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Chapter 2

Compound Interest

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 2.1

Question 1

Find the amount and the compound interest on ₹8000 at 5% per annum for 2 years.

Answer

Principal for first year = ₹8000.

Interest for the first year = ₹ 8000×5×1100\dfrac{8000 \times 5 \times 1}{100} = ₹400.

Amount after one year = ₹8000 + ₹400 = ₹8400.

Principal for the second year = ₹8400.

Interest for the second year = ₹ 8400×5×1100\dfrac{8400 \times 5 \times 1}{100} = ₹420.

Amount after 2 years = ₹8400 + ₹420 = ₹8820.

Compound interest for 2 years = Final amount - Principal = ₹8820 - ₹8000 = ₹820.

Hence, the amount and compound interest on ₹8000 at 5% per annum after 2 years is ₹8820 and ₹820 respectively.

Question 2

A man invested ₹46875 at 4% per annum compound interest for 3 years. Calculate :

(i) the interest for the first year.

(ii) the amount standing to his credit at the end of the second year.

(iii) the interest for the third year.

Answer

(i) Principal for first year = ₹46875.

Interest for the first year = ₹ 46875×4×1100=187500100\dfrac{46875 \times 4 \times 1}{100} = \dfrac{187500}{100} = ₹1875.

Hence, the interest for the first year = ₹1875.

(ii) Amount after one year = ₹46875 + ₹1875 = ₹48750.

Principal for the second year = ₹48750.

Interest for the second year = ₹ 48750×4×1100=195000100\dfrac{48750 \times 4 \times 1}{100} = \dfrac{195000}{100} = ₹1950.

Amount after 2 years = ₹48750 + ₹1950 = ₹50700.

Hence, the amount standing to his credit at the end of the second year is ₹50700.

(iii) Principal for third year = ₹50700.

Interest for the third year = ₹ 50700×4×1100=202800100\dfrac{50700 \times 4 \times 1}{100} = \dfrac{202800}{100} = ₹2028.

Hence, the interest for the third year = ₹2028.

Question 3

Calculate the compound interest for the second year on ₹8000 invested for 3 years at 10% p.a.

Also find the sum due at the end of third year.

Answer

Principal for first year = ₹8000.

Interest for the first year = ₹ 8000×10×1100\dfrac{8000 \times 10 \times 1}{100} = ₹800.

Amount after one year = ₹8000 + ₹800 = ₹8800.

Principal for the second year = ₹8800.

Interest for the second year = ₹ 8800×10×1100\dfrac{8800 \times 10 \times 1}{100} = ₹880.

Amount after 2 years = ₹8800 + ₹880 = ₹9680.

Principal for the third year = ₹9680.

Interest for the third year = ₹ 9680×10×1100\dfrac{9680 \times 10 \times 1}{100} = ₹968.

Amount after 3 years = ₹9680 + ₹968 = ₹10648.

Hence, the compound interest for the second year on ₹8000 invested for 3 years at 10% p.a. is ₹880 and the sum due at the end of third year is ₹10648.

Question 4

Ramesh invested ₹12800 for three years at the rate of 10% per annum compound interest. Find :

(i) the sum due to Ramesh at the end of the first year.

(ii) the interest he earns for the second year.

(iii) the total amount due to him at the end of three years.

Answer

(i) Principal for first year = ₹12800.

Interest for the first year = ₹ 12800×10×1100=128000100\dfrac{12800 \times 10 \times 1}{100} = \dfrac{128000}{100} = ₹1280.

Amount after first year = ₹12800 + ₹1280 = ₹14080

Hence, the sum due to Ramesh at the end of first year = ₹14080.

(ii) Principal for the second year = ₹14080.

Interest for the second year = ₹ 14080×10×1100=140800100\dfrac{14080 \times 10 \times 1}{100} = \dfrac{140800}{100} = ₹1408.

Hence, the interest Ramesh earns for second year = ₹1408.

(iii) Amount after 2 years = ₹14080 + ₹1408 = ₹15488.

Interest for the third year = ₹ 15488×10×1100=154880100\dfrac{15488 \times 10 \times 1}{100} = \dfrac{154880}{100} = ₹1548.80

Amount after 3 years = ₹15488 + ₹1548.80 = ₹17036.80

Hence, the total amount due to Ramesh at the end of third year is ₹17036.80

Question 5

The simple interest on a sum of money for 2 years at 12% per annum is ₹1380. Find :

(i) the sum of money.

(ii) the compound interest on this sum for one year payable half-yearly at the same rate.

Answer

Given, S.I. = ₹1380, rate = 12% p.a. and time = 2 years.

(i) Let the sum of money be P, then

S.I.=P×R×T100⇒1380=P×12×2100⇒138000=24P⇒P=13800024⇒P=5750.S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1380 = \dfrac{P \times 12 \times 2}{100} \\[1em] \Rightarrow 138000 = 24P \\[1em] \Rightarrow P = \dfrac{138000}{24} \\[1em] \Rightarrow P = 5750.

Hence, the sum of money is ₹5750.

(ii) Since, the rate of interest is 12% per annum, therefore, the rate of interest half-yearly = 6%.

Principal for first half-year = ₹5750.

Interest for first half-year = ₹ 5750×6×1100=34500100\dfrac{5750 \times 6 \times 1}{100} = \dfrac{34500}{100} = ₹345.

∴ Amount after first half-year = ₹5750 + ₹345 = ₹6095.

Principal for the second half-year = ₹6095.

Interest for the 2nd half-year = ₹ 6095×6×1100=36570100\dfrac{6095 \times 6 \times 1}{100} = \dfrac{36570}{100} = ₹365.70

∴ Compound interest on the above sum for one year payable half-yearly = ₹345 + ₹365.70 = ₹710.70

Hence, the compound interest on ₹5750 for one year payable half-yearly at the same rate is ₹710.70

Question 6

A person invests ₹10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹11200. Calculate :

(i) the rate of interest per annum.

(ii) the amount at the end of second year.

Answer

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹11200.

Interest = Amount - Principal = ₹11200 - ₹10000 = ₹1200.

Interest = P×R×T100\dfrac{P \times R \times T}{100}

⇒1200=10000×R×1100⇒1200=100R⇒R=1200100⇒R=12\Rightarrow 1200 = \dfrac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100R \\[1em] \Rightarrow R = \dfrac{1200}{100} \\[1em] \Rightarrow R = 12%.

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹11200.

Interest for second year = 11200×12×1100=134400100\dfrac{11200 \times 12 \times 1}{100} = \dfrac{134400}{100} = ₹1344.

Amount at the end of second year = ₹11200 + ₹1344 = ₹12544.

Hence, the amount at the end of second year = ₹12544.

Question 7

Mr. Lalit invested ₹5000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹5325. Calculate :

(i) the rate of interest.

(ii) the amount at the end of second year, to the nearest rupee.

Answer

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹5325.

Interest = Amount - Principal = ₹5325 - ₹5000 = ₹325.

Interest = P×R×T100\dfrac{P \times R \times T}{100}

⇒325=5000×R×1100⇒325=50R⇒R=32550⇒R=6.5\Rightarrow 325 = \dfrac{5000 \times R \times 1}{100} \\[1em] \Rightarrow 325 = 50R \\[1em] \Rightarrow R = \dfrac{325}{50} \\[1em] \Rightarrow R = 6.5%.

Hence, the rate of interest is 6.5% per annum.

(ii) Amount after first year = ₹5325.

Interest for second year = 5325×6.5×1100=34612.5100\dfrac{5325 \times 6.5 \times 1}{100} = \dfrac{34612.5}{100} = ₹346.125.

Amount at the end of second year = ₹5325 + ₹346.125 = ₹5671.125.

Hence, the amount at the end of second year to the nearest rupee = ₹5671.

Question 8

A man invests ₹5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹5600. Calculate :

(i) the rate of interest per annum.

(ii) the interest accrued in the second year.

(iii) the amount at the end of the third year.

Answer

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹5600.

Interest = Amount - Principal = ₹5600 - ₹5000 = ₹600.

Interest = P×R×T100\dfrac{P \times R \times T}{100}

⇒600=5000×R×1100⇒600=50R⇒R=60050⇒R=12\Rightarrow 600 = \dfrac{5000 \times R \times 1}{100} \\[1em] \Rightarrow 600 = 50R \\[1em] \Rightarrow R = \dfrac{600}{50} \\[1em] \Rightarrow R = 12%.

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹5600.

Principal for second year = ₹5600.

Interest for second year = 5600×12×1100=67200100\dfrac{5600 \times 12 \times 1}{100} = \dfrac{67200}{100} = ₹672.

Hence, the interest accrued in second year = ₹672.

(iii) Amount after second year = ₹5600 + ₹672 = ₹6272

Principal for third year = ₹6272.

Interest for third year = 6272×12×1100=75264100\dfrac{6272 \times 12 \times 1}{100} = \dfrac{75264}{100} = ₹752.64

Amount after third year = ₹6272 + ₹752.64 = ₹7024.64

Hence, the amount at the end of third year = ₹7024.64

Question 9

Find the amount and the compound interest on ₹2000 at 10% p.a. for 2122\dfrac{1}{2} years, compounded annually.

Answer

Principal for first year = ₹2000.

Interest for the first year = ₹ 2000×10×1100\dfrac{2000 \times 10 \times 1}{100} = ₹200.

Amount after one year = ₹2000 + ₹200 = ₹2200.

Principal for the second year = ₹2200.

Interest for the second year = ₹ 2200×10×1100\dfrac{2200 \times 10 \times 1}{100} = ₹220

Amount after 2 years = ₹2200 + ₹220 = ₹2420.

Principal for next 12\dfrac{1}{2} year = ₹2420.

Interest for the next 12\dfrac{1}{2} year = ₹ 2420×10×12100=12100100\dfrac{2420 \times 10 \times \dfrac{1}{2}}{100} = \dfrac{12100}{100} = ₹121.

Amount after 2122\dfrac{1}{2} years = ₹2420 + ₹121 = ₹2541.

Compound interest for 2122\dfrac{1}{2} years = Final amount - Principal = ₹2541 - ₹2000 = ₹541.

Hence, the amount and compound interest on ₹2000 at 10% per annum after 2122\dfrac{1}{2} years is ₹2541 and ₹541 respectively.

Question 10

Find the amount and the compound interest on ₹50000 for 1121\dfrac{1}{2} years at 8% per annum, the interest being compounded semi-annually.

Answer

Since, the rate of interest is 8% per annum, therefore, the rate of interest half-yearly = 12\dfrac{1}{2} of 8% = 4%.

Principal for first half-year = ₹50000.

Interest for first half-year = 50000×4×1100=200000100\dfrac{50000 \times 4 \times 1}{100} = \dfrac{200000}{100} = ₹2000.

Amount after first half-year = ₹50000 + ₹2000 = ₹52000.

Principal for second half-year = ₹52000.

Interest for the second half-year = 52000×4×1100=208000100\dfrac{52000 \times 4 \times 1}{100} = \dfrac{208000}{100} = ₹2080.

Amount after one year = ₹52000 + ₹2080 = ₹54080.

Principal for third half-year = ₹54080.

Interest for the third half-year = 54080×4×1100=216320100\dfrac{54080 \times 4 \times 1}{100} = \dfrac{216320}{100} = ₹2163.20

Amount after 1121\dfrac{1}{2} year = ₹54080 + ₹2163.20 = ₹56243.20

Compound interest for 1121\dfrac{1}{2} year = Final amount - principal = ₹56243.20 - ₹50000 = ₹6243.20

Hence, the amount and the compound interest on ₹50000 for 1121\dfrac{1}{2} years at 8% per annum, the interest being compounded semi-annually are ₹56243.20 and ₹6243.20 respectively.

Question 11

Calculate the amount and the compound interest on ₹5000 in 2 years when the rate of interest for successive years is 6% and 8% respectively.

Answer

Principal for first year = ₹5000.

Interest for the first year = ₹ 5000×6×1100=30000100\dfrac{5000 \times 6 \times 1}{100} = \dfrac{30000}{100} = ₹300.

Amount after one year = ₹5000 + ₹300 = ₹5300.

Principal for the second year = ₹5300.

Interest for the second year = ₹ 5300×8×1100=42400100\dfrac{5300 \times 8 \times 1}{100} = \dfrac{42400}{100} = ₹424.

Amount after 2 years = ₹5300 + ₹424 = ₹5724.

Compound interest = Final amount - Principal = ₹5724 - ₹5000 = ₹724.

Hence, the amount and the compound interest are ₹5724 and ₹724 respectively.

Question 12

A sum of ₹9600 is invested for 3 years at 10% per annum at compound interest.

(i) What is the sum due at the end of the first year?

(ii) What is the sum due at the end of the second year?

(iii) Find the compound interest earned in 2 years.

(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.

(v) Hence, write down the compound interest for the third year.

Answer

(i) Principal for first year = ₹9600.

Interest for the first year = ₹ 9600×10×1100=96000100\dfrac{9600 \times 10 \times 1}{100} = \dfrac{96000}{100} = ₹960.

Amount after one year = ₹9600 + ₹960 = ₹10560.

