Question 15 Marks
Ashok borrowed $Rs. 12,000$ at some rate on compound interest. After a year, he paid back $Rs.4,000$. If the compound interest for the second year is $Rs. 920$, find:
- The rate of interest charged
- The amount of debt at the end of the second year
Answer$(i)$ Let $\mathrm{X}\ \% \ $ be the rate of interest charged.
For $1^{\text {st }}$ year :
$P=\text { Rs. } 12,000, R=X\ \%\ $ and $T=1$
$ \Rightarrow$ Interest $(I)=\frac{12,000 \times X \times 1}{100}=120 X$
For $2^{\text {nd }}$ year:
After a year, Ashok paid back $Rs. 4,000.$
$P=\text { Rs. } 12,000+\text { Rs. } 120 X-\text { Rs. } 4,000=\text { Rs. } 8,000+\text { Rs. } 120X$
$\Rightarrow$ Interest$ (I)=\frac{(8000+120 X) \times 1}{100}=\left(80 X+1.20 X^2\right)$
The compound interest for the second year is $Rs. 920 .$
$\text { Rs. }\left(80 X+1.20 X^2\right)=\text { Rs. } 920$
$ \Rightarrow 1.20 X^2+80 X-920=0$
$ \Rightarrow 3 X^2+200 X-2300=0$
$ \Rightarrow 3 X^2+230 X-30 X-2300=0$
$ \Rightarrow X(3 X+230)-10(3 X+230)=0$
$ \Rightarrow(3 X+230)(X-10)=0$
$ \Rightarrow X=-230 / 3 \text { or } X=10$
As rate of interest cannot be negative so $x=10$.
Therefore the rate of interest charged is $10\ \%\ $.
$(ii)$For $1^{\text {st }}$ year :
Interest $=\text { Rs. } 120 \mathrm{X}=\text { Rs. } 1200$
For $2^{\text {nd }}$ year :
Interest $=\text { Rs. }\left(80 X+1.20 X^2\right)=\text { Rs. } 920$
The amount of debt at the end of the second year is equal to the addition of principal of the second year and interest for the two years.
Debt $=\text { Rs. } 8,000+\text { Rs. } 1200+\text { Rs. } 920=\text { Rs. } 10,120$
View full question & answer→Question 25 Marks
Find the sum on which the difference between the simple interest and compound interest at the rate of $8\ \%\ $ per annum compounded annually would be $Rs. 64$ in $2$ years.
AnswerLet $Rs, X$ be the sum.
Simple Interest$(I)=\frac{X \times 8 \times 2}{100}=0.16 X$
Compound interest
For $1^{\text {st }}$ year :
$P=\text { Rs. } X_i R=8 \%\ $ and $T=1$
$ \Rightarrow$ Interest $(I)=\frac{X \times 8 \times 1}{100}=0.08 X$
And amount $=\text{₹}(x+0.08 x)$
$\text { = ₹ } 1.08 \mathrm{x}$
For $2^{\text {nd }}$ year :
$P=\text { Rs. } X+\text { Rs. } 0.08 X=\text { Rs. } 1.08 X$
$ \Rightarrow$ Interest $(I)=\frac{1.08 \mathrm{X} \times 8 \times 1}{100}=0.0864 X$
And, amount $=\text{₹}(1.08 x+0.0864 x)$
$=\text{₹} 1.1664 \mathrm{x}$
So,
$C.I =$ Amount $-p$
$=\text{₹}(1.1664 x-x)$
$=\text{₹} 0.1664 x$
The difference between the simple interest and compound interest at the rate of $8\ \%\ $ per annum compounded annually should be $Rs. 64$ in $2$ years.
$\text{₹} 0.1664 x-\text{₹} 0.16 x=\text{₹} 64$
$\text{₹} 0.0064 \mathrm{x}=\text{₹} 64$
$x=\text{₹} 10000$
Therefore, the sum is $\text{₹ 10,000}$.
View full question & answer→Question 35 Marks
On a certain sum of money, invested at the rate of $10$ percent per annum compounded annually, the interest for the first year plus the interest for the third year is $Rs. 2,652.$ Find the sum.
