Question 13 Marks
A sum of $Rs. 13,500$ is invested at $16\%$ per annum compound interest for $5$ years. Calculate :
$(i)$ the interest for the first year.
$(ii)$ the amount at the end of first year.
$(iii)$ the interest for the second year, correct to the nearest rupee.
AnswerFor $1^{\text {st }}$ year :
$P= Rs. 13,500 ; R=16 \%$ and $T=1$ year
Interest $= Rs.\frac{13,500 \times 16 \times 1}{100}=Rs.2,160$
Amount $= Rs. 13,500+ Rs. 2,160= Rs. 15,660$
For $2^{\text {nd }}$ year :
$P= Rs. 15,660 ; R=16 \%$ and $T=1$ year
Interest $=Rs.\frac{15,660 \times 16 \times 1}{100}= Rs. 2,505.60$
$= Rs. 2,506$
View full question & answer→Question 23 Marks
During every financial year, the value of a machine depreciates by $12\%.$ Find the original cost of a machine which depreciates by $Rs. 2,640$ during the second financial year of its purchase.
AnswerLet original value of machine $= Rs. 100$
For $1^{\text {st }}$ year
$P= Rs. 100 ; R=12 \%$ and $T=1$ year
Depreciation in $1^{\text {st }}$ year $= Rs \frac{100 \times 12 \times 1}{100}= Rs. 12$
Value at the end of $1^{\text {st }}$ year $= Rs. 100- Rs. 12= Rs. 88$
For $2^{\text {nd }}$ year
$P= Rs. 88 ; R=12 \%$ and $T=1$ year
Depreciation in $2^{\text {nd }}$ year $= Rs. \frac{88 \times 12 \times 1}{100}= Rs. 10.56$
When depreciation in $2^{\text {nd }}$ year is $Rs. 10.56 ,$ original cost is $Rs. 100$
When depreciation in $2^{\text {nd }}$ year is $Rs.2,640,$ original cost $=\frac{100 \times 2640}{10.56}$
$= Rs. 25,000$
View full question & answer→Question 33 Marks
The value of a machine depreciated by $10\%$ per year during the first two years and $15\%$ per year during the third year. Express the total depreciation of the machine, as percent, during the three years.
AnswerLet the value of the machine in the beginning $= Rs. 100$
For $1^{st}$ year depreciation $= 10\%$ of $Rs. 100 = Rs. 10$
Value of machine for second year $= 100 - 10 = Rs. 90$
For $2^{nd}$ year depreciation $= 10\%$ of $90 = Rs. 9$
Value of machine for third year $= 90 - 9 = Rs. 81$
For $3^{rd}$ year depreciation $= 15\%$ of $81 = Rs. 12.15$
Value of machine at the end of third year $= 81 - 12.15= Rs. 68.85$
Net depreciation $= Rs. 100 - Rs. 68.85$
$= Rs. 31.15$ or $31.15\%.$
View full question & answer→Question 43 Marks
What sum will amount of $Rs. 6,593.40$ in $2$ years at C.I. , if the rates are $10$ per cent and $11$ per cent for the two successive years ?
AnswerLet principal $(p)= Rs. 100$
For $1^{\text {st }}$ year :
$P= Rs.100$
$ R=10 \%$
$ T=1$ year
$ I=\frac{100 \times 10 \times 1}{100}=\text { Rs. } 10 .$
$A=100+10=\text { Rs. } 110$
For $2^{\text {nd }}$ year :
$P= Rs. 110$
$ R=11 \%$
$ T=1$ year
$ I=\frac{110 \times 11 \times 1}{100}= Rs. 12.10 .$
$A=110+12.10= Rs. 122.10$
If Amount is $Rs. 122.10$ on a sum of $Rs. =100$
If amount is $Rs. 1 ,$ sum $=\frac{100}{122.10}$
If amount is $Rs. 6593.40 ,$ sum $=\frac{100}{122.10} \times 6593.40$
$= Rs. 5400$
View full question & answer→Question 53 Marks
$Rs. 8,000$ is lent out at $7\%$ compound interest for $2$ years. At the end of the first year $Rs. 3,560$ are returned. Calculate :
$(i)$ the interest paid for the second year.
