1. Shweta deposits Rs 300 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
Answer

Deposit per month = Rs 350,

Rate of interest = 8% p.a.

Period (x) = 1 year = 12 months

∴ Total principal for one month 350 = x(x + 1)

= Rs 350 × (12 × 13)/12

= Rs 350 × 78

= Rs 27300

∴ Interest = prt/100 = (27300 × 8 × 1)/(100 × 12)

= Rs 182

∴ Amount of Maturity = Rs 350 × 12 + Rs 182

= Rs 4200 + 182

= Rs 4382.





2. Saloni deposited Rs 150 per month ina 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?



Answer

Deposit per month = Rs 150

Rate of interest = 8% per

Period (x) = 8 month

∴ Total principal for one month = Rs 150 × (x + 1)/2

= Rs 150 × 8(8 +1)/2

= (150 × 8 × 9)

= Rs 5400

∴ Interest = prt/100 = (5400 × 8 × 1)/(100 × 12) = Rs 36

Amount of maturity = Rs 150 × 8 + Rs 36

= Rs 1200 + Rs 36

= Rs 1236





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.



Answer

Deposit per month (P) = Rs 1000

Period = 3 years = 36 months

Rate = 8%

Total Principal = 36(36 + 1)/2 × 1000

Interest = PRT/100 = (36 × 37 × 1000 × 8)/(2 × 12 × 100)

= 12 × 37 × 10

= 4440

Matured price = P × n × S.I.

= 1000 × 36 + 4440

= 36000 + 4440

= Rs 40440






4. Sonia had a deposit account in a bank and deposited Rs600 per month for 21/2 years. If the rate of interest was 10% p.a., find the maturity value of this account.



Answer

Principal (n) = 2.5 years = 2.5×12 month = 30 months

Principal (P) = Rs 600

Rate = 10% p.a.

Maturity Value = ....

S.I. = P×n(n+1)×r×1(2×100×12)

S.I. = 600×30×31/2×100×12

Simple Interest = Rs 2335

Now Maturity Value = P×n + S.I.

Maturity Value = 600×30 + 2325

= Rs 20325





5. 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?



Answer

Amount deposited month (P) = Rs 200

Period (n) = 36 months,

Rate (R) = 11% p.a.

Now amount deposited in 36 months = Rs 200 × 36

= Rs 7200

Simple interest (S.I.) = P(n(n +1)/2) × 1/12 × R/100

= 200(36(36+1))/2 × 1/12 × 11/110

= (200 × 36 × 37 × 11)/(2 × 12 × 100)

= 1221

∴ Kiran will get maturity value

= Rs 7200 + 1221

= Rs 8421





6.
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.



Answer :

Interest = Rs 58800

Monthly deposit (P) = Rs 600

Period (n) = 4 years or 48 months

∴ Deposit for 1 month = {P(n)(n+1)}/2

= (600 × 48 × 49)/2

= Rs. 705600

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

Interest = Prt/100

⇒ 58800 = (705600 × r × 1)/(100 × 12)

5880 = 588r

∴ r = 5880/588 = 10
∴ Rate of interest = 10% p.a.






7. 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.



Answer

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 = P ×n(n +1)/2

= (300 × 24)(24 +1)/2 = (300 × 24 × 25)/2 = Rs 90000

∴ Interest earned = PRT/100 = (9000 × R × 1)/(100 × 12)

= 75R

Now 300 × 24 + 75R = 7725

⇒ 7200 + 75R = 7725

⇒ 75R = 7725 - 7200 = 525

⇒ R = 525/75 = 7

∴ Rate of Interest = 7% p.a.






8. 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.



Answer

Deposit per month = Rs 2500

Period = 2 years = 24 months

Maturity value = Rs 67500

∴ Total principal for 1 month = (P × n(n +1))/2

= ₹ (2500 × 24 × 25)/2

= ₹ 750000

∴ Interest = ₹ 67500 - 24 × 2500

= ₹ 67500 - 60000

= ₹ 7500

Period = 1 month = 1/12 year

∴ Rate of interest = S.I. × 100)/(7500 × 100 × 12)/(75000 × 1)

= 12%






9. Shahrukh opened a Recurring Deposit Account in a Bank and deposited Rs 800 per month for 1.1/2 years. If he received Rs 15084 at the time of maturity, find the rate of interest per annum.



