Selina Solutions Concise Maths Class 7 Chapter 10 Simple Interest contains problems on finding the interest in the given exercise problems. The main objective of providing solutions is to improve the analytical and problem solving abilities among students. For a better hold on the topics, students are advised to download Selina Solutions Concise Maths Class 7 Chapter 10 Simple Interest PDF, from the links given below.
Chapter 10 provides the basic idea about simple interest, principal, rate of interest, time period and amount. The method of determining them and the important formulas used in the process are explained in brief in the solutions PDF.
Selina Solutions Concise Maths Class 7 Chapter 10: Simple Interest Download PDF
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Exercise 10 page: 116
1. Find the S.I. and the amount on:
(i) ₹ 150 for 4 years at 5% per year.
(ii) ₹ 350 for 3 ½ years at 8% p.a.
(iii) ₹ 620 for 4 months at 8 p per rupee per month.
(iv) ₹ 3,380 for 30 months at 4 ½ % p.a.
(v) ₹ 600 from July 12 to Dec. 5 at 10% p.a.
(vi) ₹ 850 from 10th March to 3rd August at 2 ½ % p.a.
(vii) ₹ 225 for 3 years 9 months at 16% p.a.
Solution:
(i) ₹ 150 for 4 years at 5% per year
We know that
P = ₹ 150
R = 5% per year
T = 4 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (150 × 5 × 4)/ 100
= ₹ 30
Amount = P + S.I
Substituting the values
= 150 + 30
= ₹ 180
(ii) ₹ 350 for 3 ½ years at 8% p.a.
We know that
P = ₹ 350
R = 8% p.a.
T = 3 ½ years = 7/2 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (350 × 8 × 7)/ (100 × 2)
= ₹ 98
Amount = P + S.I
Substituting the values
= 350 + 98
= ₹ 448
(iii) ₹ 620 for 4 months at 8 p per rupee per month
We know that
P = ₹ 620
R = 8 p per rupee per month = 8% p.m.
T = 4 months
Here
S.I = (P × R × T)/ 100
Substituting the values
= (620 × 8 × 4)/ 100
= ₹ 198.40
Amount = P + S.I
Substituting the values
= 620 + 198.40
= ₹ 818.40
(iv) ₹ 3,380 for 30 months at 4 ½ % p.a.
We know that
P = ₹ 3,380
R = 4 ½ % p.a. = 9/2 %
P = 30 months = 30/12 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (3380 × 9 × 30)/ (100 × 2 × 12)
= ₹ 380.25
Amount = P + S.I
Substituting the values
= 3380 + 380.25
= ₹ 3760.25
(v) ₹ 600 from July 12 to Dec. 5 at 10% p.a.
We know that
P = ₹ 600
R = 10% p.a.
T = July 12 to Dec 5
July = 19 days
Aug = 31 days
Sep = 30days
Oct = 31 days
Nov = 30 days
Dec = 05 days
Total = 146 days
T = 146/365 years = 2/5 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (600 × 10 × 2)/ (100 × 5)
= ₹ 24
Amount = P + S.I
Substituting the values
= 600 + 24
= ₹ 624
(vi) ₹ 850 from 10th March to 3rd August at 2 ½ % p.a.
We know that
P = ₹ 850
R = 2 ½% = 5/2% p.a.
T = 10th March to 3rd August
March = 21 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Aug = 3 days
Total = 146 days
T = 146/365 = 2/5 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (850 × 5 × 2)/ (100 × 2 × 5)
= ₹ 8.50
Amount = P + S.I
Substituting the values
= 850 + 8.50
= ₹ 858.50
(vii) ₹ 225 for 3 years 9 months at 16% p.a.
We know that
P = ₹ 225
R = 16% p.a.
T = 3 years 9 months = 3 and 9/12 years = 3 ¾ years = 15/4 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (225 × 16 × 15)/ (100 × 4)
= ₹ 135
Amount = P + S.I
Substituting the values
= 225 + 135
= ₹ 360
2. On what sum of money does the S.I. for 10 years at 5% become ₹ 1,600?
Solution:
It is given that
S.I = ₹ 1,600
R = 5% p.a.
T = 10 years
We know that
P = (S.I × 100)/ (R × T)
Substituting the values
= (1600 × 100)/ (5 × 10)
So we get
= ₹ 3,200
3. Find the time in which ₹ 2,000 will amount to ₹ 2,330 at 11% p.a.
Solution:
It is given that
A = ₹ 2,330
P = ₹ 2,000
We know that
S.I = A – P
Substituting the values
= 2330 – 2000
= ₹ 330
Here
Time = (S.I × 100)/ (P × R)
Substituting the values
= (330 × 100)/ (2000 × 11)
So we get
= 3/2
= 1 ½ years
4. In what time will a sum of money double itself at 8% p.a.
Solution:
Consider the principal
P = ₹ 100
It is given that
A = 100 × 2 = ₹ 200
We know that
S.I = A – P
Substituting the values
= 200 – 100
= ₹ 100
R = 8% p.a.
Here
Time = (S.I × 100)/ (P × R)
Substituting the values
= (100 × 100)/ (100 × 8)
So we get
= 25/2
= 12 ½ years
5. In how many years will ₹ 870 amount to ₹ 1,044, the rate of interest being 2 ½% p.a.?
Solution:
It is given that
P = ₹ 870
A = ₹ 1044
We know that
S.I = A – P
Substituting the values
= 1044 – 870
= ₹ 174
R = 2 ½ = 5/2 % p.a.