$(ii)$ the total interest paid in two years.
$(iii)$ the total amount of money paid in two years to clear the debt.
Answer$(i)$ For $1^{\text {st }}$ year :
$P= Rs. 8,000 ; R=7 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{8,000 \times 7 \times 1}{100}=\text { Rs. } 560 .$
Amount $=Rs.8,000+ Rs. 560=Rs. 8,560$
Money returned $=Rs. 3,560$
Balance money for $2^{\text {nd }}$ year$= Rs. 8,560 - Rs. 3,560= Rs. 5,000$
For $2^{\text {nd }}$ year :
$P= Rs. 5,000 ; R=7 \%$ and $T=1$ year.
Interest paid for the second year $= Rs. \frac{5,000 \times 7 \times 1}{100}$
$\text { = Rs. } 350$
$(ii)$ The total interest paid in two years$= Rs. 350 + Rs. 560= Rs. 910$
$(iii)$ The total amount of money paid in two years to clear the debt
$=\text { Rs. } 8,000+\text { Rs. } 910$
$=\text { Rs. } 8,910$
View full question & answer→Question 63 Marks
Ramesh invests $Rs. 12,800$ 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 the third year.
AnswerFor $1^{\text {st }}$ year :
$P= Rs. 12,800 ; R=10 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{12,800 \times 10 \times 1}{100}=\text { Rs. } 1,280 .$
Amount $= Rs.12,800+ Rs.1,280= Rs. 14,080 .$
For $2^{\text {nd }}$ year :
$P= Rs. 14,080 ; R=10 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{14,080 \times 10 \times 1}{100}=\text { Rs. } 1,408. $
Amount $= Rs.14,080+Rs.1,408=Rs. 15,488$
For $3^{\text {rd }}$ year :
$P= Rs. 15,488 ; R=10 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{15,488 \times 10 \times 1}{100}=\text { Rs. } 1,548.80$
Amount $= Rs. 15,488+ Rs. 1,548.80= Rs. 17,036.80$
View full question & answer→Question 73 Marks
Mohit invests $Rs. 8,000$ for $3$ years at a certain rate of interest, compounded annually. At the end of one year it amounts to $Rs. 9,440.$ Calculate :
$(i)$ the rate of interest per annum.
$(ii) $the amount at the end of the second year.
$(iii)$ the interest accrued in the third year.
AnswerFor $1^{\text {st }}$ year
$P=\text { Rs. } 8,000 ; A=9,440$ and $T=1$ year
Interest$=\text { Rs. } 9,440-\text { Rs. } 8,000=\text { Rs. } 1,440$
Rate$=\frac{I \times 100}{P \times T} \%$
$ =\frac{1,440 \times 100}{8,000 \times 1} \%=18 \%$
For $2^{\text {nd }}$ year
$P= Rs. 9,440 ; R=18 \%$ and $T=1$ year
Interest $=\text { Rs } \frac{9,440 \times 18 \times 1}{100}=\text { Rs. } 1,699.20$
Amount $= Rs. 9,440+ Rs. 1,699 \cdot 20= Rs.11,139 \cdot 20$
For $3^{\text {rd }}$ year
$P= Rs. 11,139.20 ; R=18 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{11,139.20 \times 18 \times 1}{100}$
$=\text { Rs. } 2,005.06$
View full question & answer→Question 83 Marks
The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. 1,089$ and for the third year it is $Rs. 1,197.90.$ Calculate the rate of interest and the sum of money.
