Simple Interest Test

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Simple Interest Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

At simple interest a sum becomes (3/2) of itself in 3 years. The rate of interest per annum is

A
60%

B
7½

C
55/6%

D
50/3%

Solution

Let th required sum be ₹ x
Rate of interest = r %
Time = 3 years
Amount = (3/2) × sum
Amount = principal + SI
(3/2) × x = x + [(P × R × T)/100]
= (3/2) x = x + [(x × r × 3)/ (100)]
= (3/2) = (1 + (r×3/100))
= r×3/100=1/2
= r = 50/3%
Question 2
1 / -0

Principal = 5200, SI = 975 and time = 2 ½ years.

A
10 (2/3) %

B
13 (3/4) %

C
9 (1/4) %

D
11 (2/3) %

Solution

Given: – P = 4800, SI = 840, t = 1 ½ years = 3/2
We know that,
R = (100 × SI) / (P × T)
= (100 × 840)/ (4800 × (3/2))
= (100 × 840 × 2) / (4800 × 3)
= (100 × 280) / (2400)
= (280) / (24)
= (35/3)
= 11 (2/3)
= 11 (2/3) % p.a.
Question 3
1 / -0

At what rate percent per annum simple interest will a sum double itself in 100 years?

A
8%

B
1%

C
12%

D
12 ½ %

Solution

Let the sum be ₹ x and the rate be r%
Then,
Amount = 2x
= P + SI = 2x
= P + [(P × R × T)/100] = 2x
= x (1 + _{r × 100)/ 100} = 2x
= (100 + (100 × r))/100 = 2
= 100 × r = 200 – 100
= r = 100/100
= r = 1 %
Question 4
1 / -0

Principal = 3000, Amount = 4350 and rate = 15% p.a.

A
3 years 6 months

B
2 years

C
3 years

D
4 years 2 months

Solution

Given: – P = 3000, Amount = 4350, R = 15%, T =?
We Know That,
SI = A – P
= 4350 – 3000
= 1350
T = (100 × SI) / (P × R)
= (100 × 1350) / (3000 × 15)
= (1 × 135) / (3 × 15)
= (9/ 3)
= 3 years
Question 5
1 / -0

Shashi borrowed 12000 from the State Bank of India for 2 years 4 months at 8% per annum. What amount will clear off her debt?

A
15360

B
14880

C
14240

D
13980

Solution

From the question,
Shashi borrowed 12000 from the State Bank of India (Principal)
Time = 2 years 4 months
We know that, 1 year = 12 months
∴ 2 years 4 months = (28/12) = (7/3)
SI = 8 % .p.a.
First we have to find Simple Interest,
SI = (P × R × T)/100
= (12000 × 8 × (7/3))/ 100
= 12000 × 8 × (7/3) × (1/100)
= (12000 × 8 × 7 × 1)/ (3 × 100)
= (40 × 8 × 7 × 1)/ (1 × 1)
= 2240
Amount = (principal + SI)
= (12000 + 2240)
= 14240
∴The amount will clear off her debt is 14240
Question 6
1 / -0

Principal = ₹ 5300, rate = 4% p.a. and time = 2 ½ years. S.I.=?

A
530

B
470

C
490

D
510

Solution

Given: – P = ₹ 5300, R = 4% p.a. and time = 2 ½ years = (5/2)
If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest.
SI = (P × R × T)/100
= (5300 × 4 × (5/2))/ 100
= 5300 × 4 × (5/2) × (1/100)
= (5300 × 4 × 5 × 1)/ (2 × 100)
= (2650 × 4 × 1 × 1)/ (1 × 20)
= (2650 × 1 × 1 × 1)/ (1 × 5)
= (2650 / 5)
= ₹ 530
Question 7
1 / -0

P=₹ 8000, amount = ₹ 8360, R=6% per annum, time =?

A
8 months

B
9 months

C
1 ¼ years

D
1½ years

Solution

Because,
Given: – P = ₹ 8000, A = ₹ 8368, R = 6%
We Know That,
Amount = Principal (1+ _{Rate × time)/100}
= (8360/ 8000) = 1 + ((6 × t)/100)
= (8360/ 8000) -1 = ((6 × t)/100)
= t = [(8360 – 8000)/ 8000] × (100/6)
= (360/8000) × (100/6)
= (6/8) × 12 months
= 9 months
Question 8
1 / -0

Principal = ₹ 19080, SI = ₹ 1908 and rate = 4% p.a., T=?

A
3

B
3.5

C
2

D
2.5

Solution

Given: – P = ₹ 19080. SI = ₹ 1908, R = 4%, T =?
We Know That,
T = (100 × SI) / (P × R)
= (100 × 1908) / (19080 × 4)
= (25 ×1908) / (19080)
= (47700/ 19080)
= 2.5years
= 2 ½ years
Question 9
1 / -0

Principal = ₹ 4800, rate = 15 % p.a. and time = 5 months. Amount=?

