Concise Mathematics Class 10 ICSE Solutions for Chapter 2 - Banking (Recurring Deposit Accounts)

Exercise - 2(A)

1. Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits Rs 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.

Ans: Installment per month (P) = Rs. 600

Number of months (n) = 20

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=600\times \dfrac{20~\left( ~20~+~1 \right)}{2~\times ~12}\times \dfrac{10}{100}$

$\Rightarrow \left( S.I. \right)~=100\times \dfrac{10~\left( ~21 \right)}{~2}\times \dfrac{1}{10}$

$\Rightarrow \left( S.I. \right)~=50\times 21$

$\Rightarrow \left( S.I. \right)~=1050$

The amount that Manish will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (600 $\times $ 20) + Rs. 1,050

= Rs. 12,000 + Rs. 1,050

= Rs. 13,050

2. Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposit Rs 640 per month for 4$\dfrac{1}{2}$years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per annum.

Ans: Installment per month (P) = Rs. 640

Number of months (n) = 4.5 $\times $ 12 = 54

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=640\times \dfrac{54~\left( ~54~+~1 \right)}{2~\times ~12}\times \dfrac{12}{100}$

$\Rightarrow \left( S.I. \right)~=64\times \dfrac{54~\left( ~55 \right)}{~2}\times \dfrac{1}{10}$

$\Rightarrow \left( S.I. \right)~=32\times 27\times 11$

$\Rightarrow \left( S.I. \right)~=9504$

The amount that Mrs. Mathew will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (640 $\times $ 54) + Rs. 9,504

= Rs. 34,560 + Rs. 9,504

= Rs. 44,064

3. Each of A and B opened a Recurring Deposit Account in a bank. If A deposits Rs 1200 per month for 3 year and B deposits Rs 1500 per month for 2$\dfrac{1}{2}$ year, find on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10 % per annum.

Ans:  For A:

Installment per month (P) = Rs. 1200

Number of months (n) = 3 $\times $ 12 = 36

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=1200\times \dfrac{36~\left( ~36~+~1 \right)}{2~\times ~12}\times \dfrac{10}{100}$

$\Rightarrow \left( S.I. \right)~=100\times \dfrac{36~\left( ~37 \right)}{~2}\times \dfrac{1}{10}$

$\Rightarrow \left( S.I. \right)~=5\times 37\times 36$

$\Rightarrow \left( S.I. \right)~=6,660$

The amount that A will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (1200 $\times $ 36) + Rs. 6,660

= Rs. 43,200 + Rs. 6,660

= Rs. 49,860

For B:

Installment per month (P) = Rs. 1500

Number of months (n) = 2.5 $\times $ 12 = 30

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=1500\times \dfrac{30~\left( ~30~+~1 \right)}{2~\times ~12}\times \dfrac{10}{100}$

$\Rightarrow \left( S.I. \right)~=15\times \dfrac{15~\left( ~31 \right)}{~~12}\times 10$

$\Rightarrow \left( S.I. \right)~=225\times 31\times \dfrac{5}{6}$

$\Rightarrow \left( S.I. \right)~=5,812.5$

The amount that B will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (1500 $\times $ 30) + Rs. $5,812.5$

= Rs. 45,000 + Rs. $5,812.5$

= Rs. 50,812.5

Difference between both amounts = Rs. 50.812.50 - Rs. 49,800

= Rs. 952.50

Hence, B will get more money than A by Rs. 952.50

4. Ashish deposits a certain sum of money every month in a recurring deposit account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets Rs 12,715 as the maturity value of this account, what sum of money did he pay every month?

Ans: Let Installment per month (P) = Rs. P

Number of months (n) = 12 

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

Maturity value of this account = 12,715

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{12~\left( ~12~+~1 \right)}{2~\times ~12}\times \dfrac{11}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{13}{~2}\times \dfrac{11}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{143}{200}$

$\Rightarrow \left( S.I. \right)~=0.715P$

The amount that Ashish will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (P $\times $ 12) + Rs. $0.715P$

= Rs. (P $\times $ 12.715)

$\Rightarrow Rs.~P\times 12.715=Rs.~12,715$

$\Rightarrow P=1,000$

So Ashish paid Rs 1,000 every month for 12 months.