Answer$(i)$ C.I. for second year $= Rs. 1,089$
C.I. for third year $= Rs. 1,197.90$
$\because$ Difference between the C.I. of two successive years
$= Rs. 1,197.90 - Rs. 1089 = Rs. 108.90$
$\Rightarrow Rs. 108.90$ is the interest of one year on $Rs. 1089 .$
$ \therefore$ Rate of interest $=\text { Rs. } \frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{T}} \%$
$ =\frac{100 \times 108.90}{1089 \times 1} \%=10 \%$
$(ii)$ Let the sum of money $= Rs. 100$
$\therefore$ Interest on it for $1^{\text {st }}$ year $=10 \%$ of $Rs.100= Rs. 10$
$\Rightarrow$ Amount in one year $= Rs.100+ Rs.10= Rs. 110$
Similarly, C.I. for $2^{\text {nd }}$ year $=10 \%$ of $Rs. 110= Rs. 11$
When C.I. for $2^{\text {nd }}$ year is $Rs. 11 , sum = Rs. 100$
When C.I. for $2^{\text {nd }}$ year is $Rs. 1089 ,$ sum $= Rs. \frac{100 \times 1089}{11}$
$= Rs. 9,900 .$
View full question & answer→Question 93 Marks
A man borrows $Rs.10,000$ at $10\%$ compound interest compounded yearly. At the end of each year, he pays back $30\%$ of the sum borrowed. How much money is left unpaid just after the second year ?
AnswerFor $1^{\text {st }}$ year :
$P= Rs. 10,000 ; R=10 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{10,000 \times 10 \times 1}{100}=\text { Rs. } 1,000$
Amount at the end of $1^{\text {st }}$ year $= Rs. 10,000+ Rs. 1,000= Rs. 11,000$
Money paid at the end of $1^{\text {st }}$ year $=30%$ of $Rs. 10,000=Rs. 3,000$
$\therefore$ Principal for $2^{\text {nd }}$ year $= Rs. 1,000 - Rs. 3,000= Rs. 8,000$
For $2^{\text {nd }}$ year :
$P= Rs. 8,000 ; R=10 \%$ and $T=1$ year
Interest$=\text { Rs. } \frac{8,000 \times 10 \times 1}{100}=\text { Rs. } 800$
Amount at the end of $2^{\text {nd }}$ year $= Rs. 8,000+ Rs. 800= Rs. 8,800$
Money paid at the end of $2^{\text {nd }}$ year $=30 \%$ of $Rs. 10,000= Rs. 3,000$
$\therefore$ Principal for $3^{\text {rd }}$ year $= Rs. 8,800- Rs. 3,000= Rs. 5,800 .$
View full question & answer→Question 103 Marks
Find the sum, invested at $10\%$ compounded annually, on which the interest for the third year exceeds the interest of the first year by $Rs. 252.$
AnswerLet the sum of money be $Rs. 100 .$
Rate of interest $=10 \%$ p.a.
Interest at the end of $1^{\text {st }}$ year $=10 \%$ of $Rs. 100= Rs. 10$
Amount at the end of $1^{\text {st }}$ year $= Rs.100+ Rs.10= Rs. 110$
Interest at the end of $2^{\text {nd }}$ year $=10 \%$ of $Rs. 110= Rs. 11$
Amount at the end of $2^{\text {nd }}$ year $=Rs.110+Rs.11= Rs. 121$
Interest at the end of $3^{\text {rd }}$ year $=10 \%$ of $Rs.121= Rs.12 \cdot 10$
Difference between interest of $3^{\text {rd }}$ year and $1^{\text {st }}$ year
$\text { = Rs. } 12.10 \text { - Rs. } 10 \text { = Rs. } 2 \cdot 10$
When difference is $Rs. 2.10 ,$ principal is $Rs. 100$
When difference is $Rs. 252 ,$ principal $=\frac{100 \times 252}{2.10}$
$= Rs. 12,000 .$
View full question & answer→Question 113 Marks
A man invests $Rs. 5,600$ at $14\%$ per annum compound interest for $2$ years. Calculate: The interest for the second year, correct to the nearest rupee.
