Banking MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions

Banking MCQs Class 10 RS Aggarwal Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2. Step by step solutions of Banking MCQs questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE  for detail information about ICSE Board Class-10

Banking MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions

Banking MCQs Class 10 RS Aggarwal Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2

oard ICSE
Publications Goyal Brothers Prakashan
Subject Maths
Class 10th
Chapter-2 Banking
Writer RS Aggarwal
Book Name Foundation
Topics Solution of Banking MCQs
Edition 2024-2025

Page- 20,21

Multiple Choice Questions :

Banking MCQs Class 10 RS Aggarwal Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2

Que-1: A recurring deposit is also known as :
(a) maturity deposit  (b) cumulative time deposit  (c) regular saving deposit   (d) investment fund deposit

Solution- (b) cumulative time deposit

Que-2: In a recurring deposit (R.D.) :
(a) a person gets the same interest every month

(b) a person gets the same maturity amount every year
(c) a person deposits the same amount every year
(d) the government deposits an amount equal to the interest every year 

Solution- (c) a person deposits the same amount every year

Que-3: In a recurring deposit, the maturity value is given by : (a) (Pxn) + I   (b) P x n x I    (c) [PxnxI]/100   (d) [(Pxn) + I]/100

Solution- (a) (Pxn) + I

Que-4:  If Ramesh Kumar has an R.D. in a post office, he has to deposit : (a) an amount only once   (b) the same amount every month   (c) a decreasing amount every month   (d) an increasing amount every month

Solution- (b) the same amount every month

Que-5: In an R.D., the maturity value is the sum of the total amount deposited and the interest. If P is the same amount deposited every month for n months and R is the rate of interest, then Interest I is equal to :
(a) P x (n/12) x (R/100)   (b) [P x n(n+1)]/12 x (R/100) (c) [P x n(n+1)] / (2×12) x R/100   (d) [P x n]/(2×12) x (R/100)

Solution- (c) [P x n(n+1)]/(2×12) x R/100

Que-6: Mohit opened a recurring deposit account in a bank for 2 years. He deposited Rs1000 every month and receives Rs25500 on maturity. The interest he earned in 2 years is :
(a) Rs13500   (b) Rs3000   (c) Rs24000   (d) Rs1500

Solution- (d) Rs1500

Reason:  Mohit deposited a total of 1000 x 24 = 24000 in the recurring deposit account over two years. If he received 25500 on maturity, then the interest he earned in 2 years is 25500 – 24000 = 1500.

Que-7: Naveen deposits Rs800 every month in a recurring deposit account for 6 months. If he receives Rs4884 at the time of maturity, then the interest he earns is : (a) Rs84   (b) Rs42   (c) Rs24   (d) Rs284

Solution- (a) Rs84

Reason: Maturity value = ₹ 4884
Deposited value = ₹ 800 × 6 = ₹ 4800
∴ Interest earns by him = ₹ 4884 – ₹ 4800
= ₹ 84

Directions (Q8 to Q11): study the following information and answer the questions that follow :

Joseph has a recurring deposit account in a bank for two years at the rate of 8% per annum simple interest.
Que-8: If at the time of maturity Joseph receives Rs2000 as interest, then the monthly installment is : (a) Rs1200   (b) Rs600    (c) Rs1000   (d) Rs1600

Solution- (c) Rs1000

Reason: n = 2 years
r = 8%
I = Rs2000
I = [Pxn(n+1)]/(2×12) x r/100
2000 = [Px24x25]/(2×12) 8/100
P = (2000×100)/(25×8)
P = Rs1000.

Que-9: The total amount deposited in a bank is :
(a) Rs25000   (b) Rs24000   (c) Rs26000   (d) Rs 23000

Solution- (b) Rs24000

Reason: Total amount deposit = P x n
= 1000 x 24 = 24000

Que-10: The amount Joseph receives on maturity is : (a) Rs27000   (b) Rs25000   (c) Rs26000   (d) Rs28000

Solution- (c) Rs26000

Reason: MV = (Pxn) + I
MV = 24000 + 2000 = 26000.

Que-11: If the monthly installment is Rs100 and the rate of interest is 8%, in how many months Joseph will receive Rs52 as interest?
(a) 18   (b) 30   (c) 12   (d) 6

Solution- (c) 12

Reason: P = Rs100
r = 8%
I = Rs52
I = [Pxn(n+1)]/(2×12) x r/100
52 = [100 x n(n+1)]/(2×12) x 8/100
n(n+1) = (52 x 100 x 2 x 12)/(100 x 8)
n(n+1) = 52 x 3
n(n+1) = 13 x 4 x 3
n(n+1) = 13 x 12
n(n+1) = 12 x (12+1)
n = 12.

— : End of Banking MCQs Class 10 RS Aggarwal Goyal Brothers Prakashan ICSE Foundation Maths Solutions  : —

Return to :–  RS Aggarwal ICSE Class 10 Solutions Goyal Brothers

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