AnswerLet Principal $= Rs. 100$
For $1^{\text {st }}$ year
$P= Rs. 100 ; R=10\ \%\ $ and $T=1$ year
Interest $=\text { Rs. } \frac{100 \times 10 \times 1}{100}=\text { Rs. } 10$
Amount $= Rs. 100+ Rs. 10= Rs. 110$
For $2^{\text {nd }}$ year
$P= Rs. 110 ; R=10\ \%\ $ and $T=1$ year
Interest $=\text { Rs. } \frac{110 \times 10 \times 1}{100}=\text { Rs. } 11$
Amount $= Rs. 110+ Rs. 11= Rs. 121$
For $3^{\text {rd }}$ year
$P= Rs. 121 ; R=10\ \%\ $ and $T=1$ year
Interest $=\text { Rs } \frac{121 \times 10 \times 1}{100}=\text { Rs. } 12.10$
Sum of $C.I.$ for $1^{\text {st }}$ year and $3^{\text {rd }}$ year $= Rs. 10+ Rs. 12 \cdot 10= Rs. 22 \cdot 10$
When sum is $Rs. 22.10$ , principal is $Rs. 100$
When sum is $ Rs. 2,652$, principal $= Rs. \frac{100 \times 2652}{22.10}$
$=\text { Rs. } 12,000 \text {. }$
View full question & answer→Question 45 Marks
Rachna borrows $Rs. 12,000$ at $10$ percent per annum interest compounded half$-$yearly. She repays $Rs. 4,000$ at the end of every six months. Calculate the third payment she has to make at end of $18$ months in order to clear the entire loan.
AnswerFor $1^{\text {st }}$ half $-$ year :
$P=\text { Rs. } 12,000 ; R=10\ \%\ $ and $T=\frac{1}{2} \text { year }$
Interest $=\text { Rs. } \frac{12,000 \times 10 \times 1}{100 \times 2}=\text { Rs. } 600$
Amount $= Rs. 12,000+ Rs. 600= Rs. 12,600$
Money paid at the end of $1^{\text {st }}$ half year $= Rs. 4,000$
Balance money for $2^{\text {nd }}$ half$-$year $=Rs. 12,600- Rs. 4,000= Rs. 8,600$
For $2^{\text {nd }}$ half $-$ year :
$P= Rs. 8,600 ; R=10\ \%\ $ and $T=\frac{1}{2}$ year
Interest $= Rs. \frac{8,600 \times 10 \times 1}{100 \times 2}= Rs. 430$
Amount $= Rs. 8,600+ Rs. 430= Rs. 9,030$
Money paid at the end of $2^{\text {nd }}$ half$-$year $= Rs. 4,000$
Balance money for $3^{\text {rd }}$ half $-$ year $= Rs. 9,030- Rs. 4,000= Rs. 5,030$
For $3^{\text {rd }}$ half$-$year
$P= Rs. 5,030 ; R=10 \%$ and $T=\frac{1}{2}$ year
Interest $=\text { Rs. } \frac{5,030 \times 10 \times 1}{100 \times 2}=\text { Rs. } 251.50$
Amount $= Rs. 5,030+ Rs. 251 \cdot 50= Rs. 5,281 \cdot 50$
View full question & answer→Question 55 Marks
The cost of a machine depreciated by $Rs. 4,000$ during the first year and by $Rs. 3,600$ during the second year. Calculate :
- The rate of depreciation.
- The original cost of the machine.
- Its cost at the end of the third year.
Answer$(i)$ Difference between depreciation in value between the first and second years $Rs. 4,000 - Rs. 3,600= Rs. 400.$
$\Rightarrow$ Depreciation of one year on $Rs. 4,000= Rs. 400$
$\Rightarrow$ Rate of depreciation $=\frac{400}{4000} \times 100 \ \%\ =10\ \%\ $
$(ii)$ Let $Rs. 100$ be the original cost of the machine.