A
4850

B
4950

C
5100

D
5000

Solution

Given: – P = ₹ 4800, R = 15 % p.a. and time = 5 months = (5/12) years
If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest.
SI = (P × R × T)/100
= (4800 × (15) × (5/12))/ 100
= 4800 × (15) × (5/12) × (1/100)
= (4800 × 15 × 5 × 1)/ ( 12 × 100)
= (48 × 15 × 5 × 1)/ (12)
= (3600)/ (12)
= ₹ 300
Amount = (principal + SI)
= (4800 + 300)
= ₹ 5100
Question 10
1 / -0

Find the rate when: Principal = 4280, SI = 900 and time = 3 years.

A
5.65%

B
4.56%

C
6.23%

D
5.63%

Solution

Given: – P = 4280, SI = 800, t = 3 years.
We know that,
R = (100 × SI) / (P × T)
= (100 × 800)/ (4280 × 3)
= (100 × 800) / (4280 × 3)
= (80000/ 4280)
= 6.23 %
Question 11
1 / -0

Principal = 4650, amount = 5424.20 and time = 4 years.

A
4.16 %

B
5.16 %

C
6.18%

D
5 .25%

Solution

Given: – P = 4650, amount = 5424.20, t = 4 years.
We know that,
SI = A – P
= 5424.20 – 4650
= 774.2
R = (100 × SI) / (P × T)
= (100 × 774.20)/ (4650 × 4)
= (100 × 77420) / (4650 × 4 ×100)
= (3871 /465 × 2)
= (3871/930)
= 4.16 % p.a.
Question 12
1 / -0

The simple interst at x % per annum for x years will be ₹ x on a sum of

A
₹ X

B
₹ 100X

C
₹ (100/X)

D
₹ (100/x2)

Solution

From the question,
SI = ₹ x
Rate = x % .p.a.
Time = x years
Then,
SI = (P × R × T)/ 100
X = (P × X× X)/ 100
P = (100X)/ (X × X)
P = ₹ (100/X)
Question 13
1 / -0

Principal = ₹ 3000, rate = 6% p.a. and time = 3 years 3 months. Amount=?

A
3375

B
3585

C
3786

D
2875

Solution

Given: – P = ₹ 3000, R = 6% p.a. and
Time = 3 years 3 months
We know that, 1 year = 12 months
∴ 3 years 3 months = (39/12) = (13/4)
If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest.
SI = (P × R × T)/100
= (3000 × 6 × (13/4))/ 100
= 3000 × 6 × (13/4) × (1/100)
= (3000 × 6 × 13 × 1)/ (4 × 100)
= (15 × 3 × 13 × 1)/ (1 × 1)
= (15 × 3 × 13 × 1)
= ₹ 585
Amount = (principal + SI)
= (3000 + 585)
= ₹ 3585
Question 14
1 / -0

Find the time when: Principal = ₹ 3200, SI = ₹ 1152 and rate = 12% p.a.

A
3

B
2.5

C
2

D
3.5

Solution

Given: – P = ₹ 3200. SI = ₹ 1152, R = 12%, T =?
We Know That,
T = (100 × SI) / (P × R)
= (100 × 1152) / (3200 × 12)
= (1152) / (32 × 12)
= (1152) / (384)
= 3 years
Question 15
1 / -0

Principal = ₹ 3200, rate = 12% p.a. and time = 2 years. S.I=?

A
768

B
648

C
750

D
848

Solution

Given: – P = ₹ 3200, R = 12% p.a. and time = 2years.
If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest.
SI = (P × R × T)/100
= (3200 × 12 × 2)/ 100
= (32 × 12 × 2)/ 1
= ₹768

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Simple Interest Test - 6

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Simple Interest Test - 6

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SHARING IS CARING

Question 1

60%

7½

55/6%

50/3%

Solution

Question 2

10 (2/3) %

13 (3/4) %

9 (1/4) %

11 (2/3) %

Solution

Question 3

8%

1%

12%

12 ½ %

Solution

Question 4

3 years 6 months

2 years

3 years

4 years 2 months

Solution

Question 5

15360

14880

14240

13980

Solution

Question 6

530

470

490

510

Solution

Question 7

8 months

9 months

1 ¼ years

1½ years

Solution

Question 8

3

3.5

2

2.5

Solution

Question 9

4850

4950

5100

5000

Solution

Question 10

5.65%

4.56%

6.23%

5.63%

Solution

Question 11

4.16 %

5.16 %

6.18%

5 .25%

Solution

Question 12

₹ X

₹ 100X

₹ (100/X)

₹ (100/x2)

Solution