Simple Interest ICSE Class-7th Concise Selina Maths Solutions Chapter-10 . We provide step by step Solutions of Exercise / lesson-10 Simple Interest for ICSE Class-7 Concise Selina Mathematics . Our Solutions contain all type Questions to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

## Simple Interest ICSE Class-7th Concise Selina Maths Solutions Chapter-10

### Exercise – 10

#### Question 1.

Find the S.I. and amount on :
(i) Rs. 150 for 4 years at 5% per year.
(ii) Rs. 350 for 3$\frac {1 }{ 2 }$ years at 8% p.a.
(iii) Rs. 620 for 4 months at 8 p. per rupee per month.
(iv) Rs. 3,380 for 30 months at 4 $\frac { 1 }{ 2 }$ % p.a.
(v) 600 from July 12 to Dec. 5 at 10% p.a.
(vi) Rs. 850 from 10th March to 3rd August at 2 $\frac { 1 }{ 2 }$ % p.a.
(vii) Rs. 225 for 3 years 9 months at 16% p.a.

(i)

P = Rs. 150, R = 5% per year T = 4 years

∴ S.I. = (P.R.T)/100

= (150×5×4)/100

= Rs. 30

and amount = P + S.I.

= Rs. 150 + Rs. 30

= Rs. 180

(ii)

P = Rs. 350, R = 8% p.a. T = 3 (1/2) years = 7/2 years

∴ S.I. = (P.R.T)/100

=(350×8×7)/(100×2)

= Rs. 98

and amount = P + S.I.

= Rs. 350 + Rs. 98

= Rs. 448

(iii)

P = Rs. 620, R = 8 p. per rupee per month = 8% p.m..

T = 4 months

∴ S.I. = (P.R.T)/100

=(620×8×4)/100

= Rs. 19840/100

= Rs. 198.40

and amount = P + S.I.

= Rs. 620 + Rs. 198.40

= Rs. 818.40

(iv)

Principal (P) = Rs. 3380

Rate = 4 (1/2)% p.a. = 9/2%

Period = 30 months = 30/12 years

S.I. = (PRT/100)

=(3380×9×30)/(100×2×12)

= Rs. 1521/4

= Rs. 380.25

Amount = P + S.I.

= Rs. 3380 + 380.25

= Rs. 3760.25

(v)

P = Rs. 600, R = 10% p.a.

T = July 12 to Dec. 5

 July = 19 Days Aug. = 31 Days Sep. = 30 Days Oct. = 31 Days Nov. = 30 Days Dec. = 05 Days Total 146 Days

= (146/365) years = 2/5 years

∴ S.I. = (P.R.T)/100

=(600×10×2)/(100×5)

= Rs. 24

∴ Amount = P + S.I.

= Rs. 600 + Rs. 24

= Rs. 624

(vi)

P = Rs. 850, R = 2 (1/2)%

=5/2% p.a.

T = 10th march to 3rd Aug.

 March = 21 Days April = 30 Days May = 31 Days June = 30 Days July = 31 Days Aug. = 03 Days Total 146 Days

=146/365 years = 2/5 years

∴ S.I. = (P.R.T)/100

=(850×5×2)/(100×2×5)= 850/100

∴ Amount = P + S.I.

= Rs. 850 + Rs. 8.50

= Rs. 858.50

(vii)

P = Rs. 225, R = 16% p.a. T = 3 years 9 months

=3 (9/12)

=3 (3/4) years

=15/4 years

∴ S.I. = (P.R.T./100)

=(225×16×15)/100×4

= Rs. 135

∴ Amount = P + S.I.

= Rs. 225 + Rs. 135

= Rs. 360

#### Question 2.

On what sum of money does the S.I. for 10 years at 5% become Rs. 1,600 ?

S.I. = Rs. 1600, R = 5% p.a. T = 10 years

∴ P = (S.I.×100)/(R×T)

=(1600×100)/(5×10) = Rs. 3200

#### Question 3.

Find the time in which Rs. 2,000 will amount to Rs. 2,330 at 11% p.a. ?

Amount (A) = Rs. 2330

Principal (P) = Rs. 2000

∴ S.I. = A – P

= Rs. 2330 – Rs. 2000

= Rs. 330

R = 11 % p.a.

∴ Time = (S.I.×100)/(P×R)

=(330×100)/(2000×11)

=3/2

=1 (1/2) years

#### Question 4.

In what time will a sum of money double it self at 8% p.a ?

Let the principal (P) = ₹ 100

∴ Amount (A) = ₹ 100 × 2 = ₹ 200

∴ S.I. = A – P = ₹ 200 – ₹ 100 = ₹ 100

Rate (R) = 8 % p.a.

∴ Time = (S.I.×100)/(P×R)

=(100×100)/(100×8)

=25/2

=12 (1/2) years

#### Question 5.

In how many years will be ₹870 amount to ₹1,044, the rate of interest being 2$\frac { 1 }{ 2 }$ % p.a ?

Principal (P) = ₹ 870

Amount (A) = ₹ 1044

∴ S.I. = P – A = ₹ 1044 – ₹ 870 = ₹ 174

Rate (R) = 2 (1/2)=5/2% p.a.

∴ Time = (S.I.×100)/(P×R )

= (174×100×2)/(870×5)

=8 years

#### Question 6.

Find the rate percent if the S.I. on ₹275 is 2 years is ₹22.

Principal (P) = ₹ 275, S.I. = ₹ 22

Time = 2 years

∴ Rate = (S.I.×100)/P×T

= (22×100)/(275×2)

= 4 % p.a.

#### Question 7.

Find the sum which will amount to ₹700 in 5 years at 8% rate p.a.

Amount = ₹ 700, Rate (R) = 8 % p.a.

