Question 514 Marks
The value of a machine depreciates by $15\%$ in the first year and by $12\%$ in the second year. Find the value of the machine if its depreciation in the second year was $Rs.1,632.$
AnswerLet value of the machine be $Rs\ x$.
$ V_0=Rs\ x ; n=2 ; r=15 \% $
Depreciation in the first year $=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs R \times\left(1-\frac{15}{100}\right)$
$\Rightarrow V _{ t }= Rs R \times \frac{17}{20}$
$\Rightarrow V _{ t }= Rs 0.85 x $
Depreciation in the second year when $r$ is $12 \%=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs 0.85 x \times\left(1-\frac{12}{100}\right)$
$\Rightarrow V _{ t }= Rs 0.85 x \times \frac{22}{25}$
$\Rightarrow V _{ t } Rs 0.748 x $
Depreciation in the value of machine in the second year
$ =\operatorname{Rs}(0.85 x-0.748 x)=\operatorname{Rs} 1,632$
$\Rightarrow 0.102 x=\operatorname{Rs} 1,632$
$\Rightarrow x=\operatorname{Rs} 16,000 $
The original value of the machine was $Rs.16,000 .$
View full question & answer→Question 524 Marks
The value of a refrigerator depreciates by 8% of its value at the beginning of the year. Find the original value of the refrigerator if it depreciated by Rs 2,392 in the second year.
AnswerLet value of the refrigerator be Rs x.
$ V_0=R s x ; n=2 ; r=8 \% $
Depreciation in the first year $=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs x \times\left(1-\frac{8}{100}\right)$
$\Rightarrow V _{ t }= Rs x \times \frac{23}{25}$
$\Rightarrow V _{ t }= Rs 0.92 x $
Depreciation in the second year $=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs 0.92 x \times\left(1-\frac{8}{100}\right)$
$\Rightarrow V _{ t }= Rs 0.92 x \times \frac{23}{25}$
$\Rightarrow V _{ t }= Rs 0.8464 x $
Depreciation in the value of refrigerator in the second year
$ =\operatorname{Rs}(0.92 x-0.8464 x)=\operatorname{Rs} 2,392$
$\Rightarrow 0.0736 x=\operatorname{Rs} 2,392$
$\Rightarrow x=\operatorname{Rs} 32,500 $
The original value of the refrigerator was Rs 32,500 .
View full question & answer→Question 534 Marks
The value of a scooter depreciates by 12% of its value at the beginning of the year. Find the original value of the scooter if it depreciated by Rs 2,640 in the second year.
AnswerLet value of the soooter be Rs $x$.
$ V_0=\text { Rs } x ; n=2 ; r=12 \% $
Depreciation in the first year $=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs x \times\left(1-\frac{12}{100}\right)$
$\Rightarrow V _{ t }= Rs x \times \frac{22}{25}$
$\Rightarrow V _{ t }= Rs 0.88 x $
Depreciation in the second year $=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs 0.88 x \times\left(1-\frac{12}{100}\right)$
$\Rightarrow V _{ t }= Rs 0.88 x \times \frac{22}{25}$
$\Rightarrow V _{ t }= Rs 0.7744 x $
Depreciation in the value of soooter in the second year
$ =\operatorname{Rs}(0.88 x-0.7744 x)=\operatorname{Rs} 2,640$
$\Rightarrow 0.1056 x=\operatorname{Rs} 2,640$
$\Rightarrow x=\operatorname{Rs} 25,000 $
The original value of the soooter was Rs 25,000 .
View full question & answer→Question 544 Marks
Meera borrowed $Rs.12,500$ on compound interest from Rajeev for $2$ years when the rates of interest for successive years were $8\%$ and $10\%.$ If Meera returned $Rs.7,500$ at the end of the first year, find the amount she has to return at the end of the second year.
Answer$ P=\text { Rs. } 12,500, R=8 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 12500 \times 8 \times 1}{100} $
$ =\text { Rs } 1,000 $
Amount due after $1$ st year =Rs. $12,500+$ Rs. $1,000= Rs 13,500$
Amount paid after $1$ st year= Rs. 7,500
Balance amount $= Rs. 13,500 - Rs. 7,500= Rs. 6,000$
Interest for second year when $r=10 \%$ p.a.
$ =\frac{\operatorname{Rs} 6000 \times 10 \times 1}{100}$
$ =\operatorname{Rs} 600 $
Amount due after $2$nd year
$ =\text { Rs. } 6,000+\text { Rs. } 600=\text { Rs } 6.600 $
Meera has to return $Rs.6,600$ to Rajeev at the end of second year
View full question & answer→Question 554 Marks
Prakash borrowed $Rs.10,000$ from Rajesh for $2$ years at $6\%$ and $8\%$ p.a. compound interest for successive years. If Prakash returns $Rs.5,600$ at the end of the first year, how much does he have to give to Rajesh at the end of the second year to clear the loan?
Answer$ P=\text { Rs. } 10,000, R=6 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 10000 \times 6 \times 1}{100} $
$= Rs.600$
Amount due after $1$ st year
$ \text { =Rs. } 10,000+\text { Rs. } 600=\text { Rs } 10,600 $
Amount paid after $1$ st year $= Rs. 5,600$
Balance amount $=$ Rs. $10,600-$ Rs. $5,600= Rs. 5,000$ Interest for second year when $r=8 \%$ p.a.
$ =\frac{\operatorname{Rs} 5000 \times 8 \times 1}{100}$
$=\operatorname{Rs} 400 $
Amount due after $2$ nd year
$ =\text { Rs. } 5,000+\text { Rs. } 400 $
$Rs.5,400$
Prakash has to return $Rs.5,400$ to Rajesh at the end of second year.
View full question & answer→Question 564 Marks
Manoj borrowed $Rs.25,000$ from Sohan at $8.4\%$ p.a. compound interest. After $2$ years Manoj cleared $Rs.17,500$ and a motorcycle. Find the cost of the motorcycle.
AnswerHere, $P=Rs.25,000 ; r=8.4 \%$ p.a.; $t=2$ years
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$=\text { Rs } 25000\left(1+\frac{8.4}{100}\right)^2$
$=\text { Rs } 25000\left(1+\frac{84}{100 \times 10}\right)^2$
$=\text { Rs } 25000\left(\frac{271}{250}\right)^2$
$=\text { Rs } 25000 \times \frac{271}{250} \times \frac{271}{250}$
$\therefore A =\text { Rs.29,376.40 } $
Hence, amount due after $2$ years $=$ Rs $29,376.40$
Amount paid after $2$ years $=$ Rs $17,500$
Balance amount $=$ Amount due after $2$ years $-$ amount paid after $2$ years $=$ cost of the motorcycle $ =\operatorname{Rs}(29,376.40-17,500) $
Cost of the motorcycle $=$ Rs $11.876 .40$
View full question & answer→Question 574 Marks
Rajeev borrowed $Rs.15,000$ from Sanjay at $12\%$ p.a. compound interest. After $2$ years Rajeev gave $Rs.7,500$ and a scooter to clear the account. Find the cost of the scooter.