Answer

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

r = ?

No. of months (n) = 1.1/2 = 3/2 × 12

= 18 months

∴ Interest = P × n(n +1)/(2 × 12) × r/100

= ₹ 800 × 18(18 + 1)/(2 × 12) × r/100

= ₹ 800 × (18 × 19)/(2 × 12) × r/100

= 114r

∴ Maturity amount = 114r + 800 × 18

= ₹ 15084 = 114r + ₹ 14400

⇒ ₹ 15084 - ₹ 14400 = 114r

⇒ 684 = 114r

r = 684/114 = 6%





10. 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



Answer

Interest = Rs 1200

Period (n) = 2 years = 24 months

Let monthly deposit = ₹ P p.m.

∴ Interest = P × n(n +1)/(2 × 12) × r/100

1200 = (P × 24 × 25)/24 × 6/100

⇒ 1200 = 6/4.P

∴ P = (1200 × 4)/6 = 800

∴ Monthly deposit = ₹ 800

And maturity value = P × n + Interest

= ₹ 800 × 24 + ₹ 1200

= ₹ 19200 + ₹ 1200

= ₹ 20400





11. 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.



Answer

Let monthly instalment is Rs P

Here n = 1 year = 12 months

n = 12

∵ M.V. = n(n +1)/(2 × 12) × (P × R)/100 + P.n

⇒ ₹ 6455 = 12(12 + 1)/(2 × 12) × (P × 14)/100 + P.12

₹ 6455 = (13 × P × 7)/100 + P.12

⇒ ₹ 6455 = (91P + 1200P)/100

⇒ ₹ 645500 = 1291 P

⇒ P = 645500/1291 = ₹500




12. Samita has a recurring deposit account in a bank of Rs2000 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.



Answer

Deposit per month = Rs 2000,

Rate of interest = 10%,

Let period = n months

Then principal for one month = 2000 × n(n + 1)/2 = 1000n(n + 1) and interest

= {1000n(n + 1)×10×1}/(100 × 12)

= 100n(n + 1)/12

∴ Maturity value = 2000 × n + 100n(n + 1)/12

∴ 2000n + 100n(n + 1)/12 = 83100

⇒ 24000n + 100n² + 100n = 83100 × 2

⇒ 240n + n² + n = 831 × 12

⇒ n² + 241n - 9972 = 0

⇒ n² + 277n - 36n - 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





MCQs of Chapter 2 Banking


1. If Shahrukh opened a recurring account in a bank and deposited Rs 800 per month for 1.1/2 years, then the total money deposited in the account is



(a) Rs 11400



(b) Rs 14400



(c) Rs 13680



(d) none of these



Answer


(b) Rs 14400

Monthly deposit = Rs 800

Period (n) = 1.1/2 years = 18 months

∴ Total money deposit = Rs 800 × 18

= Rs 14400
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2. Mrs. 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



Answer


(c) Rs 130

Deposit per month (P) = Rs 250

Period (n) = 1 year = 12 months

Rate (r) = 8% p.a.

∴ Interest = (P × n × (n + 1))/(2 × 12) × r/100

= (250 × 12 + 13)/(2 × 12) × 8/100

= ₹ 130






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



Answer


(d) Rs 12875

Deposit (P) = Rs 500 per month

Period (n) = 2 years = 24 months

Rate (r) = 7% p.a.

∴ Interest = P × n × (n + 1)/(2 × 12) × r/100

= (500 × 24 × 25 × 7)/(2 × 12 × 100)

= ₹ 875

∴ Maturity value = P × 24 + Interest

= ₹ 500 × 24 + 875

= ₹ 12000 × 875

= ₹ 12875





4. John deposited Rs 400 every month in a bank’s recurring deposit for 2.1/2 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%



Answer


(b) 7%

Deposit (P) = Rs 400 per month

Period (n) = 2.1/2 years = 3 months

Interest = Rs 1085

Let r% be the rate of interest

∴ Interest = P × n × (n + 1)/(2 × 12) × r/100

1085 = ₹ (400 × 30 × 31 × r)/(2 × 12 × 100)

1085 = 155r ⇒ r = 1085/155 = 7

∴ Rate 7% p.a.