Hence, the amount due at the end of first year = ₹10560.

(ii) Principal for second year = ₹10560.

Interest for the second year = ₹ 10560×10×1100=105600100\dfrac{10560 \times 10 \times 1}{100} = \dfrac{105600}{100} = ₹1056.

Amount after 2 years = ₹10560 + ₹1056 = ₹11616.

Hence, the amount due at the end of second year = ₹11616.

(iii) Compound interest = Final amount - Principal = ₹11616 - ₹9600 = ₹2016.

Hence, the compound interest earned in 2 years = ₹2016.

(iv) Difference between (ii) and (i) = ₹11616 - ₹10560 = ₹1056.

Interest on the above sum for 1 year = 1056×10×1100=10560100\dfrac{1056 \times 10 \times 1}{100} = \dfrac{10560}{100} = ₹105.60

Hence, the difference = ₹1056 and interest earned on it = ₹105.60

(v) Principal for third year = ₹11616.

Interest for the third year = ₹ 11616×10×1100=116160100\dfrac{11616 \times 10 \times 1}{100} = \dfrac{116160}{100} = ₹1161.60

Hence, the compound interest for third year = ₹1161.60

Question 13

The simple interest on a certain sum of money for 2 years at 10% per annum is ₹1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.

Answer

Let the sum of money be ₹x.

Given, simple interest = ₹1600.

∴1600=P×R×T100⇒1600=x×10×2100⇒1600=20x100⇒160000=20x⇒x=16000020⇒x=₹8000.\therefore 1600 = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1600 = \dfrac{x \times 10 \times 2}{100} \\[1em] \Rightarrow 1600 = \dfrac{20x}{100} \\[1em] \Rightarrow 160000 = 20x \\[1em] \Rightarrow x = \dfrac{160000}{20} \\[1em] \Rightarrow x = ₹8000.

Principal for first year = ₹8000.

Interest for the first year = ₹ 8000×10×1100=80000100\dfrac{8000 \times 10 \times 1}{100} = \dfrac{80000}{100} = ₹800.

Amount after one year = ₹8000 + ₹800 = ₹8800.

Principal for the second year = ₹8800.

Interest for the second year = ₹ 8800×10×1100=88000100\dfrac{8800 \times 10 \times 1}{100} = \dfrac{88000}{100} = ₹880.

Amount after 2 years = ₹8800 + ₹880 = ₹9680.

Principal for the third year = ₹9680.

Interest for the third year = ₹ 9680×10×1100=96800100\dfrac{9680 \times 10 \times 1}{100} = \dfrac{96800}{100} = ₹968

Amount after 3 years = ₹9680 + ₹968 = ₹10648

Compound interest = Final amount - Principal = ₹10648 - ₹8000 = ₹2648.

Hence, the amount and compound interest due = ₹10648 and ₹2648 respectively.

Question 14

Vikram borrowed ₹20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after 2122\dfrac{1}{2} years.

Answer

Simple interest = P×R×T100.\dfrac{P \times R \times T}{100}.

∴ S.I. = 20000×10×2.5100=500000100\dfrac{20000 \times 10 \times 2.5}{100} = \dfrac{500000}{100} = ₹5000.

Calculating, compound interest :

Principal for first year = ₹20000.

Interest for first year = 20000×10×1100=200000100\dfrac{20000 \times 10 \times 1}{100} = \dfrac{200000}{100} = ₹2000.

Amount after first year = ₹20000 + ₹2000 = ₹22000.

Principal for second year = ₹22000.

Interest for second year = 22000×10×1100=220000100\dfrac{22000 \times 10 \times 1}{100} = \dfrac{220000}{100} = ₹2200.

Amount after second year = ₹22000 + ₹2200 = ₹24200.

Principal for next 12\dfrac{1}{2} year = ₹24200.

Interest for next 12\dfrac{1}{2} year = 24200×10×12100=242000200\dfrac{24200 \times 10 \times \dfrac{1}{2}}{100} = \dfrac{242000}{200} = ₹1210.

Amount after 2122\dfrac{1}{2} year = ₹24200 + ₹1210 = ₹25410.

Compound interest = Final amount - principal = ₹25410 - ₹20000 = ₹5410.

Difference between C.I. and S.I. = ₹5410 - ₹5000 = ₹410.

Hence, Venkat gained ₹410 after 2122\dfrac{1}{2} years.

Question 15

A man borrows ₹6000 at 5% compound interest. If he repays ₹1200 at the end of each year, find the amount outstanding at the beginning of the third year.

Answer

Principal for first year = ₹6000, rate = 5%.

Interest for first year = 6000×5×1100=30000100\dfrac{6000 \times 5 \times 1}{100} = \dfrac{30000}{100} = ₹300.

Amount after first year = ₹6000 + ₹300 = ₹6300.

Money refunded at the end of first year = ₹1200.

Principal for second year = ₹6300 - ₹1200 = ₹5100.

Interest for second year = 5100×5×1100=25500100\dfrac{5100 \times 5 \times 1}{100} = \dfrac{25500}{100} = ₹255.

Amount after second year = ₹5100 + ₹255 = ₹5355.

Money refunded at the end of second year = ₹1200.

Principal for third year = ₹5355 - ₹1200 = ₹4155.

Hence, the amount outstanding at the beginning of third year = ₹4155.

Question 16

Mr. Raina deposits ₹ 1,600 in a bank every year in the beginning of the year, at 5% per annum compound interest. Calculate the amount due to him at the end of 2 years. Also find his gain in two years.

Answer

For 1st year,

P = ₹ 1,600

R = 5%

T = 1 year

Using formula,

I = P×R×T100\dfrac{P \times R \times T}{100}

Substituting the values, we get

⇒I=1,600×5×1100=16×5=80.\Rightarrow I = \dfrac{1,600 \times 5 \times 1}{100}\\[1em] = 16 \times 5 \\[1em] = 80.

A = P + I = ₹ 1,600 + ₹ 80 = ₹ 1,680

For 2nd year,

P = ₹ 1,680 + ₹ 1,600 (Amount deposited at beginning of every year) = ₹3,280

R = 5%

T = 1 year

⇒I=3,280×5×1100=16,400100=164.\Rightarrow I = \dfrac{3,280 \times 5 \times 1}{100}\\[1em] = \dfrac{16,400}{100} \\[1em] = 164.

A = P + I = ₹ 3,280 + ₹ 164 = ₹ 3,444.

Gain = Total balance - money invested

= ₹ 3,444 - (₹ 1,600 x 2)

= ₹ 3,444 - ₹ 3,200

= ₹ 244.

Hence, at the end of 2 years amount = ₹ 3,444 and the gain = ₹ 244.

Question 17

Mr. Dubey borrows ₹100000 from State Bank of India at 11% per annum compound interest. He repays ₹41000 at the end of first year and ₹47700 at the end of second year. Find the amount outstanding at the beginning of the third year.

Answer

Principal for first year = ₹100000, rate = 11%.

Interest for first year = 100000×11×1100=1100000100\dfrac{100000 \times 11 \times 1}{100} = \dfrac{1100000}{100} = ₹11000.

Amount after first year = ₹100000 + ₹11000 = ₹111000.

Money refunded at the end of first year = ₹41000.

Principal for second year = ₹111000 - ₹41000 = ₹70000.

Interest for second year = 70000×11×1100=770000100\dfrac{70000 \times 11 \times 1}{100} = \dfrac{770000}{100} = ₹7700.

Amount after second year = ₹70000 + ₹7700 = ₹77700.

Money refunded at the end of second year = ₹47700.

Principal for third year = ₹77700 - ₹47700 = ₹30000.

Hence, the amount outstanding at the beginning of third year = ₹30000.

Question 18

Jaya borrowed ₹50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.

Answer

Principal for first year = ₹50000, rate = 12%.

Interest for first year = 50000×12×1100=600000100\dfrac{50000 \times 12 \times 1}{100} = \dfrac{600000}{100} = ₹6000.

Amount after first year = ₹50000 + ₹6000 = ₹56000.

Money refunded at the end of first year = ₹33000.

Principal for second year = ₹56000 - ₹33000 = ₹23000, rate = 15%.

Interest for second year = 23000×15×1100=345000100\dfrac{23000 \times 15 \times 1}{100} = \dfrac{345000}{100} = ₹3450.

Amount after second year = ₹23000 + ₹3450 = ₹26450.

Hence, Jaya must pay ₹26450 at the end of second year to clear her debt.

Exercise 2.2

Question 1

Find the amount and the compound interest on ₹5000 for 2 years at 6% per annum, interest payable yearly.

Answer

Formula for calculating amount,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Using formula we get,

A=₹5000(1+6100)2=₹5000×(106100)2=₹5000×(5350)2=₹5000×5350×5350=₹140450002500=₹5618.A = ₹5000\Big(1 + \dfrac{6}{100}\Big)^2 \\[1em] = ₹5000 \times \Big(\dfrac{106}{100}\Big)^2 \\[1em] = ₹5000 \times \Big(\dfrac{53}{50}\Big)^2 \\[1em] = ₹5000 \times \dfrac{53}{50} \times \dfrac{53}{50} \\[1em] = ₹\dfrac{14045000}{2500} \\[1em] = ₹5618.

Compound interest = Final amount - Principal = ₹5618 - ₹5000 = ₹618.

Hence, the amount and the compound interest on ₹5000 for 2 years at 6% per annum is ₹5618 and ₹618 respectively.

Question 2

Find the amount and the compound interest on ₹8000 for 4 years at 10% per annum, interest reckoned yearly.

Answer

Formula for calculating amount,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Using formula we get,

A=₹8000(1+10100)4=₹8000×(110100)4=₹8000×(1110)4=₹8000×1110×1110×1110×1110=₹11712800010000=₹11712.80A = ₹8000\Big(1 + \dfrac{10}{100}\Big)^4 \\[1em] = ₹8000 \times \Big(\dfrac{110}{100}\Big)^4 \\[1em] = ₹8000 \times \Big(\dfrac{11}{10}\Big)^4 \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] = ₹\dfrac{117128000}{10000} \\[1em] = ₹11712.80

Compound interest = Final amount - Principal = ₹11712.80 - ₹8000 = ₹3712.80

Hence, the amount and the compound interest on ₹8000 for 4 years at 10% per annum is ₹11712.80 and ₹3712.80 respectively.

Question 3

If the interest is compounded half-yearly, calculate the amount when the principal is ₹7400, the rate of interest is 5% and the duration is one year.

Answer

Since rate of interest is 5% per annum, therefore rate of interest per conversion period (half-yearly) = 2.5%.

As the money is invested for one year, therefore,

n (the number of conversion periods) = 2.

∴A=P(1+r100)n=₹7400×(1+2.5100)2=₹7400×(102.5100)2=₹7400×(10251000)2=₹7400×(4140)2=₹7400×4140×4140=₹124394001600=₹7774.625\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹7400 \times \Big(1 + \dfrac{2.5}{100}\Big)^2 \\[1em] = ₹7400 \times \Big(\dfrac{102.5}{100}\Big)^2 \\[1em] = ₹7400 \times \Big(\dfrac{1025}{1000}\Big)^2 \\[1em] = ₹7400 \times \Big(\dfrac{41}{40}\Big)^2 \\[1em] = ₹7400 \times \dfrac{41}{40} \times \dfrac{41}{40} \\[1em] = ₹ \dfrac{12439400}{1600} \\[1em] = ₹7774.625

Hence, amount = ₹7774.625

Question 4

Find the amount and the compound interest on ₹5000 at 10% p.a. for 1121\dfrac{1}{2} years, compound interest reckoned semi-annually.

Answer

Since, compound interest is reckoned semi-annually,

rate = 102\dfrac{10}{2} % = 5%.

n (no. of conversion periods) = 3 half-years.

∴A=P(1+r100)n=₹5000×(1+5100)3=₹5000×(105100)3=₹5000×(2120)2=₹5000×2120×2120×2120=₹5000×92618000=₹5788.125\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹5000 \times \Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹5000 \times \Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹5000 \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = ₹5000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹5000 \times \dfrac{9261}{8000} \\[1em] = ₹5788.125

C.I. = Amount - Principal = ₹5788.125 - ₹5000 = ₹788.125

Hence, amount = ₹5788.125 and compound interest = ₹788.125

Question 5

Find the amount and the compound interest on ₹100000 compounded quarterly for 9 months at the rate of 4% p.a.

Answer

Since rate of interest is 4% per annum, therefore rate of interest per conversion period (quarterly) = 14×4\dfrac{1}{4} \times 4 = 1%.

As the money is invested for 9 months, therefore,

n (the number of conversion periods) = 93\dfrac{9}{3} = 3.

