Simple Interest Class 8 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Simple Interest Class 8 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8
| Board | ICSE |
| Publications | Goyal Brothers Prakashan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-8 | Simple Interest and Compound Interest |
| Exe-8A | Problems on Simple Interest |
| Academic Session | 2024-2025 |
How to Solve Simple Interest Problems
( Class 8 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8 )
The formula for calculating simple interest is: = (P x R x T) / 100, Where P = Principal amount (the beginning balance). R = Interest rate per year , T = Number of Year
Note If Time is given in month then divide it by 12 to make in year and if time is given in days then divide it by 365 to make it in year.
Final Amount = Principal (beginning balance) + Simple interest
Exercise- 8A
( Simple Interest and Compound Interest Class 8 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-8 )
Que-1: Find the simple interest and amount on :
(i) Rs4500 for 2*(1/2) years at 7*(2/3)% per annum
(ii) Rs6360 for 6 years 3 months at 8% per annum
(iii) Rs19200 for 11 months at 9*(3/4)% per annum
(iv) Rs58400 for 75 days at 6*(1/2)% per annum
Sol: (i) SI = (P×R×T)/100
= (4500×(5/2)×(23/3))/100
= 5175/6 = Rs862.50
Amount = S.I. + P
= 862.50 + 4500
= Rs5362.50
(ii) Given principle (P) = Rs 6360
time(T) = 6 years 3 months = 6+ 3/12 = 25/4 years
Rate of interest (R) = 8%
S.I = P*T*R/100
S.I = (6360*25*8) / 100*4
S.I = Rs 3180
Amount = principle + simple interest
= 6360 + 3180 = Rs 9540
(iii) Given principle (P) = Rs19200
time(T) = 11 months = 11/12 years
Rate of interest (R) = 9*(3/4) = 39/4 %
S.I = P*T*R/100
S.I = (19200*11*39) / 100*4*12
S.I. = Rs1716
Amount = principle + simple interest
= 1716 + 19200
= Rs20916
(iv) Given principle (P) = Rs58400
time(T) = 75 days = 75/365 year
Rate of interest (R) = 6*(1/2) = 13/2 %
S.I = P*T*R/100
S.I = (58400*75*13) / 100*365*2
S.I. = Rs780
Amount = principle + simple interest
= 58400 + 780
= Rs59180.
Que-2: Find the simple interest on Rs8600 from 18th October, 2006 to 13th March, 2007 at 8% per annum. Also find the amount.
Sol: Given : Rs.8600 from 18th October,2006 to 13th march ,2007 at 8%per annum.
We have :
Time = 146 days
Principal = ₹8600
Rate = 8%
S.I. = P×R×T/100
= ₹ 8600×8×146 / 100×365
= ₹275.20
Amount = Principal + S.I.
= ₹ 8600+₹275.2
= 8875.20
Que-3: Ashish lent Rs10500 to Sunidhi at 7% per annum simple interest. After 5 years, Sunidhi discharged the debt by giving a watch and Rs13000 in cash. What is the value of the watch ?
Sol: Principal = P = 10500
rate of interest = R = 7%
time = 5 years
simple interest = PTR/100
= 10500 x 5 x 7 /100
= 105 x 35
= 3675.
Amount to be returned after 5 years = P + R
= 10500 + 3675
= 14175
Cash given = 13000
remaining balance = 14175 – 13000
= 1175
Que-4: In what will the simple interest on Rs7560 be Rs1102.50 at 6*(1/4)% per annum ?
Sol: I = p×r×t/100
1102.50 = 7560×6×t/100
t = 1102.50×100/7560×6
t = 110250/45360
t = 2.43 years
t = 2 years 4 months.
Que-5: In how much time will Rs25600 amount to Rs35664, when money is worth 9*(1/4)% per annum simple interest ?
Sol: Amount (A) = 35,664
Principal (P) = 25,600
Simple Interest = Amount – Principal
= 35664 – 25600
= 10064
Rate = 37/4 %
Simple Interest = P×R×T/100
10064 = 25,600 × 37/4 × T /100
1006400= 6400×37×T
1006400 = 236800 × T
T = 1006400/236800
T = 4.25 Year!
T = 4 years 3 months
Que-6: At what rate percent per annum will Rs1625 amount to Rs2080 in 3*(1/2)% years?
Sol: Principle = 1625
Amount = 2080
Time = 3 1/2 = 7/2 years = 3.5
Interest = Amount – Principal
Interest = 2080 – 1625
Interest = 455.
I = Interest, P = Principal, T = Time.
Rate percentage = (Ix100)/(PxT)
= (455×100)/(1625×3.5)
= 45500/5687.5
= 8%
Que-7: At what rate percent per annum will the simple interest on Rs6720 be Rs1911 in 3 years 3 months ?
Sol: The simple interest given is Rs. 1911.
Principal amount is Rs. 6720.
