The DAV Class 8 Maths Book Solutions Pdf and **DAV Class 8 Maths Chapter 6 Brain Teasers **Solutions of Compound Interest offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 6 Brain Teasers Solutions

Question 1A.

Tick (✓) the correct option.

(i) For a given rate of interest and time, what is more profitable to the depositor?

(a) compound interest

(b) simple interest

(c) both are equally profitable

(d) cannot be determined

Solution:

(a) compound interest

(ii) A sum of ₹ 10,000 at 8% per annum for six months compounded quarterly amounts

(a) ₹ 400

(b) ₹ 404

(c) ₹ 408

(d) ₹ 10,404

Solution:

(d) ₹ 10,404

Amount = 10000 \(\left(1+\frac{2}{100}\right)^2\)

= 10000 \(\left(\frac{100+2}{100}\right)^2\)

= 10000 \(\left(\frac{102}{100}\right)^2\)

= 10000 \(\left(\frac{102}{100}\right) \times \frac{102}{100}\)

= ₹ 10,404

(iii) In A = P \(\left(1+\frac{\mathrm{R}}{100}\right)^n\), A stands for

(a) time period

(b) principal

(c) principal + interest

(d) interest

Solution:

(c) principal + interest

(iv) If the number of conversion periods is greater than or equal to 2, then compound interst is

(a) less than simple interest

(b) greater than simple interest

(c) less than or equal to simple interst

(d) greater than or equal to simple interest

Solution:

(b) greater than simple interest

(v) Reema wants to do a one-year deposit. She should opt for

(a) a simple interst of 10%

(b) a compound interest of 10% compounded quarterly

(c) a compound interest of 10% compounded annually

(d) a compound interst of 10% compounded half-yearly

Solution:

(b) a compound interest of 10% compounded quarterly

Question 1B.

Answer the following questions.

(i) Find the compound interest on ₹ 1,000 at 10% per annum for two years.

(ii) If ₹ 6,000 is deposited for two years at 4% per annum compounded quarterly, then find the time period and rate to compute compound interest.

(iii) The value of a machine worth ₹ 5,00,000 depreciates at the rate of 10% every year. In how many years will its value be ₹ 3,64,500?

(iv) Find the difference between the compound interest and simple interest on ₹ 5,000 for two years at 5% per annum.

(v) If ₹ 20,000 is deposited for three years at 5% compounded annually, then what will be the principal for the second year?

Solution:

(i) P = ₹ 1,000,

r = 10%,

n = 2 years,

C.I.= ?

A = P \(\left(1+\frac{r}{100}\right)^n\)

= 1000 \(\left(1+\frac{10}{100}\right)^2\)

= 1000 \(\left(\frac{100+10}{100}\right)^2\)

= 1000 \(\left(\frac{110}{100}\right)^2\)

= 1000 × \(\frac{11}{10} \times \frac{11}{10}\)

= ₹ 1210

C.I. = A – P

= ₹ 1210 – ₹ 1000 = ₹ 210.

(ii) P = ₹ 6000,

r = 4% p.a.

= \(\frac{1}{4}\) × 4 quarterly

= 1% quarterly (p.q.)

n = 2 years

= 2 × 4 = 8 quarters

(iii) Initial value = ₹ 5,00,000

Final value = ₹ 3,64,500

Rate of depreciation = 10%

Time period = n years

∴ A = P \(\left(1-\frac{r}{100}\right)^n\)

⇒ 3,64,500 = 5,00,000 \(\left(1-\frac{10}{100}\right)^n\)

⇒ \(\frac{364500}{500000}=\left(\frac{100-10}{100}\right)^n\)

⇒ \(\frac{729}{1000}=\left(\frac{90}{100}\right)^n\)

⇒ \(\left(\frac{9}{10}\right)^3=\left(\frac{9}{10}\right)^n\)

⇒ n = 3 years

(iv) P = ₹ 5000,

n = 2 years,

r = 5% p.a.,

C.I. = ?