Time (T) = 5 years

Let principal (P) = ₹ 100

then S.I. = (P.R.T./100)

=(100×8×5)/100

=₹40

∴ Amount (A) = P + S.I.

= ₹ 100 + 40 = ₹ 140

If amount is ₹ 140, then principal = ₹ 100

and, if amount is Rs. 700, then principal

= ₹ (100×700)/140

=₹500

#### Question 8.

What is the rate of interest, if ₹3,750 amounts to ₹4,650 in 4 years ?

Principal (P) = ₹ 3750

Amount (A) = ₹ 4650

∴ S.I. = A – P = ₹ 4650 – 3750 = ₹ 900

Time (T) = 4 years

∴ Rate = (S.I×100)/P×T

= (900×100)/(3750×4)

= 6 % p.a.

#### Question 9.

In 4 years, ₹6,000 amount to ₹8,000. In what time will ₹525 amount to ₹700 at the same rate ?

In first case, principal (P) = ₹ 6000

Amount (A) = ₹ 8000

∴ S.I. = A – P = ₹ 8000 – ₹ 6000 = ₹ 2000

Time (T) = 4 years

∴ R = (S.I.×100)/P×T

=(2000×100)/(6000×4)

=25/3%

=8 (1/3)% p.a.

In second case, Principal (P) = ₹ 525

Amount (A) = ₹ 700

∴ S.I. = A – P = ₹ 700 – ₹ 525 = ₹ 175

Rate (R) = 25/3% of p.a.

∴ Time = (S.I.×100)/P×R

= (Rs.175×100×3)/(525×25)

= 4 years

#### Question 10.

The interest on a sum of money at the end of 2$\frac { 1 }{ 2 }$ years is $\frac { 4 }{ 5 }$ of the sum. What is the rate percent ?

Let the sum (P) = Rs. 100

∴ S.I. = Rs. 100×(4/5) = Rs. 80

Period (T) = 2 (1/2)

=5/2 years.

∴ Rate = (S.I.×100)/P×T

=(80×100×2)/(100×5)

= 32 % p.a.

#### Question 11.

What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years ?

In second case, Principal (P) = Rs. 900

Rate (R) = 4%, Time (T) = 5 years

∴ S.I. = (P×R×T)/100

= (900×4×5)/100 = Rs. 180

In first case, S.I. = Rs. 180

Rate = 5 %, Time = 3 years

∴ Sum = (S.I.×100)/(R×T)

= (180×100)/(5×3)

= Rs. 1200

#### Question 12.

A sum of Rs. 1,780 become Rs. 2,136 in 4 years,
Find :
(i) the rate of interest.
(ii) the sum that will become Rs. 810 in 7 years at the same rate of interest ?

(i) In first case, Principal (P) = Rs. 1780

Amount (A) = Rs. 2136

∴ S.I. = A – P = Rs. 2136 – 1780 = Rs. 356

Time (T) = 4 years

∴ Rate = (S.I.×100)/(P×T)

= (356×100)/(1780×4)

=5%p.a

(ii) In second case, Let principal (P) = Rs. 100

Rate (R) = 5 % p.a., Time (T) = 7 years

∴ S.I. = (P×R×T)/100

= (100×5×7)/100

= Rs. 35

∴ Amount = P + S.I. = Rs. 100 + 35 = Rs. 35

If amount is Rs. 135, then principal = Rs. 100and if amount is Rs. 810, then principal

= Rs. (100×810)/135

=Rs.600

#### Question 13.

A sum amounts to Rs. 2,652 in 6 years at 5% p.a. simple interest.
Find :
(i) the sum
(ii) the time in which the same sum will double itself at the same rate of interest.

(i) Case First , Let principal (P) = rs. 100

Rate (R) = 5 % p.a., Time (T) = 6 years

∴  S.I. = (P×R×T)/100

= (100×5×6)/100

= Rs. 30

and, amount = Rs. 100 + Rs. 30 = Rs. 130

If amount is Rs. 130, then principal = Rs. 100

and, if amount is Rs. 2652, then principal

= (100×2652)/130

= Rs. 2040

(ii) Case second , Let sum (P) = Rs. 100

Amount (A) = Rs. 100 × 2 = Rs. 200

S.I. = A – P = Rs. 200 – 100 = Rs. 100

Rate = 5 % p.a.

Time = (S.I.×100)/(P×R)

= (100×100)/(100×5)= 20 years

#### Question 14.

P and Q invest Rs. 36,000 and Rs. 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets Rs. 3,080 more interest than Q; find the rate of interest.

P’s investment (P1) = Rs. 36000

and Q’s investment (P2) = Rs. 25000

Period (T) = 4 years, Let rate of interest = x %

Q’s interest = Rs. (36000×x×4)/100

= Rs. 1440 x    (∵ S.I.=PRT/100)

and Q’s interest = (25000 × x × 4)/100

= Rs. 1000 x

Difference in their interest

= Rs. (1440 – 1000) x = Rs. 440 x

But difference = Rs. 3080

∴ 440 x = 3080 ⇒ x = 3080/440

⇒ x = 7%

∴ Rate of interest = 7 % p.a.

#### Question 15.

A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.

Let the sum (P) = ₹ 100

∴ S.I. = ₹ (1/4) × ₹100 = ₹25

Period (T) = 5 years

∴ Rate % = (S.I.×100)/P×T

(25×100)/(100×5)

=5%

#### Question 16.

The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.

Let the sum (P) = ₹ 100

S.I. = (30/100)× ₹100 = ₹ 30

Period (T) = 5 years

∴ Rate = (S.I.×100)/(P×T)

= (30×100)/(100×5)

=6%

— End of Simple Interest Solutions :–

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