5. A man has a recurring deposit account in a bank for 3$\dfrac{1}{2}$ year. If the rate of interest is 12 % per annum and the man gets Rs 10,206 on maturity, find the value of monthly instalments.

Ans: Let Installment per month (P) = Rs. P

Number of months (n) = $3.5\times 12=42$ 

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

Maturity value of this account = 10,206

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{42~\left( ~42~+~1 \right)}{2~\times ~12}\times \dfrac{12}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{21~\times ~43}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{903}{100}$

$\Rightarrow \left( S.I. \right)~=9.03P$

The amount that man will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (P $\times $ 42) + Rs. $9.03P$

= Rs. (P $\times $ 51.03)

$\Rightarrow Rs.~P\times 51.03=Rs.~10,206$

$\Rightarrow P=200$

Hence monthly installment is Rs 200.

6. (i) Puneet has a recurring deposit account in the Bank of Baroda and deposits Rs 140 per month for 4 year. If he gets Rs 8,092 on maturity, find the rate of interest given by the bank.

Ans: Installment per month (P) = Rs. 140

Number of months (n) = $4\times 12=48$

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

Maturity value of this account = 8,092

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=140\times \dfrac{48~\left( ~48~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=r\times \dfrac{140~\times ~2~\times ~49}{100}$

$\Rightarrow \left( S.I. \right)~=r\times \dfrac{13,720}{100}$

$\Rightarrow \left( S.I. \right)~=137.2~r$

The amount that Puneet will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (140 $\times $ 48) + Rs. $137.2~r$

= Rs. (6,720 + $137.2~r$)

$\Rightarrow Rs.~\left( 6,720~+~137.2~r \right)=Rs.~8,092$

$\Rightarrow 137.2~r=8,092-6,720$

$\Rightarrow 137.2~r=1,372$

$\Rightarrow r=10$

Hence the rate of interest is given by the bank 10 % p.a.

(ii) David opened a recurring deposit account in a bank and deposited Rs 300 per month for two year. If he received Rs 7,725 on maturity, find the rate of interest per annum.

Ans: Installment per month (P) = Rs. 300

Number of months (n) = $2\times 12=24$

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

Maturity value of this account = 7,725

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=300\times \dfrac{24~\left( ~24~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=r\times 3\times 25$

$\Rightarrow \left( S.I. \right)~=75r$

The amount that David will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (300 $\times $ 24) + Rs. $75r$

= Rs. (7,200 + $75r$)

$\Rightarrow Rs.~\left( 7,200~+~75r \right)=Rs.~7,725$

$\Rightarrow 75r=7,725-7,200$

$\Rightarrow 75r=525$

$\Rightarrow r=7$

Hence the rate of interest is given by the bank 7 % p.a.

7. Amit deposited Rs 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 interest is calculated at the end of every month?

Ans: Installment per month (P) = Rs. 150

Number of months (n) = $8$

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

Maturity value of this account = ?

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=150\times \dfrac{8~\left( ~8~+~1 \right)}{2~\times ~12}\times \dfrac{8}{100}$

$\Rightarrow \left( S.I. \right)~=150\times 3\times \dfrac{2}{25}$

$\Rightarrow \left( S.I. \right)~=6\times 6$

$\Rightarrow \left( S.I. \right)~=Rs.~36$

The amount that Amit will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (150 $\times $ 8) + Rs. $36$

= Rs. (1,200 + $36$)

= Rs. 1.236

8. Mrs. Geeta deposited Rs 350 per month in a bank for 1 year and 3 months under the recurring deposit scheme. If the maturity value of her deposits is Rs 5,565; find the rate of interest per annum.

Ans:  Installment per month (P) = Rs. 350

Number of months (n) = $12+3=15$

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

Maturity value of this account = 5,565

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)=350\times \dfrac{15~\left( ~15~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)=35\times r\times \dfrac{15~\times ~4}{2~\times ~3~\times ~10}$

$\Rightarrow \left( S.I. \right)=35r$

$\Rightarrow \left( S.I. \right)=Rs.~35r$

The amount that Mrs. Geeta will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (350 $\times $ 15) + Rs. $35r$

= Rs. (5,250 + $35r$)

According to question:

$\Rightarrow Rs.~\left( 5,250~+~35r \right)=5,565$

$\Rightarrow 35r=315$

$\Rightarrow r~=9$

Hence the rate of interest per annum is 9 % p.a.