∴A=P(1+r100)n=₹100000×(1+1100)3=₹100000×(101100)3=₹100000×(101100)3=₹100000×101100×101100×101100=₹100000×101×101×1011000000=₹103030110=₹103030.10\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹100000 \times \Big(1 + \dfrac{1}{100}\Big)^3 \\[1em] = ₹100000 \times \Big(\dfrac{101}{100}\Big)^3 \\[1em] = ₹100000 \times \Big(\dfrac{101}{100}\Big)^3 \\[1em] = ₹100000 \times \dfrac{101}{100} \times \dfrac{101}{100} \times \dfrac{101}{100} \\[1em] = ₹\dfrac{100000 \times 101 \times 101 \times 101}{1000000} \\[1em] = ₹\dfrac{1030301}{10} \\[1em] = ₹103030.10

Compound interest = Final amount - Principal = ₹103030.10 - ₹100000 = ₹3030.10

Hence, amount = ₹103030.10 and compound interest = ₹3030.10.

Question 6

Find the difference between C.I. and S.I. on sum of ₹4800 for 2 years at 5% per annum payable yearly.

Answer

S.I. = P×R×T100\dfrac{P \times R \times T}{100}.

Putting values in formula we get,

S.I.=4800×5×2100=48000100=₹480.S.I. = \dfrac{4800 \times 5 \times 2}{100} \\[1em] = \dfrac{48000}{100} \\[1em] = ₹480.

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=P[(1+r100)n−1]=₹4800×[(1+5100)2−1]=₹4800×[(105100)2−1]=₹4800×[(2120)2−1]=₹4800×[441400−1]=₹4800×[441−400400]=₹4800×[41400]=₹196800400=₹492.C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(1 + \dfrac{5}{100}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(\dfrac{105}{100}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(\dfrac{21}{20}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\dfrac{441}{400} - 1\Big] \\[1em] = ₹4800 \times \Big[\dfrac{441 - 400}{400}\Big] \\[1em] = ₹4800 \times \Big[\dfrac{41}{400}\Big] \\[1em] = ₹\dfrac{196800}{400} \\[1em] = ₹492.

C.I. - S.I. = ₹492 - ₹480 = ₹12.

Hence, the difference between C.I. and S.I. = ₹12.

Question 7

Find the difference between the simple interest and compound interest on ₹2500 for 2 years at 4% per annum, compound interest being reckoned semi-annually.

Answer

Since interest is calculated half yearly, hence rate = 42\dfrac{4}{2} % = 2%

Time = 2 years or 4 half-years.

S.I.= P×R×T100\dfrac{P \times R \times T}{100}.

Putting values in formula we get,

S.I.=₹2500×2×4100=₹20000100=₹200.S.I. = \dfrac{₹2500 \times 2 \times 4}{100} \\[1em] = ₹\dfrac{20000}{100} \\[1em] = ₹200.

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=P[(1+r100)n−1]=₹2500×[(1+2100)4−1]=₹2500×[(102100)4−1]=₹2500×[(5150)4−1]=₹2500×[67652016250000−1]=₹2500×[6765201−62500006250000]=₹2500×[5152016250000]=₹5152012500=₹206.084C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(1 + \dfrac{2}{100}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(\dfrac{102}{100}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(\dfrac{51}{50}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\dfrac{6765201}{6250000} - 1\Big] \\[1em] = ₹2500 \times \Big[\dfrac{6765201 - 6250000}{6250000}\Big] \\[1em] = ₹2500 \times \Big[\dfrac{515201}{6250000}\Big] \\[1em] = ₹\dfrac{515201}{2500} \\[1em] = ₹206.084

C.I. - S.I. = ₹206.084 - ₹200 = ₹6.084

Hence, the difference between C.I. and S.I. = ₹6.084

Question 8

Find the amount and the compound interest on ₹2000 in 2 years if the rate is 4% for the first year and 3% for the second year.

Answer

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

For first year, P = ₹2000 and rate = 4%.

Using formula,

A=P(1+r100)n=₹2000×(1+4100)1=₹2000×(104100)=₹2000×2625=₹80×26=₹2080.A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹2000 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] = ₹2000 \times \Big(\dfrac{104}{100}\Big) \\[1em] = ₹2000 \times \dfrac{26}{25} \\[1em] = ₹80 \times 26 \\[1em] = ₹2080.

For second year, P = ₹2080 and rate = 3%.

Using formula,

A=P(1+r100)n=₹2080×(1+3100)1=₹2080×(103100)=₹214240100=₹2142.40A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹2080 \times \Big(1 + \dfrac{3}{100}\Big)^1 \\[1em] = ₹2080 \times \Big(\dfrac{103}{100}\Big) \\[1em] = ₹\dfrac{214240}{100} \\[1em] = ₹2142.40

C.I. = Final Amount - Principal = ₹2142.40 - ₹2000 = 142.40

Hence, the amount = ₹2142.40 and compound interest = ₹142.40

Question 9

Find the compound interest on ₹3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.

Answer

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

For first year, P = ₹3125 and rate = 4%.

Using formula,

A=P(1+r100)n=₹3125×(1+4100)1=₹3125×(104100)=₹3125×2625=₹8125025=₹3250.A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹3125 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] = ₹3125 \times \Big(\dfrac{104}{100}\Big) \\[1em] = ₹3125 \times \dfrac{26}{25} \\[1em] = ₹\dfrac{81250}{25} \\[1em] = ₹3250.

For second year, P = ₹3250 and rate = 5%.

Using formula,

A=₹3250×(1+5100)1=₹3250×(105100)=₹3250×2120=₹6825020=₹3412.50A = ₹3250 \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] = ₹3250 \times \Big(\dfrac{105}{100}\Big) \\[1em] = ₹3250 \times \dfrac{21}{20} \\[1em] = ₹\dfrac{68250}{20} \\[1em] = ₹3412.50

For third year, P = ₹3412.50 and rate = 6%.

Using formula,

A=₹3412.50×(1+6100)1=₹3412.50×(106100)=₹3412.50×5350=₹180862.550=₹3617.25A = ₹3412.50 \times \Big(1 + \dfrac{6}{100}\Big)^1 \\[1em] = ₹3412.50 \times \Big(\dfrac{106}{100}\Big) \\[1em] = ₹3412.50 \times \dfrac{53}{50} \\[1em] = ₹\dfrac{180862.5}{50} \\[1em] = ₹3617.25

C.I. = Final Amount - Principal = ₹3617.25 - ₹3125 = ₹492.25

Hence, compound interest = ₹492.25

Question 10

What sum of money will amount to ₹9261 in 3 years at 5% per annum compound interest?

Answer

Let principal = P.

Given, A = ₹9261, rate = 5%, n = 3.

We know,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

9261=P(1+5100)39261=P(105100)39261=P(2120)39261=P×92618000P=9261×80009261P=₹8000.9261 = P\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] 9261 = P\Big(\dfrac{105}{100}\Big)^3 \\[1em] 9261 = P\Big(\dfrac{21}{20}\Big)^3 \\[1em] 9261 = P \times \dfrac{9261}{8000} \\[1em] P = \dfrac{9261 \times 8000}{9261} \\[1em] P = ₹8000.

Hence, ₹8000 will amount to ₹9261 in 3 years at 5% per annum compound interest.

Question 11

What sum invested at 4% per annum compounded semi-annually amounts to ₹7803 at the end of one year?

Answer

Let principal = P.

Given, A = ₹7803.

Since interest is compounded semi-annually,

rate = 42\dfrac{4}{2}% = 2%.

n (the number of conversion periods) = 2.

We know,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

7803=P(1+2100)27803=P(102100)27803=P(5150)27803=P×26012500P=7803×25002601P=₹7500.7803 = P\Big(1 + \dfrac{2}{100}\Big)^2 \\[1em] 7803 = P\Big(\dfrac{102}{100}\Big)^2 \\[1em] 7803 = P\Big(\dfrac{51}{50}\Big)^2 \\[1em] 7803 = P \times \dfrac{2601}{2500} \\[1em] P = \dfrac{7803 \times 2500}{2601} \\[1em] P = ₹7500.

Hence, ₹7500 will amount to ₹7803 in 1 year at 4% per annum compounded semi-annually.

Question 12

What sum invested for 1121\dfrac{1}{2} years compounded half-yearly at the rate of 4% p.a. will amount to ₹132651?

Answer

Let principal = P.

Given, A = ₹132651.

Since interest is compounded half-yearly,

rate = 42\dfrac{4}{2}% = 2%.

n (the number of conversion periods) = 3.

We know,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

132651=P(1+2100)3132651=P(102100)3132651=P(5150)3132651=P×132651125000P=132651×125000132651P=₹125000.132651 = P\Big(1 + \dfrac{2}{100}\Big)^3 \\[1em] 132651 = P\Big(\dfrac{102}{100}\Big)^3 \\[1em] 132651 = P\Big(\dfrac{51}{50}\Big)^3 \\[1em] 132651 = P \times \dfrac{132651}{125000} \\[1em] P = \dfrac{132651 \times 125000}{132651} \\[1em] P = ₹125000.

Hence, ₹125000 will amount to ₹132651 in 1121\dfrac{1}{2} year at 4% per annum compounded semi-annually.

Question 13

On what sum will the compound interest for 2 years at 4% per annum be ₹5712?

Answer

Let principal = P,

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

⇒5712=P[(1+4100)2−1]⇒5712=P[(104100)2−1]⇒5712=P[(2625)2−1]⇒5712=P[676625−1]⇒5712=P[676−625625]⇒5712=P×51625⇒P=5712×62551⇒P=112×625⇒P=₹70000.\Rightarrow 5712 = P\Big[\Big(1 + \dfrac{4}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\Big(\dfrac{104}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\Big(\dfrac{26}{25}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\dfrac{676}{625} - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\dfrac{676 - 625}{625}\Big] \\[1em] \Rightarrow 5712 = P \times \dfrac{51}{625} \\[1em] \Rightarrow P = \dfrac{5712 \times 625}{51} \\[1em] \Rightarrow P = 112 \times 625 \\[1em] \Rightarrow P = ₹70000.

Hence, principal = ₹70000.

Question 14

A man invests ₹1200 for two years at compound interest. After one year the money amounts to ₹1275. Find the interest for the second year correct to the nearest rupee.

Answer

Let rate of interest be r% per annum.

Given, ₹1200 amounts to ₹1275 after one year.

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get,

⇒1275=1200(1+r100)1⇒12751200=1+r100⇒12751200−1=r100⇒1275−12001200=r100⇒751200=r100⇒r=75001200⇒r=254⇒r=614\Rightarrow 1275 = 1200\Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{1275}{1200} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1275}{1200} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1275 - 1200}{1200} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{75}{1200} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{7500}{1200} \\[1em] \Rightarrow r = \dfrac{25}{4} \\[1em] \Rightarrow r = 6\dfrac{1}{4}%.

Principal for second year = ₹1275.

Interest for second year = P×R×T100\dfrac{P \times R \times T}{100}.

Substituting value we get,

Interest =₹1275×254×1100=₹1275×25400=₹31875400=₹79.6875≈₹80.\text{Interest } = \dfrac{₹1275 \times \dfrac{25}{4} \times 1}{100} \\[1em] = \dfrac{₹1275 \times 25}{400} \\[1em] = ₹\dfrac{31875}{400} \\[1em] = ₹79.6875 \approx ₹80.

Hence, the interest for the second year = ₹80.

Question 15

At what rate percent per annum compound interest will ₹2304 amount to ₹2500 in 2 years?

Answer

Let rate of interest = r.

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Given, A = ₹2500, P = ₹2304, n = 2.

Putting values in formula we get,

⇒2500=2304(1+r100)2⇒25002304=(1+r100)2⇒(5048)2=(1+r100)2⇒5048=(1+r100)⇒5048−1=r100⇒50−4848=r100⇒248=r100⇒r=248×100⇒r=20048=416\Rightarrow 2500 = 2304\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{2500}{2304} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{50}{48}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{50}{48} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \dfrac{50}{48} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{50 - 48}{48} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{2}{48} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{2}{48} \times 100 \\[1em] \Rightarrow r = \dfrac{200}{48} = 4\dfrac{1}{6}%.

Hence, rate of interest = 4164\dfrac{1}{6}%.

Question 16

A sum compounded annually becomes 2516\dfrac{25}{16} times of itself in two years. Determine the rate of interest per annum.

Answer

Let rate of interest = r.

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Let Principal = P.

Given, sum becomes 2516\dfrac{25}{16} times of itself in two years,

∴ A = 2516\dfrac{25}{16}P.

Putting values in formula we get,

⇒2516P=P(1+r100)2⇒2516=(1+r100)2⇒(54)2=(1+r100)2\Rightarrow \dfrac{25}{16}P = P\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{25}{16} = \Big(1 + \dfrac{r}{100}\Big)^2\\[1em] \Rightarrow \Big(\dfrac{5}{4}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2\\[1em]

Taking square root on both sides,

⇒54=1+r100⇒54−1=r100⇒5−44=r100⇒14=r100⇒r=1004⇒r=25\Rightarrow \dfrac{5}{4} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{5}{4} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{5 - 4}{4} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{4} \\[1em] \Rightarrow r = 25%.

Hence, rate of interest = 25%.

Question 17

At what rate percent will ₹2000 amount to ₹2315.25 in 3 years at compound interest?

Answer

Let rate of interest = r.

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Given, A = ₹2315.25, P = ₹2000, n = 3.

