Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn how to solve Multiple Choice Questions on Compound Interest very easily. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Board ICSE
Publications Goyal brothers Prakshan
Subject Maths
Class 9th
Chapter-2 Compound Interest
Writer RS Aggrawal
Book Name Foundation
Topics Solution of MCQs
Academic Session 2024-2025

How to Solve MCQs on Compound Interest  Easily

To solve mcqs on  Compound Interest use formula or without formula and calculate unknown value Also focus on growth and deprecitaion formula in some cases. At last check the option available which is most suitable after calculation.

Multiple Choice Questions 

Compound Interest Class 9 RS Aggarwal Goyal Brothers ICSE Foundation Maths Solutions.

Page- 41,42

Que-1: The compound interest on Rs 3750 for 2 years at 8% p.a., compounded annually is :

(a) Rs604   (b) Rs614   (c) Rs624   (d) Rs642

Solution- (c) Rs624

Reason:  A = P * (1+R/100)^n
A = 3750 * (1+8/100)^2
CI = A- Principal = 624Rs.

Que-2: A man invests Rs46875 at 4% p.a. compound interest for 3 years. The interest for the first year will be :

(a) Rs1785   (b) Rs1587   (c) Rs1875   (d) Rs1758

Solution- (c) Rs1875

Reason: It is given that
Principal= 46875
Rate of interest = 4% p.a.
(i) Interest for the first year= Prt/100
Substituting the values
=(46875×4×1)/100
= 1875.

Que-3: A man deposits Rs10000 in a cooperative bank for 3 years at 9% p.a. If interest is compounded annually, then the amount he will get from the bank after 3 years is :

(a) Rs12950.29   (b) Rs12905.29   (c) Rs12059.29   (d) Rs12095.29

Solution- (a) Rs12950.29

Reason: Principal amount (P) is Rs.10,000
rate (r) is 9% or 0.09
years (t) is 3
A = 10000(1 + 0.09/1)^(1*3)
= 10000(1 + 0.09)^3
= 10000(1.09)^3
≈ 10000(1.295029)
≈ Rs.12,950.29.

Que-4: Rs16000 is deposited in a bank for three years. The rate of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes Rs21384. The rate of interest for third year will be :

(a) 7%   (b) 10%   (c) 11%   (d) 12%

Solution- (b) 10%

Reason: Initial deposit (P) = Rs16,000
Interest rate for the first year = 8%
Interest rate for the second year = 12%
Final amount after three years = Rs. 21,384
A1​ = P(1+r1/100​​)
where r1​ is the interest rate for the first year.
𝐴1 = 16000(1+8/100)
𝐴1 = 16000(1+0.08)
𝐴1 = 16000×1.08
A1 ​= 17280
A2 ​= A1​(1+r2/100​​)
where r2​ is the interest rate for the second year.
𝐴2 = 17280(1+12/100)
𝐴2 = 17280(1+0.12)
𝐴2 = 17280×1.12
𝐴2 = 19353.60
A3​ = A2​(1+r3/100​​)
We know that A3 ​= 21384.
So, 21384 = 19353.60(1+𝑟3/100)
To find 1+(𝑟3/100)​​:
1+(𝑟3/100) = 21384/19353.60
1+(𝑟3/100) = 1.105
Solving for r3​:
𝑟3/100 = 1.105 − 1
𝑟3/100 = 0.105
𝑟3 = 0.105 × 100
𝑟3 = 10%.

Que-5: A man borrows Rs5000 at 12% compound interest p.a., interest payable every six months. He pays back Rs1800 at the end of every six months. The third payment he has to make at the end of 18 months in order to clear the entire loan will be :

(a) Rs2024.60   (b) Rs2204.60   (c) Rs2240.60   (d) Rs2402.60

Solution- (a) Rs2024.60

Reason:  For first year,
P = 5000
R =12%
T = 1/2 year
I = PTR/100
I = 5000×12×1/100×1/2
A = P + I
A = 5000+ 300
A = 5300
Man repaid = 5300 – 1800
= 3500
For next six month,
P = 3500
R = 12 %
T = 1/2
I = PTR/100
= 3500×12×1/100×1/2
= 210
Now we know that
A = P + I
= 3500 + 210
= 3710
Man repaid = 3710- 1800
= 1910
For last six months ,
P = 1910
R = 12
T = 1/2
I = PTR/100
= 114.6
A = P + I
= 1910+ 114.6
= 2024.6

Que-6: The compound interest for the second year on Rs8000 invested for 3 years at 10% p.a. is :

(a) Rs780   (b) Rs880   (c) Rs980   (d) Rs1080

Solution- (b) Rs880

Reason:For the first year
P = Rs8,000
N = 1year
R = 10 %
We have S.I. = PNR/100 = (8,000×1×10)/100 = Rs800
And Amount at the end of first year P+S.I. = Rs8,000 + Rs800 = Rs8,800
Now, for the second year
P = Rs8,800
N = 1year
R = 10 %
We have S.I. = PNR/100 = (8,800×1×10)/100 = Rs880
Thus, Compound interest for the second year = Rs880.

