Simple Interest Test

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Simple Interest Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

The interest on a sum of money at the end of 4 years is 3/5 of the sum. What is the rate percent?

A
16%

B
14%

C
12%

D
15%

Solution

Consider the sum P = ₹ 100
S.I = 100 × 3/5 = ₹ 60
T = 4 years
We know that
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (60 × 100 )/ (100 × 4)
So we get
= 15% p.a.
Question 2
1 / -0

Amount of Rs. 10600 was invested by Mr Lalit dividing it into two different investment schemes A and B at a simple interest rate of 12% and 15%. What was the amount in plan B if the amount of interest earned in two years was Rs. 2820.

A
2560.33

B
2380

C
2466.67

D
2655

Solution

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (10600 – x).
Then, [(x) × 15 × 2]/100 + [(10600 – x) × 12 × 2]/100 = 2820
30x – 24x = 282000 – (10600 x 24)
6x = 48800
x = 8133.33
So, sum invested in Scheme B = Rs. (10600 – 8133.33)
= Rs. 2466.67
Question 3
1 / -0

A sum of ₹ 1,400 becomes ₹ 2,240 in 3 years. Find: the rate of interest

A
15%

B
14%

C
12%

D
20%

Solution

It is given that
P = ₹ 1400
A = ₹ 2240
We know that
S.I = A – P
Substituting the values
= 2240 – 1400
= ₹ 840
T = 3 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (840 × 100)/ (1400 × 3)
= (280/14)
So we get
= 20% p.a.
Question 4
1 / -0

Find the simple interest, when: Principal = Rs 21000, Rate of Interest = 16% per annum and Time = 3 months.

A
Rs 840

B
Rs 960

C
Rs 780

D
Rs 860

Solution

Given Principal = Rs 21000, Rate of Interest = 16% per annum and Time = 3 months = (3/12) = (1/4) years
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (21000 × (1/4) × 16)/100
SI = (21000 × 1 × 16)/100 × 4
= Rs 840
Question 5
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 6
1 / -0

A sum amounts to ₹ 3648 in 4 years at 5% p.a. simple interest. Find the time in which the same sum will double itself at the same rate of interest.

A
20 years

B
10 years

C
15 years

D
8 years

Solution

Consider P = ₹ 100
R = 5% p.a.
T = 4 years
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (100 × 5 × 4)/ 100
= ₹ 20
Here amount = 100 + 20 = ₹ 120
If the amount is ₹ 120 then principal is ₹ 100
If the amount is ₹ 3648 then principal = (100 × 3648)/ 120
= ₹ 3040
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
Question 7
1 / -0

What sum of money lent out at 8% for 5 years will produce the same interest as ₹ 500 lent out at 5% for 6 years?

A
425

B
375

C
385

D
415

Solution

It is given that
P = ₹ 500
R = 5%
T = 6 years
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (500 × 5 × 6)/ 100
= ₹ 150
It is given that
S.I = ₹ 150
R = 8%
T = 5 years
We know that
Sum P = (S.I × 100)/ (R × T)
Substituting the values
= (150 × 100)/ (8 × 5)
So we get
= ₹ 375
Question 8
1 / -0

Find the simple interest, when: Principal = Rs 7500, Rate of Interest = 8% per annum and Time = 6 months.

A
250

B
350

C
300

D
450

Solution

Given Principal = Rs 7500, Rate of Interest = 8% per annum and Time = 6 months = ½ years
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (7500 × ½ × 8)/100
SI = (7500 × 1 × 8)/100 × 2
= Rs 300
Question 9
1 / -0

A sum of money is lent for 8 years at R% simple interest per annum. If the interest earned be one-fifth of the money lent, find the value of R.

A
4 %

B
2 %

C
3 %

D
2.5%

Solution

Consider the sum P = ₹ 100
We know that
S.I = 1/5 × 100 = ₹ 20
T = 8 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (20 × 100)/ (100 × 8)
So we get
= 2.5%
Question 10
1 / -0

Find the simple interest, when: Principal = Rs 4000, Rate of Interest = 12% per annum and Time = 73 days.

