Simple Interest Test

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Simple Interest Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

Find the S.I. and the amount on: ₹ 300 for 6 years at 9% per year.

A
SI = 162, Amount = 462

B
SI = 152, Amount = 452

C
SI – 262, Amount = 562

D
SI = 62, Amount = 362

Solution

₹ 300 for 6 years at 9% per year
We know that
P = 300
R = 9% per year
T = 6 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (300 × 9 × 6)/ 100
= 162
Amount = P + S.I
Substituting the values
= 300 + 162
= 462
Question 2
1 / -0

Find the sum which will amount to ₹ 1200 in 5 years at 9% p.a.

A
1100

B
800

C
1000

D
1050

Solution

It is given that
Amount = ₹ 1200
R = 9% p.a.
T = 5 years
Consider P = ₹ 100
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (100 × 5 × 4)/ 100
= ₹ 45
Here
A = P + S.I
Substituting the values
= 100 + 20
= ₹ 120
If the amount is ₹ 120 then the principal is ₹ 100
If the amount is ₹ 1200 then the principal = (100 × 1200)/ 120 = ₹ 1000
Question 3
1 / -0

Find the S.I. and the amount on 480 for 6 months at 9 % per rupee per month

A
SI = 298.40, Amount = 918.40

B
SI = 98.40, Amount = 718.40

C
SI = 398.40, Amount = 1018.40

D
SI = 198.40, Amount = 818.40

Solution

₹ 480 for 6 months at 9 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
Question 4
1 / -0

In how many years will ₹ 390 amount to ₹ 1,066, the rate of interest being 6 ½% p.a.?

A
3 (2/3) years

B
2 (2/3) years

C
4 (2/5) years

D
3 (1/4) years

Solution

It is given that
P = ₹ 390
A = ₹ 1066
We know that
S.I = P – A
Substituting the values
= 1066 – 390
= ₹ 676
R = 6 ½ = 13/2 % p.a.
We know that
Time = (S.I × 100)/ (P × R)
Substituting the values
= (676 × 100 × 2)/ (390 × 13)
So we get
= (8/3) years
= 2 (2/3) years.
Question 5
1 / -0

In 5 years, ₹ 7,000 amounts to ₹ 11,200. In what time will ₹ 720 amount to ₹ 900 at the same rate?

A
2 (1/12) years

B
3 (5/12) years

C
2 (5/12) years

D
None of the above

Solution

It is given that
P = ₹ 7,000
A = ₹ 11,200
We know that
S.I = A – P
Substituting the values
= 11200 – 7000
= ₹ 4200
T = 5 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (4200 × 100)/ (7000 × 5)
So we get
= 12% p.a.
It is given that
P = ₹ 720
A = ₹ 900
We know that
S.I = A – P
Substituting the values
= 900 – 720
= ₹ 180
R = 12% p.a.
Here
Time = (S.I × 100)/ (P × R)
Substituting the values
= (180 × 100 )/ (720 × 12)
So we get
= (25/12)
= 2 (1/12) years
Question 6
1 / -0

Find the S.I. and the amount on: ₹ 540 from 23rd March to 16th August at 3 ½ % p.a.

A
SI = 8.56, AMOUNT = 548.56

B
SI = 7.56, AMOUNT = 547.56

C
SI = 14, AMOUNT = 554

D
SI = 39.2, AMOUNT = 579.2

Solution

₹ 540 from 23rd March to 16th August at 2 ½ % p.a.
We know that
P = 540
R = 3 ½% = 7/2% p.a.
T = 23rd March to 16th August
March = 8 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Aug = 16 days
Total = 146 days
T = 146/365 = 2/5 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (540 × 7 × 2)/ (100 × 2 × 5)
= 7.56
Amount = P + S.I
Substituting the values
= 540 + 7.56
= 547.56
Question 7
1 / -0

Find the S.I. and the amount on: ₹ 720 for 4 ½ years at 8% p.a.

A
SI = 259.2, AMOUNT = 979.2

B
SI = 249.2, AMOUNT = 949.2

C
SI = 159.2, AMOUNT = 879.2

D
SI = 279.8, AMOUNT = 969.8

Solution

₹ 720 for 4 ½ years at 8% p.a.
We know that
P = 720
R = 8% p.a.
T = 4 ½ years = 9/2 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (720 × 8 × 9)/ (100 × 2)
= (36 × 4 × 9) / (5)
= (1296/5)
= (259.2)
Amount = P + S.I
Substituting the values
= 720 + 259.2
= ₹ 979.2
Question 8
1 / -0

In what time will a sum of money double itself at 16% p.a.

A
6 (1/4) years

B
5 years

C
5 (1/4) years

D
6 years

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 = 16% p.a.
Here
Time = (S.I × 100)/ (P × R)
Substituting the values
= (100 × 100)/ (100 × 16)
So we get
= 25/4
= 6 (1/4) years
Question 9
1 / -0

On what sum of money does the S.I. for 15 years at 10% become 4,500?

