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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 2 - Banking (Recurring Deposit Accounts)

## Selina Textbook Solutions Chapter 2 - Banking (Recurring Deposit Accounts)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 2 - Banking (Recurring Deposit Accounts).

All Selina textbook questions of Chapter 2 - Banking (Recurring Deposit Accounts) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 2 - Banking (Recurring Deposit Accounts) Page/Excercise 2(A)

Solution 1

Instalment per month(P) = Rs600

Number of months(n) = 20

Rate of interest(r)= 10%p.a.

The amount that Manish will get at the time of maturity

=Rs(600x20)+ Rs1,050

=Rs12,000+ Rs1,050

= Rs13,050 Ans.

Solution 2

Instalment per month(P) = Rs640

Number of months(n) = 54

Rate of interest(r)= 12%p.a.

The amount that Manish will get at the time of maturity

=Rs(640x54)+ Rs9,504

=Rs34,560+ Rs9,504

= Rs44,064 Ans.

Solution 3

For A

Instalment per month(P) = Rs1,200

Number of months(n) = 36

Rate of interest(r)= 10%p.a.

The amount that A will get at the time of maturity

=Rs(1,200x36)+ Rs6,660

=Rs43,200+ Rs6,660

= Rs49,860

For B

Instalment per month(P) = Rs1,500

Number of months(n) = 30

Rate of interest(r)= 10%p.a.

The amount that B will get at the time of maturity

=Rs(1,500x30)+ Rs5,812.50

=Rs45,000+ Rs5,812.50

= Rs50,812.50

Difference between both amounts= Rs50,812.50 - Rs49,860

= Rs952.50

Then B will get more money than A by Rs952.50 Ans.

Solution 4

Let Instalment per month(P) = Rs y

Number of months(n) = 12

Rate of interest(r)= 11%p.a.

Maturity value= Rs(y x12)+Rs0.715y= Rs12.715y

Given maturity value= Rs12,715

Then Rs12.715y = Rs12,715

Ans.

Solution 5

Let Instalment per month(P) = Rs y

Number of months(n) = 42

Rate of interest(r)= 12%p.a.

Maturity value= Rs(y x42)+Rs9.03y= Rs51.03y

Given maturity value= Rs10,206

Then Rs51.03y = Rs10206

Ans.

Solution 6

(a)

Instalment per month(P) = Rs140

Number of months(n) = 48

Let rate of interest(r)= r %p.a.

Maturity value= Rs(140 x48)+Rs(137.20)r

Given maturity value= Rs8,092

Then Rs(140 x48)+Rs(137.20)r = Rs8,092

137.20r = Rs8,092 -; Rs6,720

r =

(b)

Instalment per month(P) = Rs300

Number of months(n) = 24

Let rate of interest(r)= r %p.a.

Maturity value= Rs(300 x24)+Rs(75)r

Given maturity value= Rs7,725

Then Rs(300 x24)+Rs(75)r = Rs7,725

75 r = Rs7,725 -; Rs7,200

r =

Solution 7

Instalment per month(P) = Rs150

Number of months(n) = 8

Rate of interest(r)= 8%p.a.

The amount that Manish will get at the time of maturity

=Rs(150x8)+ Rs36

=Rs1,200+ Rs36

= Rs1,236 Ans.

Solution 8

Instalment per month(P) = Rs350

Number of months(n) = 15

Let rate of interest(r)= r %p.a.

Maturity value= Rs(350 x15)+Rs(35)r

Given maturity value= Rs5,565

Then Rs(350 x15)+Rs(35)r = Rs5,565

35r = Rs5,565 -; Rs5,250

r =

Solution 9

Instalment per month(P) = Rs1,200

Number of months(n) = n

Let rate of interest(r)= 8 %p.a.

Maturity value= Rs(1,200x n)+Rs4n(n+1)= Rs(1200n+4n2+4n)

Given maturity value= Rs12,440

Then 1200n+4n2+4n = 12,440

Then number of months = 10 Ans.

Solution 10

Instalment per month(P) = Rs300

Number of months(n) = n

Let rate of interest(r)= 12 %p.a.

Maturity value= Rs(300x n)+Rs1.5n(n+1)

= Rs(300n+1.5n2+1.5n)

Given maturity value= Rs8,100

Then 300n+1.5n2+1.5n = 8,100

Then time= 2years

Solution 11

(i)

Maturity value = Rs67,500

Money deposited= Rs2,500 x 24= Rs60,000

Then total interest earned= Rs67,500 - Rs60,000= Rs7,500 Ans.

(ii)

Instalment per month(P) = Rs2,500

Number of months(n) = 24

Let rate of interest(r)= r %p.a.

Then 625 r= 7500

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 2 - Banking (Recurring Deposit Accounts) Page/Excercise 2(B)

Solution 1

Instalment per month(P) = Rs600

Number of months(n) = 48

Rate of interest(r)= 8%p.a.

The amount that Manish will get at the time of maturity

=Rs(600x48)+ Rs4,704

=Rs28,800+ Rs4,704

= Rs33,504Ans.

Solution 2

Installment per month(P) = Rs80

Number of months(n) = 18

Let rate of interest(r)= r %p.a.

Maturity value= Rs(80 x18)+Rs(11.4r)

Given maturity value= Rs1,554

ThenRs(80 x18)+Rs(11.4r) = Rs1,554

11.4r = Rs1,554 - Rs1,440

Solution 3

Installment per month(P) = Rs400

Number of months(n) = n

Let rate of interest(r)= 8 %p.a.

Maturity value= Rs(400x n)+

Given maturity value= Rs16,176

ThenRs(400x n)+= Rs16,176

1200n +4n2+4n= Rs48,528

4n2+1204n = Rs48,528

n2+301n - 12132= 0

(n+337)(n-36)=0

n= -337 or n=36

Then number of months = 36months= 3yearsAns.

Solution 4

Let installment per month= Rs P

Number of months(n) = 24

Rate of interestÂ®= 8%p.a.

Maturity value= Rs(P x 24)+ Rs 2P= Rs26P

Given maturity value= Rs30,000

Solution 5

Let Installment per month= Rs P

Number of months(n) = 36

Rate of interest(r)= 8%p.a.

Given interest = Rs9,990

(ii)Maturity value= Rs(2,250x 36)+ Rs9,990=Rs90,990Ans.

Solution 6

Installment per month(P) = Rs900

Number of months(n) = 48

Let rate of interest(r)= r %p.a.

Maturity value= Rs(900 x48)+Rs(882)r

Given maturity value= Rs52,020

ThenRs(900 x48)+Rs(882)r = Rs52,020

882r = Rs52,020 - Rs43,200

r =

Solution 7

Installment per month(P) = Rs1,800

Number of months(n) = 48

Let rate of interest(r)= r %p.a.

Maturity value= Rs(1,800 x48)+Rs(1,764)r

Given maturity value= Rs1,08,450

ThenRs(1,800 x48)+Rs(1764)r = Rs1,08,450

1764r = Rs1,08,450 - Rs86,400

r =

Solution 8

Solution 9

Solution 10

Solution 11

Interest, I = Rs. 1,200

Time, n = 2 years = 2 × 12 = 24 months

Rate, r = 6%

(i) To find: Monthly instalment, P

Now,

So, the monthly instalment is Rs. 800.

(ii) Total sum deposited = P × n = Rs. 800 × 24 = Rs. 19,200

Amount of maturity = Total sum deposited + Interest on it

= Rs. (19,200 + 1,200)

= Rs. 20,400

## Browse Study Material

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 2 - Banking (Recurring Deposit Accounts)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.

# Text Book Solutions

ICSE X - Mathematics

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