Putting values in formula we get,

⇒2315.25=2000(1+r100)3⇒2315.252000=(1+r100)3⇒231525200000=(1+r100)3⇒92618000=(1+r100)3⇒(2120)3=(1+r100)3\Rightarrow 2315.25 = 2000\Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{2315.25}{2000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{231525}{200000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em]

Taking cube root on both sides we get,

⇒2120=1+r100⇒2120−1=r100⇒21−2020=r100⇒120=r100⇒r=10020⇒r=5\Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{20} \\[1em] \Rightarrow r = 5%.

Hence, rate of interest = 5%.

Question 18

If ₹40000 amounts to ₹48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum.

Answer

Let rate of interest per annum be r% per annum, i.e. r2\dfrac{r}{2}% half-yearly.

n = 2 years or 4 half-years.

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Given, A = ₹48620.25 and P = ₹40000.

Putting values in formula we get,

⇒48620.25=40000(1+r2100)4⇒48620.2540000=(1+r200)4⇒48620254000000=(1+r200)4⇒194481160000=(1+r200)4⇒(2120)4=(1+r200)4⇒2120=1+r200⇒2120−1=r200⇒21−2020=r200⇒120=r200⇒r=20020⇒r=10\Rightarrow 48620.25 = 40000\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^4 \\[1em] \Rightarrow \dfrac{48620.25}{40000} = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \dfrac{4862025}{4000000} = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \dfrac{194481}{160000} = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^4 = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{1}{20} = \dfrac{r}{200} \\[1em] \Rightarrow r = \dfrac{200}{20} \\[1em] \Rightarrow r = 10%.

Hence, rate of interest = 10% per annum.

Question 19

Determine the rate of interest for a sum that becomes 216125\dfrac{216}{125} times of itself in 1121\dfrac{1}{2} years, compounded semi-annually.

Answer

Let rate of interest per annum be r% per annum, i.e. r2\dfrac{r}{2}% half-yearly.

n = 1121\dfrac{1}{2} years or 3 half-years.

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Let principal be P,

∴ A = 216125P\dfrac{216}{125}P.

Putting values in formula we get,

⇒216125P=P(1+r2100)3⇒216125=(1+r200)3⇒(65)3=(1+r200)3\Rightarrow \dfrac{216}{125}P = P\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{216}{125} = \Big(1 + \dfrac{r}{200}\Big)^3\\[1em] \Rightarrow \Big(\dfrac{6}{5}\Big)^3 = \Big(1 + \dfrac{r}{200}\Big)^3\\[1em]

Taking cube root on both sides,

65=1+r20065−1=r2006−55=r20015=r200r=2005r=40\dfrac{6}{5} = 1 + \dfrac{r}{200} \\[1em] \dfrac{6}{5} - 1 = \dfrac{r}{200} \\[1em] \dfrac{6 - 5}{5} = \dfrac{r}{200} \\[1em] \dfrac{1}{5} = \dfrac{r}{200} \\[1em] r = \dfrac{200}{5} \\[1em] r = 40%.

Hence, rate of interest = 40% per annum.

Question 20

At what rate percent p.a. compound interest would ₹80000 amount to ₹88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.

Answer

Let the rate of interest be r% per annum.

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting value we get,

⇒88200=80000(1+r100)2⇒8820080000=(1+r100)2⇒441400=(1+r100)2⇒(2120)2=(1+r100)2⇒1+r100=2120⇒r100=2120−1⇒r100=21−2020⇒r100=120⇒r=120×100⇒r=5\Rightarrow 88200 = 80000\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{88200}{80000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{441}{400} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{21}{20} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{21}{20} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{21 - 20}{20} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{20} \\[1em] \Rightarrow r = \dfrac{1}{20} \times 100 \\[1em] \Rightarrow r = 5%.

After 3 years,

A=₹80000(1+5100)3=₹80000×(105100)3=₹80000×(2120)3=₹80000×2120×2120×2120=₹80000×92618000=₹92610.A = ₹80000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹80000 \times \Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹80000 \times \Big(\dfrac{21}{20}\Big)^3 \\[1em] = ₹80000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹80000 \times \dfrac{9261}{8000} \\[1em] = ₹92610.

Hence, the rate of interest = 5% per annum and amount after 3 years = ₹92610.

Question 21

A certain sum amounts to ₹5292 in 2 years and to ₹5556.60 in 3 years at compound interest. Find the rate and the sum.

Answer

Let the rate of interest be r% per annum.

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

As sum amounts to ₹5292 in 2 years, substituting value we get,

∴5292=P(1+r100)2\therefore 5292 = P\Big(1 + \dfrac{r}{100}\Big)^2 .....(Eq. 1)

As sum amounts to ₹5556.60 in 3 years, substituting value we get,

∴5556.60=P(1+r100)3\therefore 5556.60 = P\Big(1 + \dfrac{r}{100}\Big)^3 .....(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

⇒1+r100=5556.605292⇒1+r100=555660529200⇒r100=555660529200−1⇒r100=26460529200⇒r=26460529200×100⇒r=264605292⇒r=5\Rightarrow 1 + \dfrac{r}{100} = \dfrac{5556.60}{5292} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{555660}{529200} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{555660}{529200} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{26460}{529200} \\[1em] \Rightarrow r = \dfrac{26460}{529200} \times 100 \\[1em] \Rightarrow r = \dfrac{26460}{5292} \\[1em] \Rightarrow r = 5%.

Substituting value of r in Eq. 1 we get,

⇒5292=P(1+5100)2⇒5292=P(1+120)2⇒5292=P(2120)2⇒5292=P×441400⇒P=5292×400441⇒P=12×400⇒P=₹4800.\Rightarrow 5292 = P\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P \times \dfrac{441}{400} \\[1em] \Rightarrow P = 5292 \times \dfrac{400}{441} \\[1em] \Rightarrow P = 12 \times 400 \\[1em] \Rightarrow P = ₹4800.

Hence, sum = ₹4800 and rate = 5%.

Question 22

A certain sum amounts to ₹798.60 after 3 years and ₹878.46 after 4 years. Find the interest rate and the sum.

Answer

Let the rate of interest be r% per annum.

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

As sum amounts to ₹798.60 in 3 years substituting value we get,

∴798.60=P(1+r100)3\therefore 798.60 = P\Big(1 + \dfrac{r}{100}\Big)^3 .....(Eq. 1)

As sum amounts to ₹878.46 in 4 years substituting value we get,

∴878.46=P(1+r100)4\therefore 878.46 = P\Big(1 + \dfrac{r}{100}\Big)^4 .....(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

⇒1+r100=878.46798.60⇒1+r100=8784679860⇒r100=8784679860−1⇒r100=87846−7986079860⇒r100=798679860⇒r100=110⇒r=10010⇒r=10\Rightarrow 1 + \dfrac{r}{100} = \dfrac{878.46}{798.60} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{87846}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{87846}{79860} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{87846 - 79860}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{7986}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{10} \\[1em] \Rightarrow r = \dfrac{100}{10} \\[1em] \Rightarrow r = 10%.

Substituting value of r in Eq. 1 we get,

⇒798.60=P(1+10100)3⇒798.60=P(1+110)3⇒798.60=P(1110)3⇒798.60=P×13311000⇒P=798.60×10001331⇒P=7986001331⇒P=₹600.\Rightarrow 798.60 = P\Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(1 + \dfrac{1}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P \times \dfrac{1331}{1000} \\[1em] \Rightarrow P = 798.60 \times \dfrac{1000}{1331} \\[1em] \Rightarrow P = \dfrac{798600}{1331} \\[1em] \Rightarrow P = ₹600.

Hence, sum = ₹600 and rate = 10%.

Question 23

In what time will ₹15625 amount to ₹17576 at 4% per annum compound interest?

Answer

Let the time be n years.

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting values we get,

⇒17576=15625(1+4100)n⇒17576=15625(104100)n⇒1757615625=(2625)n⇒(2625)3=(2625)n⇒n=3 years.\Rightarrow 17576 = 15625\Big(1 + \dfrac{4}{100}\Big)^n \\[1em] \Rightarrow 17576 = 15625\Big(\dfrac{104}{100}\Big)^n \\[1em] \Rightarrow \dfrac{17576}{15625} = \Big(\dfrac{26}{25}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{26}{25}\Big)^3 = \Big(\dfrac{26}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.

Hence, ₹15625 will amount to ₹17576 in 3 years at 4% per annum compound interest.

Question 24(i)

In what time will ₹1500 yield ₹496.50 as compound interest at 10% per annum compounded annually?

Answer

Let the time be n years.

C.I.=P[(1+r100)n−1].C.I. = P \Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big].

Substituting values we get,

⇒496.50=1500[(1+10100)n−1]⇒496.50=1500(110100)n−1500⇒496.50+1500=1500(110100)n⇒1996.501500=(1110)n⇒199650150000=(1110)n⇒13311000=(1110)n⇒(1110)3=(1110)n⇒n=3 years.\Rightarrow 496.50 = 1500\Big[\Big(1 + \dfrac{10}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 496.50 = 1500\Big(\dfrac{110}{100}\Big)^n - 1500 \\[1em] \Rightarrow 496.50 + 1500 = 1500\Big(\dfrac{110}{100}\Big)^n \\[1em] \Rightarrow \dfrac{1996.50}{1500} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \dfrac{199650}{150000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.

Hence, in 3 years ₹1500 yield ₹496.50 as compound interest at 10% per annum compounded annually.

Question 24(ii)

Find the time (in years) in which ₹12500 will produce ₹3246.40 as compound interest at 8% per annum, interest compounded annually.

Answer

Let the time be n years.

C.I.=P[(1+r100)n−1].C.I. = P \Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big].

Substituting values we get,

⇒3246.40=12500[(1+8100)n−1]⇒3246.40=12500(108100)n−12500⇒3246.40+12500=12500(108100)n⇒15746.4012500=(2725)n⇒15746401250000=(2725)n⇒1968315625=(2725)n⇒(2725)3=(2725)n⇒n=3 years.\Rightarrow 3246.40 = 12500\Big[\Big(1 + \dfrac{8}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 3246.40 = 12500\Big(\dfrac{108}{100}\Big)^n - 12500 \\[1em] \Rightarrow 3246.40 + 12500 = 12500\Big(\dfrac{108}{100}\Big)^n \\[1em] \Rightarrow \dfrac{15746.40}{12500} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \dfrac{1574640}{1250000} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \dfrac{19683}{15625} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^3 = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.

Hence, in 3 years ₹12500 yield ₹3246.40 as compound interest at 8% per annum compounded annually.

Question 25

₹16000 invested at 10% p.a., compounded semi-annually, amounts to ₹18522. Find the time period of investment.

Answer

Rate = 10% p.a. i.e. \dfrac{10%}{2} = 5% compounded semi-annually.

Let time be n half-years.

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting values we get,

⇒18522=16000(1+5100)n⇒18522=16000(105100)n⇒1852216000=(2120)n⇒92618000=(2120)n⇒(2120)3=(2120)n⇒n=3 half-years.\Rightarrow 18522 = 16000\Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow 18522 = 16000\Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{18522}{16000} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow n = 3 \text{ half-years}.

n = 3 half-years i.e. 1121\dfrac{1}{2} years.

Hence, time = 1121\dfrac{1}{2} years.

Question 26

What sum will amount to ₹2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years?

Answer

If the rates are different then formula is,

A=P(1+r1100)(1+r2100)A = P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)

Substituting values we get,

A=P(1+5100)(1+6100)⇒2782.50=P(1+120)(1+350)⇒2782.50=P×2120×5350⇒P=2782.50×20×5021×53⇒P=₹2500.A = P\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{6}{100}\Big) \\[1em] \Rightarrow 2782.50 = P\Big(1 + \dfrac{1}{20}\Big)\Big(1 + \dfrac{3}{50}\Big) \\[1em] \Rightarrow 2782.50 = P \times \dfrac{21}{20} \times \dfrac{53}{50} \\[1em] \Rightarrow P = \dfrac{2782.50 \times 20 \times 50}{21 \times 53} \\[1em] \Rightarrow P = ₹2500.

Hence, sum = ₹2500.

Question 27

A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹225 and ₹240. Find :

(i) the rate of interest.

(ii) the original sum.

(iii) the interest earned in the third year.

Answer

Given,

Interest for first year = ₹225

Interest for second year = ₹240.

Difference = ₹15.

Here, ₹15 is the interest on ₹225 for 1 year.

(i) We know that,

Rate=S.I.×100P×T=15×100225×1=203=623\text{Rate} = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \dfrac{15 \times 100}{225 \times 1} \\[1em] = \dfrac{20}{3} \\[1em] = 6\dfrac{2}{3}%.

Hence, rate of interest = 6236\dfrac{2}{3}%.

(ii) We know that,

P=S.I.×100R×T=225×100203×1=22500203=6750020=₹3375.\text{P} = \dfrac{S.I. \times 100}{R \times T} \\[1em] = \dfrac{225 \times 100}{\dfrac{20}{3} \times 1} \\[1em] = \dfrac{22500}{\dfrac{20}{3}} \\[1em] = \dfrac{67500}{20} \\[1em] = ₹3375.