A
2500

B
3000

C
3500

D
3200

Solution

It is given that
S.I = 4,500
R = 10% p.a.
T = 15 years
We know that
P = (S.I × 100)/ (R × T)
Substituting the values
= (4500 × 100)/ (10 × 15)
So we get
= 3,000
Question 10
1 / -0

Find the rate percent, if the S.I. on ₹ 350 in 2 years is ₹ 34

A
5%

B
6%

C
7%

D
4%

Solution

It is given that
P = ₹ 350
S.I = ₹ 28
T = 2 years
We know that
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (28 × 100)/ (350 × 2)
So we get
= 4% p.a.
Question 11
1 / -0

What is the rate of interest, if ₹ 4,500 amounts to ₹ 5,400 in 4 years?

A
6%

B
5%

C
7%

D
4.5%

Solution

It is given that
P = ₹ 4,750
A = ₹ 5,650
We know that
S.I = A – P
Substituting the values
= 5400 – 4500
= ₹ 900
T = 4 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (900 × 100)/ (4500 × 4)
So we get
= 5% p.a.
Question 12
1 / -0

Find the total amount on: ₹ 4480 for 20 months at 6 ½ % p.a

A
3458.25

B
4965.33

C
5469.22

D
3644.25

Solution

₹ 4480 for 20 months at 6 ½ % p.a.
We know that
P = 4480
R = 6 ½ % p.a. = 13/2 %
P = 20 months = 20/12 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (4480 × 13 × 20)/ (100 × 2 × 12)
= (448 × 13) / (12)
= 485.33
Amount = P + S.I
Substituting the values
= 4480 + 485.33
= 4965.33
Question 13
1 / -0

Find the S.I. and the amount on: ₹ 375 for 4 years 6 months at 12% p.a.

A
SI = 153.2, AMOUNT = 478.2

B
SI = 256.2, AMOUNT = 576.2

C
SI = 159.2, AMOUNT = 479.2

D
SI = 202.5, AMOUNT = 577.5

Solution

₹ 375 for 4 years 6 months at 12% p.a.
We know that
P = 375
R = 12% p.a.
T = 4 years 6 months = 4 6/12 = 4 ½ years = 9/2 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (375 × 12 × 9)/ (100 × 2)
= 202.5
Amount = P + S.I
Substituting the values
= 375 + 202.5
= 577.5
Question 14
1 / -0

Find the time in which ₹ 2,500 will amount to ₹ 2,860 at 12% p.a.

A
2 years

B
1 (1/5) years

C
2 (3/5) years

D
1 (2/5) years

Solution

It is given that
A = ₹ 2,880
P = ₹ 2,500
We know that
S.I = A – P
Substituting the values
= 2860 – 2500
= ₹ 360
Here
Time = (S.I × 100)/ (P × R)
Substituting the values
= (360 × 100)/ (2500 × 12)
So we get
= (30/25)
= (6/5)
= 1 (1/5) years
Question 15
1 / -0

Find the S.I. and the amount on: ₹ 900 from July 15 to Dec. 10 at 10% p.a.

A
SI = 40, Amount = 940

B
SI = 98, Amount = 998

C
SI = 18, Amount = 918

D
SI = 36, Amount = 936

Solution

₹ 900 from July 15 to Dec. 8 at 10% p.a.
We know that
P = 900
R = 10% p.a.
T = July 15 to Dec 7
July = 16 days
Aug = 31 days
Sep = 30days
Oct = 31 days
Nov = 30 days
Dec = 8 days
Total = 146 days
T = 146/365 years = 2/5 years
Here
S.I = (P × R × T)/ 100
Substituting the values
= (900 × 10 × 2)/ (100 × 5)
= 36
Amount = P + S.I
Substituting the values
= 900 + 36
= 936

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Re-Attempt Weekly Quiz Competition

Simple Interest Test - 7

Result

Simple Interest Test - 7

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SHARING IS CARING

Question 1

SI = 162, Amount = 462

SI = 152, Amount = 452

SI – 262, Amount = 562

SI = 62, Amount = 362

Solution

Question 2

1100

800

1000

1050

Solution

Question 3

SI = 298.40, Amount = 918.40

SI = 98.40, Amount = 718.40

SI = 398.40, Amount = 1018.40

SI = 198.40, Amount = 818.40

Solution

Question 4

3 (2/3) years

2 (2/3) years

4 (2/5) years

3 (1/4) years

Solution

Question 5

2 (1/12) years

3 (5/12) years

2 (5/12) years

None of the above

Solution

Question 6

SI = 8.56, AMOUNT = 548.56

SI = 7.56, AMOUNT = 547.56

SI = 14, AMOUNT = 554

SI = 39.2, AMOUNT = 579.2

Solution

Question 7

SI = 259.2, AMOUNT = 979.2

SI = 249.2, AMOUNT = 949.2

SI = 159.2, AMOUNT = 879.2

SI = 279.8, AMOUNT = 969.8

Solution

Question 8

6 (1/4) years

5 years

5 (1/4) years

6 years

Solution

Question 9

2500

3000

3500

3200

Solution

Question 10

5%

6%

7%

4%

Solution

Question 11

6%

5%

7%

4.5%

Solution

Question 12

3458.25

4965.33

5469.22

3644.25

Solution