9. A recurring deposit account of Rs 1,200 per month has a maturity value of Rs 12,440. If the rate of interest is 8 % and the interest is calculated at the end of every month; find the time ( in months) of this recurring deposit account.

Ans: Installment per month (P) = Rs. 1,200

Number of months (n) = $n$

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

Maturity value of this account = 12,440

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=1,200\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{8}{100}$

$\Rightarrow \left( S.I. \right)~=4\times n\left( n+1 \right)$

$\Rightarrow \left( S.I. \right)~=Rs.~4{{n}^{2}}+4n$

Maturity value of this account:

= Total sum deposited + Simple interest

= Rs. (1,200 $\times $ n) + Rs. $4{{n}^{2}}+4n$

= Rs. ($4{{n}^{2}}+1,204n$)

According to question:

$\Rightarrow \left( 4{{n}^{2}}+1,204n \right)=12,440$

$\Rightarrow {{n}^{2}}+301n-3,110=0$

$\Rightarrow {{n}^{2}}+311n-10n-3,110=0$

$\Rightarrow n\left( n+311 \right)-10\left( n+311 \right)=0$

$\Rightarrow \left( n+311 \right)\left( n-10 \right)=0$

$\Rightarrow n~=-311~\And ~10$

Months can’t be negative so n = 10

Hence the time of this recurring deposit account is 10 months.

10. Mr. Gulati has a recurring deposit account of Rs 300 per month. If the rate of interest is 12 % and the maturity value of this account is Rs 8,100; find the time (in years) of this recurring deposit account.

Ans: Installment per month (P) = Rs. 300

Number of months (n) = $n$

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

Maturity value of this account = 8,100

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=300\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{12}{100}$

$\Rightarrow \left( S.I. \right)~=\dfrac{3}{2}\times n\left( n+1 \right)$

$\Rightarrow \left( S.I. \right)~=Rs.~\dfrac{3}{2}{{n}^{2}}+\dfrac{3}{2}n$

Maturity value of this account:

= Total sum deposited + Simple interest

= Rs. (300 $\times $ n) + Rs. $\dfrac{3}{2}{{n}^{2}}+\dfrac{3}{2}n$

= Rs. ($\dfrac{3}{2}{{n}^{2}}+\dfrac{603}{2}n$)

According to question:

$\Rightarrow \left( \dfrac{3}{2}{{n}^{2}}+\dfrac{603}{2}n \right)=8,100$

$\Rightarrow {{n}^{2}}+201n-5,400=0$

$\Rightarrow {{n}^{2}}+225n-24n-5,400=0$

$\Rightarrow n\left( n+225 \right)-24\left( n+225 \right)=0$

$\Rightarrow \left( n+225 \right)\left( n-24 \right)=0$

$\Rightarrow n~=-225~\And ~24$

Months can’t be negative so n = 24

Hence the time of this recurring deposit account is 24 months which is equal to 2 years.

11. Mr. Gupta opened a recurring deposit account in a bank. He deposited Rs 2,500 per month for two year. At the time of maturity he got Rs 67,500. Find:

i) the total interest earned by Mr. Gupta

Ans: Maturity value = Rs. 67,500

Money deposited = Rs. 2,500 $\times $ 24

= Rs. 60,000

Total interest = Rs. 67,500 - Rs. 60,000

Total interest = Rs. 7,500 

ii) the rate of interest per annum.

Ans: Installment per month (P) = Rs. 2,500

Number of months (n) = $2\times 12=24$

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

Total interest = Simple Interest = Rs. 7,500 

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)=2,500\times \dfrac{24~\left( ~24~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)=25\times 25\times r$

$\Rightarrow \left( S.I. \right)=Rs.~625r$

$\Rightarrow 625r=7,500$

$\Rightarrow r=12$

Hence the rate of interest per annum is 12 % p.a.

