SChand CLASS 10 Chapter 2 Banking Exercise 2

 Exercise 2

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Question 1

Mr. Rajiv Anand has opened a recurring deposit account of Rs. 400 per month for 20 month in a bank. Find the amount he will get at the tie of maturity, if the rate of interest is 8.5% p. a., if the interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 400, n = 20 months, r = 8.5%.

Putting these values in SI formula, we get

SI $= \frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= 400 \times 20 \times (20 + 1) \times \frac{ 8.5 }{ 2400}$

⇒ SI = 595

Thus, the maturity value is MV = Pn + SI = (400×20) + 595 = Rs. 8595

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Question 2

Mr. Savita Khosla deposits Rs 900 per month in a recurring account for 2 years. If she gets Rs. 1800 as interest at the time of maturity, find the rate of interest if the interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 900, n = 24 months, SI = Rs. 1800

Putting these values in SI formula, we get

SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒$1800 = 900 \times 24 \times (24 + 1) \times \frac{ r}{2400}$

⇒ r = 8

Thus, the required rate of interest is 8% p.a.

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Question 3

Mr. Brown deposit Rs. 1100 per month in a cumulative time deposit account in a bank for 16 months. If at the end of maturity he gets Rs. 19096, find the rate of interest if interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 1100, n = 16 months, MV = 19096,

So, SI = MV – Pn = 19096 – (1100×16) = Rs. 1496

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ $1496 = 1100 \times 16 \times (16 + 1) \times \frac{r}{2400}$

⇒ r = 12

Thus, the required rate of interest is 12% p.a.

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Question 4

Sandhya has a recurring deposit account in Vijya Bank and deposits Rs 400 per month for 3 years. If she gets Rs 16176 on maturity, find the rate of interest given by the bank.

Sol :

From the question, we have P = Rs. 400, n = 36 months, MV = 16176,

So, SI = MV – Pn = 16176 – (400×36) = Rs. 1777

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒$ 1777 = 400 \times 36 \times (36 + 1) \times \frac{r}{2400}$

⇒ r = 8

Thus, the required rate of interest is 8% p.a.

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Question 5

A man deposits Rs 600 per month in a bank for 12 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits if the rate of interest is 8 % p.a. and interest is calculated at the end of every month?

Sol :

From the question, we have P = Rs. 600, n = 12 months, r = 8%.

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= 600 \times 12 \times (12 + 1) \times \frac{8}{2400}$

⇒ SI = 312

Thus, the maturity value is MV = Pn + SI = (600×12) + 312 = Rs. 7512

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Question 6

Anil deposits Rs 300 per month in a recurring deposit account for 2 years. If the rate of interest is 10% per year, calculate the amount that Anil will receive at the end of 2 years, i.e at the time of maturity.

Sol :

From the question, we have P = Rs. 300, n = 24 months, r = 10%.

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= 300 \times 24 \times (24 + 1) \times \frac{10}{2400}$

⇒ SI = 750

Thus, the maturity value is MV = Pn + SI = (300×24) + 750 = Rs. 7950

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Question 7

Sudhir opened a recurring deposit account with a bank for 1 ½ years. If the rate of interest is 10% and the bank pays Rs 1554 on maturity, find how much did Sudhir deposit per month?

Sol :

From the question, we have n = 18 months, r = 10%, MV = Rs. 1554

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= P\times 18 \times (18 + 1) \times \frac{10 }{2400}$

⇒ SI $= \frac{57P}{40}$

Since the maturity value is MV = Pn + SI

⇒ $1554 = 18P + \frac{57P}{40}$

⇒ P = Rs. 80 per month.

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Question 8

Renu has a cumulative deposit account of Rs. 200 per month at 10% per annum. If she gets Rs 6775 at the time of maturity, find the total time for which the account was held.

Sol :

From the question, we have P = Rs. 200, r = 10%, MV = Rs. 6775

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= 200 \times n(n + 1) \times \frac{10}{2400}$

⇒ SI $= \frac{\text{5n(n + 1)}}{6}$

Since the maturity value is MV = Pn + SI

⇒$ 6775 = 200n + \frac{5n(n + 1)}{6}$

⇒ n² + 241n – 8130 = 0

⇒ (n – 30)(n + 271) = 0

⇒ n = 30 months $= 2\frac{1}{2}$ years.