Hence, the sum = ₹3375.

(iii) Here,

Amount after second year = ₹3375 + ₹225 + ₹240 = ₹3840.

Interest earned in third year = 3840×203×1100=3840×20×1300\dfrac{3840 \times \dfrac{20}{3} \times 1}{100} = \dfrac{3840 \times 20 \times 1}{300} = ₹256.

Hence, interest earned in third year = ₹256.

Question 28

On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹25 if the rate of interest charged for both is 5% p.a.?

Answer

Given,

Let Sum (P) = ₹x,

Rate (R) = 5% p.a.

Period (n) = 2 years.

We know that,

S.I.=PRT100\text{S.I.} = \dfrac{\text{PRT}}{100}

Substituting values we get,

S.I.=x×5×2100=₹x10.\text{S.I.} = \dfrac{x \times 5 \times 2}{100} \\[1em] = ₹\dfrac{x}{10}.

When interest is compounded annually,

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting values,

A=x(1+5100)2=x(1+120)2=x×(2120)2=x×441400=₹441x400.A = x\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] = x\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = x \times \dfrac{441}{400} \\[1em] = ₹\dfrac{441x}{400}.

C.I. = A - P = ₹441x400−₹x=₹41x400.₹\dfrac{441x}{400} - ₹x = ₹\dfrac{41x}{400}.

C.I.−S.I.=₹41x400−₹x10=₹41x−40x400=₹x400\text{C.I.} - \text{S.I.} = ₹\dfrac{41x}{400} - ₹\dfrac{x}{10} \\[1em] = ₹\dfrac{41x - 40x}{400} \\[1em] = ₹\dfrac{x}{400}

Given,

Difference = ₹25.

∴x400=25⇒x=25×400⇒x=₹10000.\therefore \dfrac{x}{400} = 25 \\[1em] \Rightarrow x = 25 \times 400 \\[1em] \Rightarrow x = ₹10000.

Hence, sum of money = ₹10000.

Question 29

The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is ₹15. Find the sum of money lent out.

Answer

Let Sum (P) = ₹x.

Given,

Rate = 10% p.a. or 5% half-yearly.

Period = 1 year or 2 half-years.

We know that,

A=P(1+r100)n.A = P\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting values,

A=x(1+5100)2=x(1+120)2=x×(2120)2=x×441400=₹441x400.A = x\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] = x\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = x \times \dfrac{441}{400} \\[1em] = ₹\dfrac{441x}{400}.

C.I. = A - P = ₹441x400−₹x=₹41x400.₹\dfrac{441x}{400} - ₹x = ₹\dfrac{41x}{400}.

We know that,

S.I.=PRT100\text{S.I.} = \dfrac{\text{PRT}}{100}

Substituting values we get,

S.I.=x×10×1100=₹x10.\text{S.I.} = \dfrac{x \times 10 \times 1}{100} \\[1em] = ₹\dfrac{x}{10}.

C.I.−S.I.=₹41x400−₹x10=₹41x−40x400=₹x400\text{C.I.} - \text{S.I.} = ₹\dfrac{41x}{400} - ₹\dfrac{x}{10} \\[1em] = ₹\dfrac{41x - 40x}{400} \\[1em] = ₹\dfrac{x}{400}

Given,

Difference = ₹15,

∴x400=15⇒x=15×400=₹6000.\therefore \dfrac{x}{400} = 15 \\[1em] \Rightarrow x = 15 \times 400 = ₹6000.

Hence, sum of money = ₹6000.

Question 30

The amount at compound interest which is calculated yearly on a certain sum of money is ₹1250 in one year and ₹1375 after two years. Calculate the rate of interest.

Answer

Given,

Amount after one year = ₹1250,

Amount after second year = ₹1375.

Difference = ₹1375 - ₹1250 = ₹125.

So, ₹125 is the interest on ₹1250 for 1 year.

We know that,

R=S.I.×100P×T=125×1001250×1=10R = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \dfrac{125 \times 100}{1250 \times 1} \\[1em] = 10%.

Hence, rate of interest = 10%.

Question 31

The simple interest on a certain sum for 3 years is ₹225 and the compound interest on the same sum at the same rate for 2 years is ₹153. Find the rate of interest and principal.

Answer

Let principal be ₹P and rate of interest be R% p.a.

According to the first condition of the question,

S.I. on ₹P for 3 years = ₹225.

∴P×R×T100=225⇒P×R×3100=225⇒P×R=225×1003⇒P×R=7500⇒P=7500R.......(i)\therefore \dfrac{P \times R \times T}{100} = 225 \\[1em] \Rightarrow \dfrac{P \times R \times 3}{100} = 225 \\[1em] \Rightarrow P \times R = \dfrac{225 \times 100}{3} \\[1em] \Rightarrow P \times R = 7500 \\[1em] \Rightarrow P = \dfrac{7500}{R} .......(i)

According to the second condition of the question,

C.I. on ₹P for 2 years at R% p.a. = ₹153.

∴P[(1+R100)n−1]=153⇒P[(1+R100)2−1]=153⇒P[(100+R100)2−1]=153⇒P[(100+R)21002−1]=153⇒P[(100+R)2−10021002]=153⇒P[1002+R2+200R−10021002]=153⇒P[R2+200R1002]=153\therefore P\Big[\Big(1 + \dfrac{R}{100}\Big)^n - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(1 + \dfrac{R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(\dfrac{100 + R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{(100 + R)^2}{100^2} - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{(100 + R)^2 - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{100^2 + R^2 + 200R - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{R^2 + 200R}{100^2}\Big] = 153 \\[1em]

Using value of P from Eq 1 above:

⇒7500R×R(R+200)1002=153⇒7500(R+200)1002=153⇒R+200=153×100×1007500⇒R+200=1530075⇒R+200=204⇒R=204−200=4\Rightarrow \dfrac{7500}{R} \times \dfrac{R(R + 200)}{100^2} = 153 \\[1em] \Rightarrow \dfrac{7500(R + 200)}{100^2} = 153 \\[1em] \Rightarrow R + 200 = \dfrac{153 \times 100 \times 100}{7500} \\[1em] \Rightarrow R + 200 = \dfrac{15300}{75} \\[1em] \Rightarrow R + 200 = 204 \\[1em] \Rightarrow R = 204 - 200 = 4%.

Substituting value of R in (i) we get,

P=75004=₹1875.P = \dfrac{7500}{4} = ₹1875.

Hence, rate of interest = 4% and sum = ₹1875.

Question 32

Find the difference between compound interest on ₹8000 for 1121\dfrac{1}{2} years at 10% p.a. when compounded annually and semi-annually.

Answer

Case 1:

When compounded annually,

rate for first year = 10%, rate for next 12\dfrac{1}{2} year = 5%.

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

On substituting values,

A=₹8000(1+10100)(1+5100)=₹8000(110100)(105100)=₹8000×1110×2120=₹(40×11×21)=₹9240.A = ₹8000\Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{5}{100}\Big) \\[1em] = ₹8000\Big(\dfrac{110}{100}\Big)\Big(\dfrac{105}{100}\Big) \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{21}{20} \\[1em] = ₹(40 \times 11 \times 21) \\[1em] = ₹9240.

C.I. = A - P = ₹9240 - ₹8000 = ₹1240.

Case 2:

When compounded semi-annually,

rate = 5%, n = 3.

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

On substituting values,

A=₹8000(1+5100)3=₹8000(105100)3=₹8000×2120×2120×2120=₹(21×21×21)=₹9261.A = ₹8000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹8000\Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹8000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹(21 \times 21 \times 21) \\[1em] = ₹9261.

C.I. = A - P = ₹9261 - ₹8000 = ₹1261.

Difference between two C.I. = ₹1261 - ₹1240 = ₹21.

Question 33

A sum of money is lent out at compound interest for two years at 20% p.a., C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, C.I. being reckoned half-yearly, it would have fetched ₹482 more by way of interest. Calculate the sum of money lent out.

Answer

Let the sum lent out be ₹x.

When C.I. is reckoned yearly,

Rate = 20%

∴A=P(1+r100)n=x(1+20100)2=x(1+15)2=x(65)2=36x25\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x\Big(1 + \dfrac{20}{100}\Big)^2 \\[1em] = x\Big(1 + \dfrac{1}{5}\Big)^2 \\[1em] = x\Big(\dfrac{6}{5}\Big)^2 \\[1em] = \dfrac{36x}{25}

C.I. = A - P = 36x25−x=11x25.\dfrac{36x}{25} - x = \dfrac{11x}{25}.

When C.I. is reckoned half-yearly,

Rate = 202\dfrac{20}{2} % = 10%.

n (no. of conversion periods) = 4 half-years.

∴A=P(1+r100)n=x(1+10100)4=x(1+110)4=x(1110)4=14641x10000\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x\Big(1 + \dfrac{10}{100}\Big)^4 \\[1em] = x\Big(1 + \dfrac{1}{10}\Big)^4 \\[1em] = x\Big(\dfrac{11}{10}\Big)^4 \\[1em] = \dfrac{14641x}{10000}

C.I. = A - P = 14641x10000−x=4641x10000.\dfrac{14641x}{10000} - x = \dfrac{4641x}{10000}.

Given, difference between C.I. in both cases = ₹482.

∴4641x10000−11x25=482⇒4641x−4400x10000=482⇒241x10000=482⇒x=482×10000241⇒x=₹20000.\therefore \dfrac{4641x}{10000} - \dfrac{11x}{25} = 482 \\[1em] \Rightarrow \dfrac{4641x - 4400x}{10000} = 482 \\[1em] \Rightarrow \dfrac{241x}{10000} = 482 \\[1em] \Rightarrow x = \dfrac{482 \times 10000}{241} \\[1em] \Rightarrow x = ₹20000.

Hence, sum lent out = ₹20000.

Question 34

A sum of money amounts to ₹13230 in one year and to ₹13891.50 in 1121\dfrac{1}{2} years at compound interest, compounded semi-annually. Find the sum and rate of interest per annum.

Answer

Let sum be ₹x and rate be r%.

Since C.I. is reckoned half-yearly, rate = r2\dfrac{r}{2}%

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

In one year,

n = 2

A = ₹13230

rate = r2\dfrac{r}{2}% as interest is calculated half-yearly.

Substituting values in formula,

13230=x(1+r2100)213230=x(1+r200)2........(i)13230 = x\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^2 \\[1em] 13230 = x\Big(1 + \dfrac{r}{200}\Big)^2 ........(i)

In one and half year,

n = 3

A = ₹13891.50

rate = r2\dfrac{r}{2}% as interest is calculated half-yearly.

Substituting values in formula,

13891.50=x(1+r2100)313891.50=x(1+r200)3........(ii)13891.50 = x\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^3 \\[1em] 13891.50 = x\Big(1 + \dfrac{r}{200}\Big)^3 ........(ii)

Dividing eqn. (ii) by (i),

⇒13891.5013230=x(1+r200)3x(1+r200)2⇒138915132300=1+r200⇒r200=138915132300−1⇒r200=138915−132300132300⇒r200=6615132300⇒r=6615×200132300⇒r=1323000132300⇒r=10\Rightarrow \dfrac{13891.50}{13230} = \dfrac{x\Big(1 + \dfrac{r}{200}\Big)^3}{x\Big(1 + \dfrac{r}{200}\Big)^2} \\[1em] \Rightarrow \dfrac{138915}{132300} = 1 + \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{138915}{132300} - 1 \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{138915 - 132300}{132300} \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{6615}{132300} \\[1em] \Rightarrow r = \dfrac{6615 \times 200}{132300} \\[1em] \Rightarrow r = \dfrac{1323000}{132300} \\[1em] \Rightarrow r = 10%.

Putting value of r in eq. (i) we get,

⇒13230=x(1+10200)2⇒13230=x(1+120)2⇒13230=x(2120)2⇒13230=x×441400⇒x=13230×400441⇒x=₹12000.\Rightarrow 13230 = x\Big(1 + \dfrac{10}{200}\Big)^2 \\[1em] \Rightarrow 13230 = x\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] \Rightarrow 13230 = x\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow 13230 = x \times \dfrac{441}{400} \\[1em] \Rightarrow x = \dfrac{13230 \times 400}{441} \\[1em] \Rightarrow x = ₹12000.

Hence, sum = ₹12000 and rate = 10% per annum.

Exercise 2.3

Question 1

The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years.

Answer

By formula,

V = V0(1+r1100)(1+r2100)V_0\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)

Putting values in formula we get,

V=200000(1+10100)(1+15100)=200000×110100×115100=20×110×115=253000.V = 200000\Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big) \\[1em] = 200000 \times \dfrac{110}{100} \times \dfrac{115}{100} \\[1em] = 20 \times 110 \times 115 \\[1em] = 253000.

Hence, the population of the town at the end of two years is 253000.

Question 2

The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years?

Answer

By formula,

V = V0(1+r100)nV_0\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

V=15625(1+4100)3=15625×(104100)3=15625×(2625)3=15625×2625×2625×2625=15625×1757615625=17576.V = 15625\Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{104}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{26}{25}\Big)^3 \\[1em] = 15625 \times \dfrac{26}{25} \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] = \dfrac{15625 \times 17576}{15625} \\[1em] = 17576.

