Rd Sharma 2020 2021 for Class 7 Maths Chapter 13 - Simple Interest

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Page No 13.7:

Question 1:

Find the simple interest, when:
(i) Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.
(ii) Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.
(iii) Principal = Rs 4500, Rate of Interest = 4% per annum and Time = months.
(iv) Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months.
(v) Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days.

Answer:

(i) Principal (P) = Rs 2000
Rate of interest (R) = 5% p.a.
Time (T) = 5 years
Simple interest = $\frac{P \times R \times T}{100} = \frac{2000 \times 5 \times 5}{100} = Rs 500$

(ii) Principal (P) = Rs 500
Rate of interest (R) = 12.5% p.a.
Time (T) = 4 years
Simple interest = $\frac{P \times R \times T}{100} = \frac{500 \times 12.5 \times 4}{100} = Rs 250$

(iii) Principal (P) = Rs 4500
Rate of interest (R) = 4% p.a.
Time (T) = 6 months

T = $\frac{6}{12} = \frac{1}{2}$ year (1 year = 12 months)
Simple interest = $\frac{P \times R \times T}{100} = \frac{4500 \times 4 \times \frac{1}{2}}{100} = \frac{4500 \times 4 \times 1}{100 \times 2} = Rs 90$

(iv) Principal (P) = Rs 12000
Rate of interest (R) = 18% p.a.
$Time (T) = 4 months = \frac{4}{12} = \frac{1}{3} y ear$ (1 year = 12 months)

Simple interest = $\frac{P \times R \times T}{100} = \frac{12000 \times 18 \times 1}{100 \times 3} = Rs 720$

(v) Principal (P) = Rs 1000
Rate of interest (R) = 10% p.a.
$Time (T) = 73 days = \frac{73}{365} year$ (1 year = 365 days)

Simple interest = $\frac{P \times R \times T}{100} = \frac{1000 \times 10 \times 73}{100 \times 365} = Rs 20$

Page No 13.8:

Question 2:

Find the interest on Rs 500 for a period of 4 years at the rate of 8% per annum. Also, find the amount to be paid at the end of the period.

Answer:

Principal amount (P) = Rs 500
Time period (T) = 4 years
Rate of interest (R) = 8% p.a.

Interest = $\frac{P \times R \times T}{100} = \frac{500 \times 4 \times 8}{100} = Rs 160$

Total amount paid = Principal amount + Interest = Rs 500 + 160
= Rs 660

Page No 13.8:

Question 3:

A sum of Rs 400 is lent at the rate of 5% per annum. Find the interest at the end of 2 years.

Answer:

Principal amount (P) = Rs 400
Time period (T) = 2 years
Rate of interest (R) = 5% p.a.

Interest paid after 2 years $= \frac{P \times R \times T}{100} = \frac{400 \times 5 \times 2}{100} = Rs 40$

Page No 13.8:

Question 4:

A sum of Rs 400 is lent for 3 years at the rate of 6% per annum. Find the interest.

Answer:

Principal amount (P) = Rs 400
Time period (T) = 3 years
Rate of interest (R) = 6% p.a.
Interest after 3 years = $\frac{P \times R \times T}{100} = \frac{400 \times 6 \times 3}{100}$ = Rs 72

Page No 13.8:

Question 5:

A person deposits Rs 25000 in a firm who pays an interest at the rate of 20% per annum. Calculate the income he gets from it annually.

Answer:

Principal amount (P) = Rs 25000
Time period (T) = 1 year
Rate of interest (R) = 20% p.a.
Annual interest = $\frac{P \times R \times T}{100} = \frac{25000 \times 20 \times 1}{100}$ = Rs 5000

Page No 13.8:

Question 6:

A man borrowed Rs 8000 from a bank at 8% per annum. Find the amount he has to pay after $4 \frac{1}{2}$ years.