AnswerHere, $P=Rs.15,000 ; r=12^{\circ} /$ p.a.; $t=2$ years
$ \therefore A = P \left(1+\frac{ r }{100}\right)^n$
$=\operatorname{Rs} 15000\left(1+\frac{12}{100}\right)^2$
$=\operatorname{Rs} 15000\left(\frac{112}{100}\right)^2$
$=\operatorname{Rs} 15000\left(\frac{28}{25}\right)^2$
$=\operatorname{Rs} 15000 \times \frac{28}{25} \times \frac{28}{25}$
$\therefore A =\operatorname{Rs} 18816 $
Hence, amount due after $2$ years $=$ Rs.$18,816$
Amount paid after $2$ years $=$ Rs $7,500$
Balance amount $=$ Amount due after $2$ years $-$ amount paid after $2$ years $=$ cost of the scooter $=$ Rs $(18,816- 7,500)$ Cost of the scooter $=$ Rs $11,316$
View full question & answer→Question 584 Marks
Pooja borrowed Rs 15,000 from Sonali at 11% p.a. compound interest. If she repays Rs 7,550 at the end of first year and Rs 6,101 at the end of second year, find the amount Pooja needs to give to Sonali at the end of third year to clear her debt.
Answer$ P=\text { Rs. } 15,000, R=11 \% \text { p.a. }$
$\text { Interest for first year }$
$=\frac{\text { Rs } 15000 \times 11 \times 1}{100}$
$=\text { Rs, } 1650 $
Amount due after 1 st year
$ =\text { Rs. } 15,000+\text { Rs. } 1,650=\text { Rs } 16,650 $
Amount paid after 1 st year= Rs. 7,550
Balance amount= Rs. 16,650 - Rs. 7,550 =Rs. 9, 100 Interest for second year
$ =\frac{\operatorname{Rs} 9100 \times 11 \times 1}{100} $
=Rs 1001
Amount paid after 2nd year = Rs. 61101
Balance amount = Rs. 10, 101 - Rs. $61101=$ Rs. 4,000
Interest for third year
$ =\frac{\operatorname{Rs} 4000 \times 11 \times 1}{100} $
$ =\text { Rs } 440 $
Amount due after 3rd year
$ =\text { Rs. } 4,000+\text { Rs. } 440=\text { Rs 4,440 } $
Pooja needs to pay Rs 4,440 to Sonali at the end of third vear to clear her debt.
View full question & answer→Question 594 Marks
Calculate the arnount and the cornpound interest for the following:
$Rs.12,500$ for $3$ years at $12\%$ for the first year, $15\%$ for the second year and $17\%$ for the third year.
Answer$ P=\operatorname{Rs} 12500 ; $
(i) Interest for the first year
$ T=1 \text { year, } R=12 \% \text { for first year }$
$=\frac{\text { Rs } 12500 \times 12 \times 1}{100}$
$=\text { Rs } 1500 $
(ii) Principal for the second year
$=$ Amount after one year
$=$ Rs $12500+$ Rs $1500$
$=$ Rs $14000$
(iii) Interest for the second year
$T=1$ year,$R=15 \%$ for second year
$ =\frac{\operatorname{Rs} 14000 \times 15 \times 1}{100}$
$=\text { Rs } 2100 $
(iv) Principal for the third year
$ =\text { Amount after second year }$
$=\text { Rs } 14000+\text { Rs } 2100$
$=\text { Rs } 16100 $
(v) Interest for the third year
$ T=1 \text { year, } R=17 \% \text { for second year }$
$=\frac{\operatorname{Rs} 16100 \times 17 \times 1}{100}$
Therefore Amount at the end of $3$ rd year
$ =\text { Rs } 16100+\text { Rs } 2737$
$=\text { Rs } 18837 $
Amount $=$ Rs 18837
$ \text { C.I. }=\text { A - P }$
$=\operatorname{Rs}(18837-12500)$
$\text { C.I. }=6337 $
View full question & answer→Question 604 Marks
Calculate The Amount and the Cornpound Interest for the Following:
Rs 15,000 for 2 years at 6°/o for the first year and 7°/o for tl1e second year.
Answer$ P=\operatorname{Rs} 15000 \text {; } $
(i) Interest for the first year
$T=1$ year, $R=6 \%$ for first year
$ =\frac{\operatorname{Rs} 15000 \times 6 \times 1}{100}$
$=\operatorname{Rs} 900 $
(ii) Principal for the second year
= Amount After one year
$=$ Rs $15000+$ Rs 900
$=$ Rs 15900
(iii) Interest for the second year
$T=1$ year,$R=7 \%$ for second year
$ =\frac{\operatorname{Rs} 15900 \times 7 \times 1}{100}$
$=\operatorname{Rs} 1113 $
Therefore Amount at the end of 2nd year
$ =\text { Rs } 15900+\text { Rs } 1113$
$=\text { Rs } 17013 $
Amount $=$ Rs 17013
$ \text { C.I. = A - P }$
$=\operatorname{Rs}(17013-15000)$
$\text { C.I. = Rs } 2013 $
View full question & answer→Question 614 Marks
Calculate the amount and the compound interest for the following :
Rs. 12500 for 2 years at 8% for the first year and 10% for the second year.
Answer$ P=\text { Rs. 12500; } $
(i) Interest for the first year
$ T =1 \text { year, } R =8 \% \text { for first year } $
$ =\frac{\operatorname{Rs} 12500 \times 8 \times 1}{100}$
$=\operatorname{Rs} 1000 $
(ii) Principal for the second year
$=$ Amount after one year
$ =\text { Rs } 12500+\text { Rs } 1000$
$=\text { Rs } 13500 $
(iii) Interest for the second year
$ T =1 \text { year, } R =10 \% \text { for second year }$
$=\frac{\operatorname{Rs~} 13500 \times 10 \times 1}{100}$
$=\operatorname{Rs~} 1350 $
Therefore ,Amount at the end of 2nd year
$ =\text { Rs } 13500+\text { Rs } 1350 $
Amount $=$ Rs 14850
$ \text { C.I. }=\text { A - P }$
$=\operatorname{Rs}(14850-12500)$
$\text { C.I. }=\operatorname{Rs} 2350 $$ P=\text { Rs. 12500; } $
(i) Interest for the first year
$ T =1 \text { year, } R =8 \% \text { for first year } $
$ =\frac{\operatorname{Rs} 12500 \times 8 \times 1}{100}$
$=\operatorname{Rs} 1000 $
(ii) Principal for the second year
$=$ Amount after one year
(iii) Interest for the second year
Therefore ,Amount at the end of 2nd year
$ =\text { Rs } 13500+\text { Rs } 1350 $
Amount $=$ Rs 14850
View full question & answer→Question 624 Marks
Mr. Chatterjee borrowed Rs 50,000 in compound interest from Mr. Patel for 2 years when the rates of interest for the successive years were $7 \frac{1}{2} \%$ and $9 \frac{1}{4} \%$. If Mr. Chatterjee returned Rs 27,750 at the end of the first year, find the amount he needs to return at the end of the seoond year to clear the loan.