S.I. = ?

A = P \(\left(1+\frac{r}{100}\right)^n\)

= 5000 \(\left(1+\frac{5}{100}\right)^2\)

= 5000 \(\left(1+\frac{1}{20}\right)^2\)

= 5000 \(\left(\frac{21}{20}\right)^2\)

= ₹ 5512.50

C.I. = A – P

= 5512.50 – 5000

= ₹ 512.50

S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{5000 \times 5 \times 2}{100}\)

= ₹ 500

Difference = C.I. – S.I.

= ₹ 512.50 – ₹ 500

= ₹ 12.50

(v) P = 20,000,

R = 5% p.a.,

n = 1 year

S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{20,000 \times 5 \times 1}{100}\)

= ₹ 1000

Thus, the principal for second year = ₹ 20000 + ₹ 1000

= ₹ 21,000

Question 2.

Preeti invested ₹ 50,000 at 8% p.a. for 3 years and the interest is compounded annually. Calculate:

(i) The amount standing to her credit at the end of the second year.

(ii) The interest for the third year.

Solution:

Here, P = ₹ 50,000,

R = 8% p.a.,

T = 3 years

Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{50000 \times 8 \times 1}{100}\)

= ₹ 4000

Principal for the second year = ₹ 50,000 + ₹ 4000

= ₹ 54000

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{54000 \times 8 \times 1}{100}\)

= ₹ 4320

Principal for the third year = ₹ 54000 + ₹ 4320

= ₹ 58320

Interest for the third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{58320 \times 8 \times 1}{100}\)

= ₹ 4665.60

∴ Amount standing to her credit at the end of second year = ₹ 58320

and the interest for the third year = ₹ 4665.60.

Question 3.

A man had ₹ 75000. He invested ₹ 35000 in a company which pays him 9% interest per annum and he invested rest of the money in another company which pays him 9.5% interest p.a. Find the total compound interest received by him after 2 years.

Solution:

For the first company:

C.I. = P \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

= 35000 \(\left[\left(1+\frac{9}{100}\right)^2-1\right]\)

= 35000 \(\left[\left(\frac{109}{100}\right)^2-1\right]\)

= 35000 [(1.09)^{2} – 1]

= 35000 [1.1881 – 1]

= 35000 × 0.1881

= 6583.50

For the second company:

C.I. = P \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

= 40000 \(\left[\left(1+\frac{9.5}{100}\right)^2-1\right]\)

= 40000 \(\left[\left(\frac{219}{200}\right)^2-1\right]\)

= 40000 \(\left[\frac{47961}{40000}-1\right]\)

= 40000 \(\left[\frac{47961-40000}{40000}\right]\)

= 40000 × \(\frac{7961}{40000}\)

= ₹ 7961

Hence total C.I. = ₹ 6583.50 + ₹ 7961

= ₹ 1454450

Question 4.

A certain sum of money at compound interest becomes ₹ 7396 in 2 years and ₹ 7950.70 in 3 years. Find the rate of interest.

Solution:

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

7396 = P \(\left[1+\frac{\mathrm{R}}{100}\right]^2\) ………….(i)

and 7950.70 = P \(\left[1+\frac{\mathrm{R}}{100}\right]^3\) …………..(ii)

Dividing (ii) by (i), we get

\(\frac{7950.70}{7396}=\frac{P\left[1+\frac{R}{100}\right]^3}{P\left[1+\frac{R}{100}\right]^2}\)

\(\frac{795070}{739600}=\left(1+\frac{\mathrm{R}}{100}\right)\)

1.075 = 1 +\(\frac{R}{100}\)

\(\frac{R}{100}\) = 1.075 – 1

∴ R = 0.075 × 100 = 7.5%

Hence, the required rate of interest = 7.5%.

Question 5.

Pooja started a business by investing ₹ 2,00,000. During the first three successive years, she earned a profit of 5%, 8% and 12% p.a. respectively. If in each year, the profit was | added on the capital at the end of the previous year, calculate her total profit after 3 years.