Answer:

Principal amount (P) = Rs 8000
Time period (T) = $4 \frac{1}{2} = \frac{9}{2} years$
Rate of interest (R) = 8% p.a.
Interest = $\frac{P \times R \times T}{100} = \frac{8000 \times 8 \times 9}{100 \times 2}$ = Rs 2880

Total amount paid after $4 \frac{1}{2}$ years = Principal amount + Interest = Rs 8000 + Rs 2880
= Rs 10880

Page No 13.8:

Question 7:

Rakesh lent out Rs 8000 for 5 years at 15% per annum and borrowed Rs 6000 for 3 years at 12% per annum. How much did he gain or lose?

Answer:

Principal amount lent out by Rakesh (P) = Rs 8000
Time period (T) = 5 years
Rate of interest (R) = 15% p.a.
Interest = $\frac{P \times R \times T}{100} = \frac{8000 \times 15 \times 5}{100}$ = Rs 6000

Principal amount borrowed by Rakesh (P) = Rs 6000
Time period (T) = 3 years
Rate of interest (R) = 12% p.a.
Interest = $\frac{P \times R \times T}{100} = \frac{6000 \times 12 \times 3}{100}$ = Rs 2160

Amount gained by Rakesh = Rs 6000 − Rs 2160 = Rs 3840

Page No 13.8:

Question 8:

Anita deposits Rs 1000 in a savings bank account. The bank pays interest at the rate of 5% per annum. What amount can Anita get after one year?

Answer:

Principal amount (P) = Rs 1000
Time period (T) = 1 year
Rate of interest (R) = 5% p.a.

Interest = $\frac{P \times R \times T}{100} = \frac{1000 \times 5 \times 1}{100}$ = Rs 50

Total amount paid after 1 year = Principal amount + Interest = Rs 1000 + Rs 50
= Rs 1050

Page No 13.8:

Question 9:

Nalini borrowed Rs 550 from her friend at 8% per annum. She returned the amount after 6 months. How much did she pay?

Answer:

Principal amount (P) = Rs 550
Time period (T) = 6 months = $\frac{6}{12} = \frac{1}{2}$ year (1 year = 12 months)
Rate of interest (R) = 8% p.a.

Interest = $\frac{P \times R \times T}{100} = \frac{550 \times 8 \times 1}{100 \times 2} = Rs 22$

Total amount paid after 6 months = Principal amount + Interest = Rs 550 + Rs 22
= Rs 572

Page No 13.8:

Question 10:

Rohit borowed Rs 600000 from a bank at 9% per annum for 2 years. He lent this sum of money to Rohan at 10% per annum for 2 years. How much did Rohit earn from this transaction?

Answer:

Principal amount lent out by Rohit (P) = Rs. 60000
Time period (T)        = 2 years
Rate of interest (R) = 10% p.a.

Interest = $\frac{P \times R \times T}{100} = Rs. \frac{60000 \times 10 \times 2}{100} = Rs. 12000$

Principal amount borrowed by Rohit from the bank (P) = Rs. 60000
Time period (T)                                         = 2 years
Rate of interest (R)                                   = 9% p.a.
Interest = $\frac{P \times R \times T}{100} = Rs. \frac{60000 \times 9 \times 2}{100} = Rs. 10800$

Amount gained by Rohit = Rs. 12000 - 10800 = Rs. 1200

Page No 13.8:

Question 11:

Romesh borrowed Rs 2000 at 2% per annum and Rs 1000 at 5% per annum. He cleared his debt after 2 years by giving Rs 2800 and a watch. What is the cost of the watch?

Answer:

Principal amount borrowed by Romesh (P) = Rs. 2000
Time period (T) = 2 years
Rate of interest (R) = 2% p.a.

Interest = $\frac{P \times R \times T}{100} = Rs . \frac{2000 \times 2 \times 2}{100} = Rs . 80$

Principal amount borrowed by Romesh (P) = Rs. 1000
Time period (T) = 2 years
Rate of interest (R) = 5% p.a.