Depreciation during the $1^{\text {st }}$ year $=10 \%$ of R$s. 100= Rs. 10$
When the values depreciates by $Rs. 10$ during the $1^{\text {st }}$ year, Original cost $= Rs. 100$
$\Rightarrow$ When the depreciation during $1^{\text {st }}$ year $= Rs. 4,000$
Original Cost $=\frac{100}{10} \times 4000= Rs. 40,000$
The original cost of the machine is $Rs. 40,000 .$
$(iii)$ Total depreciation during all the three years
$=$ Depreciation in value during $(1^{\text {st }}$ year $+2^{\text {nd }}$ year $+3^{\text {rd }}$ year$)$
$ =\text { Rs. } 4,000+\text { Rs. } 3,600+10\ \%\ $ of $(\text {Rs. } 40,000-\text { Rs. } 7,600)$
$ =\text { Rs. } 4,000+\text { Rs. } 3,600+\text { Rs. } 3,240$
$ =\text { Rs. } 10,840$
The cost of the machine at the end of the third year
$=\text { Rs. } 40,000-\text { Rs. } 10,840=\text { Rs. } 29,160$
View full question & answer→Question 65 Marks
On a certain sum of money, the difference between the compound interest for a year, payable half$-$yearly, and the simple interest for a year is $Rs. 180/-$ Find the sum lent out, if the rate of interest in both the cases is $10\ \%\ $ per annum.
AnswerLet principal $p= Rs. 100 ; R=10\ \%\ ; T=1$ year
$\text { SI }=\frac{100 \times 10 \times 1}{100}=\text { Rs. } 10 \text {. }$
$\mathrm{Cl}$ payable at every 6 months
So, $R=\frac{10}{2}=5 \%$
$I=\frac{100 \times 5 \times 1}{100}=\text { Rs. } 5$
$A=100+5=\text { Rs. } 105$
For second year
$P=\text { Rs. } 105$
$I=\frac{105 \times 5 \times 1}{100}=\text { Rs. } 5.25$
Total compound interest $=5+5.25= Rs. 10.25$
Difference of $\mathrm{Cl}$ and $\mathrm{SI}=10.25-10= Rs. 0.25$
When difference in interest is $Rs. 0.25$ , sum $= Rs. 100 .$
If the difference is $Rs. 1$ then
sum$=\frac{100}{0.25}$
If the difference is $Rs. =180$ then
sum $=\frac{100}{0.25} \times 180=\text { Rs. } 72,000$
View full question & answer→Question 75 Marks
A man borrows $Rs.5,000$ at $12$ percent compound interest payable every six months. He repays $Rs.1,800$ at the end of every six months. Calculate the third payment he has to make at the end of $18$ months in order to clear the entire loan.
AnswerFor $1^{\text {st }}$ six months :
$P= Rs. 5,000, R=12\ \%\ $ and $T=6$ months $=\frac{1}{2}$ year
$\therefore$ Interest $=\frac{5,000 \times 12 \times 1}{2 \times 100}= Rs. 300 .$
And, Amount $= Rs. 5,000+ Rs.300= Rs. 5,300$
Since, money repaid $= Rs. 1,800$
Balance $= Rs. 5,300- Rs. 1,800= Rs. 3,500$
For $2^{\text {nd }}$ six months :
$P= Rs. 3,500, R=12\ \%\ $ and $T=6$ months $=\frac{1}{2}$ year
$\therefore$ Interest $=\frac{3,500 \times 12 \times 1}{2 \times 100}=\text { Rs. } 210 \text {. }$
And, Amount $= Rs. 3,500+ Rs. 210= Rs. 3,710 .$
Again money repaid $= Rs. 1,800$
Balance $= Rs. 3,710- Rs. 1,800= Rs. 1,910 .$
For $3^{\text {rd }}$ six months :
$P= Rs. 1,910, R=12\ \%\ $ and $T=6$ months $=\frac{1}{2}$ year
$\therefore$ Interest $=\frac{1,910 \times 12 \times 1}{2 \times 100}= Rs. 114.60 .$
And, Amount $= Rs. 1,910+ Rs. 114.60= Rs. 2,024.60$
Thus, the $3^{\text {rd }}$ payment to be made to clear the entire loan is $2,024.60$.