ML Aggarwal Solution Class 10 Chapter 2 Banking Test

Q1 Q2 Q3 Q4 Q5

Test

Question 1

Mr. Dhruv deposits Rs 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.

Sol :

Deposit per month = Rs 600

Rate of interest = 10% p.a.

Period (n) = 5 years 60 months.

Total principal for one month

$=₹ 600 \times \frac{n(n+1)}{2}=₹ 600 \times \frac{60(60+1)}{2}$

$=₹ \frac{600 \times 60 \times 61}{2}=₹ 1098000$

Interest $=\frac{PRT}{100}$ $=\frac{1098000 \times 10 \times 1}{100 \times 12}=₹ 9150$

∴Amount of maturity $=₹ 600 \times 60+₹ 9+50$

=₹ 36000+₹ 9150=₹ 45150

Question 2

Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a $2 \frac{1}{2}$ years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?

Sol :

In case of Ankita,

Deposit per month = Rs 400

Period (n) = 3 years = 36 months

Rate of interest = 10%

Total principal for one month

$=400 \times \frac{n(n+1)}{2}=400 \times \frac{36(36+1)}{2}$

$=₹ \frac{400 \times 36 \times 37}{2}=₹ 266400$

Interest $=\frac{PRT}{100}$ $=\frac{266400 \times 10 \times 1}{100 \times 12}=₹ 2220$

∴Amount of maturity =₹ 400 \times 36+₹ 2220

=₹ 14400+₹ 2220=₹ 16620

In case of Anshul, 

Deposit p.m.=₹ 500

Rate of interest =10 %

Period $(n)=2 \frac{1}{2}$ years =30 months

∴Total principal for one month

$=₹ 500 \times \frac{n(n+1)}{2}=500 \times \frac{30(30+1)}{2}$

$=₹ \frac{500 \times 30 \times 31}{2}=₹ 232500$

Interest $=\frac{232500 \times 10 \times 1}{100 \times 12}=₹ 1937.50$

Amount of maturity =₹ 500 × 30+₹ 1937.50

=₹ 15000+₹ 1937.50=₹ 16937.50

At maturity Anshul will get more amount

Difference =₹ 16937.50-₹ 16620.00

=₹ 317.50

Question 3

Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find

(i) the rate of simple interest,

(ii) the total interest earned by Shilpa

Sol :

Deposit per month (P) = Rs 800

Amount of maturity = Rs 48200

Period (n)=4 years =48 months

Let rate of interest be R % p.a.

Total principal for one month

$=\frac{P(n)(n+1)}{2}=\frac{800 \times 48 \times(48+1)}{2}$ $=₹ \frac{800 \times 48 \times 49}{2}=₹ 940800$

Total deposit =₹ 800 \times 48=₹ 38400

∴Interest earned =₹ 48200-₹ 38400=₹ 9800

(i) Rate of interest $=\frac{\text { S.I. } \times 100}{P \times T}$

$=\frac{9800 \times 100 \times 12}{940800 \times 1}=12.5 \%$

(ii) Total interest earned by Shilpa $=₹ 9800$

Question 4

Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for $4 \frac{1}{2}$ years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.

Sol :

Let each monthly instalment = Rs x

Rate of interest = 11 %

Period $(n)=4 \frac{1}{2}$ years or 54 months,

Total principal for one month

$=₹ x \times \frac{n(n+1)}{2}=₹ x \times \frac{54(54+1)}{2}$

$=x \times \frac{54 \times 55}{2}=1485 x$

Interest $=\frac{1485 x \times 11 \times 1}{100 \times 12}=13 \cdot 6125 x$

∴Total amount of maturity $=54 x+13 \cdot 6125 x$

=67.6125x

∴67.6125 x=101418.75

$x=\frac{101418 \cdot 75}{67 \cdot 6125}=₹ 1500$

∴Deposit per month =₹ 1500

Question 5

Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.

Sol :

Deposit during the month (P) = Rs 600

Rate of interest = 7% p.a.