Solution:

A = P \(\left[1+\frac{R_1}{100}\right]\left[1+\frac{R_2}{100}\right]\left[1+\frac{R_3}{100}\right]\)

= 200,000 \(\left[1+\frac{5}{100}\right]\left[1+\frac{8}{100}\right]\left[1+\frac{12}{100}\right]\)

= 200,000 × \(\frac{21}{20} \times \frac{27}{25} \times \frac{28}{25}\)

= \(\frac{200,000 \times 15876}{12500}\)

Hence the total profit = ₹ 254,061 – ₹ 200,000 = ₹ 54016.

Question 6.

Mahesh borrowed a certain sum for 2 years at simple interest from Bhim. Mahesh lent this sum to Vishnu at the same rate for 2 years compound interest. At the end of 2 years, Mahesh received ₹ 410 as compound interest but paid ₹ 400 as simple interest. Find the sum and rate of interest.

Solution:

S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

400 = \(\frac{P \times R \times 2}{100}\)

P × R = 20000 …………..(i)

C.I. = \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

410 = \(\left[\left(1+\frac{R}{100}\right)^2-1\right]\) …………….(ii)

Dividing (ii) by (i), we get

∴ Putting the value of R in eq.(i), we get

P × 5 = 20000

P = \(\frac{20000}{5}\)

∴ P = ₹ 4000

Hence the sum is ₹ 4000

and rate of interest = 5% p.a.

Question 7.

A man invested ₹ 1000 for three years at 11% simple interest p.a. and another ₹ 1000 is invested at 10% compound interest p.a. compounded annually for 3 years. Find which investment is better?

Solution:

Here C.I. > S.I.

Hence, the second option is better.

Question 8.

M/s Heera Associates let out ₹ 4,00,000 for 1 year at 16% p.a. compounded annually. How much they could earn if the interest is compounded half-yearly?

Solution:

If the interest is compounded half-yearly,

∴ R = \(\frac{16}{2}\)

= 8% per half-year

and T = 1 × 2

= 2 half-years

∴ A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 400000 \(\left[1+\frac{8}{100}\right]^2\)

= 400000 × \(\left[\frac{27}{25}\right]^2\)

= 400000 × \(\frac{729}{625}\)

= ₹ 466560

Hence, the required amount = ₹ 466560.

Question 9.

Sirish borrowed a sum of ₹ 1,63,840 at 12.5% p.a. compounded annually. On the same day, he lent out the same amount to Sahej at the same rate of interest but compounded half-yearly. Find his gain after 2 years.

Solution:

Case I:

When the interest is compounded annually,

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 163840 \(\left[1+\frac{12.5}{100}\right]^2\)

= 163840 \(\left[1+\frac{1}{8}\right]^2\)

= 163840 × \(\left(\frac{9}{8}\right)^2\)

= 163840 × \(\frac{81}{64}\)

= ₹ 207360

Case II:

When the interest is compounded half-yearly,

∴ R = \(\frac{12.5}{2}\) %

= 6.25% per half year

and T = 2 × 2 = 4 half years

A = P \(\left[1+\frac{R}{100}\right]^{\mathrm{T}}\)

= 163840 \(\left[1+\frac{6.25}{100}\right]^4\)

= 163840 × \(\left(\frac{17}{16}\right)^4\)

= 163840 × \(\frac{83521}{65536}\)

= ₹ 208802.50

Hence, his gain = ₹ 208802.50 – ₹ 207360

= ₹ 1442.50

Question 10.

The annual rate of growth in population of a certain city is 8%. If its present population is ₹ 196830, what was the population 3 years ago?

Solution:

Present population = Population 3 year ago \(\left[1+\frac{8}{100}\right]^3\)

196830 = P × \(\left[\frac{27}{25}\right]^3\)

196830 = P × \(\frac{19683}{15625}\)

P = \(\frac{196830 \times 15625}{19683}\)

= 156250

Hence, the required population is 156250.