Interest = $\frac{P \times R \times T}{100} = Rs . \frac{1000 \times 5 \times 2}{100} = Rs . 100$

Total amount that he will have to return = Rs. 2000 + 1000 + 80 + 100 = Rs. 3180

Amount repaid = Rs. 2800
Value of the watch = Rs. 3180 - 2800 = Rs. 380

Page No 13.8:

Question 12:

Mr Garg lent Rs 15000 to his friend. He charged 15% per annum on Rs 12500 and 18% on the rest. How much interest does he earn in 3 years?

Answer:

Principal amount (P) = Rs 12500
Time period (T) = 3 years
Rate of interest (R) = 15% p.a.

Interest = $\frac{P \times R \times T}{100} = \frac{12500 \times 15 \times 3}{100}$ = Rs 5625

Rest of the amount lent = Rs 15000 − Rs 12500 = Rs 2500
Rate of interest = 18 % p.a.
Time period = 3 years

Interest = $\frac{P \times R \times T}{100} = \frac{2500 \times 18 \times 3}{100}$ = Rs 1350

Total interest earned = Rs 5625 + Rs 1350 = Rs 6975

Page No 13.8:

Question 13:

Shikha deposited Rs 2000 in a bank which pays 6% simple interest. She withdrew Rs 700 at the end of first year. What will be her balance after 3 years?

Answer:

Principal amount deposited (P) = Rs 2000
Time period (T) = 1 year
Rate of interest (R) = 6% p.a.
Interest after 1 year = $\frac{P \times R \times T}{100} = \frac{2000 \times 6 \times 1}{100} = Rs 120$
So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120
After 1 year, amount withdrawn = Rs 700
Principal amount left (P₁) = Rs 2120 − Rs 700 = Rs 1420
Time period (T) = 2 years
Rate of interest (R) = 6% p.a.
Interest after 2 years = $\frac{P_{1} \times R \times T}{100} = \frac{1420 \times 6 \times 2}{100} = Rs 170.40$

Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40

Page No 13.8:

Question 14:

Reema took a loan of Rs 8000 from a money lender, who charged interest at the rate of 18% per annum. After 2 years, Reema paid him Rs 10400 and wrist watch to clear the debt. What is the price of the watch?

Answer:

Principal amount (P) = Rs 8,000
Rate of interest (R) = 18%
Time period (T) = 2 years
Interest after 2 years = $\frac{P \times R \times T}{100} = \frac{8000 \times 18 \times 2}{100}$ = Rs 2,880
Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880 = Rs 10,880
Amount paid = Rs 10,400
Value of the watch = Rs 10,880 − Rs 10,400 = Rs 480

Page No 13.8:

Question 15:

Mr Sharma deposited Rs 20000 as a fixed deposit in a bank at 10% per annual. If 30% is deducted as income tax on the interest earned, find his annual income.

Answer:

Amount deposit (P) = Rs 20,000
Rate of interest (R) = 10% p.a.
Time period (T) = 1 year

Interest after 1 year = $\frac{P \times R \times T}{100} = \frac{20000 \times 10 \times 1}{100}$ = Rs 2,000

Amount deducted as income tax = $30 % of Rs 2, 000 = \frac{30 \times 2000}{100} = Rs 600$

Annual interest after tax deduction = Rs 2,000 − Rs 600 = Rs 1,400

Page No 13.8:

Question 1:

If the simple interest on a certain sum for 2 years at the rate of 5% per annum is ₹4000, then the sum is

(a) ₹46,000
(b) ₹44,000
(c) ₹40,000
(d) ₹48,000

Answer:

We know, $I = \frac{P \times T \times R}{100}$

It is given that,
T = 2 years
R = 5%
I = ₹4000

Then,
$4000 = \frac{P \times 5 \times 2}{100} \Rightarrow 4000 = \frac{10 P}{100} \Rightarrow P = 40000$

Thus, P = ₹40,000

Hence, the correct option is (c).

Page No 13.9:

Question 2:

In how many years will a certain sum become 3 times itself at 25% per annum under simple interest?