Amount of maturity = Rs 15450

Let time = n months

∴Total principal $=\frac{\mathrm{P}(n)(n+1)}{2}$

$=\frac{600 \times n(n+1)}{2}=\frac{600\left(n^{2}+n\right)}{2}=300\left(n^{2}+n\right)$

∴Interest $=\frac{\text { PRT }}{100}=\frac{300\left(n^{2}+n\right) \times 7 \times 1}{100 \times 12}$

$=\frac{7}{4}\left(n^{2}+n\right)$

∴$600 n+\frac{7}{4}\left(n^{2}+n\right)=15450$

⇒2400n+7n²+7n=61800

⇒7n²+2407 n-61800=0

⇒7n²-168 n+2575 n-61800=0

⇒7 n(n-24)+2575(n-24)=0

⇒(n-24)(7 n+2575)=0

Either n-24=0, then n=24

or 7 n+2575=0, then

$7 n=-2575 \Rightarrow n=\frac{-2575}{7}$

Which is not possible being negative.

∴n=24

∴Period=24 months or 2 years

ML Aggarwal Solution Class 10 Chapter 2 Banking MCQs

Q1 Q2 Q3 Q4

MCQs

Question 1

If Sharukh opened a recurring deposit account in a bank and deposited Rs 800 per month for $1 \frac{1}{2}$ years,  then the total money deposited in the account is

(a) Rs 11400

(b) Rs 14400

(c) Rs 13680

(d) none of these

Sol :

Monthly deposit = Rs 800

Period $(n)=1 \frac{1}{2}$ years $=18$ months

∴ Total money deposit = Rs 800 × 18

= Rs 14400

Ans : (b)

Question 2

Mrs. Asha Mehta deposit Rs 250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is

(a) Rs 65

(b) Rs 120

(c) Rs 130

(d) Rs 260

Sol :

Deposit per month (P) = Rs 250

Period (n) = 1 year = 12 months

Rate (r) = 8% p.a.

∴Interest $=\frac{\mathrm{P} \times n \times(n+1)}{2 \times 12} \times \frac{r}{100}$

$=\frac{250 \times 12+13}{2 \times 12} \times \frac{8}{100}=₹ 130$

Ans : (c)

Question 3

Mr. Sharma deposited Rs 500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7% per annum, then the amount he gets on maturity is

(a) Rs 875

(b) Rs 6875

(c) Rs 10875

(d) Rs 12875

Sol :

Deposit (P) = Rs 500 per month

Period (n) = 2 years = 24 months

Rate (r) = 7% p.a.

∴Interest $=\frac{P \times n \times(n+1)}{2 \times 12} \times \frac{r}{100}$

$=\frac{500 \times 24 \times 25 \times 7}{2 \times 12 \times 100}=₹ 875$

∴Maturity value =P×24+Interest

=₹ 500 \times 24+875=₹ 12000 × 875

=₹ 12875

Ans : (d)

Question 4

John deposited Rs 400 every month in a bank’s recurring deposit account for $2 \frac{1}{2}$ years . If he gets Rs 1085 as interest at the time of maturity, then the rate of interest per annum is

(a) 6%

(b) 7%

(c) 8%

(d) 9%

Sol :

Deposit (P) = Rs 400 per month

Period (n) $=2 \frac{1}{2}$ years $=3$ months

∴Interest = Rs 1085

Let r% be the rate of interest

Interest $=\frac{P \times n \times(n+1)}{2 \times 12} \times \frac{r}{100}$

$1085=₹ \frac{400 \times 30 \times 31 \times r}{2 \times 12 \times 100}$

$1085=155 r \Rightarrow r=\frac{1085}{155}=7$

∴Rate 7 % p.a.

Ans : (b)

ML Aggarwal Solution Class 10 Chapter 2 Banking Exercise 2

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Exercise 2

Question 1

Shweta deposits Rs. 350 per month in a recurring deposit account for one year at the rate of 8% p.a. Find the amount she will receive at the time of maturity.

Sol :

Deposit per month = Rs 350,

Rate of interest = 8% p.a.