Increase in population = 17576 - 15625 = 1951.

Hence, the increase in population in next 3 years = 1951.

Question 3

The population of a city increases each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find :

(i) its population 2 years hence

(ii) its population 2 years ago.

Answer

(i) By formula,

V = V0(1+r100)nV_0\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

V=6760000(1+4100)2=6760000×(104100)2=6760000×(2625)2=6760000×2625×2625=6760000×676625=10816×676=7311616.V = 6760000\Big(1 + \dfrac{4}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\dfrac{104}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\dfrac{26}{25}\Big)^2 \\[1em] = 6760000 \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] = \dfrac{6760000 \times 676}{625} \\[1em] = 10816 \times 676 \\[1em] = 7311616.

Hence, the population after 2 years = 7311616.

(ii) Let population 2 years ago be P and present population will be the final population.

By formula,

V = V0(1+r100)nV_0\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

⇒6760000=P×(1+4100)2⇒6760000=P×(104100)2⇒6760000=P×(2625)2⇒6760000=P×2625×2625⇒6760000=P×676625⇒P=6760000×625676⇒P=625×10000⇒P=6250000.\Rightarrow 6760000 = P \times \Big(1 + \dfrac{4}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\dfrac{104}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\dfrac{26}{25}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] \Rightarrow 6760000 = P \times \dfrac{676}{625} \\[1em] \Rightarrow P = \dfrac{6760000 \times 625}{676} \\[1em] \Rightarrow P = 625 \times 10000 \\[1em] \Rightarrow P = 6250000.

Hence, the population 2 years ago was 6250000.

Question 4

The cost of a refrigerator is ₹9000. Its value depreciates at the rate of 5% every year. Find the total depreciation in its value at the end of 2 years.

Answer

Depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Putting values in formula we get,

V=₹9000(1−5100)2=₹9000×(95100)2=₹9000×(1920)2=₹9000×1920×1920=₹9000×361400=₹8122.5V = ₹9000\Big(1 - \dfrac{5}{100}\Big)^2 \\[1em] = ₹9000 \times \Big(\dfrac{95}{100}\Big)^2 \\[1em] = ₹9000 \times \Big(\dfrac{19}{20}\Big)^2 \\[1em] = ₹9000 \times \dfrac{19}{20} \times \dfrac{19}{20} \\[1em] = ₹\dfrac{9000 \times 361}{400} \\[1em] = ₹8122.5

Depreciation = ₹9000 - ₹8122.5 = ₹877.5

Hence, the total depreciation in its value at the end of 2 years = ₹877.5

Question 5

Dinesh purchased a scooter for ₹24000. The value of scooter is depreciating at the rate of 5% per annum. Calculate its value after 3 years.

Answer

Depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Putting values in formula we get,

V=₹24000(1−5100)3=₹24000×(95100)3=₹24000×(1920)3=₹24000×1920×1920×1920=₹24000×68598000=₹(3×6859)=₹20577.V = ₹24000\Big(1 - \dfrac{5}{100}\Big)^3 \\[1em] = ₹24000 \times \Big(\dfrac{95}{100}\Big)^3 \\[1em] = ₹24000 \times \Big(\dfrac{19}{20}\Big)^3 \\[1em] = ₹24000 \times \dfrac{19}{20} \times \dfrac{19}{20} \times \dfrac{19}{20} \\[1em] = ₹\dfrac{24000 \times 6859}{8000} \\[1em] = ₹(3 \times 6859) \\[1em] = ₹20577.

Hence, the value of scooter after 3 years = ₹20577.

Question 6

A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?

Answer

Let yearly produce 2 years ago be P. Then present production will be the final production.

By formula,

V = V0(1+r100)nV_0\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

⇒2187=P×(1+8100)2⇒2187=P×(108100)2⇒2187=P×(2725)2⇒2187=P×2725×2725⇒2187=P×729625⇒P=2187×625729⇒P=3×625⇒P=1875.\Rightarrow 2187 = P \times \Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \dfrac{27}{25} \times \dfrac{27}{25} \\[1em] \Rightarrow 2187 = P \times \dfrac{729}{625} \\[1em] \Rightarrow P = \dfrac{2187 \times 625}{729} \\[1em] \Rightarrow P = 3 \times 625 \\[1em] \Rightarrow P = 1875.

Hence, the yearly produce of wheat two years ago was 1875 quintals.

Question 7

The value of a property decreases every year at the rate of 5%. If its present value is ₹411540, what was its value three years ago?

Answer

Depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Let value three years ago be V0.

Putting values in formula we get,

⇒₹411540=V0(1−5100)3⇒₹411540=V0(95100)3⇒₹411540=V0(1920)3⇒₹411540=V0×1920×1920×1920⇒V0=₹411540×20×20×2019×19×19⇒V0=₹411540×80006859⇒V0=₹(60×8000)⇒V0=₹480000.\Rightarrow ₹411540 = V_0\Big(1 - \dfrac{5}{100}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0\Big(\dfrac{95}{100}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0\Big(\dfrac{19}{20}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0 \times \dfrac{19}{20} \times \dfrac{19}{20} \times \dfrac{19}{20} \\[1em] \Rightarrow V_0 = ₹\dfrac{411540 \times 20 \times 20 \times 20}{19 \times 19 \times 19} \\[1em] \Rightarrow V_0 = ₹\dfrac{411540 \times 8000}{6859} \\[1em] \Rightarrow V_0 = ₹(60 \times 8000) \\[1em] \Rightarrow V_0 = ₹480000.

Hence, the value of property 3 years ago was ₹480000.

Question 8

Ahmed purchased an old scooter for ₹16000. If the cost of the scooter after 2 years depreciates to ₹14440, find the rate of depreciation.

Answer

Let rate of depreciation be r%.

Depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Putting values in formula we get,

⇒14440=16000(1−r100)21444016000=(1−r100)214441600=(1−r100)2(3840)2=(1−r100)21−r100=3840r100=1−3840r100=240r=20040r=5\Rightarrow 14440 = 16000\Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] \dfrac{14440}{16000} = \Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] \dfrac{1444}{1600} = \Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] \Big(\dfrac{38}{40}\Big)^2 = \Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] 1 - \dfrac{r}{100} = \dfrac{38}{40} \\[1em] \dfrac{r}{100} = 1 - \dfrac{38}{40} \\[1em] \dfrac{r}{100} = \dfrac{2}{40} \\[1em] r = \dfrac{200}{40} \\[1em] r = 5%.

Hence, the rate of depreciation = 5%.

Question 9

A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-2015. Find the annual rate of growth of production of cars.

Answer

Let rate of growth be r% per annum.

Growth formula,

V = V0(1+r100)nV_0\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

⇒92610=80000(1+r100)39261080000=(1+r100)392618000=(1+r100)3(2120)3=(1+r100)31+r100=2120r100=2120−1r100=21−2020r100=120r=10020r=5\Rightarrow 92610 = 80000\Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \dfrac{92610}{80000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \dfrac{9261}{8000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Big(\dfrac{21}{20}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] 1 + \dfrac{r}{100} = \dfrac{21}{20} \\[1em] \dfrac{r}{100} = \dfrac{21}{20} - 1 \\[1em] \dfrac{r}{100} = \dfrac{21 - 20}{20} \\[1em] \dfrac{r}{100} = \dfrac{1}{20} \\[1em] r = \dfrac{100}{20} \\[1em] r = 5%.

Hence, the annual rate of growth of production of cars is 5%.

Question 10

The value of a machine worth ₹500000 is depreciating at the rate of 10% every year. In how many years will its value be reduced to ₹364500?

Answer

Let value be depreciated from ₹500000 to ₹364500 in n years.

Depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Putting values in formula we get,

⇒364500=500000(1−10100)n⇒364500500000=(100−10100)n⇒7291000=(90100)n⇒(910)3=(910)n∴n=3.\Rightarrow 364500 = 500000\Big(1 - \dfrac{10}{100}\Big)^n \\[1em] \Rightarrow \dfrac{364500}{500000} = \Big(\dfrac{100 - 10}{100}\Big)^n \\[1em] \Rightarrow \dfrac{729}{1000} = \Big(\dfrac{90}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{9}{10}\Big)^3 = \Big(\dfrac{9}{10}\Big)^n \\[1em] \therefore n = 3.

Hence, the value of machine will be depreciated from ₹500000 to ₹364500 in 3 years.

Question 11

Mahindra set up a factory by investing ₹2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, calculate his total profit.

Answer

When rate of interest are different then total amount is given by,

A=P(1+r1100)(1+r2100)A = P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)

Substituting values we get,

A=₹2500000(1+5100)(1+10100)=₹2500000(1+120)(1+110)=₹2500000×2120×1110=₹2500000×21×1120×10=₹12500×21×11=₹2887500.A = ₹2500000\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{10}{100}\Big) \\[1em] = ₹2500000\Big(1 + \dfrac{1}{20}\Big)\Big(1 + \dfrac{1}{10}\Big) \\[1em] = ₹2500000 \times \dfrac{21}{20} \times \dfrac{11}{10} \\[1em] = \dfrac{₹2500000 \times 21 \times 11}{20 \times 10} \\[1em] = ₹12500 \times 21 \times 11 \\[1em] = ₹2887500.

Profit = Amount - Principal = ₹2887500 - ₹2500000 = ₹387500

Hence, profit = ₹387500.

Question 12

The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?

Answer

Let initial value of property be V0.

By growth formula,

V=V0(1+r100)n.V = V_0\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting values we get,

V=V0(1+25100)3=V0(1+14)3=V0(54)3=V0×12564=125V064.V = V_0\Big(1 + \dfrac{25}{100}\Big)^3 \\[1em] = V_0\Big(1 + \dfrac{1}{4}\Big)^3 \\[1em] = V_0\Big(\dfrac{5}{4}\Big)^3 \\[1em] = V_0 \times \dfrac{125}{64} \\[1em] = \dfrac{125V_0}{64}.

Change in value (C) = Final value - Initial value.

∴C=125V064−V0=125V0−64V064=61V064.\therefore C = \dfrac{125V_0}{64} - V_0 \\[1em] = \dfrac{125V_0 - 64V_0}{64} \\[1em] = \dfrac{61V_0}{64}.

Percentage increase = Change in valueOriginal value×100.\dfrac{\text{Change in value}}{\text{Original value}} \times 100.

Substituting values we get,

Percentage increase=61V064V0×100=61V064V0×100=610064=152516=95516\text{Percentage increase} = \dfrac{\dfrac{61V_0}{64}}{V_0} \times 100 \\[1em] = \dfrac{61V_0}{64V_0} \times 100 \\[1em] = \dfrac{6100}{64} \\[1em] = \dfrac{1525}{16} \\[1em] = 95\dfrac{5}{16}%.

Hence, after 3 years the value of property will increase by 9551695\dfrac{5}{16}%.

Question 13

Mr. Durani bought a plot of land for ₹180000 and a car for ₹320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?

Answer

Since, value of land grows uniformly at the rate of 30% p.a. hence, by growth formula,

V=V0(1+r100)nV = V_0\Big(1 + \dfrac{r}{100}\Big)^n.

Substituting values we get value of land after 3 years,

V=₹180000(1+30100)3=₹180000(130100)3=₹180000(1310)3=₹180000×1310×1310×1310=₹180000×21971000=₹180×2197=₹395460.V = ₹180000\Big(1 + \dfrac{30}{100}\Big)^3 \\[1em] = ₹180000\Big(\dfrac{130}{100}\Big)^3 \\[1em] = ₹180000\Big(\dfrac{13}{10}\Big)^3 \\[1em] = ₹180000 \times \dfrac{13}{10} \times \dfrac{13}{10} \times \dfrac{13}{10} \\[1em] = ₹180000 \times \dfrac{2197}{1000} \\[1em] = ₹180 \times 2197 \\[1em] = ₹395460.

Given, the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter,

Hence, value of car after 3 years = V=V0(1−r1100)(1−r2100)(1−r3100)V = V_0\Big(1 - \dfrac{r_1}{100}\Big)\Big(1 - \dfrac{r_2}{100}\Big)\Big(1 - \dfrac{r_3}{100}\Big)

Substituting values, we get value of car after 3 years,

V=₹320000(1−20100)(1−15100)(1−15100)=₹320000×80100×85100×85100=₹(32×4×17×85)=₹184960.V = ₹320000\Big(1 - \dfrac{20}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)\Big(1 - \dfrac{15}{100}\Big) \\[1em] = ₹320000 \times \dfrac{80}{100} \times \dfrac{85}{100} \times \dfrac{85}{100} \\[1em] = ₹(32 \times 4 \times 17 \times 85) \\[1em] = ₹184960.

Present value of land and car = ₹180000 + ₹320000 = ₹500000.

Total value of land and car after 3 years = ₹395460 + ₹184960 = ₹580420.

Profit = Amount - Initial value = ₹580420 - ₹500000 = ₹80420.

After 3 years profit of Mr. Durani would be ₹80420.