Number of years = 3 years and 3 months = 3×12 + 3 months
= 39/12 years = 3.25 years
Simple Interest = P×R×T/100
1911 = (6720*3.25*R)/100
R = (1911*100)/(6720*3.25)
R = 8.75% = 8*(3/4)%
Que-8: At what rate percent of simple interest will a sum of money double itself in 12 years?
Sol: Let say Sum of Money = P
Sum of Money Doubles = 2P
Interest Earned = 2P – P = P
Rate of interest = R
Time = 12 Years
Simple interest = P * R * T /100
= P = P * R * 12 /100
= R = 100/12
= R = 25/3
= R = 8.33% = 8*(1/3)%
Que-9: Simple interest on a certain sum is (9/16) of the sum. Find the rate percent and the time if both are numerically equal.
Sol: Let the sum of money be x
Simple interest = 9x/16
Let the rate percent = R%
Time = R years
As we know,
Simple Interest = P * R * T/100
∵ (x * R * R)/100 = 9x/100
R²/100 = 9/100
R² = 900/16
R = 30/4
R = 7*(1/2)% p.a.
T = R = 7*(1/2) years
Que-10: What sum will yield Rs406 as simple interest in 1 year 2 months at 6*(1/4)% per annum ?
Sol: Let the principle be Rs. x .
SI = Rs. 406
Rate = 25/4 %
Time = 1 year and 2 months
= 1 + 2/12 = 7/6 years.
ATQ,
406 = x × 25/400 × 7/6
x =[ 406 ×400×6]÷[25×7]
= 5568
Que-11: What sum will amount to Rs1748 in 2*(1/2) years at 7*(1/2)% per annum simple interest ?
Sol: S.I. = 1748
R = 7*(1/2)% = 15/2 %
T = 2*(1/2) = 5/2 years
Simple Interest = P * R * T/100
1748 = p+(prt/100)
p = 1748×100/(100+7.5×2.5)
= 1472
Que-12: A sum of money becomes (8/5) of itself in 5 years at a certain rate of simple interest. Find the rate of interest.
Sol: Given: Time, T=5 years
Let the principal be P
Rate of interest =R
Amount, A = 8P/5
Simple Interest, S.I= Amount − Principal
= (8P/5) − P
= 3P/5
∴ S.I = 3P/5
Again, S.I = P×R×T/100
⇒ 3P/5 = P×R×5/100
⇒ 3/5 = 5R/100
⇒ 3/5 = R/20
⇒ R = (3×20)/5
⇒ R = 3×4
∴ R = 12%.
Que-13: What sum of money lent at 12*(1/2)% per annum will produce the same interest in 4 years as Rs8560 produces in 5 years at 12% per annum ?
Sol: Interest = (Principal x Rate x Time) / 100
Interest = (8560 x 12 x 5) / 100 = 5136
So, the interest earned by 8560 in 5 years is 5136.
Interest = (Principal x Rate x Time) / 100
Let the principal be P.
The interest earned by P in 4 years at 12.5% per annum will be:
Interest = (P x 12.5 x 4) / 100 = 0.5P
We want the interest earned by P in 4 years to be equal to 5136, so:
0.5P = 5136
P = 10272
Que-14: If Rs1250 amount to Rs1550 in 3 years at simple interest, what will Rs3200 amount to in 4 years at the same rate ?
Sol: Interest = (Principal x Rate x Time) / 100
(1550-1250) = (1250x3xR)/100
R = (300×100)/(1250×3)
R = 8%
I = (3200x4x8)/100
I = 1024
Now we can add the above interest and the principle amount.
= 1024 + 3200
= 4224.
Que-15: A sum of money lent at simple interest amounts to Es3224 in 2 years and Rs4160 in 5 years. Find the sum and rate of interest.
Sol: x = sum of money to be lent
y = % rate per annum
a/q
when time= 2 years, Amount = 3224
so, 3224 = x + x*y*2/100. ——— (I)
and when time= 5 years, Amount = 4160
so, 4160 = x + x*y*5/100. ——— (II)
solving equation I and ii we get..
x = 2600
and y = 12%
Que-16: A lend Rs 2500 to B and a certain sum to C at the same time at 7% per annum simple interest. If after 4 years, A altogether receives Rs1120 as interest from B and C, find the sum lent to C.
Sol: Let the sum lent to C be x.
Given the rate of interest is 7%.
Now according to the question,
[(2500×7×4)/100] + [(x×7×4)/100] = 1120
⇒ 2500×28+28x = 112000
⇒ 2500+x = 4000
⇒ x = 4000−2500 = 1500
Hence the sum lent to C is Rs.1500.
Que-17: The simple interest on a certain sum for 3 years at 8% per annum is Rs96 more than the simple interest on the same sum for 2 years at 9% per annum. Find the sum.
Sol: Condition I:- For 3 years at the rate of 8% per annum.