Answer$P=$ Rs. $50,000, R=7 \frac{1}{2} \%$ p.a. $=\frac{15}{2} \%$ p.a.
$ \text { Interest for first year }$
$=\frac{\operatorname{Rs} 50000 \times \frac{15}{2} \times 1}{100}$
$=\frac{\operatorname{Rs} 50000 \times 15 \times 1}{2 \times 100}$
$=\operatorname{Rs} 3,750 $
Amount due after 1 st year
$ \text { =Rs. } 50,000+\text { Rs. } 3,750=\text { Rs 53,750 } $
Amount paid after 1 st year $=$ Rs. 27,750
Balance amount= Rs. 53,750 - Rs. $27,750=$ Rs. 26,000
Interest for second year when $r=9 \frac{1}{4} \%$ p.a. $=\frac{37}{4} \%$ p.a.
$ =\frac{\operatorname{Rs} 26000 \times \frac{37}{4} \times 1}{100}$
$=\frac{\operatorname{Rs} 26000 \times 37 \times 1}{4 \times 100}$
$=\text { Rs 2, } 405 $
Amount due after 2 nd year
$=$ Rs. $26,000+$ Rs. $2,405=$ Rs 28,405
Mr. Chatterjee has to return Rs 28,405 to Mr. Patel at the end of second year to clear his loan.
View full question & answer→Question 634 Marks
Gayatri invested $Rs.25,000$ for $3$ years and 6 months in a bank which paid
$10\%$ p.a- compound interest. Calculate the amount to the nearest.Ts $-10,$ that she
received at the end of the period.
Answer$C _1=\frac{25000 \times 1 \times 10}{100}=2500 $
$ P _1=27500 $
$ C _2=\frac{27500 \times 10}{100}=2750 $
$P _2=30250$
$ C _3=\frac{30250 \times 1 \times 10}{100}=3025 $
$ P _3=33275$
$ C _4=\frac{33275 \times 10 \times 1}{100}=1663.75$
$ P _4=34940$
View full question & answer→Question 644 Marks
Calculate the amount and the compound interest for the following:
$Rs. 76,000$ at $10 \%$ p.a. in $2 \frac{1}{2}$ years
AnswerHere, $P=$ Rs.$76,000 ; r=10 \%$ p.a.; $t=2 \frac{1}{2}$ years
For the first year: $t=1$ year
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 76000 \times 10 \times 1}{100}$
$\text { S.I. }=\text { Rs7, } 600 $
$ A=P+S . I .$
$=\operatorname{Rs}(76,000+7,600)=\operatorname{Rs} 83,600=\text { new principal } $
For the second year: $t=1$ year; $P=R s 83,600$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 83600 \times 10 \times 1}{100}$
$\text { S.I. }=\text { Rs8, } 360 $
$ A=P+S . I .$
$A=\operatorname{Rs}(83,600+8,360)=\operatorname{Rs} 91960=\text { new principal } $
For the third year: $t=1 / 2$ year; $P=\operatorname{Rs} 91,960$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 91960 \times 10 \times 1}{100 \times 2}$
$\text { S.I. }=\text { Rs4, 598 } $
$ A=P+S . I .$
$A=\operatorname{Rs}(91,960+4,598)=\operatorname{Rs} 96,558 $
C.I. = Interest in first year + interest in second year+ interest in third year
$ \text { C.I. }=\operatorname{Rs}(7,600+8,360+4,598)=\operatorname{Rs} 20,558 $
View full question & answer→Question 654 Marks
Calculate the amount and the compound interest for the following:
$Rs. 22,500$ at $12 \%$ p.a. in $1 \frac{3}{4}$ years
AnswerHere, $P= Rs. 22,500 ; r=12 \%$ p.a.; $t=1 \frac{3}{4}$ years
For the first year: $t=1$ year
$\text { S.I. }=\frac{ P \times r \times t }{100} $
$ \text { S.I. }=\frac{ Rs 22500 \times 12 \times 1}{100} $
$ \text { S.I. }=\text { Rs2, } 700 $
$ A=P+S . I . $
$ =\operatorname{Rs}(22,500+2,700)=\operatorname{Rs} 25,200=\text { new principal }$
For the second year: $t=3 / 4$ year; $P=Rs.25,200$
$\text { S.I. }=\frac{P \times r \times t}{100} $
$ \text { S.I. }=\frac{\operatorname{Rs~} 25200 \times 12 \times 3}{100 \times 4} $
$ \text { S.I. }=\text { Rs } 2268$
$A=P+S . I .$
$ A=R s(25,200+2,268)=\operatorname{Rs} 27,468 $
$ \text { C.I. }=\text { Interest in first year + interest in second year }$
$\text { C.I. }=\operatorname{Rs}(2,700+2,268)=\operatorname{Rs} 4,968$
View full question & answer→Question 664 Marks
Calculate the amount and the compound interest for the following:
$Rs. 25,000$ at $8 \frac{2}{5} \%$ p.a. in $1 \frac{1}{3}$ years
AnswerHere, $P=$ Rs. 25,$000 ; r=8 \frac{2}{5} \%$ p.a. $=42 / 5 \%$;
$t =1 \frac{1}{3}$ years
For the first year: $t =1$ year
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 25000 \times 42 \times 1}{100 \times 5}$
$\text { S.I. }=\text { Rs2, } 100 $
$ A=P+S . I .$
$=\operatorname{Rs}(25,000+2,100)=\operatorname{Rs} 27,100=\text { new principal } $
For the second year: $t=1 / 3$ year; $P=R s 27,100$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 27100 \times 42 \times 1}{100 \times 5 \times 3}$
$\text { S.I. }=\text { Rs } 758.80 $
$A=P+S . I .$
$A=\operatorname{Rs}(27,100+758.80)=\operatorname{Rs} 27,858.80$
$\text { C.I. }=\text { Interest in first year + interest in second year }$
$\text { C.I. }=\operatorname{Rs}(2,100+758.80)=\operatorname{Rs} 2,858.80$
View full question & answer→Question 674 Marks
Calculate the amount and the compound interest for the following:
$\text { Rs. } 20,000 \text { at } 9 \% \text { p.a. in } 2 \frac{1}{3} \text { years }$
AnswerHere, $P=$ Rs. 20,$000 ; r=9^{\circ} \%$ p.a. $; t=2 \frac{1}{3}$ years
For the first year: $t=1$ year
$ A=P+S . I$
$=\operatorname{Rs}(20,000+1,800)=\operatorname{Rs} 21,800=\text { new principal } $
For the second year: $t=1$ year; $P=R s 21,800$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\text { Rs } 21800 \times 9 \times 1}{100}$
$\text { S.I. }=\text { Rs } 1,962 $
$ A=P+S . I .$
$A=\operatorname{Rs}(21,800+1,962)=\operatorname{Rs} 23,762=\text { new principal } $
For the third year: $t=1 / 3$ year; $P=R s 23,762$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 23762 \times 9 \times 1}{100 \times 3}$
$\text { S.I. }=\text { Rs712. } 86 $
$ A=P+S . I .$
$A=\operatorname{Rs}(23,762+712.86)=\operatorname{Rs} 24,474.86 $
C.I. = Interest in first year + interest in second year + interest in third year
$ \text { C.I. }=\operatorname{Rs}(1,800+1,962+712.86)=\operatorname{Rs} 4,474.86 $
View full question & answer→Question 684 Marks