A
108

B
96

C
88

D
72

Solution

Given
Principal = Rs 4000, Rate of Interest = 12% per annum and
Time = 73 days = (73/365) days
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (4000 × (73/365) × 12)/100
SI = (4000 × 73 × 12)/100 × 365
= Rs 96
Question 11
1 / -0

A money lender claims to be lending at simple interest, but he adds the interest every 6 months in the calculation of principal. The rate of interest charged by him is 16%. What will be the effective rate of interest?

A
16.64 %

B
14.48 %

C
15.45 %

D
12.25 %

Solution

Let the sum be Rs. 100.
Then,
Simple interest for 1st 6 months = Rs. [100 × 16 × 1]/[100 × 2] = Rs. 8
Simple interest for last 6 months = Rs. [108 × 16 × 1]/[100 x 4] = Rs.8.64
So, amount at the end of 1 year = Rs. (100 + 8 + 8.64) = Rs. 116.64
Effective rate = (116.64 – 100) = 16.64%
Question 12
1 / -0

A sum of ₹ 1,600 becomes ₹ 2560 in 3 years. Find the rate of interest.

A
15%

B
20%

C
12%

D
16%

Solution

SI = ( A – P )
SI = 2560 – 1600
SI = 960
R = ( 960 × 100 ) / (1600 × 3)
R = 20%
Question 13
1 / -0

P and Q invest ₹ 28,000 and ₹ 22,000 respectively at the same rate of interest per year. If at the end of 3 years, P gets ₹ 2160 more interest than Q, find the rate of interest.

A
15%

B
12%

C
9%

D
8%

Solution

It is given that
P’s investment (P1) = ₹ 28,000
Q’s investment (P2) = ₹ 22,000
T = 3 years
Consider the rate of interest = x %
So we get
P’s interest (S.I) = (P × R × T)/ 100
Substituting the values
= (28000 × x × 3)/ 100
= ₹ 840x
Q’s interest = (P × R × T)/ 100
Substituting the values
= (22000 × x × 3)/ 100
= ₹ 660x
Here the difference in their interest = 840x – 660x = ₹ 180x
The difference given = ₹ 2160
So we get
180x = 2160
x = 2160/180
x = 12%
So the rate of interest = 12% p.a.
Question 14
1 / -0

The simple interest earned on a certain sum in 4 years is 20% of the sum. Find the rate of interest.

A
5%

B
4%

C
6%

D
4.5%

Solution

Consider sum P = ₹ 100
We know that
SI = 20/100 × 100 = ₹ 20
T = 4 years
Here
Rate = (S.I × 100)/ (P × T)
Substituting the values
= (20 × 100)/ (100 × 4)
So we get
= 5%
Question 15
1 / -0

A sum amounts to ₹ 2882 in 3 years at 5% p.a. simple interest. Find the sum.

A
2700

B
2506.08

C
2406

D
2308.4

Solution

Consider P = ₹ 100
R = 5% p.a.
T = 3 years
We know that
S.I = (P × R × T)/ 100
Substituting the values
= (100 × 5 × 3)/ 100
= ₹ 15
Amount = Rs 100+15 = 115
If amount is 2882 then principal is
(100 x 2882)/115
= 2506.08

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Re-Attempt Weekly Quiz Competition

Simple Interest Test - 8

Result

Simple Interest Test - 8

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SHARING IS CARING

Question 1

16%

14%

12%

15%

Solution

Question 2

2560.33

2380

2466.67

2655

Solution

Question 3

15%

14%

12%

20%

Solution

Question 4

Rs 840

Rs 960

Rs 780

Rs 860

Solution

Question 5

₹ x

₹ 100x

₹ (100/x)

₹ (100/x2)

Solution

Question 6

20 years

10 years

15 years

8 years

Solution

Question 7

425

375

385

415

Solution

Question 8

250

350

300

450

Solution

Question 9

4 %

2 %

3 %

2.5%

Solution

Question 10

108

96

88

72

Solution

Question 11

16.64 %

14.48 %

15.45 %

12.25 %

Solution

Question 12

15%

20%

12%

16%

Solution