S.I. = (P×8×3)/100…..(1)
condition II:- For 2 years at the rate of 9% per annum.
S.I. = (P×9×2)/100…..(2)
Now according to the question-
(P×8×3)/100 = [(P×9×2)/100] + 96
⇒ (P×24)/100 = [(P×18)+9600]/100
⇒ 24P = 18P = 9600
⇒ 24P−18P = 9600
⇒ P = 9600/6 = 1600
Hence the sum is Rs.1600.
Que-18: Two equal sums of money lent at simple interest at 11% p.a., for 3*(1/2) years and 4*(1/2) years respectively. If the difference in interests for two periods was Rs412.50, find each sum.
Sol: Let each sum be x.
Given that rate of interest is 11%
Now according to the question-
[(x×11×9)/(100×2)] − [(x×11×7)/(100×2)] = 412.5
99x−77x = 412.5×200
⇒ 22x = 82500
⇒ x = 82500/22 = 3750
Hence the value of each sum is Rs.3750.
Que-19: Divide Rs6000 into two parts so that the simple interest on the first part for 9 months at 12% per annum is equal to the simple interest on the second part for 1*(1/2) years at 10% per annum.
Sol: Let one part is x and other part is (6000−x)
Now,
S.I on first part = S.I. on second part
[x×(12/100)×(9/12)] = [(6000−x)×(10/100)×(3/2)]
∴ x = 3750
and (6000−x)
⇒ 6000−3750 = 2250
Therefore two parts of 6000 are 3750 and 2250.
Que-20: Divide a sum of Rs13500 into two parts such that if one part be lent at 8*(1/3)% per annum for 2 years 9 months and the other at 7*(1/2)% per annum for 1 year 8 months, the total interest received is Rs2375.
Sol: Let the one part be x and other be (13500−x)
[(x/100)×(25/3)×(2+(9/12)) + [(13500−(x/100))×(15/2)×(1+(8/12))
(x/12)(11/4) + [(13500−x)/100]×(15/2)×(5/3) = 2375
5x = 114000−81000
x = 6600
13500−6600 = 6900
So, two parts are Rs6600 and Rs6900
Que-21: In what time will a sum of money lent at 8*(1/3)% simple interest become 4 times of itself.
Sol: as per question A = 4P
You also know that the rate (R) is 8 1/3%, which is equivalent to 25/3%.
SI = (P × R × T) / 100
SI = (P × (25/3) × T) / 100
SI = A – P
A – P = (P × (25/3) × T) / 100
4P – P = (P × (25/3) × T) / 100
3P = (P × (25/3) × T) / 100
3 = (25/3) × T / 100
T = (3 × 100) / (25/3)
T = 300 / (25/3)
T = 36 years
Que-22: A certain sum of money lent out 6*(2/3)% p.a., produces the same simple interest in 6 years as Rs3200 lent out at 8*(2/5)% p.a. for 7 years, Find the sum.
Sol: TO KNOW THE SUM,
P: 3200, R: 8 2/5% = 42/5, T=7
SI = PRT/100
= (3200×42×7)/100×5
= 9408/5 = Rs.1881.6
In second case, P=?
P = SI× 100/R×T
= (18816×100×3)/(10×20×6)
= Rs.4704.
Que-23: Naveen and Praveen borrowed Rs42000 and Rs55000 respectively for 3*(1/2) years at the rate of interest. If Praveen has to pay Rs3640 more than Naveen, find the rate of interest.
Sol: Given that Naveen borrowed Rs.42000 and Praveen borrowed Rs.55000.
Time period = 72 years
SI = PRT/100
S.I. paid by Naveen = [42000×r×(7/2)]/100 = 210×r×7
S.I. paid by Praveen = [55000×r×(7/2)]/100 = 225×r×7
Given that the difference in the interest paid by Praveen and Naveen is Rs.3640
Therefore,
225×r×7 − 210×r×7 = 3640
r×7×(275−210) = 3640
⇒ r = 3640/(7×65)
= 8%
Hence the rate of interest is 8%.
Que-24: A sum of money was put at simple interest at a certain rate for 2 years. If this sum had been put at 3% higher rate, it would have earn Rs720 more as interest. Find the sum.
Sol: So 3% produces 720/2 for 1 year
which is 360 per year
Interest = P*T*R/100
360 = P*1*3 / 100
On Solving P = 360*100/3
P = 12,000
Que-25: A sum of money invested at 6% p.a., simple interest for a certain periods of time yields Rs960 as interest. If this sum had been invested for 5 years more, it would have yielded Rs2160 as interest. Find the sum.
Sol: 5 years interest = 2160-960 = 1200
int for one year = 1200/5= 240
int for 4 years = 960
let sum be p, time be n, rate of int be r
simple int = pnr/100
960= p* 4* 6/100
p= 960*100/4*6
= Rs. 4000
–: Simple Interest Class 8 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Ch-8 Solutions :–
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