Period (x) = 1 year

= 12 months

∴Total principal for one month $350 \times \frac{x(x+1)}{2}=\operatorname{Rs} 350 \times \frac{12 \times 13}{2}$

= Rs. 350×78

= Rs. 27300

∴Interest $=\frac{PRT}{100}=\frac{27300 \times 8 \times 1}{100 \times 12}=$ Rs. 182

∴Amount of Maturity=Rs 350×12+Rs 182

=Rs 4200+182=Rs 4382 

Question 2

Salom deposited Rs 150 per month in a bank for 8 months under the Recurring Deposit Scheme. ‘What will be the maturity value of his deposit if the rate of interest is 8% per annum ?

Sol :

Deposit per month = Rs. 150

Rate of interest = 8% per

Period (x) = 8 month

∴Total principal for one month$=\operatorname{Rs.} 150 \times \frac{x(x+1)}{2}=\operatorname{Rs} 150 \times \frac{8(8+1)}{2}$ $=\frac{150 \times 8 \times 9}{2}=\operatorname{Rs} .5400 .$

∴Interest $=\frac{p r t}{100}=\frac{5400 \times 8 \times 1}{100 \times 12}=$ Rs. 36

∴Amount of Maturity=Rs 150×8+Rs 36

=Rs 1200+36=Rs 1236 

Question 3

Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value. (2009)

Sol :

Deposit per month (P) = Rs. 1000

Period = 3 years = 36 months

Rate = 8%

Total principal $=\frac{36(36+1)}{2} \times 1000$

Intercst $=\frac{\mathrm{PR} \mathrm{T}}{100}=\frac{36 \times 37 \times 1000 \times 8}{ 2 \times 12 \times 100}$

=12×37×10=4440

Matured value $=P \times n+S . I=1000 \times 36+4440=36000+4440=$ Rs. 40440

Question 4

Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the banks pays interest at the rate of 11% per annum, find the amount she gets on maturity ?

Sol :

Amount deposited month (P) = Rs. 200

Period (n) = 36 months,

Rate (R) = 11% p.a.

Now amount deposited in 36 months = Rs. 200 x 36 = Rs 7200

Simple Interest $(\mathrm{S.I})=\mathrm{P}\left(\frac{n(n+1)}{2}\right) \times \frac{1}{12} \times \frac{R}{100}$

$=200\left(\frac{36(36+1)}{2}\right) \times \frac{1}{12} \times \frac{11}{100}$

$=\frac{200 \times 36 \times 37 \times 11}{2 \times 12 \times 100}=1221$

∴Kiran will get maturity value=7200+1210=8421

Question 5

Haneef has a cumulative bank account and deposits Rs. 600 per month for a period of 4 years. If he gets Rs. 5880 as interest at the time of maturity, find the rate of interest.

Sol :

Interest = Rs. 58800

Monthly deposit (P) = Rs. 600

Period (n)=4 years or 48 months

∴Deposit for 1 month $=\frac{P(n)(n+1)}{2}$

$=\frac{600 \times 48 \times 49}{2}=\mathrm{Rs} .705600$

Let, rate of interest =r % p.a.

Interest $=\frac{\operatorname{PRT}}{100}$

$5880=\frac{705600 \times r \times 1}{100 \times 12}$

5880=588r

∴$r=\frac{5880}{588}=10$

∴Rate of interest =10 % p.a.

Question 6

David opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest per annum. (2008)

Sol :

Deposit during one month (P) = Rs. 300

Period = 2 years = 24 months.

Maturity value = Rs. 7725

Let R be the rate percent, then

Now principal for 1 month $=\frac{P \times n(n+1)}{2}$

$=\frac{300 \times 24(24+1)}{2}=\frac{300 \times 24 \times 25}{2}=$ Rs. 90000

∴Interest earned $=\frac{\mathrm{PRT}}{100}=\frac{90000 \times \mathrm{R} \times 1}{100 \times 12}$

=75R

Now 300×24+75R=7725

7200+75 R=7725

75 R=7725-7200=525

$\mathrm{R}=\frac{525}{75}=7$

∴Rate of Interest =7 % p.a.