View full question & answer→Question 85 Marks
A sum of money is lent at $8\ \%\ $ per annum compound interest. If the interest for the second year exceeds that for the first year by $Rs. 96$, find the sum of money.
AnswerLet money be $Rs 100$
For $1^{\text {st }}$ year
$P=\text { Rs. } 100 ; R=8\ \%\ $ and $T=1 \text { year. }$
Interest for the first year $=\text { Rs. } \frac{100 \times 8 \times 1}{100}=\text { Rs. } 8$
Amount $= Rs. 100+ Rs. 8 = Rs. 108$
For $2^{\text {nd }}$ year
$P=\text { Rs. } 108 ; R=8\ \%\ $ and $T=1 \text { year. }$
Interest for the second year $= Rs. \frac{108 \times 8 \times 1}{100}=\text { Rs. } 8.64$
Difference between the interests for the second and first year $= Rs. 8.64 - Rs. 8= Rs. 0.64$
Given that interest for the second year exceeds the first year by $Rs. 96 .$
When the difference between the interests is $Rs. 0.64$ , principal is $Rs. 100$
When the difference between the interests is $Rs.96$, principal $= Rs. \frac{96 \times 100}{0.64}= Rs. 15,000 .$
View full question & answer→Question 95 Marks
A man lends $Rs. 12,500$ at $12\ \%\ $ for the first year, at $15\ \%\ $ for the second year and at $18\ \%\ $ for the third year. If the rates of interest are compounded yearly ; find the difference between the $C.I.$ fo the first year and the compound interest for the third year.
AnswerFor $1^{\text {st }}$ year
$P= Rs. 12500 ; R=12\ \%\ ; R=1$ year
$\mathrm{I}=\frac{12500 \times 12 \times 1}{100}=\text { Rs. } 1500$
$A=12500+1500=\text { Rs. } 14000$
For $2^{\text {nd }}$ year
$P=$ Rs. $14000 ; R=15 \ \%\ ; T=1$ year
$I=\frac{14000 \times 15 \times 1}{100}=\text { Rs. } 2100$
$A=14000+2100=\text { Rs. } 16100$
For $3^{\text {rd }}$ year
$P= Rs. 16100 ; R=18\ \%\ ; T=1$ year
$\text { I }=\frac{16100 \times 18 \times 1}{100}=\text { Rs. } 2898$
$A=16100+2898=\text { Rs. } 18,998$
Difference between the compound interest of the third year and first year
$=\text { Rs. } 2898-\text { Rs. } 1500$
$ =\text { Rs. } 1398$
View full question & answer→Question 105 Marks
Calculate the difference between the simple interest and the compound interest on $Rs. 4,000$ in $2$ years at $8\ \%\ $ per annum compounded yearly.
AnswerFor $1^{\text {st }}$ year
$P=\text { Rs. } 4000$
$ R=8$
$ T=1$ year
$\mathrm{T}=1$ year
$I=\frac{4000 \times 8 \times 1}{100}=320$
$A=4000+320=\text { Rs. } 4320$
For $2^{\text {nd }}$ year
$P=\text { Rs. } 4320$
$ R=8 \ \%\ $
$ T=1$ year
$ I=\frac{4320 \times 8 \times 1}{100}=\text { Rs. } 345.60$
$A=4320+345.60=4665.60$
Compound interest$ =\text { Rs. } 4665.60-\text { Rs. } 4000=\text { Rs. } 665.60$
Simple interest for $2$ years $=\frac{4000 \times 8 \times 2}{100}=\text { Rs. } 640$
Difference of $\mathrm{Cl}$ and $\mathrm{SI}=665.60-640= Rs 25.60 .$
View full question & answer→Question 115 Marks
A man borrows $Rs. 10,000$ at $5\ \%\ $ per annum compound interest. He repays $35\ \%\ $ of the sum borrowed at the end of the first year and $42\ \%\ $ of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt ?
