Ch- Test Compound Interest Class 9 OP Malhotra ICSE Maths Solutions Ch-2.  We Provide Step by Step Solutions / Answer of Compound Interest OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Compound Interest Class 9 OP Malhotra Ch-Test ICSE Maths Solutions Ch-2

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-2	Compound Interest
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Compound Interest

Compound Interest Class 9 OP Malhotra ICSE Maths Solutions Ch-2

Que-1: Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At end of first year it amounts to ₹ 6,720. Calculate:
(i) the rate of interest
(ii) the amount at the end of the second year.

Sol: Nikita invests (P) = ₹ 6000
Period (T) = two years
and after 1 year she got back ₹ 6720
∴ Interest for 1 year = ₹ (6720 – 6000) = ₹ 720
(i) ∴ Rate of interest compounded annually
= (720×100)/6000 = 12%

(ii) Interest at end of second year
= (6720×12×1)/100
∴ Amount at end of second year = ₹ (6720 + 806.40) = ₹ 7,526.40

Que-2: Rohit borrows ₹ 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit’s profit in the transaction at the end of two years.

Sol: Rohit borrows ₹ 86,000 for 2 years at simple interest at 5% per annum
∴ Interest = (86,000×2×5)/100 = ₹ 8600
Rohit gets ₹ 8600 in 2 years as simple interest Rohit lends money to Akshay at 5% p.a. compounded annually for 2 years
Amount he gets back after 2 years is
A = P [1+(r/100)^n] = 86,000 [1+(5/100)]²
= 86,000 [1+(1/20)]² [t = 2, r = 5%]
A = 86,000 x (21/20) x (21/20)
A = ₹ 94,815
C.I. = A – P
= 94,815 – 86,000
C.I. = ₹ 8815
Rohit’s profit in the transaction at the end of two years = ₹(8815 – 8600) = ₹ 215

Que-3: Mr. Kumar borrowed ₹ 15,000 for two years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays ₹ 6,200 at the end of the first year, find the outstanding amount at the end of the second year.

Sol: For 1st year :
S.I = (15000×8×1)/100 = ₹ 1200
Amount = ₹ (15000 + 1200) = ₹ 16200
Remaining amount after repayment
= ₹ (16200 – 6200) = ₹ 10000
For 2nd year :
P = ₹ 10000
S.I.= (10000×10×1)/100 = ₹ 1000
Amount at the end of 2nd year = ₹ 10000 + ₹ 1000 = ₹ 11000

Que-4: In what period of time will ₹ 12,000 yield ₹ 3,972 as compound interest at 10% per annum, if compounded on an yearly basis?

Sol: P = ₹ 12000
A = ₹ 12000 + ₹ 3972 = ₹ 15972
R = 10% p.a., n = ?
A = P [1+(r/100)^n]
15972 = 12,000 [1+(10/100)]^n
15972/12000 = [11/10]^n
1331/1000 = [11/10]^n
(11/10)³ = [11/10]^n
Comparing,
n = 3
∴ Time = 3 years

Que-5: On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹ 25 if the rate of interest charged for both is 5%?

Sol: Difference between C.I. and S.l. = ₹25
Let principal (P) = ₹ 100
Rate (R) = 5% p.a. and period (n) = 2 years
Then S.I. = PRT/100 = (100×5×2)/100 = ₹ 10
and when interest is compounded annually
A = P[1+(R/100)]^n = 100[1+(5/100)]^2
= ₹100 x (21/20) × (21/20) = ₹441/4
and C.I. = A – P = ₹ (441/4) – 100
= ₹ (441−400)/4 = ₹ 41/4
Now difference between C.I. and S.l.
= ₹ (41/4) – 10
= ₹ (41−40)/4 = ₹ 1/4
If difference is ₹ 1/4, then principal = ₹ 100
and if difference is ₹ 1, then principal
= ₹ (100×4)/1
and if difference is ₹ 25, the principal
= (100×4×25)/1 = ₹ 10000

Multiple Choice Questions (MCQs)