Exercise - 2(B)

1. Pramod deposits Rs 600 per month in a Recurring Deposit Account for 4 years. If the rate of interest is 8% per annum; calculate the maturity value of his account.

Ans: Installment per month (P) = Rs. 600

Number of months (n) = $4\times 12=48$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=600\times \dfrac{48~\left( ~48~+~1 \right)}{2~\times ~12}\times \dfrac{8}{100}$

$\Rightarrow \left( S.I. \right)~=6\times 2\times 49\times 8$

$\Rightarrow \left( S.I. \right)~=96\times 49$

$\Rightarrow \left( S.I. \right)~=4,704$

The amount that Pramod will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (600 $\times $ 48) + Rs. $4,704$

= Rs. 28,800 + Rs. $4,704$

= Rs. 33,504

Hence the maturity value of his account is Rs. 33,504.

2. Ritu has a Recurring Deposit Account in a bank and deposits Rs 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of this account is Rs 1,554.

Ans: Installment per month (P) = Rs. 80

Number of months (n) = $18$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=80\times \dfrac{18~\left( ~18~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=r\times 6\times 19\times \dfrac{1}{10}$

$\Rightarrow \left( S.I. \right)~=11.4r$

The amount that Ritu will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (80 $\times $ 18) + Rs. $11.4r$

= Rs. 1,440 + Rs. 11.4r

According to question:

$\Rightarrow 1,440~+~11.4r=1,554$

$\Rightarrow 11.4r=114$

$\Rightarrow r=10$

Hence the rate of interest paid by the bank is 10 % p.a.

3. The maturity value of a R.D. the account is Rs 16,176. If the monthly installment is Rs 400 and the rate of interest is 8 %, find the time (period) of this R.D. account.

Ans: Installment per month (P) = Rs. 400

Number of months (n) = $n$

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

Maturity value = Rs 16,176

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=400\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{8}{100}$

$\Rightarrow \left( S.I. \right)~=\dfrac{8}{6}\times n\left( n+1 \right)$

The maturity value = Total sum deposited + Simple interest

= Rs. (400 $\times $ n) + Rs. $\dfrac{8}{6}\times n\left( n+1 \right)$

= Rs. $\dfrac{8}{6}{{n}^{2}}+\dfrac{2408}{6}n$

According to question:

$\Rightarrow \dfrac{8}{6}{{n}^{2}}+\dfrac{2408}{6}n=16,176$

$\Rightarrow {{n}^{2}}+301n-12,132=0$

$\Rightarrow {{n}^{2}}+337n-36n-12,132=0$

$\Rightarrow n\left( n+337 \right)-36\left( n+337 \right)=0$

$\Rightarrow \left( n+337 \right)\left( n-36 \right)=0$

$\Rightarrow n~=-337~\And ~36$

Months can’t be negative so n = 36

Hence the time of this recurring deposit account is 36 months.

4. Mr. Bajaj needs Rs 30,000 after 2 years. What least money (in multiple of Rs 5) must he deposit every month in a recurring deposit account to get required money at the end of 2 years, rate of interest being 8 % p.a.?

Ans: Let Installment per month (P) = Rs. P

Number of months (n) = $2\times 12=24$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{24~\left( ~24~+~1 \right)}{2~\times ~12}\times \dfrac{8}{100}$

$\Rightarrow \left( S.I. \right)~=P\times 25\times 8\times \dfrac{1}{100}$

$\Rightarrow \left( S.I. \right)~=2P$

The amount that Mr. Bajaj will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (P $\times $ 24) + Rs. $2P$

= Rs. 26P

According to question:

$\Rightarrow 26P=30,000$

$\Rightarrow P=\dfrac{30,000}{26}$

$\Rightarrow P=1153.84$

P = 1155 (Multiple of 5)

5. Mr. Richard has had a recurring deposit account in a post office for 3 years at 7.5% p.a. simple interest. If he gets Rs 8,325 as interest at the time of maturity, find:

i) the monthly instalments.