Chapter Test for Banking


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.



Answer

Deposit per month = Rs 600

Rate of interest= 10% p.a.

Period (n) = 5 years 60 months

Total principal for one month = ₹ 600 × n(n + 1)/2

= 600 × 60(60 + 1)/2

= ₹ (600 × 60 × 61)/2

= ₹ 109800

= ₹ (600 × 60 × 61)/2

= ₹ 1098000

Interest = prt/100

= (1098000 × 10 × 1)/(100 × 12)

= ₹ 9150

∴ Amount of maturity = ₹ 600 × 60 + ₹ 9150

= ₹ 36000 + ₹ 9150
= ₹ 45150





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 2.1/2 years recurring deposit. The Bank paid 10% p.a. simple interest for both. At maturity who will get more and by how much?



Answer :

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 × n(n + 1)/2

= 400 × 36(36 + 1)/2

= ₹ (400 × 36 × 37)/2

= ₹ 266400

Interest = prt/100

= (266400 × 10 × 1)/(100 × 12)

= ₹ 2220

∴ Amount of maturity = ₹ 400 × 36 + ₹ 2220

= ₹ 14400 + ₹ 2220

= ₹ 16620

In case of Anshul,

Deposit p.m. = ₹ 500

Rate of Interest = 10%

Period (n) = 2.1/2 years= 30months

∴ Total principal for one month = ₹ 500 × n(n + 1)/2

= 500 × 30(30 + 1)/2

= ₹ (500 × 30 × 31)/2

= ₹ 232500

Interest = (232500 × 10 × 1)/(100 × 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





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



Answer

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 = P(n)(n + 1)/2

= {(800 × 48 × (48 × 1)}/2

= ₹ (800 × 48 × 49)/2

= ₹ 940800

Total deposit = ₹ 800 × 48

= ₹ 38400

And amount of maturity = ₹ 48200

∴ Interest earned = ₹ 48200 - ₹ 38400

= ₹ 9800





4. Mr. Chaturvedi has a recurring deposit account in Gindlay’s Bank for 4.1/2 years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.



Answer

Let each monthly instalment = Rs x

Rate of interest = 11%

Period (n) = 4.1/2 years or 54 months,

Total principal for one month = ₹ x × n(n + 1)/2

= ₹ x × 54(54 + 1)/2

= ₹ x (54 × 55)/2

= 1485x

Interest = (1485x 11 × 1)/(100 × 12)

= 13.6125x

∴ Total amount of maturity = 54x + 13.6125x

= 67.6125x

∴ 67.6125x = 101418.75

x = 101418.75/67.6125

= ₹ 1500

∴ Deposit per month = ₹ 1500





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.



Answer

Deposit during the month (P) = Rs 600

Rate of interest = 7% p.a.

Amount of maturity = Rs 15450

Let time = n months

∴ Total principal = P(n)(n + 1)/2

= 600 × n(n + 1)/2

= 600(n² + n)/2

= 300(n² + n)

∴ Interest = PRT/100

= (300(n² + n) × 7 × 1)/(100 × 12)

= 7/4(n² + n)

∴ 600n + 7/4(n²+ n) = 15450

⇒ 2400n + 7n² + 7n = 61800

⇒ 7n² + 2407n - 61800 = 0

⇒ 7n² - 168n + 2575n - 61800 = 0

⇒ 7n(n - 24) + 2575(n - 24) = 0

⇒ (n - 24)(7n + 2575) = 0

Either n - 24 = 0, then n = 24

Or 7n + 2575 = 0, then

7n = -2575

⇒ n = -2575/7

Which is not possible being negative.

∴ n = 24
∴ Period = 24 months or 2 years