Self Evaluation and Revision Banking Class 10 ICSE Maths Solutions OP Malhotra 2026-27. We Provide Step by Step Solutions of Banking- RD , maturity value. Visit official Website CISCE  for detail information about ICSE Board Class-10 Mathematics.

Self Evaluation and Revision Banking Class 10 ICSE Maths Solutions OP Malhotra 2026-27

Board	ICSE
Publications	S Chand
Subject	Maths
Class	10th
Chapter-2	Banking
Writer	OP Malhotra
Self Evaluation	Extra Practice Questions
Edition	2026-2027

Self Evaluation on Banking

Self Evaluation and Revision Banking Class 10 ICSE Maths Solutions OP Malhotra 2026-27

Que-1: Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and the interest is calculated at the end of every month .

Sol: Deposit per month (P) = ₹ 150
Rate (R) = 8%
Period (n) = 8 months
∴ Principal for 1 month = P×n(n+1)/2
= 150×8×(8+1)/2=₹150×8×9/2
= ₹ 5400
Interest = PRT/100=₹5400×8×1/100×12 = ₹ 36
∴ Maturity value = Deposit + Interest
= ₹ 150 x 8 + 36
= ₹ 1200 + 36 = ₹ 1236

Que-2: Mr. R.K. Nair gets ₹ 6455 at the end of one year at the rate of 14% per annum in a Recurring Deposit Account. Find the monthly instalment.

Sol: Let deposit per month = ₹ x
Rate (R) = 14% p.a.
Period (n) = 1 year =12 months
Maturity value = ₹ 6455
Principal for 1 month = P x n(n+1)/2
= x × 12×(12+1)/2=x+12×13/2 = 78 x
and interest = PRT/100=78x×14×1/100×12
= 91x/100 … (i)
Interest = Maturity value – Total deposit
= ₹ 6455 – x × 12 = 6455 – 12x … (ii)
From (i) and (ii)
91×100 = 6455 – 12x
⇒ 91x = 645500 – 1200x
⇒ 91x + 1200x = 645500
⇒ 1291x = 645500
⇒ x = 645500/1291 = 500
∴ Deposit per month = ₹ 500

Que-3: Mohan deposits ₹ 80 per month in a cumulative deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is 6% per annum.

Sol: Deposit per month (P) = ₹ 80
Rate (R) = 6% p.a.
Period (n) = 6 years = 72 months
Now principal for 1 month = n(n+1)/2
= 80×72×(72+1)/2
= ₹ 80×72×73/2 = ₹ 210240
Interest = PRT/100=210240×6×1/100×12
= 105120/100 = ₹ 1051.20
∴ Maturity value = Deposit + Interest
= 72 x 80+ 1051.20
= ₹ 5760 + 1051.20 = ₹ 6811.20

Que-4: Saloni deposited ₹ 150 per month in her bank for eight months under the Recurring Deposit Scheme. What will the maturity value of her deposit, if the rate of interest is 8% per annum and the interest is calculated at the end of every month.

Sol: Deposit per month = ₹ 150
Rate (R) = 8%
Period (n) = 8 months
∴ Principal for 1 month = P×n(n+1)/2
= ₹150×8×(8+1)/2
= 150×8×9/2
= ₹ 5400
∴ Interest = PRT/100=5400×8×1/100×12 = ₹ 36
∴ Maturity value = P x n + Interest
= ₹ 150 x 8 + 36
= ₹ 1200 + 36
= ₹ 1236

Que-5: David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7725 at the time of maturity, find the rate of interest per annum.

Sol: Deposit per month (P) = ₹ 300
Period (n) = 2 years or 24 months
Let R be the rate % per annum
Maturity value = ₹ 7725
Total principal for 1 month = P×n(n+1)/2
= 300×24×(24+1)/2
= 300×24×25/2 = ₹ 90000
∴ Interest = PRT/100=90000×R×1/100×12 = ₹ 75R … (i)
Now maturity value = P x n + Interest
7725 = 300 x 24 + 75R
⇒ 7725 = 7200 + 75R
⇒ 75R = 7725 – 7200 = 525
⇒ R = 525/75 = 7
Rate of interest = 7% p.a.

Que-6: Mrs. Goswami deposits ₹ 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.

Sol: P = 36(36+1)/2 x 100
∴ Interest = 36×37×1000×8/2×12×100
= 12 x 37 x 10 = 4440
Matured value = 36000 + 4440 = ₹ 40440

Que-7: Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2500 per month for two years. At the time of maturity he got ₹ 67,500. Find :
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.