AnswerFor the first year:
$P= Rs. 5,600, N=1$ year and $R=14 \%$
We have,
$\text { S.I. }=\frac{\mathrm{PNR}}{100}=\frac{5,600 \times 1 \times 14}{100}=\text { Rs. } 784 .$
And Amount at the end of first year $P + S.I. = Rs. 5600+ Rs. 784=Rs. 6,384 .$
Now for the second year:
For $2^{\text {nd }}$ year
$\mathrm{P}=6384, \mathrm{R}=14 \%, \mathrm{~N}=1 \text { year }$
$ \text { S.I. }=\frac{\mathrm{PNR}}{100}=\frac{6,384 \times 14 \times 1}{100}=R s .893 .76$
To the nearest rupee, it is $Rs. 894 ($nearly$).$
View full question & answer→Question 123 Marks
A manufacturer estimates that his machine depreciates by $15\%$ of its value at the beginning of the year. Find the original value $($cost$)$ of the machine, if it depreciates by $\text{Rs}. 5,355$ during the second year.
AnswerLet the original cost of the machine $= Rs. 100 .$
$\therefore$ Depreciation during the $1^{\text {st }}$ year $=15 \%$ of $Rs. 100= Rs, 15 .$
Value of the machine at the beginning of the $2^{\text {nd }}$ year
$=\text { Rs. } 100-\text { Rs. } 15=\text { Rs. } 85$
$\therefore$ Depreciation during the $2^{\text {nd }}$ year $=15 \%$ of $ Rs. 85= Rs. 12.75$
Now, when depreciation during $2^{\text {nd }}$ year $=R s, 12.75$,
original cost $= Rs. 100$
$\therefore$ when depreciation during $2^{\text{nd}}$ year $= Rs. 5,355.$
original cost $=\text{ Rs. } \frac{100}{12.75} \times 5,355=\text{ Rs. } 42,000.$
Hence, original cost of the machine is $Rs. 42,000 .$
View full question & answer→Question 133 Marks
A borrowed $Rs. 2,500$ from $B$ at $12\%$ per annum compound interest. After $2$ years$, A$ gave $Rs. 2,936$ and a watch to $B$ to clear the account. Find the cost of the watch.
AnswerFor $1^{\text {st }}$ year,
$P=\text { Rs. } 2500$
$ R=12 \%$
$ T=1 \text { year }$
$ I=\frac{2500 \times 12 \times 1}{100}=\text { Rs. } 300 .$
Amount $=2500+300= Rs. 2800$
For $2^{\text {nd }}$ year,
$P=\text { Rs. } 2800$
$ R=12 \%$
$ T=1$ year
$I=\frac{2800 \times 12 \times 1}{100}=\text { Rs. } 336$
Amount $=2800+336= Rs. 3136$
Amount repaid by $A$ to $B = Rs. 2936$
The amount of watch $=Rs. 3136 - Rs. 2936$
$= Rs. 200$
View full question & answer→Question 143 Marks
Calculate the compound interest for the second year on $Rs. 8,000/-$ invested for $3$ years at $10\%$ per annum.
AnswerFor $1^{\text {st }}$ year
$P=\text { Rs. } 8000$
$ R=10 \%$
$ T=1$ year
$I=\frac{8000 \times 10 \times 1}{100}=800$
$A=8000+800=\text { Rs. } 8800$
For $2^{\text {nd }}$ year
$P=\text { Rs. } 8800$
$ R=10 \%$
$ T=1$ year
$I=\frac{8800 \times 10 \times 1}{100}$
Compound interest for $2^{\text {nd }}$ years $= Rs. 880$
View full question & answer→Question 153 Marks
Calculate the amount and the compound interest on :$₹ 4,600$ in $2$ years when the rates of interest of successive years are $10\%$and $12\%$ respectively.
AnswerFor $1^{\text {st }}$ year
$P=\text { Rs. } 4600$
$ R=10 \%$
$ T=1$ year.
$ I=\frac{4600 \times 10 \times 1}{100}=\text { Rs. } 460$
$A=4600+460=\text { Rs. } 5060$
For $2^{\text {nd}}$ year
$P=\text { Rs. } 5060$
$ R=12 \%$
$ T=1$ year.
$ I=\frac{5060 \times 12 \times 1}{100}=\frac{60720}{100}=607.20$
$ A=5060+607.20=\text { Rs. } 5667.20$
Compound interest $=5667.20-4600=\text { Rs. } 1067.20$
Amount after $2$ years $=\text { Rs. } 5667.20$
View full question & answer→Question 163 Marks
Calculate the amount and the compound interest on $:Rs. 8,000$ in $2 \frac{1}{2}$ years at $15 \%$ per year.