### DAV Class 8 Maths Chapter 6 HOTS

Question 1.

A certain sum of money is invested at the rate of 10% per annum compound interest, the interest compounded annually. If the difference between the interests of third year and first year is ₹ 1,105, fmd the sum invested.

Solution:

Consider principal = P

Rate of interest = 10%,

time = 3 years

### DAV Class 8 Maths Chapter 6 Enrichment Questions

Question 1.

A builder employed 4,000 workers to work on a residential project. At the end of the first year, 10% workers were removed, at the end of the second year, 5% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 15% at the end of the third year. How many workers were working during the fourth year?

Solution:

Number of workers in the initial = 4000

R_{1} = 10%,

R_{2} = 5%,

R_{3} = 15%

Number of workers working in the fourth year = Present worker \(\left(1-\frac{R_1}{100}\right)\left(1-\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)\)

= 4000 \(\left(1-\frac{10}{100}\right)\left(1-\frac{5}{100}\right)\left(1+\frac{15}{100}\right)\)

= 4000 × \(\left(\frac{100-10}{100}\right)\left(\frac{100-5}{100}\right)\left(\frac{100+15}{100}\right)\)

= 4000 × \(\frac{90}{100} \times \frac{95}{100} \times \frac{115}{100}\)

= 3933 workers

Additional Questions:

Question 1.

Calculate the amount of ₹ 65,536 for 1\(\frac{1}{2}\) years at 12\(\frac{1}{2}\)% p.a., the interest being compounded semi-annually.

Solution:

Here, P = ₹ 65,536

R = 12\(\frac{1}{2}\)% p.a.

= \(\frac{25}{2 \times 2}\)

= \(\frac{25}{4}\) % semi-annual

and T = 1\(\frac{1}{2}\) years

= \(\frac{3}{2}\) × 2

= 3 half years

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 65536 \(\left[1+\frac{25}{4 \times 100}\right]^3\)

= 65536 × \(\left[\frac{17}{16}\right]^3\)

= 65536 × \(\frac{17}{16} \times \frac{17}{16} \times \frac{17}{16}\)

= ₹ 78608

Hence, the required amount is ₹ 78608.

Question 2.

Calculate the compound interest on ₹ 24000 for 6 months if the interest is payable quarterly at the rate of 8% p.a.

Solution:

Here, P = ₹ 24000,

R = 8% p.a.

= \(\frac{8}{4}\) = 2% per quarter

and T = \(\frac{6}{12}\) × 4 = 2 years

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 24000 \(\left[1+\frac{2}{100}\right]^2\)

= 24000 \(\left[\frac{51}{50}\right]^2\)

= 24000 × \(\frac{51}{50} \times \frac{51}{50}\)

= ₹ 24969.30

∴ C.I. = A – P

= ₹ 24969.60 – ₹ 24000

= ₹ 969.60

Hence, the required C.I. = ₹ 969.60

Question 3.

In what time will ₹ 5400 amount to ₹ 6773.76 at 12% p.a. compound interest?

Solution:

Here, P = ₹ 5400,

A = ₹ 6773.76,

R = 12%

∴ A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

⇒ 6773.76 = 5400 \(\left[1+\frac{12}{100}\right]^{\mathrm{T}}\)

⇒ \(\frac{6773.76}{5400}=\left[\frac{28}{25}\right]^{\mathrm{T}}\)

⇒ \(\frac{677376}{540000}=\left[\frac{28}{25}\right]^{\mathrm{T}}\)

⇒ \(\frac{784}{625}=\left[\frac{28}{25}\right]^{\mathrm{T}}\)

⇒ \(\left[\frac{28}{25}\right]^2=\left[\frac{28}{25}\right]^{\mathrm{T}}\)

∴ T = 2 years

Hence, the required time = 2 years.

Question 4.

The difference between C.I. and S.I. on ₹ 8400 for 2 years is ₹ 21 at the same rate of interest per year. Find the rate of interest.