Calculate the amount and the compound interest for the following:
Rs. 10,000 at $8 \%$ p.a. in $2 \frac{1}{4}$ years
AnswerHere, $P=$ Rs. 10,$000 ; r=8 \%$ p.a. $; t=2 \frac{1}{4}$ years
For the first year: $t=1$ year
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{ Rs 10000 \times 8 \times 1}{100}$
$\text { S.I. }=\text { Rs } 800 $
$ A=P+S . I .$
$=\operatorname{Rs}(10,000+800)=\operatorname{Rs} 10,800=\text { new principal } $
For the second year: $t=1$ year; $P=R s 10,800$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 10800 \times 8 \times 1}{100}$
$\text { S.I. }=\text { Rs864 } $
$ A=P+S . I .$
$A=\operatorname{Rs}(10,800+864)=\operatorname{Rs} 11,664=\text { new principal } $
For the third year: $t=1 / 4$ year; $P=R s 11,664$
$A=P+S . I .$
$A=R s(11,664+233.28)=\operatorname{Rs} 11,897.28$
$\text { C.I. = Interest in first year + interest in second year }$
$+ \text { interest in third year }$
$\text { C.I. }=\operatorname{Rs}(800+864+233.28)=\operatorname{Rs} 1,897.28$
View full question & answer→Question 694 Marks
Calculate the amount and the compound interest for the following:
Rs. 30,000 at $8 \%$ p.a. in $2 \frac{1}{2}$ years
AnswerHere, $P=$ Rs. 30,$000 ; r=8 \%$ p.a. $; t=2 \frac{1}{2}$ years
For the first year: $t=1$ year
$\text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{Rs 3000 \times 8 \times 1}{100}$
$\text { S.I. }=\operatorname{Rs} 2,400$
$A=P+S . I .$
$=\operatorname{Rs}(30,000+2,400)=\operatorname{Rs} 32,400=\text { new principal }$
For the seoond year: $t=1$ year; $P=R s 32,400$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 32400 \times 8 \times 1}{100}$
$\text { S.I. }=\text { Rs2,592 } $
$ A=P+S . I .$
$A=\operatorname{Rs}(32,400+2,592)=\operatorname{Rs} 34,992=\text { new principal } $
For the third year: $t=1 / 2$ year ; $P=R s 34,992$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{ Rs 34992 \times 8 \times 1}{100}$
$\text { S.I }=\text { Rsi, } 399.68 $
C.I. $=$ Interest in first year + interest in second year + interest in third year
$ \text { C.I. }=\operatorname{Rs}(2,400+2,592+1,399.68)=\operatorname{Rs} 6,391.68 $
View full question & answer→Question 704 Marks
Calculate the amount and the compound interest for the following:
$Rs. 23,750$ at $12 \%$ p.a. in $2 \frac{1}{2}$ years
AnswerHere, $P=\operatorname{Rs} 23,750 ; r=12^{\circ} \%$ p.a. $; t=2 \frac{1}{2}$ years
For the first year: t=1 year
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 23750 \times 12 \times 1}{100}$
$\text { S.I. }=\operatorname{Rs~} 2850 $
$ A=P+S . I .$
$=\operatorname{Rs}(23,750+2,850)=\operatorname{Rs} 26,600=\text { new principal } $
For the seoond year: $t=1$ year; $P=R s 26,600$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 26600 \times 12 \times 1}{100}$
$\text { S.I. }=\operatorname{Rs} 3192 $
$ A=P+S . I$
$A=\operatorname{Rs}(26.600+3,192)=\operatorname{Rs} 29.792=\text { new principal } $
For the third year: $t=1 / 2$ year; $P=R s 29,792$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 29792 \times 12 \times 1}{100 \times 2}$
$\text { S.I. }=\text { Rs1, } 787.52 $
$ A=P+S . I .$
$A=\operatorname{Rs}(29,792+1,787.52)=\operatorname{Rs} 31,579.52 $
C.I. = Interest in first year + interest in second year + interest in third year
$ \text { C.I. }=\operatorname{Rs}(2,850+3,192+1,787.52)=\operatorname{Rs} 7,829.52 $
View full question & answer→Question 714 Marks
Calculate the amount and the compound interest for the following:
Rs.17,500 at 12°10 p.a. in 3 years
AnswerHere, $P=$ Rs. 17,$500 ; r=12^{\circ} 10$ p.a. ; $t=3$ years
For the first year: $t=1$ year
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 17500 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs2, } 100 $
$ A=P+S . I .$
$=\operatorname{Rs}(17,500+2,100)=\operatorname{Rs} 19,600=\text { new principal } $
For the second year: $t=1$ year; $P=\operatorname{Rs} 19,600$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 19600 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs2, } 352 $
$ A=P+S . I .$
$A=\operatorname{Rs}(19,600+2,352)=\operatorname{Rs} 21,952=\text { new principal } $
For the third year: $t=1$ year; $P=R s 21,952$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 21952 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs2, 634. } 24 $
$A=P+S . I .$
$A=R s(21,952+2,634.24)=\text { Rs } 24,586.24$
$C . I .=\text { Interest in first year }+ \text { interest in second year }+$
$\text { interest in third year }$
$\text { C.I. }=\operatorname{Rs}(2,100+2,352+2,634.24)=\operatorname{Rs} 7,086.24$
View full question & answer→Question 724 Marks
Calculate the amount and the compound interest for the following:
$Rs.13,500$ at $10^\circ 10$ p.a. in $2$ years
AnswerHere, $P=$ Rs. $13,500 ; r=10^{\circ} 10$ p.a. $; t=2$ years
For the first year: $t=1$ year
$\text { S.I. }=\frac{ P \times r \times t }{100} $
$\text { S.I. }=\frac{ Rs 13500 \times 10 \times 1}{100} $
$\text { S.I. }=\text { Rsl, } 350$
$A=P+S . I$
$=\operatorname{Rs}(13,500+1,350)=\operatorname{Rs} 14,850=\text { new principal }$
For the second year: $t=1$ year; $P=R s 14,850$
$\text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 14850 \times 10 \times 1}{100} $
$\text { S.I. }=\text { Rsl, } 485$
$A=P+S . I .$
$A=\operatorname{Rs}(14,850+1,485)=\operatorname{Rs} 16,335$
$\text { C.I. }=\text { Interest in first year +interest in second year }$
$\text { C.I. }=\operatorname{Rs}(1,350+1,485)=\operatorname{Rs} 2,835$
View full question & answer→Question 734 Marks
The population of a city is $24,000.$ In the next $3$ years it will be $27,783.$ Find the rate of growth of the population.
