ML Aggarwal Solution Class 10 Chapter 2 Banking Test
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
=₹600×60×612=₹1098000
Interest =PRT100 =1098000×10×1100×12=₹9150
∴Amount of maturity =₹600×60+₹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 212 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×n(n+1)2=400×36(36+1)2
=₹400×36×372=₹266400
Interest =PRT100 =266400×10×1100×12=₹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)=212 years =30 months
∴Total principal for one month
=₹500×n(n+1)2=500×30(30+1)2
=₹500×30×312=₹232500
Interest =232500×10×1100×12=₹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
Let rate of interest be R % p.a.
Total principal for one month
=P(n)(n+1)2=800×48×(48+1)2 =₹800×48×492=₹940800
Total deposit =₹ 800 \times 48=₹ 38400
∴Interest earned =₹ 48200-₹ 38400=₹ 9800
(i) Rate of interest = S.I. ×100P×T
=9800×100×12940800×1=12.5%
(ii) Total interest earned by Shilpa =₹9800
Question 4
Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for 412 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)=412 years or 54 months,
Total principal for one month
=₹x×n(n+1)2=₹x×54(54+1)2
=x×54×552=1485x
Interest =1485x×11×1100×12=13⋅6125x
∴Total amount of maturity =54x+13⋅6125x
=67.6125x
∴67.6125 x=101418.75
x=101418⋅7567⋅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 =P(n)(n+1)2
=600×n(n+1)2=600(n2+n)2=300(n2+n)
∴Interest = PRT 100=300(n2+n)×7×1100×12
=74(n2+n)
∴600n+74(n2+n)=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
7n=−2575⇒n=−25757
Which is not possible being negative.
∴n=24
∴Period=24 months or 2 years
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