AnswerFor the first year,
$P_1=10,000, R=5\ \%\ $
$ I_1=\frac{10,000 \times 5 \times 1}{100}$
$ I_1=500$
$A_1=10,000+500=\text { Rs. } 10,500 .$
At the end of the first year, he repays $35 \ \%\ $ of the sum borrowed so he repays the amount
$=10,500-35 \ \%\ $ of $10,000$
$ =10,500-\frac{35}{100} \times 10,000$
$ =10,500-3,500$
$ =\text { Rs. } 7,000 .$
For the second year,
$P_2=\text { Rs. } 7000, R=5 \ \%\ $
$\mathrm{I}_2=\frac{7,000 \times 5 \times 1}{100}=350$
$A_2=7000+350=\text { Rs. } 7,350$
At the end of the second year, he repays $42\ \%\ $ of the sum borrowed so he repays the amount $=
=7,350-\frac{42}{100} \times 10,000$
$ =7350-4200$
$ =\text { Rs. } 3150$
For the Third year
$P_3=\text { Rs. } 3150, R=5\ \%\ $
$I_3=\frac{3150 \times 5 \times 1}{100}=157.5$
$A_3=3150+157.5=\text { Rs. } 3307.50$
Hence he pays $Rs. 3307.50$ at the end of the third year in order to clear the debt.
View full question & answer→Question 125 Marks
Govind borrows $Rs.18,000$ at $10\ \%\ $ simple interest. He immediately invests the money borrowed at $10\ \%\ $ compound interest compounded half$-$yearly. How much money does Govind gain in one year ?
AnswerTo calculate $S.I.$
$P=\text { Rs. } 18,000 ; R=10 \ \%\ $ and $T=1 \text { year }$
$ \text { S.I. }=\text { Rs. } \frac{18,000 \times 10 \times 1}{100}=\text { Rs. } 1,800 .$
To calculate $C.I.$
For $1^{\text {st }}$ half$-$year :
$P=$ Rs. 18,$000 ; R=10 \ \%\ $ and $T=\frac{1}{2}$ year
Interest $=\text { Rs. } \frac{18,000 \times 10 \times 1}{100 \times 2}=\text { Rs. } 900$
Amount $= Rs. 18,000+ Rs. 900= Rs. 18,900$
Fo $2^{\text {nd }}$ year :
$P= Rs. 18,900 ; R=10 \%$ and $T=\frac{1}{2}$ year
Interest$= Rs. \frac{18,900 \times 10 \times 1}{100 \times 2}=\text { Rs. } 945 \text {. }$
Amount$= Rs. 18,900+ Rs. 945= Rs. 19,845$
Compound interest $= Rs. 19,845- Rs. 18,000=Rs. 1,845$
His gain $= Rs. 1,845- Rs. 1,800= Rs. 45$
View full question & answer→Question 135 Marks
Meenal lends $Rs. 75,000$ at $C.I.$ for $3$ years. If the rate of interest for the first two years is $15\ \%\ $ per year and for the third year it is $16\ \%\ $, calculate the sum Meenalwill get at the end of the third year.
AnswerInterest for the first year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{75,000 \times 15 \times 1}{100}$
$ =\text { Rs. } 11,250$
Amount for the first year $= Rs. 75,000+ Rs. 11,250= Rs. 86,250$
Interest for the second year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$=\frac{86,250 \times 15 \times 1}{100}$
$ =\text { Rs. } 12,937.5$
Amount for the second year $= Rs. 86,250+ Rs. 12,937.5= Rs. 99,187.5$
Interest for the third year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$=\frac{99,187.5 \times 16 \times 1}{100}$
$ =\text { Rs. } 15,870$
Amount for the third year $= Rs. 99,187.5+ Rs. 15,870= Rs. 1,15,057.5$
Hence, the sum Meenal will get at the end of the third year is $Rs.1,15,057.5$
View full question & answer→Question 145 Marks
How much will $Rs. 50,000$ amount to in $3$ years, compounded yearly, if the rates for the successive years are $6\ \%\ , 8\ \%\ $ and $10\ \%\ $ respectively?