Sol: Total amount deposited by Mr. Gupta in 24 months = ₹ 2500 x 24 = ₹ 60000
Total principal for 1 month
= ₹ 2500×24×(24+1)/2
= ₹ 2500 x 12 x 25
= ₹ 750000
Let rate of interest be r% p.a.
Now interest on ₹ 750000 for 1 month
= 750000 x 1/12 x r/100 = 625 r
Maturity amount = ₹ 67500
⇒ 60000 + 625r = 67500
625r = 67500 – 60000 = 7500
r = 7500/625 = 12 % p.a.

Que-8: Ahmed has a recurring deposit account in a bank. He deposits ₹ 2,500 per month for 2 years. If he get ₹ 66,250 at the time of maturity, find
(i) The interest paid by the bank.
(ii) The rate of interest.

Sol: Monthly deposit (M.D.) = ₹ 2500
n = 2 x 12 = 24 months
Total deposited amount = ₹ 2500 x 24 = ₹ 60000
Matured amount = ₹ 66250
(i) Interest paid by the bank
= Rs. (66250 – 60000) = ₹ 6250

(ii) Equivalent principle for 1 month
= M.D. ×n(n+1)/2
= 2500×24×25/2 = ₹ 750000
R = I×100/P×T=6250×100×12/750000×1 = 10 % p.a.

Que-9: Kiran deposited ₹ 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at the rate of 11 % per annum, find the amount she gets on maturity.

Sol: P (Principal) = ₹ 200, n (Time) = 36 months,
R (Rate) = 11% p.a.
Amount deposited in 36 months
= ₹ 200 x 36 = ₹ 7200
S.I. = P x n(n+1)/2×1/12×R/100
= ₹ 200×36×37×11/2×12×100 = ₹ 1221
Amount Kiran will get on maturity
= ₹ (7200 + ₹ 1221) = ₹ 8421

Que-10: Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is 8% per annum and Mr. Britto gets ₹ 8088 from the bank after 3 years, find the value of his monthly instalment.

Sol: Let monthly installment = ₹ x
n = 3 x 12 months = 36 months
Rate = 8% p.a.
∴ I = P x n(n+1)/2×12×r/100
= x × 36×37/2×12×8/100=444/100 x
Maturity value = ₹ 8088
According to the question,
I + 36 × x = 8088
⇒ 444/100 + 36x = 8088
4.44x + 36x = 8088
40.44x = 8088
x = 8088/40.44=8080×100/4044 = ₹ 200
∴ Value of monthly installment = ₹ 200

Que-11: Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 11/2 years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum.

Sol: Money deposited per month (P) = ₹ 800 r = ?
No.of months (n) = 1 (1/2) = 32 x 1/2 = 18
∴ Interest = P x n(n+1)/2×12×r/100
= 800 x 18(18+1)/2×12×r/100
= 800 x 18×19/2×12×r/100 = 114r
∴ Maturity amount = 114r + 800 x 18
According to the question,
15084 = WAr + 14400
⇒ 15084 – 14400 = 114r
⇒ 684 = WAr
r = 684/114 = 6%

Que-12: Katrina opened a recurring deposit account with a National Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly instalment is ₹ 1,000, find the:
(i) interest earned in 2 years.
(ii) matured value.

Sol: Given,
p = ₹ 1000
n = 2 years = 24 months
r = 6%
(i) Interest = p x n(n+1)/2×r/12×100
= 1000 x 24(25+1)/2=6/12×100
= ₹ 1000 x 24(25)/2×6/12×100
Thus, the interest earned in 2 years is ₹ 1500

(ii) Sum deposited in two years = 24 x ₹ 1000 = ₹ 24,000
Maturity value = Total sum deposited in two years + Interest
= ₹ 24,000 + ₹ 1,500 = ₹ 25,500
Thus, the maturity value is ₹ 25,500

Que-13: Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find :
(i) the monthly instalment
(ii) the amount of maturity

Sol: (i) I = ₹ 1200,
n = 2 x 12 = 24 months,
r = 6%
I = P x n(n+1)/2×r/12×100
⇒ 1200 = P24(24+1)/2×1/12×100
⇒ 1200 = P×24(25)/2×6/12×100
⇒ 1200 = P x 3/2
⇒ P = 1200×2/3
⇒ P = ₹ 800
So the monthly instalment is ₹ 800

(ii) Total sum deposited = P x n = ₹ 800 x 24 = ₹ 19200
∴ The amount that Mohan will get at the time of maturity
= Total sum deposited + Interest on it
= ₹ 19200 + ₹ 1200 = ₹ 20400
Hence, the amount of maturity is ₹ 20400

— : End of Self Evaluation and Revision Banking Class 10 ICSE Maths Solutions OP Malhotra :–

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