Answer$ V _{ n }=27,783 ; V _0=24,000 ; r=? ; t =3 \text { years }$
$V _{ n }= V _0\left(1+\frac{ r }{100}\right)^{ n }$
$27783=24000\left(1+\frac{ r }{100}\right)^3$
$\frac{27783}{24000}=\left(1+\frac{ r }{100}\right)^3$
$\frac{21^3}{20^3}=\left(1+\frac{ r }{100}\right)^3$
$\left(1+\frac{ r }{100}\right)=\frac{21}{20}$
$\frac{ r }{100}=\frac{21}{20}-1$
$\frac{ r }{100}=\frac{1}{20}$
$r=\frac{1}{20} \times 100$
$r=5 \% $
The rate of growth of population is $5 \%$.
View full question & answer→Question 744 Marks
In a factory the production of scooters rose to $46,305$ from $40,000$ in $3$ years. Find the annual rate of growth of the production of scooters.
Answer$ V _{ n }=46,305 ; $
$V _0=40,000 ;$
$r=? ;$
$ t =3$ years
$V _{ n }= V _0 \times\left(1+\frac{ r }{100}\right)^{ n }$
$46305=40000\left(1+\frac{ r }{100}\right)^3$
$\frac{46305}{40000}=\left(1+\frac{ r }{100}\right)^3$
$\frac{21^3}{20^3}=\left(1+\frac{ r }{100}\right)^3$
$\left(1+\frac{ r }{100}\right)=\frac{21}{20}$
$\frac{ r }{100}=\frac{21}{20}-1$
$\frac{ r }{100}=\frac{1}{20}$
$r =\frac{1}{20} \times 100$
$r ={5 \%} $
The annual rate of growth of scooters is $5 \%$
View full question & answer→Question 754 Marks
A building worth $Rs. 1, 33,100$ is constructed on a plot of land worth $Rs. 72,900.$ After how many years will the values of both be same, if the land appreciates at $10\% p.a. $ and the building depreciates at $10\% p.a.$?
AnswerFor the building:
$ V_n=V_0\left(1-\frac{r}{100}\right)^n$
$V_n=\operatorname{Rs} 133100\left(1-\frac{10}{100}\right)^n$
$V_n=\operatorname{Rs} 133,100 \times(0.9)^n $
For the plot:
$ V_n=V_0\left(1+\frac{r}{100}\right)^n$
$V_n=\operatorname{Rs} 72900\left(1+\frac{10}{100}\right)^n$
$V_n=\operatorname{Rs} 72900 \times(1.1)^n $
Since, value becomes same:
$ 1,33,100 \times(0.9)^{ n }=72,900 \times(1.1)^{ n } \frac{\lim}{x \rightarrow \infty}$
$\frac{(1.1)^{ n }}{(0.9)^{ n }}=\frac{133100}{72900}$
$\frac{(1.1)^{ n }}{(0.9)^{ n }}=\frac{1331}{729}=\frac{11^3}{9^3}$
$t =3 $
Hence, after $3$ years value of both will be same.
View full question & answer→Question 764 Marks
Find the difference between the compound interest compounded yearly and half-yearly for the following:
Rs 20,000 for $1 \frac{1}{2}$ years at $16 \%$ p.a.
Answer$P=\operatorname{Rs} 20,000 ; t=1 \frac{1}{2}$ years
When compounded yearly: $r=16 \%$ p.a.
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$A=R s 20000\left(1+\frac{16}{100}\right)\left(1+\frac{16}{100}\right)^{\frac{1}{2}}$
$=$ Rs $20000 \times 1.16 \times\left(1+\frac{1}{2} \times \frac{16}{100}\right)$
$=$ Rs $20,000 \times 1.16 \times 1.08$
$=$ Rs 25,056
C.I. $= A - P$
$=\operatorname{Rs}(25,056-20,000)$
$=$ Rs 5,056
When compounded half-yearly:
$A = P \left(1+\frac{ r }{100}\right)^{ n }$
$A=\operatorname{Rs} 20000\left(1+\frac{8}{100}\right)^3$
$=$ Rs $20,000 \times 1.08 \times 1.08 \times 1.08$ $=$ Rs $25,194.24$
C.I. $= A - P$
$=\operatorname{Rs}(25,194.24-20,000)$
$=$ Rs 5,194.24
Hence the difference in the interest=Rs $(5,194.24$ - $ 5,056)=\operatorname{Rs} 138.24 $
View full question & answer→Question 774 Marks
Find the difference betlween the compound interest compounded yearly and half$-$yearly for the following: $Rs. 15,000$ for $1 \frac{1}{2}$ years at $12 \% p.a.$
Answer$P=\operatorname{Rs} 15,000 ; t=1 \frac{1}{2}$ years
When compounded yearly : $r=12 \% p.a.$
$ A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{12}{100}\right)\left(1+\frac{12}{100}\right)^{\frac{1}{2}}$
$=\text { Rs } 15000 \times 1.12 \times\left(1+\frac{1}{2} \times \frac{12}{100}\right)$
$=\text { Rs } 15,000 \times 1.12 \times 1.06$
$=\text { Rs } 17,808$
$\text { C.I. }=A-P$
$=\text { Rs }(17,808-15,000)$
$=\text { Rs } 2808 $
When compounded half$-$yearly :
$ A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{6}{100}\right)^3$
$=\operatorname{Rs} 15,000 \times 1.06 \times 1.06 \times 1.06$
$=\operatorname{Rs} 17,865.24$
$\text {C.I.}=A-P$
$=\text { Rs }(17,865.24-15,000)$
$=\text { Rs } 2,865.24 $
Hence the difference in the interest$=Rs (2,865.24-2,808)$
$ =\operatorname{Rs} 57.24 $
View full question & answer→Question 784 Marks
A sum of money placed at compound interest compounded annually amounts to $Rs. 26,460$ in $2$ years and to $Rs. 29,172.15$ in $4$ years. Calculate the rate of interest and the sum.
