Aptitude
Topic: Simple Interest and Compound Interest
Study Material
Concept 1: Understanding important terms.
Concept 2: Understanding Simple interest & Compound interest.
Compound interest formula is given by: Where, A= Amount, P= Principal, R= Rate of interest and T= Time (in years) Derivation of compound interest formula: Concept 3: Compound Interest VS Simple Interest:
Concept 4: Understanding calculation of SI & CI though percentages:
Example 1: What is SI of Rs.800 on 5% per annum for 3 years? We can either put formula and calculate as below: SI= (800*5*3)/100 = 120 Or, As we know that in simple interest, interest is always calculated on principal so we can just calculate 15% (5% for 3 years) of 800 which is 120. Thus, in this way we don’t have to remember the formula and we can simply get answers by using concept of percentages. Example 2: What will be the CI obtained on an amount of Rs.4800 at the rate of 5% for 3 years? We can either put formula and calculate as below: Amount= 4800[1+ (5/100)]^3 = 5556.6 ; CI= (5556.6 – 4800)= 756.6 Or, We can simply use concept of multiplying factor learnt in percentages, since 4800 is increasing by 5% every year for constant 3 years, we can write it as 4800* (1.05)^3 = 5556.6 ; CI = (5556.6-4800) = 756.6. Thus, in this way we don’t have to remember the complex formula of CI and we easily solve problems based on CI through concept of multiplying factor learnt in session of percentage. |
Problems on Simple Interest and Compound Interest
Foundation
1. |
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is: |
2. |
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B? |
3. |
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum? |
4. |
How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest? |
5. |
Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest? |
MODERATE
1. |
A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest? |
2. |
An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes: |
3. |
A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is: |
4. |
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is: |
5. |
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate? |
HOTS-HIGH ORDER THINKING SKILLS
1. |
What is the difference between the compound interests on Rs. 5000 for 1 |
2. |
The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is: |
3. |
What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.? |
4. |
At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years? |
5. |
The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is: |
SOLUTIONS
FOUNDATION
1. Answer: Option C
Explanation:
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.
2. Answer: Option A
Explanation:
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).
Then, |
|
x x 14 x 2 |
|
+ |
|
(13900 – x) x 11 x 2 |
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= 3508 |
100 |
100 |
28x – 22x = 350800 – (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 – 7500) = Rs. 6400.
3. Answer: Option D
Explanation:
Principal |
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= Rs. 8925. |
4. Answer: Option B
Explanation:
Time = |
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100 x 81 |
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= 4 years. |
450 x 4.5 |
5. Answer: Option B
Explanation:
Let rate = R% and time = R years.
Then, |
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1200 x R x R |
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= 432 |
100 |
12R2 = 432
R2 = 36
R = 6.
MODERATE
1. Answer: Option D
Explanation:
S.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = |
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100 x 3000 |
|
= 6% |
12500 x 4 |
2. Answer: Option B
Explanation:
Let the sum be Rs. 100. Then,
S.I. for first 6 months = Rs. |
|
100 x 10 x 1 |
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= Rs. 5 |
100 x 2 |
S.I. for last 6 months = Rs. |
|
105 x 10 x 1 |
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= Rs. 5.25 |
100 x 2 |
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%
3. Answer: Option B
Explanation:
Amount |
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= Rs. 3321. |
C.I. = Rs. (3321 – 3200) = Rs. 121
4. Answer: Option A
Explanation:
Let the sum be Rs. x. Then,
C.I. = |
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x |
|
1 + |
4 |
|
2 |
– x |
|
= |
|
676 |
x |
– x |
|
= |
51 |
x. |
100 |
625 |
625 |
S.I. = |
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x x 4 x 2 |
|
= |
2x |
. |
100 |
25 |
|
51x |
– |
2x |
= 1 |
625 |
25 |
x = 625.
5. Answer: Option C
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
|
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100 x 60 |
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= 10% p.a. |
100 x 6 |
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
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= 3972. |
SOLUTIONS
FOUNDATION
1. Answer: Option C
Explanation:
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.
2. Answer: Option A
Explanation:
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).
Then, |
|
x x 14 x 2 |
|
+ |
|
(13900 – x) x 11 x 2 |
|
= 3508 |
100 |
100 |
28x – 22x = 350800 – (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 – 7500) = Rs. 6400.
3. Answer: Option D
Explanation:
Principal |
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= Rs. 8925. |
4. Answer: Option B
Explanation:
Time = |
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100 x 81 |
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= 4 years. |
450 x 4.5 |
5. Answer: Option B
Explanation:
Let rate = R% and time = R years.
Then, |
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1200 x R x R |
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= 432 |
100 |
12R2 = 432
R2 = 36
R = 6.
MODERATE
1. Answer: Option D
Explanation:
S.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = |
|
100 x 3000 |
|
= 6% |
12500 x 4 |
2. Answer: Option B
Explanation:
Let the sum be Rs. 100. Then,
S.I. for first 6 months = Rs. |
|
100 x 10 x 1 |
|
= Rs. 5 |
100 x 2 |
S.I. for last 6 months = Rs. |
|
105 x 10 x 1 |
|
= Rs. 5.25 |
100 x 2 |
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%
3. Answer: Option B
Explanation:
Amount |
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= Rs. 3321. |
C.I. = Rs. (3321 – 3200) = Rs. 121
4. Answer: Option A
Explanation:
Let the sum be Rs. x. Then,
C.I. = |
|
x |
|
1 + |
4 |
|
2 |
– x |
|
= |
|
676 |
x |
– x |
|
= |
51 |
x. |
100 |
625 |
625 |
S.I. = |
|
x x 4 x 2 |
|
= |
2x |
. |
100 |
25 |
|
51x |
– |
2x |
= 1 |
625 |
25 |
x = 625.
5. Answer: Option C
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
|
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100 x 60 |
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= 10% p.a. |
100 x 6 |
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
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= 3972. |
HOTS-HIGH ORDER THINKING SKILLS
1. Answer: Option A
Explanation:
C.I. when interest |
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= Rs. 5304. |
C.I. when interest is |
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= Rs. 5306.04 |
-
Difference = Rs. (5306.04 – 5304) = Rs. 2.04
2. Answer: Option A
Explanation:
Amount = Rs. (30000 + 4347) = Rs. 34347.
Let the time be n years.
Then, 30000 |
|
1 + |
7 |
|
n |
= 34347 |
100 |
|
|
107 |
|
n |
= |
34347 |
= |
11449 |
= |
|
107 |
|
2 |
100 |
30000 |
10000 |
100 |
n = 2 years.
3. Answer: Option C
Explanation:
Amount |
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= Rs. 35123.20 |
-
C.I. = Rs. (35123.20 – 25000) = Rs. 10123.20
4. Answer: Option A
Explanation:
Let the rate be R% p.a.
Then, 1200 x |
|
1 + |
R |
|
2 |
= 1348.32 |
100 |
|
|
1 + |
R |
|
2 |
= |
134832 |
= |
11236 |
100 |
120000 |
10000 |
|
|
1 + |
R |
|
2 |
= |
|
106 |
|
2 |
100 |
100 |
|
R |
= |
106 |
100 |
100 |
R = 6%
5. Answer: Option B
Explanation:
P |
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1 + |
20 |
|
n |
> 2P |
|
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6 |
|
n |
> 2. |
100 |
5 |
Now, |
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6 |
x |
6 |
x |
6 |
x |
6 |
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> 2. |
5 |
5 |
5 |
5 |
So, n = 4 years.