Question 7

Mr. Gupta-opened a recurring deposit account in a bank. He deposited Rs. 2500 per month for two years. At the time of maturity he got Rs. 67500. Find :

(i) the total interest earned by Mr. Gupta.

(ii) the rate of interest per annum.

Sol :
Deposit per month = Rs. 2500

Period = 2 years = 24 months

Maturity value = Rs. 67500

∴Total principal for 1 month $=\frac{\mathrm{P} \times n(n+1)}{2}$

$=₹ \frac{2500 \times 24 \times 25}{2}=₹ 750000$

∴Interest =₹ 67500-24×2500

=₹ 67500-60000=₹ 7500

Period $=1$ month $=\frac{1}{12}$ year

∴Rate of interest $=\frac{\mathrm{S.I.} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{7500 \times 100 \times 12}{750000 \times 1}$

=12%

Question 8

Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs 800 per month for $1 \frac{1}{2}$ yeas . If he received Rs 15084 at the time of maturity, find the rate of interest per annum.

Sol :

Money deposited by Shahrukh per month (P)= Rs 800

r = ?

No. of months $(n)=1 \frac{1}{2}=\frac{3}{2} \times 12$

=18 months

∴Interest $=\mathrm{P} \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$=₹ 800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}$

$=₹ 800 \times \frac{18 \times 19}{2 \times 12} \times \frac{r}{100}=114 r$

∴Maturity amount =114 r+800×18

₹ 15084=114 r+₹ 14400

₹ 15084-₹ 14400=114 r

684=114 r

$r=\frac{684}{114}=6 \%$

Question 9

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs 1200 as interest at the time of maturity, find:

(i) the monthly instalment

(ii) the amount of maturity. (2016)

Sol :

Interest = Rs 1200

Period (n) = 2 years = 24 months

Rate (r) = 6% p.a.

Let monthly deposit =₹ P p.m.

∴Interest $=\frac{\mathrm{P} \times n(n+1)}{2 \times 12} \times \frac{r}{100}$

$1200=\frac{P \times 24 \times 25}{24} \times \frac{6}{100}$

$1200=\frac{6}{4} \mathrm{P}$

∴$P=\frac{1200 \times 4}{6}=800$

∴Monthly deposit =₹ 800

and maturity value =P × n+ Interest

$=₹ 800 \times 24+₹ 1200$

$=₹ 19200+₹ 1200=₹ 20400$

Question 10

Mr. R.K. Nair gets Rs 6,455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly instalment.

Sol :

Let monthly instalment is Rs P

here n = 1 year = 12 months

n = 12

∴$\mathrm{M.V.}=\frac{n(n+1)}{2 \times 12} \times \frac{\mathrm{P} \times \mathrm{R}}{100}+\mathrm{P} . \mathrm{n}$

$₹ 6455=\frac{12(12+1)}{2 \times 12} \times \frac{P \times 14}{100}+P .12$

$₹ 6455=\frac{13 \times \mathrm{P} \times 7}{100}+\mathrm{P.} 12$

$₹ 6455=\frac{91 P+1200 P}{100}$

₹ 645500=1291 P

$P=\frac{645500}{1291}=₹ 500$

Question 11

Samita has a recurring deposit account in a bank of Rs 2000 per month at the rate of 10% p.a. If she gets Rs 83100 at the time of maturity. Find the total time for which the account was held.

Sol :

Deposit per month = Rs 2000,

Rate of interest = 10%, Let period = n months

Then principal for one month

$=2000 \times \frac{n(n+1)}{2}=1000 n(n+1)$ and interest

$=\frac{1000 n(n+1) \times 10 \times 1}{100 \times 12}=\frac{100 n(n+1)}{12}$

∴$=2000 \times n+\frac{100 n(n+1)}{12}$

⇒24000 n+100n²+100 n=83100 \times 12

⇒240 n+n²+n=831 \times 12

⇒n²+241 n-9972=0

⇒n²+277 n-36 n-9972=0

⇒n(n+277)-36(n+277)=0

⇒(n+277)(n-36)=0

Either n+277=0, then n=-277, which is not possible.

or n-36=0, then x=36

∴Period=36 months or 3 years