Multiple Choice Questions

Question 1

The compound interest on ₹1000 at 10% p.a. compounded annually for 2 years is

  1. ₹190

  2. ₹200

  3. ₹210

  4. ₹1210

Answer

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=1000×[(1+10100)2−1]=1000×[(110100)2−1]=1000×[(1110)2−1]=1000×[121100−1]=1000×21100=₹210.C.I. = 1000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\dfrac{110}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\dfrac{11}{10}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\dfrac{121}{100} - 1\Big] \\[1em] = 1000 \times \dfrac{21}{100} \\[1em] = ₹210.

Hence, Option 3 is the correct option.

Question 2

If Sukriti borrows ₹8000 for two years at the rate of 10% per annum compound interest, then the amount to be paid by her at the end of two years to clear the debt is

  1. ₹8800

  2. ₹9600

  3. ₹9680

  4. ₹102400

Answer

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

A=₹8000×(1+10100)2=₹8000×(110100)2=₹8000×(1110)2=₹8000×1110×1110=₹8000×121100=₹9680.A = ₹8000 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = ₹8000 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = ₹8000 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] = ₹8000 \times \dfrac{121}{100} \\[1em] = ₹9680.

Hence, Option 3 is the correct option.

Question 3

If a man invests ₹12000 for two years at the rate of 10% per annum compound interest, then the compound interest earned by him at the end of two years is

  1. ₹2400

  2. ₹2520

  3. ₹2000

  4. ₹1800

Answer

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=₹12000×[(1+10100)2−1]=₹12000×[(110100)2−1]=₹12000×[(1110)2−1]=₹12000×[121100−1]=₹12000×21100=₹120×21=₹2520.C.I. = ₹12000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\Big(\dfrac{110}{100}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\Big(\dfrac{11}{10}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\dfrac{121}{100} - 1\Big] \\[1em] = ₹12000 \times \dfrac{21}{100} \\[1em] = ₹120 \times 21 \\[1em] = ₹2520.

Hence, Option 2 is the correct option.

Question 4

Mr. Rao bought 1-year, ₹10000 certificate of deposit that paid interest at an annual rate of 8% compounded semi-annually. The interest received by him on maturity is

  1. ₹816

  2. ₹864

  3. ₹800

  4. ₹10816

Answer

Rate = 8% i.e. 82\dfrac{8}{2}% = 4% when compounded semi-annually.

n (no. of conversion periods) = 2 half-years.

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=₹10000×[(1+4100)2−1]=₹10000×[(104100)2−1]=₹10000×[(5250)2−1]=₹10000×[27042500−1]=₹10000×2704−25002500=₹10000×2042500=₹20400002500=₹816.C.I. = ₹10000 \times \Big[\Big(1 + \dfrac{4}{100}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\Big(\dfrac{104}{100}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\Big(\dfrac{52}{50}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\dfrac{2704}{2500} - 1\Big] \\[1em] = ₹10000 \times \dfrac{2704 - 2500}{2500} \\[1em] = ₹10000 \times \dfrac{204}{2500} \\[1em] = ₹\dfrac{2040000}{2500} \\[1em] = ₹816.

Hence, Option 1 is the correct option.

Question 5

The compound interest on ₹5000 at 20% per annum for 1121\dfrac{1}{2} years compounded half-yearly is

  1. ₹6655

  2. ₹1655

  3. ₹1500

  4. ₹1565

Answer

Rate = 20% i.e. 202\dfrac{20}{2}% = 10% when compounded semi-annually.

n (no. of conversion periods) = 3 half-years.

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=₹5000×[(1+10100)3−1]=₹5000×[(110100)3−1]=₹5000×[(1110)3−1]=₹5000×[13311000−1]=₹5000×1331−10001000=₹5000×3311000=₹1655.C.I. = ₹5000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\Big(\dfrac{110}{100}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\Big(\dfrac{11}{10}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\dfrac{1331}{1000} - 1\Big] \\[1em] = ₹5000 \times \dfrac{1331 - 1000}{1000} \\[1em] = ₹5000 \times \dfrac{331}{1000} \\[1em] = ₹1655.

Hence, Option 2 is the correct option.

Question 6

If the number of conversion periods ≥ 2, then the compound interest is

  1. less than simple interest

  2. equal to simple interest

  3. greater than or equal to simple interest

  4. greater than simple interest

Answer

When the number of conversion periods ≥ 2, then the compound interest is greater than simple interest.

As, in compound interest the interest is always calculated on the compounded principal whereas in simple interest, the interest is calculated on the initial principle so for conversion periods ≥ 2 compound interest will be greater than simple interest.

Hence, Option 4 is the correct option.

Question 7

The present population of a city is 12,00,000. If it increases at the rate of 8% every year then the population of the city after 2 years is

  1. 199680

  2. 1399680

  3. 1500000

  4. 1299680

Answer

By growth formula,

V=V0(1+r100)nV = V_0\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values in formula,

V=1200000(1+8100)2=1200000(108100)2=1200000(108100)2=1200000(2725)2=1200000×2725×2725=1200000×729625=1920×729=1399680.V = 1200000\Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] = 1200000\Big(\dfrac{108}{100}\Big)^2 \\[1em] = 1200000\Big(\dfrac{108}{100}\Big)^2 \\[1em] = 1200000\Big(\dfrac{27}{25}\Big)^2 \\[1em] = 1200000 \times \dfrac{27}{25} \times \dfrac{27}{25} \\[1em] = 1200000 \times \dfrac{729}{625} \\[1em] = 1920 \times 729 \\[1em] = 1399680.

Hence, Option 2 is the correct option.

Question 8

Consider the following two statements.

Statement 1: A sum of ₹ 1,000 invested at 10% p.a. rate of interest will earn ₹ 100 interest in first year.

Statement 2: Simple and compound interest is the same for first conversion period at the same rate of interest.

Which of the following is valid?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and Statement 2 is false.

  4. Statement 1 is false, and Statement 2 is true.

Answer

Given,

P = ₹ 1,000

R = 10%

T = 1 year

Using C.I. formula,

C.I. = P[(1+R100)n−1]P\Big[\Big(1 + \dfrac{R}{100}\Big)^n - 1\Big]

Substituting the values, we get

C.I.=1,000×[(1+10100)1−1]=1,000×[(1+110)−1]=1,000×(1+110−1)=1,000×(110)=100.C.I. = 1,000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^1 - 1\Big]\\[1em] = 1,000 \times \Big[\Big(1 + \dfrac{1}{10}\Big) - 1\Big]\\[1em] = 1,000 \times \Big(1 + \dfrac{1}{10} - 1\Big)\\[1em] = 1,000 \times \Big(\dfrac{1}{10}\Big)\\[1em] = 100.

Calculating S.I.,

P = ₹ 1,000

R = 10%

T = 1 year

Using formula,

S.I. = P×R×T100\dfrac{P \times R \times T}{100}

Substituting the values, we get

⇒S.I.=1,000×10×1100=10×10=100.\Rightarrow S.I. = \dfrac{1,000 \times 10 \times 1}{100}\\[1em] = 10 \times 10 \\[1em] = 100.

Since,

C.I. = S.I. = ₹ 100.

Therefore, we can say that simple and compound interest is the same for first conversion period at the same rate of interest.

∴ Both statements are true.

Hence, option 1 is the correct option.

Assertion Reason Type Questions

Question 1

Assertion (A): The population of a town in 2015 was 10,000. It grew by 10% every year. So, in the year 2020, the population was 15,000.

Reason (R): Formula used in such problems is V = Vo (1+r100)n\Big(1 + \dfrac{r}{100}\Big)^n

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

The formula, V = Vo (1+r100)n\Big(1 + \dfrac{r}{100}\Big)^n is indeed the correct formula used for calculating future value in problems involving compound growth (like population growth, compound interest, etc.), where:

V is the final value.

Vo is the initial value.

r is the rate of growth (or interest) per period.

n is the number of periods.

This formula accurately reflects how a quantity increases by a certain percentage over multiple periods, with the growth compounding on the new total each period.

∴ Reason (R) is true.

Given,

Vo = 10,000

r = 10%.

n = 5 years

Substituting values we get :

V=10,000×(1+10100)5=10,000×(1+110)5=10,000×(10+110)5=10,000×(1110)5=10,000×1,61,0511,00,000=16,105.10V = 10,000 \times \Big(1 + \dfrac{10}{100}\Big)^5\\[1em] = 10,000 \times \Big(1 + \dfrac{1}{10}\Big)^5\\[1em] = 10,000 \times \Big(\dfrac{10 + 1}{10}\Big)^5\\[1em] = 10,000 \times \Big(\dfrac{11}{10}\Big)^5\\[1em] = 10,000 \times \dfrac{1,61,051}{1,00,000}\\[1em] = 16,105.10

Since population cannot be in fraction, the population in 2020 would be approximately 16,105.

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Question 2

Assertion (A): Two friends invest the same amount of money for the same time (> 2 years) at the same rate of interest. One earns simple interest, but the other earns compound interest. Then both will get the same amount of money back at the end of investment.

Reason (R): The principal for each conversion period increases for the compound interest calculation.

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Simple interest is calculated only on the original principal amount. The interest earned each period remains constant.

Compound interest is calculated on the original principal amount + all accumulated interest from previous periods. This means the principal for calculating interest increases with each conversion period.

The person earning compound interest will always have a higher amount back at the end of the investment period, if the conversion period is greater than one.

∴ Assertion (A) is false.

The principal for each conversion period increases for the compound interest calculation. This is true and is the fundamental principle of compound interest.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Chapter Test

Question 1

₹10000 was lent for one year at 10% per annum. By how much more will the interest be, if the sum was lent at 10% per annum, interest being compounded half-yearly?

Answer

When interest is compounded yearly,

P = ₹10000, r = 10%, T = 1.

C.I. = 10000×10×1100\dfrac{10000 \times 10 \times 1}{100} = ₹1000.

When interest is compounded half-yearly,

P = ₹10000, r = 102\dfrac{10}{2}% = 5%, T = 2.

For first half-year,

C.I. = 10000×5×1100\dfrac{10000 \times 5 \times 1}{100} = ₹500.

Amount after first half-year = ₹1000 + ₹500 = ₹1500.

Principal for second half-year = ₹1500.

C.I. = 10500×5×1100\dfrac{10500 \times 5 \times 1}{100} = ₹525.

Amount after second half-year = ₹1500 + ₹525 = ₹2025.

C.I. = Final amount - Principal = ₹2025 - ₹1000 = ₹1025.

Difference in C.I. in both cases = ₹1025 - ₹1000 = ₹25.

Hence, the interest would be ₹25 more, if the sum was lent at 10% per annum, interest being compounded half-yearly.

Question 2

A man invests ₹3072 for two years at compound interest. After one year the money amounts to ₹3264. Find the rate of interest and the amount due at the end of 2nd year.

Answer

Let the rate of interest be r% per annum.

Given, ₹3072 amounts to ₹3264 in one year.

∴ Compound interest = Final amount - Principal = ₹3264 - ₹3072 = ₹192.

∴192=3072×r×1100⇒r=192003072⇒r=6.25\therefore 192 = \dfrac{3072 \times r \times 1}{100} \\[1em] \Rightarrow r = \dfrac{19200}{3072} \\[1em] \Rightarrow r = 6.25%.

Principal for second year = ₹3264.

C.I.= 3264×6.25×1100=20400100\dfrac{3264 \times 6.25 \times 1}{100} = \dfrac{20400}{100} = ₹204.

Amount after second year = ₹3264 + ₹204 = ₹3468.

Hence, the rate of interest = 6.25% and the amount after second year = ₹3468.

Question 3

What sum will amount to ₹28090 in two years at 6% per annum compound interest? Also find the compound interest.

Answer

Let principal be ₹P.

We know,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

⇒28090=P(1+6100)2⇒28090=P(106100)2⇒28090=P(5350)2⇒28090=P×5350×5350⇒P=28090×50×5053×53⇒P=28090×25002809⇒P=₹25000.\Rightarrow 28090 = P\Big(1 + \dfrac{6}{100}\Big)^2 \\[1em] \Rightarrow 28090 = P\Big(\dfrac{106}{100}\Big)^2 \\[1em] \Rightarrow 28090 = P\Big(\dfrac{53}{50}\Big)^2 \\[1em] \Rightarrow 28090 = P \times \dfrac{53}{50} \times \dfrac{53}{50} \\[1em] \Rightarrow P = \dfrac{28090 \times 50 \times 50}{53 \times 53} \\[1em] \Rightarrow P = \dfrac{28090 \times 2500}{2809} \\[1em] \Rightarrow P = ₹25000.

C.I. = Final amount - Principal = ₹28090 - ₹25000 = ₹3090.

Hence, the principal = ₹25000 and compound interest = ₹3090.

Question 4

Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was ₹422, find :

(i) the equal sums

(ii) compound interest for each sum.

Answer

(i) Let the sum be ₹P.