Que-6: A sum of ₹ 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become

(a) ₹ 48,000
(b) ₹ 96,000
(c) ₹ 1,90,000
(d) ₹ 1,92,000

Sol: (d) ₹ 1,92,000
Sum (P) = ₹ 12000
∴ A = ₹ 12000 x 2 = ₹ 24000
Period (n) = 5 years
∴ A = P[1+(R/100)]^n
⇒ A/P = [1+(R/100)]^n
⇒ 24000/12000 = [1+(R/100)]^5
⇒ 2 = [1+(R/100)]^5
∴ (1+(R/100))^20 = 2^4 = 16
Now, amount after 20 years
= P[1+(R/100)]^20
= 12000[(1+(R/100))^5]^4
= ₹ 12000 x (2)^4 = ₹ 12000 x 16
= ₹ 1,92,000

Que-7: The difference between C.I. (compound interest) and S.I. (simple interest) on a sum of ₹ 4,000 for 2 years at 5% p.a. payable yearly is

(a) ₹ 20
(b) ₹ 10
(c) ₹ 50
(d) ₹ 60

Sol: (b) ₹ 10
Sum (P) = ₹ 4000
Rate (R) = 5% p.a.
Period (n) = 2 years
∴ S.I = PRT/100 = (4000×5×2)/100 = ₹ 400
and when interest is compounded annually,
then A = [1+(R/100)]^n
⇒ 4000 [1+(5/100)]²
= ₹ 4000 x (21/20) x (21/20) = ₹ 4410
and C.I. = A – P = ₹ 4410 – 4000 = ₹ 410
and difference between C.I. and S.I.
= ₹ 410 – 400 = ₹ 10

Que-8: If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is ₹ 31, the sum is

(a) ₹ 500
(b) ₹ 50
(c) ₹ 1000
(d) ₹ 1250

Sol: (c) ₹ 1000
Difference in C.I. and S.I. = ₹ 31
Let principal (P) = ₹ 100
Rate (R) = 10% p.a.
Period (n) = 3 years
∴ S.I. =  PRT/100 = ₹ (100×10×3)/100 = ₹ 30
When interest is compounded annually then
A = P[1+(R/100)]^n
= 100 [1+(10/100)]³
= ₹ 100 × (11/10) × (11/10) × (11/10)
= ₹ 1331/10
∴ C.I. = A – P = ₹ (1331−100)/10 = ₹331/10
and difference between C.I. and S.I.
= ₹ (1331−100)/10 = ₹ 331/10
If difference is ₹3110 then principal = ₹ 100
If difference is ₹ 1 then principal = ₹ (100×10)/31
and if difference is ₹31, then principal
= (100×10×31)/31 = ₹1000

Que-9: On what sum of money will compound interest for 2 years at 5% per annum amount to ₹164?

(a) ₹ 1600
(b) ₹ 1500
(c) ₹ 1400
(d) ₹ 1700

Sol: (a) ₹ 1600
C.I. = ₹ 164
Rate = 5% p.a.
and period (n) = 2 years
Let principal (P) = ₹ 100
Then A = P [1+(R/100)]^n
= ₹ [1+(5/100)]² = ₹100 × (21/20) × (21/20)
= ₹ 441/4
and C.I. = A – P
= ₹ (441/4) – 100 = ₹ 41/4
If C.I. is ₹ 41/4 then principal = ₹ 100
If C.I. is ₹ 1 then principal = ₹ (100×4)/41
and if C.L is ₹ 164, then principal
= ₹ (100×4×164)/41
= ₹ 1600

Que-10: A sum of money becomes eight times in 3 years. If the rate is compounded annually, in how much time will the same amount at the same compound rate becomes sixteen times?

(a) 6 years
(b) 4 years
(c) 8 years
(d) 5 years

Sol: (b) 4 years
Let principal (P) = ₹ 100
Then amount (A) = ₹ 100 x 8 = ₹ 800
Period (n) = 3 years
∴ A = P [1+(R/100)]^n
⇒ A/P = [1+(R/100)]^n
⇒ 800/100 = [1+(R/100)]³
⇒ [1+(R/100)]³ = 8 = 2³
⇒ [1+(R/100)] = 2 … (i)
In second case,
P = ₹ 100
Then A = ₹ 100 x 16 = 1600
∴ AP = [1+(R/100)]^n
⇒ 1600/100 = [1+(R/100)]^n
⇒ [1+(R/100)]^n = 16
⇒ (2)^n = 2^4                 [From (i)]
Comparing,
n = 4
∴ Period = 4 years

— : End of Compound Interest Class 9 OP Malhotra Ch-Test ICSE Maths Step by step Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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