Answer$P=x ; r=? ; t=2$ and $4$ years; $A=Rs. 26,460 \ (2$ years$)$ and $Rs. 29, 172.15 \ (4$ years$)$
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$26460= x \left(1+\frac{ r }{100}\right)^2 .......(i)$
$29172.15=x\left(1+\frac{r}{100}\right)^4 .......(ii)$
$\therefore \frac{ x \left(1+\frac{ r }{100}\right)^4}{ x \left(1+\frac{ r }{100}\right)^2}=\frac{29172.15}{26460}$
$\Rightarrow\left(1+\frac{ r }{100}\right)^2=\frac{194481}{176400}$
$\Rightarrow\left(1+\frac{ r }{100}\right)^2=\left(\frac{441}{420}\right)^2$
$\Rightarrow 1+\frac{ r }{100}=\frac{441}{420}$
$\Rightarrow \frac{ r }{100}=\frac{441}{420}-1$
$\Rightarrow \frac{ r }{100}=\frac{441-420}{420}$
$r=\frac{21}{420} \times 100$
$r=5 \%$
Using $(i)$
$x\left(1+\frac{r}{100}\right)^2=\text { Rs. } 26460$
$x\left(1+\frac{5}{100}\right)^2=\text { Rs. } 26460$
$x\left(\frac{105}{100}\right)^2=\text { Rs. } 26460$
$1.1025 x=\text { Rs. } 26,460$
$x=\text { Rs. } 24,000 $
The sum $= Rs. 24,000$ and rate of interest $=5 \%$
View full question & answer→Question 794 Marks
A sum of money placed at compound interest compounded annually amounts to $Rs. 31,360$ in $2$ years and to $Rs. 35,123.20$ in $3$ years. Calculate the rate of interest and the sum.
Answer$ P=x ; r=? ; t=2 $ and $3$ years; $A=\operatorname{Rs.} 31,360 \ (2 $ years$)$ and $Rs. 35, 123.20 .(3$ years$)$
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$31360=x\left(1+\frac{ r }{100}\right)^2 .... (i) $
$ 35123.20= x \left(1+\frac{ r }{100}\right)^3 .... (ii) $
$ \therefore \frac{ x \left(1+\frac{ r }{100}\right)^3}{x\left(1+\frac{ r }{100}\right)^2}=\frac{35123.20}{31360} $
$\Rightarrow\left(1+\frac{r}{100}\right)=\frac{35123.20}{31360}$
$ \Rightarrow \frac{ r }{100}=\frac{35123.20}{31360}-1 $
$ \Rightarrow \frac{ r }{100}=\frac{35123.20-31360}{31360} $
$ r=\frac{3763.20}{31360} \times 100$
$r=12 \% $
Using $(i)$
$ x\left(1+\frac{r}{100}\right)^2=\operatorname{Rs.} 31,360$
$x\left(1+\frac{12}{100}\right)^2=\operatorname{Rs.} 31,360$
$x\left(\frac{112}{100}\right)^2=\operatorname{Rs.} 31,360$
$1.2544 X=\operatorname{Rs.} 31,360$
$x=\text { Rs. } 25,000 $
The sum $= Rs. 25,000 $and rate of interest $=12 \%$.
View full question & answer→Question 804 Marks
A certain sum of money invested at compound interest compounded annually amounted to $Rs. 26,450$ in $2$ years and to $Rs. 30,417.50$ in $3$ years. Calculate the rate of interest and the sum invested.
AnswerHere, $r=? P=x ($say$)$
$T=2$ years and $3$ years
$A=Rs. 26,450$ in $2$ vears and $Rs.30,417.50$ in $3$ years.
$ A=P\left(1+\frac{r}{100}\right)^n $
$ 26450=x\left(1+\frac{r}{100}\right)^2 ......(i) $
$ 30417.50=x\left(1+\frac{r}{100}\right)^3 ......(ii) $
Dividing $(ii)$ by $(i)$
$ \frac{x\left(1+\frac{r}{100}\right)^3}{x\left(1+\frac{r}{100}\right)^2}=\frac{30417.50}{26450} $
$\Rightarrow 1+\frac{r}{100}=\frac{30417.50}{26450}$
$\Rightarrow \frac{ r }{100}=\frac{30417.50}{26450}-1$
$\Rightarrow \frac{ r }{100}=\frac{30417.50-26450}{26450}$
$\Rightarrow \frac{ r }{100}=\frac{3967.50}{26450}$
$\Rightarrow r =\frac{3967.50}{26450} \times 100$
$\Rightarrow r =15 \%$
$x\left(1+\frac{r}{100}\right)^2=Rs. 26450$
$x\left(1+\frac{15}{100}\right)^2=\text { Rs. } 26450$
$x \times \frac{23}{20} \times \frac{23}{20}=\text { Rs. } 26450$
$x \times \frac{529}{400}=\text { Rs. } 26450$
$x=\text { Rs. } \frac{26450 \times 400}{529}$
$x=\text { Rs. } 20,000$
Hence, rate of interest $=15 \%$ and sum invested $=Rs.20,000$
View full question & answer→Question 814 Marks
A certain sum of money invested at compound interest compounded annually amounted to $Rs. 5,082$ after $2$ years and to $Rs. 5,590.\ 20$ after $3$ years. Calculate the rate of interest and the sum invested.
AnswerHere, $r=? P=x \ ($say$)$
$T=2$ years and $3$ years
$A=Rs. 5,082$ in $2$ years and $Rs. 5,590.20$ in $3$ years.
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$5082= x \left(1+\frac{ r }{100}\right)^2 \ldots ......(i)$
$5590.20= x \left(1+\frac{ r }{100}\right)^3 ........(ii) $
Dividing $(ii)$ by $(i)$
$ \frac{x\left(1+\frac{r}{100}\right)^3}{x\left(1+\frac{r}{100}\right)^2}=\frac{5590.20}{5082}$
$\Rightarrow 1+\frac{r}{100}=\frac{5590.20}{5082}$
$\Rightarrow \frac{r}{100}=\frac{5590.20}{5082}-1$
$\Rightarrow \frac{r}{100}=\frac{5590.20-5082}{5082}$
$\Rightarrow \frac{r}{100}=\frac{508.20}{5082}$
$\Rightarrow r=\frac{508.20}{5082} \times 100$
$\Rightarrow r=10 \% $
using $(i)$
$x\left(1+\frac{r}{100}\right)^2=\operatorname{Rs.} 5082$
$x\left(1+\frac{10}{100}\right)^2=\operatorname{Rs.} 5082$
$x \times \frac{11}{10} \times \frac{11}{10}=\operatorname{Rs~} 5082$
$x \times \frac{121}{100}=\operatorname{Rs.} 5082$
$x=\operatorname{Rs.} \frac{5082 \times 100}{121}$
$x=\text { Rs. } 4200 $
Hence, rate of interest $=10^{\circ} \ 10$ and sum invested $= Rs. 4,200$.
View full question & answer→Question 824 Marks
What sum of money will amount to $Rs. 13,675. \ 20$ in $3$ years at compound interest, if the rates of interest are 1$0\%, 11\%$ and $12\%$ for the successive years?