AnswerInterest for the first year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{50,000 \times 6 \times 1}{100}$
$ =\text { Rs. } 3,000$
Amount for the first year $= Rs. 50,000+ Rs. 3,000= Rs. 53,000$
Interest for the second year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{53,000 \times 8 \times 1}{100}$
$ =\text { Rs. } 4,240$
Amount for the second year $= Rs. 53,000+ Rs. 4,240= Rs. 57,240$
Interest for the third year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{57,240 \times 10 \times 1}{100}$
$ =\text { Rs. } 5,724$
Amount for the third year $= Rs. 57,240+ Rs. 5,724 = Rs. 62,964$
Hence, the amount will be $Rs. 62,964 .$
View full question & answer→Question 155 Marks
Find the compound interest, correct to the nearest rupee, on $Rs. 2,400$ for $2 \frac{1}{2}$ years at $5$ per cent per annum.
AnswerFor $1^{\text {st }}$ years
$P=\text { Rs. } 2400$
$R=5\ \%\ $
$\mathrm{T}=1$ year
$I=\frac{2400 \times 5 \times 1}{100}=120$
$A=2400+120=\text { Rs. } 2520$
For $2^{\text {nd }}$ year
$P=\text { Rs. } 2520$
$R=5\ \%\ $
$\mathrm{T}=1$ year
$I=\frac{2520 \times 5 \times 1}{100}=\text { Rs. } 126 \text {. }$
$A=2520+126=\text { Rs. } 2646$
For final $\frac{1}{2}$ year,
$P=\text { Rs. } 2646$
$R=5\ \%\ $
$\mathrm{T}=\frac{1}{2}$ year
$I=\frac{2646 \times 5 \times 1}{100 \times 2}=\text { Rs. } 66.15$
Amount after $2 \frac{1}{2}$ years $=2646+66.15$
$=\text { Rs. } 2712.15$
Compound interest $=2712.15-2400$
$=\text { Rs. } 312.15$
$=\text { Rs. } 312$
View full question & answer→Question 165 Marks
Calculate the amount and the compound interest on$:Rs. 16,000$ in $3$ years, when the rates of the interest for successive years are $10\ \%\ , 14\ \%\ $ and $15\ \%\ $ respectively.
AnswerFor $1^{\text {st }}$ year
$P= Rs. 16000 ; R=10 \% ; T=1$ year
$I=\frac{16000 \times 10 \times 1}{100}$
$\text { I = Rs. } 1600$
$A=16000+1600=17600$
For $2^{\text {nd }}$ year,
$P= Rs. 17600 ; R=14 \% ; T=1$ year
$I=\frac{17600 \times 14 \times 1}{100}=\frac{246400}{100}$
$\text { I = Rs. } 2464 .$
$A=17600+2464=\text { Rs. } 20064$
For $3^{\text {rd }}$ year,
$P= Rs. 20064 ; R=15 \% ; T=1$ year
$I=\frac{20064 \times 15 \times 1}{100}$
$I=3009.60$
Amount after $3$ years $=20064+3009.60= Rs. 23073.60$
Compound interest $=23073.60-16000= Rs. 7073.60$
View full question & answer→Question 175 Marks
Find the compound interest on $Rs. 4,000$ accrued in three years, when the rate of interest is $8\%$ for the first year and $10\%$ per year for the second and the third years.
AnswerInterest for the first year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{4,000 \times 8 \times 1}{100}$
$ =\text { Rs. } 3,20$
Amount for the first year $= Rs. 4,000+ Rs. 3,20= Rs. 4,320$
Interest for the second year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{4,320 \times 10 \times 1}{100}$
$ =\text { Rs. } 432$
Amount for the second year $= Rs. 4,320+ Rs. 432= Rs. 4,752$
Interest for the third year $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{4,752 \times 10 \times 1}{100}$
$ =\text { Rs. } 475.20$
Amount for the third year $= Rs. 4,752+ Rs. 475.20= Rs. 5,227.20$
So, the compound interest $= Rs. 5,227.20 - Rs. 4,000= Rs.1,227.20$
Hence, the amount will get at the end of the third year is $Rs. 1,227.50.$
View full question & answer→