Solution:

S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{8400 \times R \times 2}{100}\)

= ₹ 168

C.I. = P \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

= 8400 \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^2-1\right]\)

But C.I. – S.I. = 21

⇒ 8400 \(\left[\left(1+\frac{R}{100}\right)^2-1\right]\) – 168 R = 21

Dividing by 168 bothsides, we get

Hence, the required rate of interest = 5% p.a.

Question 5.

A certain sum of money at C.I. amounts to ₹ 6600 in 1 year and ₹ 7986 in 3 years. Find the sum and the rate of interest.

Solution:

Amount in 1 year

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

Hence, the required sum = ₹ 6000

and the rate of interest is 10% p.a.

Question 6.

Arun borrowed ₹ 18000 from his friend Ramesh at 15% p.a. simple interest. He lent it to Rahul at the same rate but compounded annually. Find his gain after 3 years.

Solution:

S.I. = \(\frac{P \times R \times T}{100}\)

= \(\frac{18000 \times 15 \times 3}{100}\)

= ₹ 8100

C.I. = P \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

= 18000 \(\left[\left(1+\frac{15}{100}\right)^3-1\right]\)

= 18000 \(\left[\left(\frac{23}{20}\right)^3-1\right]\)

= 18000 \(\left[\frac{12167}{8000}-1\right]\)

= 18000 \(\left[\frac{12167-8000}{8000}\right]\)

= 18000 × \(\frac{4167}{8000}\)

= \(\frac{37503}{4}\)

= ₹ 9375.75

Gain = S.I – C.I

= ₹ 9375.75 – ₹ 8100

= ₹ 1275.75

Hence, the required gain = ₹ 1275.75

Question 7.

At what per cent per annum Ci. will Z 2000 amount to Z 2315.25 in 3 years?

Solution:

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

2315.25 = 2000 \(\left[1+\frac{\mathrm{R}}{100}\right]^3\)

⇒ \(\frac{2315.25}{2000}=\left[1+\frac{\mathrm{R}}{100}\right]^3\)

⇒ \(\frac{9261}{8000}=\left[1+\frac{R}{100}\right]^3\)

⇒ \(\left[\frac{21}{20}\right]^3=\left[1+\frac{\mathrm{R}}{100}\right]^3\)

Taking cube root on both sides, we get

⇒ \(\frac{21}{20}=1+\frac{\mathrm{R}}{100}\)

⇒ \(\frac{21}{20}-1=\frac{R}{100}\)

⇒ \(\frac{1}{20}=\frac{R}{100}\)

⇒ R = \(\frac{100}{20}\) = 5% p.a.

Hence, the required rate of interest = 5% p.a.

Question 8.

The present population of a village is 28000. If the population increases at the rate of 5% p.a. in the first year and 7% p.a. in the second year. Find the population of the village after 2 years.

Solution:

Population after 2 years = Present population × \(\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\)

= 28000 × \(\left(1+\frac{5}{100}\right)\left(1+\frac{7}{100}\right)\)

= 28000 × \(\frac{21}{20} \times \frac{107}{100}\)

= 31458

Hence, the population after 2 years will be 31458.

Question 9.

The present population of a town is 24000. If it increases at the rate of 5% p.a., then after 2 years it will be

(a) 26400

(b) 26460

(c) 24460

(d) 26640.

Solution:

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 24000 \(\left[1+\frac{5}{100}\right]^2\)

= 24000 \(\left[\frac{21}{20}\right]^2\)

= 24000 \(\frac{21}{20} \times \frac{21}{20}\)

= 60 × 441

= 26460

Hence, (b) is correct.

Question 10.

At what per cent will a sum of ₹ 3125 amount to ₹ 3645 in 2 years, if the interest is compounded annually.