AnswerFor the third year
Here $P=x ; A=R s 13,675.20 ; t=1$ year; $r=12 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs 13675.20= x \left(1+\frac{12}{100}\right)^1$
$\Rightarrow \operatorname{Rs.} 13675.20= x \left(\frac{112}{100}\right)$
$\Rightarrow x=\operatorname{Rs.} \frac{13675.20 \times 100}{112}$
$\Rightarrow x =\operatorname{Rs.} 12,210 $
The sum of money will be $Rs. 12,210$ at the end of the second year or beginnino;; of the third year.
For the second year
Here $P=x ; A=R s 12,210 ; t=1$ year ; $r=11 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$\Rightarrow \operatorname{Rs.} 12210= x \left(1+\frac{11}{100}\right)^1$
$\Rightarrow \operatorname{Rs.} 12210= x \left(\frac{111}{100}\right)$
$\Rightarrow x= Rs. \frac{12210 \times 100}{111}$
$\Rightarrow x=\operatorname{Rs.} 11,000$
The sum of money will be $Rs. 11,000$ at the end of the first year or beginning of the second year.
For the first year
Here $P=x ; A=R s 11,000 ; t=1$ year; $r=10 \%$
$p.a.$
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \Rightarrow \text { Rs. } 11000= x \left(1+\frac{10}{100}\right)^1$
$\Rightarrow \text { Rs. } 11000= x \left(\frac{11}{100}\right) $
$ \Rightarrow x=\operatorname{Rs.} \frac{11000 \times 10}{11} $
$ \Rightarrow x=\operatorname{Rs.} 10,000 $
The sum of money will be $Rs. 10,000$ at the beginning of the first year.
View full question & answer→Question 834 Marks
What sum of money will amount to $Rs. 3,326.40$ in $3$ years at compound interest, if the rates of interest are $8\%, 10\%$ and $12\%$ for the successive years?
AnswerFor the third year
Here $P=x ; A=Rs. ~ 3,326.40 ; t=1$ year ; $r=12 \% \ \ p.a.$
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs. 3326.40= x \left(1+\frac{12}{100}\right)^1$
$\Rightarrow Rs. 3326.40= x \left(\frac{112}{100}\right)$
$\Rightarrow x =\operatorname{Rs.} \frac{3326.40 \times 100}{112}$
$\Rightarrow x \operatorname{Rs} 2970 $
The sum of money will be $Rs. 2, 970$ at the end of the second year or beginning of the third year.
For the second year
Here $P=x ; A=Rs. 2,970 ; t=1$ year ; $r=10 \% \ \ p.a.$
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$\Rightarrow Rs. 2970= x \left(1+\frac{10}{100}\right)^1$
$\Rightarrow Rs. 2970= x \left(\frac{11}{10}\right)$
$\Rightarrow x = Rs. \frac{2970 \times 10}{11}$
$\Rightarrow x=\operatorname{Rs.} 2700$
The sum of money will be $Rs. 2,700$ at the end of the first year or beginning of the second year.
For the first year
Here $P=x ; A=Rs. 2,700 ; t=1$ year ; $r=8 \% \ \ p.a.$
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs. 2700= x \left(1+\frac{8}{100}\right)^1$
$\Rightarrow Rs. 2700= x \left(\frac{108}{100}\right)$
$\Rightarrow x = Rs. \frac{2700 \times 100}{108}$
$\Rightarrow x =\operatorname{Rs.} 2500 $
The sum of money will be $Rs. 2,500$ at the beginning of the first year.
View full question & answer→Question 844 Marks
Archana borrowed $Rs. 18,000$ from Ritu at $12\% \ p.a.$ compound interest. If at the end of the $1^{st}, 2^{nd},$ and $3^{rd}$ years, Archana returned $Rs. 5,250 , Rs. 5,875$ and $Rs. 6,875$ respectively, find the amount Archana has to pay Ritu at the end of the $4^{th}$ year to clear her debt.
Answer$ P=\text { Rs. } 18,000, R=12 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 18000 \times 12 \times 1}{100}$
$=\text { Rs 2,160 }$
Amount due after $1^{st}$ year
$ =\text { Rs. } 18,000+\text { Rs. } 2,160=\text { Rs. 20,160 } $
Amount paid after $1^{st}$ year $=$ $Rs. 5,250$
Balance amount $=$ $\text {Rs.} 20,$ $160-$ $\text {Rs.} 5,250 = \text {Rs.} 14,910$
Interest for seoond year
$ =\frac{\operatorname{Rs} 14910 \times 12 \times 1}{100}$
$=\text { Rs } 1789.20 $
Amount due after $2^{nd}$ year
$ =\text { Rs. } 14,910+\text { Rs. } 1,789.20$
$=\text { Rs. } 16,699.20 $
Amount paid after $2^{nd}$ year $= \text {Rs.} 5,875$
Balance amount $=$ $\text {Rs.} 16,699.20 -\text {Rs.} 5,875= \text {Rs.}10,824.20 $
Interest for third year
$ =\frac{\operatorname{Rs} 10824.20 \times 12 \times 1}{100}$
$=\text { Rs. } 1,298.904 $
Amount due after $3^{rd}$ year
$ =\text { Rs. } 10,824.20 \text { + Rs. } 1,298.904=\text { Rs. 12, } 123.10 $
Amount paid after $3^{rd}$ year $=$ $Rs. 6,875$
Balance amount $= \text {Rs.} 12, 123.10- \text {Rs.} 6,875 = \text {Rs.} 5,248.104$
Interest for fourth year
$ =\frac{\operatorname{Rs} 5248.104 \times 12 \times 1}{100}$
$=\text { Rs. } 629.7725 $
Amount due after $4^{th}$ year $=\text {Rs.} 5,248.104 +\text {Rs.} 629.7725= \text {Rs.} 5877.876 $
$=$ $Rs. 5877.87$
Archana needs to pay $\text {Rs.} 5877.87$ to Ritu at the end of $4^{th}$ year to clear
View full question & answer→Question 854 Marks
Rajan borrowed $Rs. 90,000$ at $15\% \ \ p.a$. compound interest. If he repays $Rs. 35,000$ at the end of each year, find the amount of loan outstanding at the beginning of the fourth year.