AnswerFor $1^{\text {st }}$ year,
$P=$ Rs. 8,$000 ; R=15 \%$, and $T=1$ year
$\therefore$ Interest $= Rs. \frac{8,000 \times 15 \times 1}{100}= Rs. 1200 .$
And, amount $= Rs. (8,000+1200)= Rs. 9,200$
For $2^{\text {nd }}$ year,
$P=$ Rs. 9,200; $R=15 \%$, and $T=1$ year
$\therefore$ Interest $= Rs. \frac{9,200 \times 15 \times 1}{100}= Rs. 1,380 .$
And, amount $= Rs. (9,200+1,380)= Rs. 10,580$
For the last $\frac{1}{2}$ year,
$P=\text { Rs. } 10,580 ; R=15 \%$ and $T=\frac{1}{2}$ year
$ \therefore$ Interest $=\text { Rs. } \frac{10,580 \times 15 \times 1}{100 \times 2}=\text { Rs. } 793.50$
And, Amount $= Rs. (10,580+793.50)= Rs. 11373.50$
$\therefore$ C.I. accrued $=$ Final amount $-$Intitial Principal
$=\text { Rs. }(11,373.50-8,000)$
$ =\text { Rs. } 3373.50$
View full question & answer→Question 173 Marks
Calculate the amount and the compound interest on $:Rs. 6,000$ in $3$ years at $5\%$ per year.
AnswerFor $1^{\text {st }}$ year,
$P=R s .6,000 ; R=5 \%$, and $T=1$ year
$\therefore$ Interest $= Rs. \frac{6,000 \times 5 \times 1}{100}= Rs. 300 .$
And, amount $= Rs. (6,000+300)= Rs. 6,300$
For $2^{\text{nd}}$ year,
$P=R s .6,300 ; R=5 \%,$ and $T=1$ year
$\therefore$ Interest $= Rs.\frac{6,300 \times 5 \times 1}{100}= Rs. 315$
And, amount $= Rs. (6,300+315)= Rs. 6,615$
For $3^{\text{rd}}$ year,
$ P=\text { Rs. } 6,615 ; R=5 \%$ and $T=1 \text { year }$
$ \therefore$ Interest$=\text { Rs. } \frac{6,615 \times 5 \times 1}{100}=\text { Rs. } 330.75$
And, Amount $=\text { Rs. }(6,615+330.75)=\text { Rs. } 6,945.75$
$ \therefore$ C.I. accrued $=$ Final amount $-$ Intitial Principal
$ =\text { Rs. }(6,945.75-6,000)$
$ =\text { Rs. } 945.75$
View full question & answer→Question 183 Marks
$Rs.16,000$ is invested at $5\%$ compound interest compounded per annum. Use the table, given below, to find the amount in $4$ years.
| Year$\downarrow$ |
Initial amount$(Rs.)$ |
Interest $(Rs.)$ |
Final amount $(Rs.)$ |
| $1^{st}$ |
$16,000$ |
$800$ |
$16,800$ |
| $2^{nd}$ |
$...........$ |
$...........$ |
$...........$ |
| $3^{rd}$ |
$...........$ |
$...........$ |
$...........$ |
| $4^{th}$ |
$...........$ |
$...........$ |
$...........$ |
| $5^{th}$ |
$...........$ |
$...........$ |
$...........$ |
Answer
| Year$\downarrow$ |
Initial amount$(Rs.)$ |
Interest $(Rs.)$ |
Final amount $(Rs.)$ |
| $1^{st}$ |
$16,000$ |
$800$ |
$16,800$ |
| $2^{nd}$ |
$16,800$ |
$840$ |
$17,640$ |
| $3^{rd}$ |
$17,640$ |
$882$ |
$18,522$ |
| $4^{th}$ |
$18,522$ |
$926.10$ |
$19,448.10$ |
| $5^{th}$ |
$19,448.10$ |
$972.405$ |
$20,420.505$ |
Thus, the amount in $4$ years is $Rs. 19448.10.$ View full question & answer→