[3 marks sum]

Question 13 Marks

Mr Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. It the rate of interest is of 8% per annum and Mr Britto gets Rs. 8088 from the bank after 3 years, find the value of his monthly instalment.

Answer

Let the value of monthly instalment be Rs. X.
Maturity value (M.V.) = Rs. 8088
Period = n = 3 years = 36 months
Rate of interest $= r = 8\%$ p.a.
$\text { S.I }=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$
$=X \times \frac{36 \times 37}{24} \times \frac{8}{100}$
$=4.44 X $
Total amount of maturity $=36 X +4.44 X$
$\Rightarrow 8088=40.44 X$
$\Rightarrow X=200$
Thus, the value of monthly installment is Rs. 200.

View full question & answer→

Question 23 Marks

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

Answer

Installment per month(P) = Rs $900$
Number of months(n) $= 48$
Let rate of interest(r) $= r\%$ p.a.
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$=900 \times \frac{48(48+1)}{2 \times 12} \times \frac{ r }{100}$
$=900 \times \frac{2352}{24} \times \frac{ r }{100}= Rs 882 r$
Maturity value $= Rs (900 \times 48) + Rs (882)r$
Given maturity value $= Rs.52,020$
Then $Rs (900 \times 48) + Rs.(882)r = Rs.52,020$
$\Rightarrow 882 r=\operatorname{Rs} 52,020-\operatorname{Rs} 43,200$
$\Rightarrow r=\frac{8820}{882}=10 \%$

View full question & answer→

Question 33 Marks

Mr. Richard has 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:

the monthly income
the amount of maturity

Answer

Let the monthly deposit be P
Interest $= Rs. 8,325|$
Rate of interest $= 7.5\%$
Time $= 3$ years $= 36$ months
$\text { (i) Interest }= P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$\Rightarrow 8325= P \times \frac{36(36+1)}{2 \times 12} \times \frac{7.5}{100}$
$\Rightarrow P =\text { Rs. } 2,000$
$\text { (ii) Maturity value }=\text { total sum deposited }+ \text { interest }$
$=2000 \times 36+8325$
$=72000+8325$
$=\text { Rs } 80,325$

View full question & answer→

Question 43 Marks

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

Answer

Let installment per month = Rs P
Number of months(n) = 24
Rate of interest = 8% p.a
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$= P \times \frac{24(24+1)}{2 \times 12} \times \frac{8}{100}$
$= P x \times 600 / 24 x \times 8 / 100= Rs 2 P $
$\text { Maturity value }=\operatorname{Rs}(P \times 24)+\operatorname{Rs} 2 P=R s 26 P$
Given maturity value $=$ Rs 30,000
Then $26 P =$ Rs 30000
$\Rightarrow P= $ Rs $\frac{30000}{26}=$ Rs $1153.84=$ Rs 1155 (multiple of 5)

View full question & answer→

Question 53 Marks

The maturity value of an R.D. Account is $Rs.16,176.$ If the monthly instalment is $Rs.400$ and the rate of interest is $8\%$; find the time (period) of this R.D Account.

Answer

Installment per month(P) $= Rs.400$
Number of months(n) $= n$
Let rate of interest(r) $= 8\%$ p.a.
$\therefore \text { S.I }= P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100} $
$=400 \times \frac{ n ( n +1)}{24} \times \frac{8}{100}=\operatorname{Rs} \frac{4 n ( n +1)}{3}$
$\text { Maturity value }=\operatorname{Rs}(400 \times n )+\operatorname{Rs} \frac{4 n ( n +1)}{3}$
Given maturity value $= Rs.16176$
Then Rs $(400 \times n)+\operatorname{Rs} \frac{4 n(n+1)}{3}=\operatorname{Rs~} 16.176$
$\Rightarrow 1200 n+4 n^2+4 n=R s 48,528$
$\Rightarrow 4 n^2+1204 n=R s 48,528 $
$\Rightarrow n^2+301 n-12132=0 $
$ \Rightarrow(n+337)(n-36)=0 $
$\Rightarrow n=-337 \text { or } n=36$
Then number of months $= 36$ months $= 3$ years

View full question & answer→

Question 63 Marks

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 the account is $Rs.1,554.$

Answer

Installment per month(P) $= Rs.80$
Number of months(n) $= 18$
Let rate of interest(r) $= r\%$ p.a
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$=80 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}$
$=80 \times \frac{342}{24} \times \frac{r}{100}= Rs.11.4 r$
Maturity value $= Rs (80 \times 18) + Rs (11.4r)$
Given maturity value $= Rs.1,554$
Then $Rs (80 \times 18 ) + Rs (11.4r) = Rs.1,554$
$\Rightarrow 11.4 r = Rs 1554- Rs 1440 $
$ \Rightarrow r =\frac{114}{11.4}=10 \%$

View full question & answer→

Question 73 Marks

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs. 1200 as interest at the time of maturity find:
(i) the monthly instalment
(ii) the amount of maturity.

Answer

Interest = Rs 1200
Period(n) = 2 years = 24 months
Rate(r) = 6% p.a
Let monthly deposit = ₹P p.m.
∴ Interest
$=\frac{ P \times n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$1200=\frac{ P \times 24 \times 25}{24} \times \frac{6}{100}$
$1200=\frac{6}{4} p$
$\therefore P=\frac{1200 \times 4}{6}=800$
∴ Monthly deposit = ₹800
and maturity value = P x n + Interest
= ₹800 x 24 + ₹1200
= ₹19200 + ₹1200
= ₹20400.