Answer$ P= Rs. 90,000, R=15 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs.} 90000 \times 15 \times 1}{100}$
$=\operatorname{Rs.} 13500 $
Amount due after $1^{st}$ year
$ = Rs. 90,000+ Rs. 13,500= Rs. 103,500 $
Amount paid after $1 ^{st}$ year $= Rs. 35,000$
Balance amount $= Rs. 103,500 - Rs. 35,000= Rs. 68,500 $
Interest for second year
$ =\frac{\operatorname{Rs} 68500 \times 15 \times 1}{100}$
$=\operatorname{Rs.} 10,275 $
Amount due after $2 ^{nd}$ year
$ = Rs. 68,500 + Rs. 10,275=\text { Rs } 78,775$
Amount paid after $2^{nd}$ year $=Rs. 35,000$
Balance amount $= Rs. 78,775 - Rs. 35,000=\text { Rs. } 43,775$
Interest for third year
$=\frac{\text { Rs } 43775 \times 15 \times 1}{100}$
$=Rs. 6,566.25 $
Amount due after $3^{rd}$ year
$ =\text { Rs. } 43,775 + Rs. 6,566.25 = Rs 50,341.25 $
Amount paid after $3^{rd}$ year $= Rs.35, 000 $
Balance amount $= Rs. 50,341.25 - Rs.35, 000 = Rs. 15,341.25 $
Loan outstanding at the beginning of the fourth year $ =\operatorname{Rs.} 15,341.25 $
View full question & answer→Question 864 Marks
Mohan borrowed $Rs. 25,000$ at $10\% \ p.a$. compound interest. If he pays back $Rs. 7,500$ every year, find the amount of loan outstanding at the beginning of the fourth year.
Answer$ P=\text { Rs. } 25,000, R=10 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 25000 \times 10 \times 1}{100}$
$=\text { Rs } 2,500 $
Amount due after $1^{st}$ year
$ =\text { Rs. } 25,000+\text { Rs. } 2,500$
$=\text { Rs } 27,500 $
Amount paid after $1^{st}$ year $= Rs. 7,500$
Balance am
ount $= Rs. 27,500- Rs. 7,500= Rs. 20,000$
Interest for seoond year
$ =\frac{\operatorname{Rs} 20000 \times 10 \times 1}{100}$
$=\operatorname{Rs} 2000 $
Amount due after $2 ^{nd}$ year
$ =\text { Rs. } 20,000+\text { Rs. } 2,000=\text { Rs } 22,000 $
Amount paid after $2 ^{nd}$ year $= Rs. 7,500$
Balance amount $= Rs. 22,000 - Rs. 7,500 = Rs. 14,500$
Interest for third year
$ =\frac{\operatorname{Rs} 14500 \times 10 \times 1}{100} $
Amount due after $3^{rd}$ year
$ =\text { Rs. } 14,500+\text { Rs. } 1,450=\text { Rs } 15,950 $
Amount paid after $3^{rd}$ year $= Rs. 7,500$
Balance amount $= Rs. 15, 950- Rs. 7,500= Rs. 8,450$
Loan outstanding at the beginning of the fourth year $= Rs. 8,450 $.
View full question & answer→Question 874 Marks
Calculate the arnount and the cornpound interest for the following: $Rs. 20,000$ for $3$ years at $7 \frac{1}{2} \%$ for the first year, $8 \%$ for the second year and $10 \%$ for the third year.
Answer$ P=\operatorname{Rs} 20000 ; $
$(i)$ Interest for he first year
$ T=1$ year $, R =7 \frac{1}{2} \% $ for first year $=\frac{15}{2} \%$
$=\frac{\operatorname{Rs} 20000 \times \frac{15}{2} \times 1}{100}$
$=\frac{\operatorname{Rs} 20000 \times 15 \times 1}{2 \times 100}$
$=\text { Rs } 1500 $
$(ii)$ Principal for the second year
$=$ Amount after one year
$ =\text { Rs } 20000+\text { Rs } 1500$
$=\text { Rs } 21500 $
$(iii)$ Interest for the second year
$T=1$ year, $R=18 \%$ for second year
$ =\frac{\operatorname{Rs} 21500 \times 8 \times 1}{100}$
$=\operatorname{Rs} 1720 $
$(iv)$ Principal for the third year
$=$ Amount after second year
$ =\text { Rs } 21500+\text { Rs } 1720$
$=\text { Rs } 23220 $
$(v)$ Interest for the third year
$ t=1 $ year $, R=10 \% $ for second year
$=\frac{\operatorname{Rs} 23220 \times 10 \times 1}{100}$
$\text { = Rs } 2322$
Therefore Amount at the end of $3^{rd}$ year
$ =\text { Rs } 23220+\text { Rs } 2322$
$=\text { Rs } 25542$
Amount $=\text { Rs } 25542$
$\text { C.I. }=\text { A }- \text { P }$
$=\text { Rs }(25542-20000)$
$\text { C.I. }=\text { Rs } 5542 $
View full question & answer→Question 884 Marks
Calculate the amount and the compound interest for the following : $Rs. 16,000$ at $15 \% \ \ p.a$. in $2 \frac{2}{3}$ years
AnswerHere, $P= Rs. 16,000 ; r=15 \% p.a. ; t=2 \frac{2}{3}$ years
For the first year: $t=1$ year
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 16000 \times 15 \times 1}{100}$
$\text { S.I. }=\text { Rs } 21,400 $
$ A=P+S . I .$
$=\operatorname{Rs}(16,000+2,400)=\operatorname{Rs } 18,400=$ new principal
For the second year: $t=1$ year; $P=Rs. 18,400$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 18400 \times 15 \times 1}{100}$
$\text { S.I. }=\text { Rs } 2,760 $
$ A=P+S . I .$
$A=\operatorname{Rs}(18,400+2,760)=\operatorname{Rs } 21,160=$ new principal
For the third year: $t=2 / 3$ year; $P=Rs. 21,160$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 21160 \times 15 \times 2}{100 \times 3}$
$\text { S.I. }=\text { Rs2116 } $
$ A=P+S . I .$
$A=\operatorname{Rs }(21,160+2116)=\operatorname{Rs } 23,276 $
$\text{C.I.} =$ Interest in first year $+$ interest in second year $+$ interest in third year
$ \text { C.I. }=\operatorname{Rs }(2,400+2,760+2116)=\operatorname{Rs } 7,276 $
View full question & answer→Question 894 Marks
Calculate the amount and the compound interest for the following : $Rs. 7500$ at $12\% \ p.a$. in $3$ years.
AnswerHere, $P= Rs. 7500 ; r=12 \% p.a.; t=3$ years
For the first year : $t=1$ year
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs } 7500 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs } 900 $
$ A=P+S . I .$
$=\operatorname{Rs }(7500+900)=\operatorname{Rs } 8400=$ new principal
For the seoond year: $t=1$ year; $P=Rs\ 8,400$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs } 8400 \times 12 \times 1}{100}$
$\text { S.I. }=\operatorname{Rs } 1,008 $
$ A=P+S . I .$
$A=\operatorname{Rs }(8,400+1,008)=\operatorname{Rs } 9,408=$ new principal
For the third year: $t=1$ year; $P=Rs\ 9,408$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 9408 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs } 1128.96 $
$A=P+S . I .$
$A=R s(9,408+1,128.96)=\operatorname{Rs } 10,536.96$
$\text { C.I.} =$ Interest in first year $+$ interest in second year $+$ interest in third year
$\text { C.I. }=\operatorname{Rs }(900+1,008+1,128.96)=\operatorname{Rs } 3,036.96$
View full question & answer→