Ans: Let Installment per month (P) = Rs. P

Number of months (n) = $3\times 12=36$

Rate of interest (r) = 7.5 % p.a.

i) Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow 8,325~=P\times \dfrac{36~\left( ~36~+~1 \right)}{2~\times ~12}\times \dfrac{7.5}{100}$

$\Rightarrow 8,325~=P\times 37\times 22.5\times \dfrac{1}{200}$

$\Rightarrow 8,325~=4.1625P$

$\Rightarrow P=2,000$

ii) the amount of maturity.

Ans: The amount that Mr. Richard will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (P $\times $ 36) + Rs. $4.1625P$

= Rs. $\left( 36\times 2,000+8,325 \right)$

= Rs. $\left( 72,000+8,325 \right)$

= Rs. $80,325$

6. Gopal has a cumulative deposit account and deposits Rs 900 per month for a period of 4 years. If he gets Rs 52,020 at the time of maturity, find the rate of interest.

Ans: Installment per month (P) = Rs. 900

Number of months (n) = $4\times 12=48$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=900\times \dfrac{48~\left( ~48~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=9\times 2\times 49\times r$

$\Rightarrow \left( S.I. \right)~=882r$

The amount that Gopal will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (900 $\times $ 48) + Rs. $882r$

= Rs. 43,200 + 882r

According to question:

$\Rightarrow 43,200~+~882r=52,020$

$\Rightarrow 882r=52,020-43,200$

$\Rightarrow 882r=8820$

$\Rightarrow r=10$

Hence the rate of interest is 10 % p.a.

7. Deepa has a 4-year recurring deposit account in a bank and deposits Rs 1,800 per month. If she gets Rs 1,08,450 at the time of maturity, find the rate of interest.

Ans: Installment per month (P) = Rs. 1,800

Number of months (n) = $4\times 12=48$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=1,800\times \dfrac{48~\left( ~48~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=18\times 2\times 49\times r$

$\Rightarrow \left( S.I. \right)~=1,764r$

The amount that Deepa will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (1,800 $\times $ 48) + Rs. $1,764r$

= Rs. 86,400 + $1,764r$

According to question:

$\Rightarrow 86,400~+~1,764r=1,08,450$

$\Rightarrow 1,764r=1,08,450-86,400$

$\Rightarrow 1,764r=22,050$

$\Rightarrow r=\dfrac{22,050}{1,764}$

$\Rightarrow r=12.5$

Hence the rate of interest is 12.5 % p.a.

8. 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 Rs 8,088 from the bank after 3 years, find the value of his monthly installment.

Ans: Let Installment per month (P) = Rs. P

Number of months (n) = $3\times 12=36$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{36~\left( ~36~+~1 \right)}{2~\times ~12}\times \dfrac{8}{100}$

$\Rightarrow \left( S.I. \right)=P\times 37\times 12\times \dfrac{1}{100}$

$\Rightarrow \left( S.I. \right)~=4.44P$

The amount that Mr. Britto will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (P $\times $ 36) + Rs. $4.44P$

= Rs. $\left( 40.44P \right)$

According to question:

$\Rightarrow 40.44P=8,088$

$\Rightarrow P=\dfrac{8,088}{40.44}$

$\Rightarrow P=200$

Thus, the value of his monthly installment is Rs. 200.

9. Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs 800 per month for 1$\dfrac{1}{2}$years. If he received Rs 15,084 at the time of maturity, find the rate of interest per annum.

Ans: Installment per month (P) = Rs. 800

Number of months (n) = $1.5\times 12=18$

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

Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=800\times \dfrac{18~\left( ~18~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=8\times \dfrac{3}{4}\times 19\times r$

$\Rightarrow \left( S.I. \right)~=114r$

The amount that Shahrukh will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (800 $\times $ 18) + Rs. $114r$

= Rs. 14,400 + $114r$

According to question:

$\Rightarrow 14,400~+~114r=15,084$

$\Rightarrow 114r=15,084-14,400$

$\Rightarrow 114r=684$

$\Rightarrow r=\dfrac{684}{114}$

$\Rightarrow r=6$

Hence the rate of interest is 6 % p.a.