View full question & answer→

Question 83 Marks

Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at the rate 6% per annum and the monthly instalment is Rs. 1,000, find the:

1) Interest earned in 2 years.

2) Matured value

Answer

Given,
P = Rs. 1000
n = 2 years = 24 months
r = 6%
1) Interest $=P \times \frac{n(n+1)}{2} \times \frac{r}{12 \times 100}$
$
=1000 \times \frac{24 \times 25}{2} \times \frac{6}{12 \times 100}
$
$
=1500
$
Thus, the interest earned in 2 years is Rs. 1500.
2) Sum deposited in two years 24 x 1000 = 24,000
Maturity value = Total sum deposited in two years + Interest
= 24000 + 1500
= 25500
Thus, the maturity value is Rs. 25,500.

View full question & answer→

Question 93 Marks

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.

Answer

Installment per month(P) = Rs 350
Number of months(n) = 15
Let rate of interest(r)= r% p.a.
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$=350 \times \frac{15(15+1)}{2 \times 12} \times \frac{ r }{100}$
$=350 \times \frac{240}{24} \times \frac{r}{100}=\operatorname{Rs} 35$
Maturity value= Rs (350 × 15) + Rs (35)r
Given maturity value = Rs 5,565
Then Rs (350 × 15) + Rs (35)r = Rs 5,565
$\Rightarrow 35 r=\operatorname{Rs} 5,565-\text { Rs } 5,250$
$\Rightarrow r=\frac{315}{35}=9 \%$

View full question & answer→

Question 103 Marks

David opened a Recurring Deposit Account in a bank and deposited Rs 300 per month for two years. If he received Rs 7,725 at the time of maturity, find the rate of interest per annum.

Answer

Instalment per month(P) = Rs 300
Number of months(n) = 24
Let rate of interest(r)= r% p.a
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$=300 \times \frac{24(24+1)}{2 \times 12} \times \frac{ r }{100}$
$=300 \times \frac{600}{24} \times \frac{ r }{100}=\operatorname{Rs}(75) r $
Maturity value = Rs (300 × 24) + Rs (75)r
Given maturity value = Rs 7,725
Then ₹ (300 × 24) + Rs (75)r = Rs 7,725
$\Rightarrow 75 r =\operatorname{Rs} 7,725-\operatorname{Rs} 7,200$
$\Rightarrow r =\frac{525}{75}=7 \%$

View full question & answer→

Question 113 Marks

Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits $Rs.140$ per month for $4$ years. If he gets $Rs.8,092$ on maturity, find the rate of interest given by the bank.

Answer

Installment per month(P) $= Rs.140$
Number of months(n) $= 48$
Let rate of interest(r) $= r\%$ p.a.
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$=140 \times \frac{48(48+1)}{2 \times 12} \times \frac{ r }{100}$
$=140 \times \frac{2352}{24} \times \frac{ r }{100}=\operatorname{Rs} 137.20$
Maturity value $= Rs (140 \times 48) + Rs (137.20)r$
Given maturity value $= Rs.8,092$
Then $Rs (140 \times 48) + Rs (137.20)r = Rs 8,092$
$\Rightarrow 137.20 r=\text { Rs } 8,092-\text { Rs } 6,720$
$ \Rightarrow r=\frac{1372}{137.20}=10 \%$

View full question & answer→

Question 123 Marks

A man has a Recurring Deposit Account in a bank for 3 1/2 years. If the rate of interest is 12% per annum and the man gets Rs 10,206 on maturity, find the value of monthly instalments.

Answer

Let Installment per month(P) = Rs y
Number of months(n) = 42
Rate of interest(r) = 12% p.a.
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$= y \times \frac{42(42+1)}{2 \times 12} \times \frac{12}{100}$
$= y \times \frac{1806}{24} \times \frac{12}{100}=\text { Rs } 9.03 y $
Maturity value= Rs (y × 42) + Rs 9.03y = Rs 51.03y
Given maturity value = Rs 10,206
Then Rs 51.03y = Rs 10206
$\Rightarrow y =\frac{10206}{51.03}=\operatorname{Rs} 200$

View full question & answer→

Question 133 Marks

Ashish deposits a certain sum of money every month is 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 ₹ $12,715$ as the maturity value of this account, what sum of money did money did he pay every month?

Answer

Let Installment per month $(P) = Rs. y$
Number of months $(n) = 12$
Rate of interest $(r) = 11\%$ p.a
So,
$\therefore S . I = P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}$
$= y \times \frac{12(12+1)}{2 \times 12} \times \frac{11}{100}$
$= y \times \frac{156}{24} \times \frac{11}{100}=\text { Rs } 0.715 y $
Hence,
Maturity value $= Rs. (y \times 12) + Rs. 0.715y = Rs. 12.715y$
Given maturity value $= Rs. 12,715$
So, on equating we have
$Rs. 12.715y = Rs.12,715$
$\Rightarrow y =\frac{12715}{12.715}= Rs.1000$
Therefore, the sum of money Ashish paid every month was $Rs. 1000.$

View full question & answer→

Mathematics [3 marks sum]

Test yourself on this topic