Solution:

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

3645 = 3125 \(\left[1+\frac{\mathrm{R}}{100}\right]^2\)

\(\frac{3645}{3125}=\left[1+\frac{\mathrm{R}}{100}\right]^2\)

\(\frac{729}{625}=\left[1+\frac{R}{100}\right]^2\)

\(\left[\frac{27}{25}\right]^2=\left[1+\frac{\mathrm{R}}{100}\right]^2\)

Taking square root on bothsides

\(1+\frac{\mathrm{R}}{100}=\frac{27}{25}\)

\(\frac{\mathrm{R}}{100}=\frac{27}{25}-1\)

\(\frac{\mathrm{R}}{100}=\frac{27-25}{25}\)

\(\frac{R}{100}=\frac{2}{25}\)

R = \(\frac{2 \times 100}{25}\) = 8% p.a.

Hence, the required rate is 8% p.a.

Question 11.

Madhu bought a house for ₹ 1,31,25000. If its value depreciates at the rate of 10% p.a., what will be its sale price after 3 years?

Solution:

Sale price after 3 years = Present value \(\left(1-\frac{R}{100}\right)^3\)

= 1,31,25000 \(\left(1-\frac{10}{100}\right)^3\)

= 1,31,25000 × \(\frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}\)

= 13125 × 729

= ₹ 9568125

Hence, the value of house after 3 years = ₹ 95,68,125.

Question 12.

Nidhi deposited ₹ 7500 in a bank, which pays her interest at the rate of 4% p.a. compounded annually. Find the amount and the interest received by her after 3 years.

Solution:

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 7500 \(\left[1+\frac{4}{100}\right]^3\)

= 7500 \(\left[\frac{26}{25}\right]^3\)

= 7500 × \(\frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)

= ₹ 8436.48

Interest = A – D

= ₹ 8436.48 – ₹ 7500 = ₹ 936.48

Hence, the required amount is ₹ 8436.48

and the interest is ₹ 936.48.

Question 13.

The difference between the compound interest and the simple interest on a certain sum of money at 8% p.a. compounded annually for 3 years is ₹ 308. Find the sum.

Solution:

Hence the required value is ₹ 12500.

Question 14.

The value of refrigerator, which was purchased two years ago, depreciates at 12% p.a. If its present value is ₹ 9680, for how much was it purchased?

Solution:

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\) [for depreciation]

9680 = P \(\left[1-\frac{12}{100}\right]^2\)

9680 = P \(\left[\frac{22}{25}\right]^2\)

9680 = P × \(\frac{22}{25} \times \frac{22}{25}\)

∴ P = \(\frac{9680 \times 25 \times 25}{22 \times 22}\)

= ₹ 12500

Hence the required value is ₹ 12500.

Multiple Choice Questions:

Question 1.

The number of years in which ₹ 2000 amount to ₹ 2662 at 10% p.a. compound interest is

(a) 2 years

(b) 4 years

(c) 3 years

(d) 1 \(\frac{1}{2}\) years

Solution:

(b) 4 years

Question 2.

The sum of money that will amount to ₹ 3630 in 2 years at 10% p.a. compound interest is

(a) ₹ 3500

(b) ₹ 3200

(c) ₹ 3000

(d) ₹ 3150

Solution:

(c) ₹ 3000

Question 3.

Compound interest on ₹ 4000 for 2 years at 10% p.a. interest being compounded half- yearly is

(a) ₹ 862.05

(b) ₹ 862.08

(c) ₹ 862.04

(d) ₹ 862.03

Solution:

(d) ₹ 862.03

Question 4.

A sum of money becomes times itself in 4 years compounded annually. Then the rate of interest is

(a) 33\(\frac{1}{2}\) %

(b) 33\(\frac{1}{3}\) %

(c) 33\(\frac{1}{4}\) %

(d) 33 %

Solution:

(b) 33\(\frac{1}{3}\) %

Question 5.

If the difierence between C.I. and S.I. on a certain sum of money for 2 years at 5% p.a. be ? 25, then the sum is

(a) ₹ 10,500

(b) ₹ 10,550

(c) ₹ 10000

(d) ₹ 10510.

Solution:

(c) ₹ 10000