Que-7: A person took a loan of Rs6000 from a bank and agreed to pay back the amount along with interest in 2 years. If the rate of compound interest for the first year is 10% and second year is 12%, the amount he had to pay after 2 years will be :

(a) Rs 7329   (b) Rs 7932   (c) Rs 7292   (d) Rs 7392

Solution- (d) Rs7392

Reason: Principal amount (P) = Rs. 6000
Interest rate for the first year (r1​) = 10%
𝐴1 = 6000(1+10/100)
𝐴1 = 6000(1+0.10)
𝐴1 = 6000×1.10 ​= 6600
Amount at the end of the first year (A1​) = Rs. 6600
Interest rate for the second year (r2​) = 12%
𝐴2 = 6600(1+12/100)
𝐴2 = 6600(1+0.12)
𝐴2 = 6600×1.12
𝐴2 = 7392.

Que-8: Nikita invests Rs 6000 for 2 years at a certain rate of interest compounded annually. At the end of the first year, it amounts to Rs6720. The rate of interest p.a. is :

(a) 8%  (b) 10%  (c) 12%  (d) 14%

Solution- (c) 12%

Reason: P = Rs. 6000, Amount at the end of the first year = Rs. 6720
S.I. for first year = Rs. (6720 – 6000) = Rs. 720
Let r% be the rate of interest p.a.
720 = (6000×𝑟×1)/100
r = 720/60
r = 12%.

Que-9: The compound interest on Rs8640 for 3 years at 8% p.a. is :

(a) Rs2345   (b) Rs3245   (c) Rs3425   (d) Rs3452

Solution- (a) Rs2345

Reason: 

Que-10: If the interest is compounded half-yearly, then, C.I. when the principal is Rs7400, the rate of interest is 5% p.a. and the duration is one year, is :

(a) Rs373.63   (b) Rs374.63   (c) Rs373.36   (d) Rs373

Solution- (b) Rs374.63

Reason:It is given that
Principal (P) = 7400
Rate of interest (r) = 5%
Period (n) = 1 year
We know that
A = P(1+r/(2×100))^2×n
Substituting the values
= 7400(1+5/200)^2
By further calculation
= 7400 × 205/200 × 205/200
= 7774.63
C.I. = A – P
C.I. = 7774.63 – 7400
C.I. = 374.63.

Que-11: The simple interest on a sum of money for 2 years at 4% per annum is Rs340. The compound interest on this sum for one year payable half-yearly at the same rate is :

(a) Rs170.70   (b) Rs107.70   (c) Rs171.70   (d) Rs270.70

Solution- (c) Rs171.70

Reason: Given : I = Rs340, T = 2 Years and R = 4%
∴P = (I×100)/(R×T)
= (Rs340×100)/(4×2) = Rs.4250
C.I. = P(1+r/(2×100))^n×2 − P
= Rs.4250(1+4/(2×100))^1×2 − Rs.4250
= Rs.4421.70 − Rs.4250
= Rs.171.70

Que-12: The compound interest on a certain sum of money at 5% p.a. for two years is Rs246. The simple interest on the same sum for three years at 6% p.a. will be :

(a) Rs432   (b) Rs430.50   (c) Rs432.75   (d) Rs431.75

Solution- (a) Rs432

Reason:  CI = Rs246, R = 5%, T = 2 years
CI = A – P
246 = P[1+5/100]^2 – P
246 = P|(21/20)^2 – 1|
246 = P[41/400]
P = (246×400)/41
= Rs2400
Now, P = Rs2400, R = 6%, T = 3 years
SI = (2400×6×3)/100
= Rs432.