10. Katrina opened a Recurring Deposit Account in a Nationalised bank for a period of 2 years. If the bank pays interest at the rate of 6 % per annum and the monthly installment is Rs 1,000, find the:

i) interest earned in two years

Ans: Installment per month (P) = Rs. 1,000

Number of months (n) = $2\times 12=24$

Rate of interest (r) = 6 % p.a.

i) Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=1,000\times \dfrac{24~\left( ~24~+~1 \right)}{2~\times ~12}\times \dfrac{6}{100}$

$\Rightarrow \left( S.I. \right)~=10\times 25\times 6$

$\Rightarrow \left( S.I. \right)~=1,500$

Hence the interest earned in two years is Rs 1,500.

ii) maturity value.

Ans: The amount that Katrina will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (1,000 $\times $ 24) + Rs. $1,500$

= Rs. 24,000 + $1,500$

= Rs. 25,500

Hence the maturity value is Rs. 25,500.

11. Mohan has a Recurring Deposit Account in a bank for 2 years at 6 % p. a. simple interest. If he gets Rs. 1,200 as interest at the time of maturity, find:

i) the monthly installment.

Ans: Let Installment per month (P) = Rs. P

Number of months (n) = $2\times 12=24$

Rate of interest (r) = 6 % p.a.

i) Simple Interest $\left( S.I. \right)~=P\times \dfrac{n~\left( ~n~+~1 \right)}{2~\times ~12}\times \dfrac{r}{100}$

$\Rightarrow \left( S.I. \right)~=P\times \dfrac{24~\left( ~24~+~1 \right)}{2~\times ~12}\times \dfrac{6}{100}$

$\Rightarrow \left( S.I. \right)~=P\times 25\times 0.06$

$\Rightarrow \left( S.I. \right)~=1.5P$

$\Rightarrow 1,200=1.5P$

$\Rightarrow P=800$

Hence the monthly installment is Rs 800.

ii) the amount of maturity.

Ans: The amount that Mohan will get at the time of maturity

= Total sum deposited + Simple interest

= Rs. (P $\times $ 24) + Rs. $1,200$

= Rs. $800\times 24$+ $1,200$

= Rs. 19,200+1,200

= Rs. 20,400

Hence the amount of maturity is Rs. 20,400.

PDF of Concise Mathematics For Class 10 Chapter 2 - Banking Recurring Deposits ICSE Solutions

Under Chapter 2 of Concise Selina for Class 10 Math, there are two Exercises included. However, problems based on calculating the maturity value of a Recurring Deposit (RD) account are the key focus in these Exercises. For solid Exam preparations, students can also have the access to the Concise Solutions for Class 10 Math Chapter 2 Banking (Recurring Deposit Accounts) PDF from the links provided at Vedantu. All the questions of Selina ICSE Solutions for Class 10 are answered and explained by Math maestros at Vedantu keeping into consideration the ICSE Board guidelines.

Download PDF For Concise Selina Solutions Math Banking (Recurring Deposit Accounts)

Selina Concise Math Solutions for Class 10 is ideal for preparing for the annual school Examinations. ICSE Class 10 syllabus for recurring deposit is not that vast but requires concentrated efforts from students to do well in Examinations. The Selina Concise Math Class 10 ICSE solutions PDF includes detailed information, fundamentals, and solved problems of recurring deposits from the subject. Each question is explained along with the assumptions and logic that are used to conclude.

Key Focus Of Concise Selina Solutions Class 10 Chapter 2 Math Banking (Recurring Deposit Accounts)

The provided solutions in the Selina textbook by Vedantu are as per the latest 2024-25 Concise Selina textbook. Check below, what you will learn from Vedantu’s Selina Concise Solutions Class 10 Chapter 2 - Banking Recurring Deposits:

1. The computing of maturity value.

2. Calculating the rate of interest (yearly and monthly).

3. Computing number of installments (yearly and monthly).

Concise Selina Solutions Class 10 Chapter 2 Math Banking (Recurring Deposit Accounts) ICSE Related Links

In this section, you will find all the PDF links by Vedantu for Class-wise and Chapter-wise problems. For this particular Chapter 2 Math Recurring Deposit Accounts, you will have access to:

1. ICSE Solutions Answers of Math Selina for Class 10 Chapter 2 - Recurring Deposit Accounts.

2. ICSE Solutions Class 10 Chapter 2 Maturity Value solved Problems.

3. Concise Mathematics Class 10 Chapter 2 and its Exercises ICSE Solutions PDF.

4. Concise Mathematics Class 10 Chapter 2 and Exercises ICSE Guide PDF.

5. ICSE Mathematics Class 6 to Class 10 Selina Solutions PDF Download.

6. ICSE Standard 6 to 10 Question Papers.

7. ICSE Standard 10 Chapter 2 Solved Question Papers on Maturity Value, Rate Of Interest and Installments.

Vedantu Selina Solutions ICSE Class 10 Chapter 2 Math Banking (Recurring Deposit Accounts) Book Descriptions

Vedantu has tailor-made solutions for you to easily find PDF ebooks without any struggle. By having access to these Selina solutions and online ebooks for Mathematics and science subjects, you will be able to have the right and convenient answers. You can also prepare with Selina Concise Mathematics For ICSE Solution anytime and anywhere by storing it in your laptop.

To get started with the Concise Mathematics Solutions For ICSE Math, you are just a click away from Vedantu which has a comprehensive collection of study materials listed. Vedantu e-book library is known to stack up hundreds of thousands of different study materials.

There are many books to allude to for Class 10 Mathematics ICSE. But skimming through all will be difficult as there are many subjects to deal with during your Boards. It would be extremely dreary and lumbering to get a handle on through a mix of books. Subsequently, Vedantu recommends the fundamental and proposed book that would conceal all ideas and activities according to the Class 10 ICSE question design, which is, Selina Concise. Assuming that students practice all question types from Selina Concise Class 10 Mathematics, then, at that point, it would be a piece of cake to acquire good marks. Vedantu subsequently accompanies Solutions for Class 10 Mathematics ICSE.

Advantages of Vedantu Solutions of Chapter 2 - Banking (Recurring Deposit Accounts) for Class 10 Mathematics ICSE:

With helpful solutions by Vedantu, students don't need to lose the speed of answering questions. Suppose you are stuck on a question and for the answer, you need to skim through the pages of your course books to comprehend the idea and afterwards sort out the solution, which would break the stream. In addition to the fact that it causes interference, it is tedious. Furthermore, when you have tests on your head, losing an iota of time can cause losing marks.

Tackling Mathematics questions is a bit-by-bit process and sticking over the problem is probably going to occur. Also, as an inquisitive student, you should know where you committed the error. Going through a weighty course reading isn't just a time taking process, however, you may not see precisely which place you committed the error. With Vedantu it is extremely straightforward. With Vedantu solutions of Class 10, you can see each progression minutely.

Vedantu comprehends the predicament of students when they are hindered because of a question and subsequently it extravagantly clarifies each question with satisfactory proclamations and theory so a student doesn't lose marks in view of ICSE step marking. Selina Concise is viewed as the base book for the ICSE Board planning and Vedantu assists students with endeavoring better.

Question Pattern:

The ICSE Mathematics for Class 10 paper will be of over 2 and a 1/2 hours term carrying a total of 80 marks and an inner assessment of 20 marks. The test paper of Class 10 ICSE Mathematics will be separated into two segments, 1 and 2.

Section I contains an aggregate of 40 marks,

Section  II contains an aggregate of 40 marks.

Section I will have more short answer questions and in Section II, students need to attend any four out of seven questions.

A significant segment of the test paper will have Banking (Recurring Deposit Accounts) questions and students can rehearse better with Selina section  Banking (Recurring Deposit Accounts) solutions by Vedantu for Class 10.

Exercising different sorts of questions is a vital method for planning for the test. Vedantu gives idea-savvy learning and comprehension through arrangement books. The arrangement of a question will require information to more than one section of the entire schedule. What's more, students should have this answer for part 1  Banking (Recurring Deposit Accounts) of Selina Concise helpful. With Vedantu dominating in the Class 10 ICSE with a decent rate opens approaches to new opportunities. Students get the chance to choose streams eagerly as indicated by their benefit.