left(1-\frac{y}{25}\right)^z\)

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Example 5. A gets Rs. 720 as an amount after 2 years from a Bank on a sum of Rs. 500. The rate of compound interest per annum is

  1. 20%
  2. 17\(\frac{1}{2}\)%
  3. 15%
  4. 10%

Solution: Answer is 3. 15%

Example 6. In case of compound interest the rate of compound interest per annum is

  1. Equal
  2. Unequal
  3. Both equal and unequal
  4. None of these

Solution: Answer is 3. Both equal and unequal

Example 7. At present the population of a village is p and if increase rate of population per year be 2r%, the population will be after n years

  1. \(\mathrm{P}\left(1-\frac{r}{100}\right)^n\)
  2. \(2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)
  3. \(\mathrm{P}\left(1-\frac{r}{100}\right)^{2 n}\)
  4. \(2 P\left(1-\frac{r}{100}\right)^{2 n}\)

Solution: Answer is 2. \(2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)

Class 10 Arithmetic Chapter 2 Mcqs With Answers

Example 8. The present price of a machine is ₹ 2P and if price of the machine decreased by 2r% in each year, the price of machine will be

  1. \(₹ \mathrm{P}\left(1-\frac{r}{100}\right)^n\)
  2. \(₹ 2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)
  3. \(₹ P\left(1-\frac{r}{100}\right)^{2 n}\)
  4. \(₹ 2 \mathrm{P}\left(1-\frac{r}{100}\right)^{2 n}\)

Solution: Answer is 4. \(₹ 2 \mathrm{P}\left(1-\frac{r}{100}\right)^{2 n}\)

Example 9. A person deposited ₹ 100 in a Rank and got the amount ₹ 121 for 2 years the rate of compound interest is

  1. 10%
  2. 20%
  3. 5%
  4. 10\(\frac{1}{2}\)%

Solution: Answer is 1. 10%

Class 10 Maths Arithmetic Chapter 2 MCQs

Example 10. The price of a machine decrease at a rate of x% from its previous years value. If the present price is ₹ P, the price of the machine 2 years ago was

  1. \(₹ \left(\mathrm{P}-\mathrm{P} \times \frac{2 x}{100}\right)\)
  2. \(₹ \frac{P}{\left(1-\frac{x}{100}\right)^2}\)
  3. \(₹ \frac{P}{\left(1+\frac{x}{100}\right)^2}\)
  4. \(₹ P\left(1-\frac{x}{100}\right)^2\)

Solution: \(\mathrm{P}=?\left(1-\frac{x}{100}\right)^2 \quad ?=\frac{\mathrm{P}}{\left(1-\frac{x}{100}\right)^2}\)

∴ Answer is \(₹ \frac{P}{\left(1-\frac{x}{100}\right)^2}\).

Example 11. If the compound interest in 1 year of a certain principal at a certain rate per annum be ₹ x and the simple interest for 1 year is ₹ y then

  1. x > y
  2. x < y
  3. x = y
  4. x > y

Solution: Simple interest for a certain amount for 1 year at any rate = compound interest for that amount for 1 year at the same rate of interest.

∴ x = y

∴ Answer is x = y

Example 12. If the rate of compound interest is r% per annum and the principal at the end of first year be ₹ P then the principal at the beginning, of the third year is

  1. \(₹ P\left(1+\frac{r}{100}\right)^3\)
  2. \(₹ \mathrm{P}\left(1+\frac{r}{100}\right)^2\)
  3. \(P\left(1+\frac{r}{100}\right)^2\)
  4. \(₹ P\left(1+\frac{r}{100}\right) \frac{r}{100}\)

Solution: Principal at the beginning of 2nd year = ₹ P.

∴ Principal at the beginning of 3rd year = \(₹ P\left(1+\frac{r}{100}\right)\)

∴ Answer is 3. \(₹ P\left(1+\frac{r}{100}\right)^2\)

Geometric Progression Mcqs Class 10

Example 13. In case of compound interest if the principal be ₹ P, rate of interest per annum = r, time = n years, and the final amount A, then

  1. \(\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n\)
  2. \(A=P\left(1+\frac{r}{100}\right)\)
  3. \(\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n-\mathrm{P}\)
  4. none of these

Solution: Answer is 1. \(\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n\)

Example 14. The fixed interval for which the interest is paid is called

  1. Interest
  2. Interest period
  3. Amount
  4. Time

Solution: Interest period

∴ Answer is 2. Interest period

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