We know that
Time = (S.I × 100)/ (P × R)
Substituting the values
= (174 × 100 × 2)/ (870 × 5)
So we get
= 8 years
6. Find the rate percent, if the S.I. on ₹ 275 in 2 years is ₹ 22.
Solution:
It is given that
P = ₹ 275
S.I = ₹ 22
T = 2 years
We know that
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (22 × 100)/ (275 × 2)
So we get
= 4% p.a.
7. Find the sum which will amount to ₹ 700 in 5 years at 8% p.a.
Solution:
It is given that
Amount = ₹ 700
R = 8% p.a.
T = 5 years
Consider P = ₹ 100
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (100 × 8 × 5)/ 100
= ₹ 40
Here
A = P + S.I
Substituting the values
= 100 + 40
= ₹ 140
If the amount is ₹ 140 then the principal is ₹ 100
If the amount is ₹ 700 then the principal = (100 × 700)/ 140 = ₹ 500
8. What is the rate of interest, if ₹ 3,750 amounts to ₹ 4,650 in 4 years?
Solution:
It is given that
P = ₹ 3,750
A = ₹ 4,650
We know that
S.I = A – P
Substituting the values
= 4650 – 3750
= ₹ 900
T = 4 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (900 × 100)/ (3750 × 4)
So we get
= 6% p.a.
9. In 4 years, ₹ 6,000 amounts to ₹ 8,000. In what time will ₹ 525 amount to ₹ 700 at the same rate?
Solution:
It is given that
P = ₹ 6,000
A = ₹ 8,000
We know that
S.I = A – P
Substituting the values
= 8000 – 6000
= ₹ 2000
T = 4 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (2000 × 100)/ (6000 × 4)
So we get
= 25/3%
= 8 1/3% p.a.
It is given that
P = ₹ 525
A = ₹ 700
We know that
S.I = A – P
Substituting the values
= 700 – 525
= ₹ 175
R = 25/3% p.a.
Here
Time = (S.I × 100)/ (P × R)
Substituting the values
= (175 × 100 × 3)/ (525 × 25)
So we get
= 4 years
10. The interest on a sum of money at the end of 2 ½ years is 4/5 of the sum. What is the rate percent?
Solution:
Consider the sum P = ₹ 100
S.I = 100 × 4/5 = ₹ 80
T = 2 ½ = 5/2 years
We know that
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (80 × 100 × 2)/ (100 × 5)
So we get
= 32% p.a.
11. What sum of money lent out at 5% for 3 years will produce the same interest as ₹ 900 lent out at 4% for 5 years?
Solution:
It is given that
P = ₹ 900
R = 4%
T = 5 years
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (900 × 4 × 5)/ 100
= ₹ 180
It is given that
S.I = ₹ 180
R = 5%
T = 3 years
We know that
Sum P = (S.I × 100)/ (R × T)
Substituting the values
= (180 × 100)/ (5 × 3)
So we get
= ₹ 1200
12. A sum of ₹ 1,780 becomes ₹ 2,136 in 4 years. Find:
(i) the rate of interest.
(ii) the sum that will become ₹ 810 in 7 years at the same rate of interest.
Solution:
(i) It is given that
P = ₹ 1780
A = ₹ 2136
We know that
S.I = A – P
Substituting the values
= 2136 – 1780
= ₹ 356
T = 4 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (356 × 100)/ (1780 × 4)
So we get
= 5% p.a.
(ii) Consider P = ₹ 100
R = 5% p.a.
T = 7 years
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (100 × 5 × 7)/ 100
= ₹ 35
Here amount = P + S.I
Substituting the values
= 100 + 35
= ₹ 135
If the amount is ₹ 135 then the principal is ₹ 100
If the amount is ₹ 810 then principal = (100 × 810)/ 135 = ₹ 600
13. A sum amounts to ₹ 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.
Solution:
(i) Consider P = ₹ 100
R = 5% p.a.
T = 6 years
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (100 × 5 × 6)/ 100
= ₹ 30
Here amount = 100 + 30 = ₹ 130
If the amount is ₹ 130 then principal is ₹ 100
If the amount is ₹ 2652 then principal = (100 × 2652)/ 130 = ₹ 2040
Consider sum P = ₹ 100
Amount = 100 × 2 = ₹ 200
We know that
S.I = A – P
Substituting the values
= 200 – 100
= ₹ 100
R = 5% p.a.
Here
T = (S.I × 100)/ (P × R)
Substituting the values
= (100 × 100)/ (100 × 5)
So we get
= 20 years
14. P and Q invest ₹ 36,000 and ₹ 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets ₹ 3,080 more interest than Q, find the rate of interest.
Solution:
It is given that
P’s investment (P1) = ₹ 36,000
Q’s investment (P2) = ₹ 25,000
T = 4 years
Consider the rate of interest = x%
So we get
P’s interest (S.I) = (P × R × T)/ 100
Substituting the values
= (36000 × x × 4)/ 100
= ₹ 1440x
Q’s interest = (P × R × T)/ 100
Substituting the values
= (25000 × x × 4)/ 100
= ₹ 1000x
Here the difference in their interest = 1440x – 1000x = ₹ 440x
The difference given = ₹ 3080
So we get
440x = 3080
x = 3080/440
x = 7%
So the rate of interest = 7% p.a.
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.
Solution:
Consider the sum P = ₹ 100
We know that
S.I = 1/4 × 100 = ₹ 25
T = 5 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (25 × 100)/ (100 × 5)
So we get
= 5%
16. The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.
Solution:
Consider sum P = ₹ 100
We know that
SI = 30/100 × 100 = ₹ 30
T = 5 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (30 × 100)/ (100 × 5)
So we get
= 6%
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