(a) 5
(b) 8
(c) 12
(d) 6

Answer:

Amount = 3 times the sum = 3P

Simple interest (I) = Amount − Sum = 3P − P = 2P

Let the sum (P) be x.
Then, simple interest (I) = 2x
Rate (R) = 25%
Time = T

$I = \frac{P \times R \times T}{100} \Rightarrow T = \frac{100 \times I}{P \times R} = \frac{100 \times 2 x}{x \times 25} = 4 \times 2 = 8 years$

Hence, the correct option is (b).

Page No 13.9:

Question 3:

The amount on ₹25,000 at 8% per annum for 6 years under simple interest is

(a) ₹35,000
(b) ₹37,000
(c) ₹45,000
(d) ₹47,000

Answer:

It is given that,
Sum (P) = ₹25,000
Rate (R) = 8%
Time (T) = 6 years

$I = \frac{P \times R \times T}{100} = \frac{25000 \times 8 \times 6}{100} = 12000$

Therefore, simple interest (I) = ₹12,000

Now, Amount = P + I = ₹25,000 + ₹12,000 = ₹37,000

Hence, the correct option is (b).

Page No 13.9:

Question 4:

The simple interest for ₹1500 at 8% per annum for 3 years is

(a) ₹400
(b) ₹360
(c) ₹450
(d) ₹500

Answer:

It is given that,
Sum (P) = ₹1500
Rate (R) = 8%
Time (T) = 3 years

$I = \frac{P \times R \times T}{100} = \frac{1500 \times 8 \times 3}{100} = 360$

Therefore, simple interest (I) = ₹360

Hence, the correct option is (b).

Page No 13.9:

Question 5:

The difference between the interest obtained for ₹1000 at 12% per annum for 3 years and that for ₹1500 at 8% per annum for $1 \frac{1}{2}$ years is

(a) ₹360
(b) ₹300
(c) ₹180
(d) ₹200

Answer:

It is given that,
Sum (P₁) = ₹1000
Rate (R₁) = 12%
Time (T₁) = 3 years

$I_{1} = \frac{P_{1} \times R_{1} \times T_{1}}{100} = \frac{1000 \times 12 \times 3}{100} = 360 . . . . (1)$

Sum (P₂) = ₹1500
Rate (R₂) = 8%
Time (T₂) = $1 \frac{1}{2}$ years = $\frac{3}{2}$ years

$I_{2} = \frac{P_{2} \times R_{2} \times T_{2}}{100} = \frac{1500 \times 8 \times 3}{100 \times 2} = 180 . . . . (2)$

Subtracting (2) from (1), we get
$I_{2} - I_{1} = 360 - 180 = 180$

Hence, the correct option is (c).

Page No 13.9:

Question 6:

Which of the following yields maximum interest for 2 years?

(a) ₹1500 at 8% per annum
(b) ₹1000 at 11% per annum
(c) ₹2000 at 5% per annum
(d) ₹900 at 20% per annum

Answer:

(a) It is given that,
Sum (P₁) = ₹1500
Rate (R₁) = 8%
Time (T₁) = 2 years

$I_{1} = \frac{P_{1} \times R_{1} \times T_{1}}{100} = \frac{1500 \times 8 \times 2}{100} = 240 . . . . (1)$

(b) It is given that,
Sum (P₂) = ₹1000
Rate (R₂) = 11%
Time (T₂) = 2 years

$I_{2} = \frac{P_{2} \times R_{2} \times T_{2}}{100} = \frac{1000 \times 11 \times 2}{100} = 220 . . . . (2)$

(c) It is given that,
Sum (P₃) = ₹2000
Rate (R₃) = 5%
Time (T₃) = 2 years

$I_{3} = \frac{P_{3} \times R_{3} \times T_{3}}{100} = \frac{2000 \times 5 \times 2}{100} = 200 . . . . (3)$

(d) It is given that,
Sum (P₄) = ₹900
Rate (R₄) = 20%
Time (T₄) = 2 years

$I_{4} = \frac{P_{4} \times R_{4} \times T_{4}}{100} = \frac{900 \times 20 \times 2}{100} = 360 . . . . (4)$

From (1), (2), (3) and (4),
₹900 at 20% per annum yields maximum interest for 2 years.