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

C.I. when sum is lent at 5% for 2 years,

C.I.=P[(1+r100)n−1]=P[(1+5100)2−1]=P[(1+120)2−1]=P[(2120)2−1]=P[441400−1]=P[441−400400]=P×41400=41P400.C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{5}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{1}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{21}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\dfrac{441}{400} - 1\Big] \\[1em] = P\Big[\dfrac{441 - 400}{400} \Big] \\[1em] = P \times \dfrac{41}{400} \\[1em] = \dfrac{41P}{400}.

C.I. when sum is lent at 6% for 2 years,

C.I.=P[(1+6100)2−1]=P[(1+350)2−1]=P[(5350)2−1]=P[28092500−1]=P×2809−25002500=309P2500.C.I. = P\Big[\Big(1 + \dfrac{6}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{3}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{53}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\dfrac{2809}{2500} - 1\Big] \\[1em] = P \times \dfrac{2809 - 2500}{2500} \\[1em] = \dfrac{309P}{2500}.

Given, difference in C.I. = ₹422.

∴309P2500−41P400=422⇒309P×4−41P×2510000=422⇒1236P−1025P10000=422⇒211P10000=422⇒P=422×10000211⇒P=₹20000.\therefore \dfrac{309P}{2500} - \dfrac{41P}{400} = 422 \\[1em] \Rightarrow \dfrac{309P \times 4 - 41P \times 25}{10000} = 422 \\[1em] \Rightarrow \dfrac{1236P - 1025P}{10000} = 422 \\[1em] \Rightarrow \dfrac{211P}{10000} = 422 \\[1em] \Rightarrow P = \dfrac{422 \times 10000}{211} \\[1em] \Rightarrow P = ₹20000.

Hence, equal sum = ₹20000.

(ii) C.I. when sum is lent at 5% for 2 years = 41P400.\dfrac{41P}{400}.

C.I.=41×20000400=41×50=₹2050.C.I. = \dfrac{41 \times 20000}{400} \\[1em] = 41 \times 50 \\[1em] = ₹2050.

C.I. when sum is lent at 6% for 2 years = 309P2500.\dfrac{309P}{2500}.

C.I.=309×200002500=309×8=₹2472.C.I. = \dfrac{309 \times 20000}{2500} \\[1em] = 309 \times 8 \\[1em] = ₹2472.

Hence, C.I. = ₹2050 when sum is lent at 5% for 2 years and C.I. = ₹2472 when sum is lent at 6% for 2 years.

Question 5

The compound interest on a sum of money for 2 years is ₹1331.20 and the simple interest on the same sum for the same period at the same rate is ₹1280. Find the sum and the rate of interest per annum.

Answer

Let the sum be ₹x and rate be r%.

Given, S.I. for 2 years = ₹1280.

∴P×R×T100=1280⇒x×r×2100=1280⇒x×r=1280×1002⇒x×r=64000.......(i)\therefore \dfrac{P \times R \times T}{100} = 1280 \\[1em] \Rightarrow \dfrac{x \times r \times 2}{100} = 1280 \\[1em] \Rightarrow x \times r = \dfrac{1280 \times 100}{2} \\[1em] \Rightarrow x \times r = 64000 .......(i)

Given, C.I. for 2 years = ₹1331.20

C.I.=P[(1+r100)n−1]⇒1331.20=x[(1+r100)2−1]⇒1331.20=x[(100+r100)2−1]⇒1331.20=x[(100+r)21002−1]⇒1331.20=x[(100+r)2−10021002]⇒1331.20=x[1002+r2+200r−10021002]⇒1331.20=x[r2+200r1002]⇒1331.20×1002=x×r×(r+200)⇒13312000=64000(r+200)....... (Using (i))⇒r+200=1331200064000⇒r+200=208⇒r=8C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(1 + \dfrac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(\dfrac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 1331.20 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 13312000 = 64000(r + 200) .......\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \dfrac{13312000}{64000} \\[1em] \Rightarrow r + 200 = 208 \\[1em] \Rightarrow r = 8%.

Putting value of r in Eq. (i) we get,

⇒x×8=64000⇒x=640008⇒x=₹8000.\Rightarrow x \times 8 = 64000 \\[1em] \Rightarrow x = \dfrac{64000}{8} \\[1em] \Rightarrow x = ₹8000.

Hence, sum = ₹8000 and rate = 8%.

Question 6

On what sum will the difference between the simple and compound interest for 3 years at 10% p.a. is ₹232.50?

Answer

Let the sum be ₹P.

S.I. = P×10×3100=3P10.\dfrac{P \times 10 \times 3}{100} = \dfrac{3P}{10}.

By formula,

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=P[(1+r100)n−1]=P[(1+10100)3−1]=P[(110100)3−1]=P[(1110)3−1]=P[13311000−1]=P[1331−10001000]=P×3311000=331P1000.C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{10}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{110}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{11}{10}\Big)^3 - 1\Big] \\[1em] = P\Big[\dfrac{1331}{1000} - 1\Big] \\[1em] = P\Big[\dfrac{1331 - 1000}{1000} \Big] \\[1em] = P \times \dfrac{331}{1000} \\[1em] = \dfrac{331P}{1000}.

Given, difference between C.I. and S.I. = ₹232.50

∴331P1000−3P10=232.50⇒331P−300P1000=232.50⇒31P1000=232.50⇒P=232.50×100031⇒P=23250031⇒P=₹7500.\therefore \dfrac{331P}{1000} - \dfrac{3P}{10} = 232.50 \\[1em] \Rightarrow \dfrac{331P- 300P}{1000} = 232.50 \\[1em] \Rightarrow \dfrac{31P}{1000} = 232.50 \\[1em] \Rightarrow P = \dfrac{232.50 \times 1000}{31} \\[1em] \Rightarrow P = \dfrac{232500}{31} \\[1em] \Rightarrow P = ₹7500.

Hence, sum = ₹7500.

Question 7

The simple interest on a certain sum for 3 years is ₹1080 and the compound interest on the same sum at the same rate for 2 years is ₹741.60. Find :

(i) the rate of interest

(ii) the principal.

Answer

Let the sum be ₹x and rate be r%.

Given, S.I. for 3 years = ₹1080.

∴P×R×T100=1080⇒x×r×3100=1080⇒x×r=1080×1003⇒x×r=36000.......(i)\therefore \dfrac{P \times R \times T}{100} = 1080 \\[1em] \Rightarrow \dfrac{x \times r \times 3}{100} = 1080 \\[1em] \Rightarrow x \times r = \dfrac{1080 \times 100}{3} \\[1em] \Rightarrow x \times r = 36000 .......(i)

Given, C.I. for 2 years = ₹741.60

C.I.=P[(1+r100)n−1]⇒741.60=x[(1+r100)2−1]⇒741.60=x[(100+r100)2−1]⇒741.60=x[(100+r)21002−1]⇒741.60=x[(100+r)2−10021002]⇒741.60=x[1002+r2+200r−10021002]⇒741.60=x[r2+200r1002]⇒741.60×1002=x×r×(r+200)⇒7416000=36000(r+200)....... (Using (i))⇒r+200=741600036000⇒r+200=206⇒r=6C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(1 + \dfrac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(\dfrac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 741.60 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 7416000 = 36000(r + 200) .......\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \dfrac{7416000}{36000} \\[1em] \Rightarrow r + 200 = 206 \\[1em] \Rightarrow r = 6%.

Hence, the rate of interest = 6%.

(ii) Putting value of r in (i) we get,

⇒x×r=36000⇒6x=36000⇒x=₹6000.\Rightarrow x \times r = 36000 \\[1em] \Rightarrow 6x = 36000 \\[1em] \Rightarrow x = ₹6000.

Hence, principal = ₹6000.

Question 8

In what time will ₹2400 amount to ₹2646 at 10% p.a. compounded semi-annually?

Answer

Since, interest is compounded semi-annually, rate = \dfrac{10%}{2} = 5%.

Let time be n half-years,

A=P(1+r100)nA =P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values in formula we get,

⇒2646=2400(1+5100)n⇒26462400=(1+5100)n⇒26462400=(105100)n⇒441400=(2120)n⇒(2120)2=(2120)n⇒n=2.\Rightarrow 2646 = 2400\Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{2646}{2400} = \Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{2646}{2400} = \Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{441}{400} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^2 = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow n = 2.

Time = 2 half-years or 1 year.

Hence, in 1 year ₹2400 amounts to ₹2646 at 10% p.a. compounded semi-annually.

Question 9

Sudarshan invested ₹60000 in a finance company and received ₹79860 after 1121\dfrac{1}{2} years. Find the rate of interest per annum compounded half-yearly.

Answer

Let rate of interest be r% p.a. i.e. r2\dfrac{r}{2}% if compounded half-yearly.

n = 1121\dfrac{1}{2} years or 3 half-years.

A=P(1+r100)nA =P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values in formula we get,

⇒79860=60000(1+r200)3⇒7986060000=(1+r200)3⇒13311000=(1+r200)3⇒(1110)3=(1+r200)3⇒1110=1+r200⇒1110−1=r200⇒110=r200⇒r=20010=20\Rightarrow 79860 = 60000\Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \dfrac{79860}{60000} = \Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = 1 + \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{r}{200} \\[1em] \Rightarrow r = \dfrac{200}{10} = 20%.

Hence, the rate of interest = 20% per annum.

Question 10

The population of a city is 320000. If the annual birth rate is 9.2% and the annual death rate is 1.7%, calculate the population of the town after 3 years.

Answer

Net growth rate = 9.2% - 1.7% = 7.5%.

By growth formula,

V = V0(1+r100)nV_0\Big(1 + \dfrac{r}{100}\Big)^n

Putting values in formula we get,

V=320000(1+7.5100)3=320000×(107.5100)3=320000×(10751000)3=320000×(4340)3=320000×4340×4340×4340=320000×7950764000=5×79507=397535.V = 320000\Big(1 + \dfrac{7.5}{100}\Big)^3 \\[1em] = 320000 \times \Big(\dfrac{107.5}{100}\Big)^3 \\[1em] = 320000 \times \Big(\dfrac{1075}{1000}\Big)^3 \\[1em] = 320000 \times \Big(\dfrac{43}{40}\Big)^3 \\[1em] = 320000 \times \dfrac{43}{40} \times \dfrac{43}{40} \times \dfrac{43}{40} \\[1em] \\[1em] = 320000 \times \dfrac{79507}{64000} \\[1em] = 5 \times 79507 \\[1em] = 397535.

Hence, the population of the town after 3 years = 397535.

Question 11

The cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If its present worth is ₹315600, find :

(i) its purchase price

(ii) its value after 3 years.

Answer

(i) By depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Let initial value of car be V0.

Putting values in formula we get,

⇒315600=V0(1−20100)2⇒315600=V0(1−20100)2⇒315600=V0(80100)2⇒315600=V0(45)2⇒315600=V0×45×45⇒315600=V0×1625⇒V0=315600×2516⇒V0=₹493125.\Rightarrow 315600 = V_0\Big(1 - \dfrac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(1 - \dfrac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(\dfrac{80}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(\dfrac{4}{5}\Big)^2 \\[1em] \Rightarrow 315600 = V_0 \times \dfrac{4}{5} \times \dfrac{4}{5} \\[1em] \Rightarrow 315600 = V_0 \times \dfrac{16}{25} \\[1em] \Rightarrow V_0 = 315600 \times \dfrac{25}{16} \\[1em] \Rightarrow V_0 = ₹493125.

Hence, the purchase price of car was ₹493125.

(ii) By depreciation formula,

V = V0(1−r100)nV_0\Big(1 - \dfrac{r}{100}\Big)^n

Putting values in formula we get,

V=315600(1−20100)3=315600×(80100)3=315600×(45)3=315600×(64125)=20198400125=₹161587.20V = 315600\Big(1 - \dfrac{20}{100}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{80}{100}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{4}{5}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{64}{125}\Big) \\[1em] = \dfrac{20198400}{125} \\[1em] = ₹161587.20

Hence, the value of car after 3 years = ₹161587.20

Question 12

Amar Singh started a business with an initial investment of ₹400000. In the first year, he incurred a loss of 4%. However, during the second year, he earned a profit of 5% which in third year rose to 10%. Calculate his net profit for the entire period of 3 years.

Answer

Given,

Investment = ₹400000

Loss in first year = 4%,

Profit in second year = 5%,

Profit in third year = 10%.

∴Amt. after 3 yrs.=₹400000(1−4100)(1+5100)(1+10100)=₹400000×96100×105100×110100=₹400000×2425×2120×1110=₹400000×55445000=₹(80×5544)=₹443520.\therefore \text{Amt. after 3 yrs.} = ₹400000\Big(1 - \dfrac{4}{100}\Big)\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{10}{100}\Big) \\[1em] = ₹400000 \times \dfrac{96}{100} \times \dfrac{105}{100} \times \dfrac{110}{100} \\[1em] = ₹400000 \times \dfrac{24}{25} \times \dfrac{21}{20} \times \dfrac{11}{10} \\[1em] = ₹400000 \times \dfrac{5544}{5000} \\[1em] = ₹(80 \times 5544) \\[1em] = ₹443520.

Net profit = Amount - Principal = ₹443520 - ₹400000 = ₹43520.

Hence, net profit after 3 years = ₹43520.

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