Que-13: Ramesh wants to get Rs6050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is :

(a) Rs5500   (b) Rs5000   (c) Rs5600   (d) Rs5800

Solution- (b) Rs5000

Reason:  Amount = Rs. 6,050
Time = 2 years
Rate of interest = 10% p.a.
Let assume that Principal = x
So,
[(100 + Rate of interest)/100]Time = (Amount/Principal)
⇒ [(100 + 10)/100]2 = (6050/ x)
⇒ (110/100) × (110/100) = (6050/ x)
⇒ (121/100) = (6050/x)
By cross multiplying
⇒ x = 6050 × (100/121)
⇒ x = 5,000

Que-14: The difference between the compound interest and simple interest on a certain sum deposited for 2 years at 5% p.a. is Rs12. The sum will be :

(a) Rs4500   (b) Rs4600   (c) Rs4800   (d) Rs5000

Solution- (c) Rs4800

Reason:N = 2years
R = 5 %
We have S.I. = (PNR)/100 = (P×2×5)/100 = 10P/100 = 0.1P
And on interest being compounded for 2 years and R = 5 %,
Amount = P(1+R/100)^N
= P(1+5/100)²
= P×(1.05)²
= 1.1025P
So, C.I. = A−P = 1.1025P − P = 0.1025P
Given, C.I. − S.I = Rs12
= 0.1025P−0.1P = Rs12
= 0.0025P = Rs12
P = Rs4,800.

Que-15: At what rate of compound interest p.a. will Rs20000 amount to Rs26620 in 3 years ?

(a) 4%   (b) 6%   (c) 8%   (d) 10%

Solution- (d) 10%

Reason:  Amount = Principal (1 + r/100)^n
= 26620 = 20000 (1+r/100)³
( 1+ r%)³ = 26620/20000
= ( 1 +r%)³ = 1331 / 1000
= ( 1+ r %) = ³√ 1331 / 1000
= ( 1+ r%) = 11/10
= ( 1 + r/100 ) = 11/10
(100+r)/100 = 11 /10
= (100 + r) = (11×100)/10
= 100 + r = 110
= rate = 110 – 100
= 10%

Que-16: In what time will Rs5000 amount to Rs5832 at 8% rate of compound interest p.a. ?

(a) 2 years   (b) 4 years   (c) 6 years   (d) 8 years

Solution- (a) 2 years

Reason: Principal = Rs. 5,000
Amount = Rs. 5,832
Rate = 8% p.a.
Amount = Principal × (1 + 𝑅/100)^n
⇒ 5832 = 5000 × (1 + 8/100)^n
⇒ 5832/5000 = (108/100)^n
⇒ (729/625) = (27/25)^n
⇒ (27/25)² = (27/25)^n
⇒ Time = 2 years

Que-17: A machine depreciates at the rate of 10% of its value at the beginning of the year. If the present value of the machine is Rs8000, its value after three years will be :

(a) Rs5382   (b) Rs5832   (c) Rs5238   (d) Rs5638

Solution- (b) Rs5832

Reason:  machine value = ₹8000
machine depreciation every year 10%
first year -> 10% of 8000 = 800
value is = 8000 – 800 = ₹7200
second year again 10% decrease
value is = 10% of 7200 = 720
so 7200 – 720 = 6480
final third year also 10% decrease
value is = 10% of 6480 = 648
final value is = 6480 – 648 = ₹5832

Que-18: The present population of a town is 200000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be :

(a) Rs253000   (b) Rs235000   (c) Rs203500   (d) Rs352000

Solution- (a) Rs253000

Reason:  Population after n years = Present population x (1+𝑟/100)^𝑛
Present population = 2,00,000
After first year, population = = 2,00,000 x (1+10/100)^1
= 2,00,000 × 11/10
= 2,20,000
Population after two years = 2,20,000 ×(1+15/100)^1
= 253000
Thus, the population after two years is 2,53,000.

Que-19: A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for Rs10000 and is sold for Rs7744. The number of years, that the machine was used is :

(a) 2   (b) 4   (c) 6   (d) 8

Solution- (a) 2

Reason: V = 7744
P = 10000
r = 12%
n = ?
V = P(1-r/100)^n
7744 = 10000(1-12/100)^n
7744/10000 = (100-12/100)^n
7744/10000 = (88/100)^n
7744/10000 = 0.88^n
0.7744 = 0.88^n
0.2552^2 = 0.127^n   {approx}
n = 2 years.

Que-20: The value of a machine depreciates every year at a constant rate. If the values of the machine in 2006 and 2008 are Rs 25000 and 19360 respectively, then the annual rate of depreciation is :

(a) 8%   (b) 10%   (c) 12%   (d) 14%

Solution- (c) 12%

Reason: P = 25000 in 2006
V = 19360 in 2008
n = 2008 – 2006 = 2 years
V = P(1+r/100)^n
19360 = 25000(1+r/100)^2
19360/25000 = (1+r/100)^2
0.7744 = (1+r/100)^2
√0.7744 = 1+r/100
0.88 = 1+r/100
1 – 0.88 = r/100
0.12 = r/100
r = 1.2 x 100
r = 12%.

– : End of Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions : —

Return to :- RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)

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