Hence, the correct option is (d).

Page No 13.9:

Question 7:

If a sum of ₹3000 is lent out at 3% per annum for 20 years under simple interest, then the amount at the end of 20^th year is

(a) ₹1800
(b) ₹1080
(c) ₹3600
(d) ₹4800

Answer:

It is given that,
Sum (P) = ₹3000
Rate (R) = 3%
Time (T) = 20 years

$I = \frac{P \times R \times T}{100} = \frac{3000 \times 3 \times 20}{100} = 1800$

Amount = I + P = ₹1800 + ₹3000 = ₹4800

Hence, the correct option is (d).

Page No 13.9:

Question 8:

If a sum of ₹2000 is lent out at 2% per annum for 10 years under simple interest, then the amount is

(a) ₹1400
(b) ₹2400
(c) ₹200
(d) ₹1500

Answer:

It is given that,
Sum (P) = ₹2000
Rate (R) = 2%
Time (T) = 10 years

$I = \frac{P \times R \times T}{100} = \frac{2000 \times 2 \times 10}{100} = 400$

Amount = I + P = ₹400 + ₹2000 = ₹2400

Hence, the correct option is (b).

Page No 13.9:

Question 9:

If interest on ₹x for 2 years at R% per annum is ₹80, the interest on ₹2x for one year at R% per annum is

(a) ₹160
(b) ₹40
(c) ₹80
(d) ₹120

Answer:

It is given that,
Sum (P₁) = ₹x
Rate (R₁) = R%
Time (T₁) = 2 years
Interest (I₁) = ₹80

$I_{1} = \frac{P_{1} \times R_{1} \times T_{1}}{100} \Rightarrow 80 = \frac{x \times R \times 2}{100} = \frac{2 R x}{100} . . . (1)$

Now,
Sum (P₂) = ₹2x
Rate (R₂) = R%
Time (T₂) = 1 year

$I_{2} = \frac{P_{2} \times R_{2} \times T_{2}}{100} = \frac{2 x \times R \times 1}{100} = \frac{2 R x}{100} = 80 [From (1)]$

Therefore, I₂ = ₹80

Hence, the correct option is (c).

Page No 13.9:

Question 10:

At simple interest a sum becomes $\frac{49}{40}$ of itself in $2 \frac{1}{2}$ years. The rate of interest per annum is

(a) 7%
(b) 8%
(c) 12%
(d) 9%

Answer:

Amount = $\frac{49}{40}$ times the sum = $\frac{49}{40}$ P

Simple interest (I) = Amount − Sum = $\frac{49}{40}$ P − P = $\frac{9}{40}$ P

Let the sum (P) be x.
Then, simple interest (I) = $\frac{9}{40}$ x
Rate (R) = R%
Time (T) = $2 \frac{1}{2}$ years = $\frac{5}{2}$ years

$I = \frac{P \times R \times T}{100} \Rightarrow R = \frac{100 \times I}{P \times T} = \frac{100 \times \frac{9}{40} x}{x \times \frac{5}{2}} = \frac{45}{5} = 9 %$

Hence, the correct option is (d).

Page No 13.9:

Question 11:

At what rate percent per annum simple interest will a sum double itself in 10 years?

(a) 8%
(b) 10%
(c) 12%
(d) $12 \frac{1}{2}$ %

Answer:

Amount = 2 times the sum = 2P

Simple interest (I) = Amount − Sum = 2P − P = P

Let the sum (P) be x.
Then, simple interest (I) = x
Rate (R) = R%
Time (T) = 10 years

$I = \frac{P \times R \times T}{100} \Rightarrow R = \frac{100 \times I}{P \times T} = \frac{100 \times x}{x \times 10} = 10 %$

Hence, the correct option is (b).

Page No 13.9:

Question 12:

In what time will a sum of ₹8000 amount to ₹8360 at 6% per annum simple interest?

(a) 8 months
(b) 9 months
(c) $1 \frac{1}{4}$ months
(d) $1 \frac{1}{2}$ years

Answer:

It is given that,
Amount = ₹8360
Sum = ₹8000

Simple interest (I) = Amount − Sum = ₹8360 − ₹8000 = ₹360

Also,
Rate (R) = 6%
Time (T) = T years

$I = \frac{P \times R \times T}{100} \Rightarrow T = \frac{100 \times I}{P \times R} = \frac{100 \times 360}{8000 \times 6} = \frac{3}{4} years = \frac{3}{4} \times 12 months = 9 months$

Hence, the correct option is (b).

Page No 13.9:

Question 13:

If a, b and c are three sums of money such that b is the simple interest on a and c is the simple interest on b for the same time and same rate. Which of the following is correct?

(a) abc = 1
(b) c² = ab
(c) b² = ac
(d) a² = bc

Answer:

It is given that,
Simple interest (I₁) = b
Sum (P₁) = a
Rate (R₁) = R%
Time (T₁) = T years

Now,
$I_{1} = \frac{P_{1} \times R_{1} \times T_{1}}{100} \Rightarrow b = \frac{a \times R \times T}{100} \Rightarrow R \times T = \frac{100 b}{a} . . . . (1)$

Also,
Simple interest (I₂) = c
Sum (P₂) = b
Rate (R₂) = R%
Time (T₂) = T years

Now,
$I_{2} = \frac{P_{2} \times R_{2} \times T_{2}}{100} \Rightarrow c = \frac{b \times R \times T}{100} \Rightarrow R \times T = \frac{100 c}{b} . . . . (2)$

On equating (1) and (2), we get
$\frac{100 b}{a} = \frac{100 c}{b} \Rightarrow b^{2} = a c$

Hence, the correct option is (c).

Page No 13.9:

Question 14:

The simple interest at R% per annum for n years will be ₹n on a sum of

(a) ₹n
(b) ₹100n
(c) ₹ $\frac{100}{n}$
(d) ₹ $\frac{100}{n^{2}}$

Answer:

It is given that,
Simple interest (I) = ₹n
Rate (R) = R%
Time (T) = n years

$I = \frac{P \times R \times T}{100} \Rightarrow P = \frac{100 \times I}{R \times T} = \frac{100 \times n}{R \times n} = \frac{100}{R}$

Hence, the correct option is (c).

Page No 13.9:

Question 15:

The simple interest on a certain sum is $\frac{16}{25}$ of the sum. If the rate percent per annum and the time are numerically equal, then the rate percent is

(a) 8%
(b) 4%
(c) 6%
(d) 12%

Answer:

Let the sum (P) be ₹x
Then, the simple interest (I) = ₹ $\frac{16}{25}$ x

Also,
Rate (R) = R%
Time (T) = R years (∵ the rate percent per annum and the time are numerically equal)

$I = \frac{P \times R \times T}{100} \Rightarrow R = \frac{100 \times I}{P \times T} \Rightarrow R = \frac{100 \times \frac{16}{25} x}{x \times R} \Rightarrow R \times R = \frac{64 x}{x} \Rightarrow R \times R = 8 \times 8 \Rightarrow R = 8 %$

Hence, the correct option is (a).

Page No 13.9:

Question 16:

At which rate percent per annum simple interest will a sum triple itself in 16 years?

(a) 12%
(b) 10.5%
(c) 11.5%
(d) 12.5%

Answer:

Amount = 3 times the sum = 3P

Simple interest (I) = Amount − Sum = 3P − P = 2P

Let the sum (P) be x.
Then, simple interest (I) = 2x
Rate (R) = R%
Time (T) = 16 years

$I = \frac{P \times R \times T}{100} \Rightarrow R = \frac{100 \times I}{P \times T} = \frac{100 \times 2 x}{x \times 16} = 12.5 %$

Hence, the correct option is (d).

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