AAO Exam-CT 12: Quant (Interest)

AAO Exam-CT 12-Quant (Interest)

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1. Simple interest on certain sum at the rate of 10% per annum after 2 years is Rs. 800. Find the compound interest on the sum at the rate of 10% per annum for 2 years.

Rs. 850
Rs. 900
Rs. 880
Rs. 840
None of these

Solution

Given:

Simple interest = Rs. 800, at the rate of 10% and time = 2 years

Formula used:

If P = Principal, R = Rate of Interest, and T = Time Period; then:

Simple interest = (P × R × T)/100, and

Compound interest = P(1 + R/100)t - P

Calculation:

800 = (P × 10 × 2)/100

⇒ P = 4,000

Now, C.I = 4,000[(1 + 10/100) - 1] = 4,000 × (21/100) = Rs. 840

∴ Compound interest is Rs. 840

Alternate Method

SI for 2 years = 400

[We know that the SI will be uniform for the given period]

SI for first year = 400 and SI for 2nd year = 400

We also know that, SI and CI for first year is same

∴ CI for First year = 400

CI for Second year = 400 + 400 × 10% = 440

Hence, Total CI = Rs. (400 + 440) = Rs. 840

2. Compound interest on a certain sum at the rate of 12% per annum after 2 years is Rs. 142464. Find the simple interest on the sum at the rate of 15% per annum for 7 years.

Rs. 5,00,000
Rs. 5,50,000
Rs. 5,88,000
Rs. 5,80,000
None of these

Solution

Given:

Compound interest = Rs. 142464, at the rate of 12% and time = 2 years

Formula used:

If P = Principal, R= Rate of Interest, and T = Time Period, then:

Simple interest = (P × R × T)/100, and

Compound interest = P(1 + R/100)t - P

Calculation:

C.I = P[(1 + 12/100)² - 1]

⇒ 142464 = P[(28/25)² - 1]

⇒ 142464 = P(159/625)

⇒ P = 142464 × (625/159)

⇒ P = Rs. 560000

So, S.I = (560000 × 15 × 7)/100 = Rs. 588000

∴ The simple interest on the sum at rate of 155 per annum for 7 years is Rs. 588000

3. Prathamesh borrowed certain amount from Gururaj. The difference between simple and compound interests compounded annually on that sum of money for 2 years at 4% per annum is Rs. 1. The sum is:

Rs 625
Rs 500
Rs 600
Rs 450
None of these

Solution

Given:

Compound Interest – simple interest = 1

Formula:

Compound interest = P × (1 + R/100)ⁿ − P

Simple Interest = principal × rate × time/100

Calculations:

Let the sum be Rs. x then,

Compound Interest = x(1 + 4/100)² – x

⇒ (676/625)x – x

⇒ (51/625) x

Simple Interest = x × 4 × 2/100

⇒ 2x/25

⇒ 51x/625 − 2x/25 = 1

⇒ x = 625

∴ The sum is Rs 625

4. What annual installment will discharge a debt of 22800 due in 10 years at 20% simple interest?

1500
1400
1200
800
None of the these

Solution

Given:

A debt of 22800 due in 10 years at 20% simple interest

Formula used

If borrowed amount be B and it is paid in equal installments

B = na + (ra/100 × Y) × n(n-1)/2

Where Y= number of installments per annum

Calculation

We know that,

B = na + (ra/100 × Y) × n(n-1)/2

Where Y= number of installments per annum

a = annual installment

Here M = 22800, y=1, r =20, n=10,

Put the value,

22800 = 10 a + 20a/100 × 10(10 -1)/2

22800 = 10a + 9a

22800 = a[19]

a = 1200

∴ Annual installment is Rs. 1200

Alternate Method

Given:

Rate = 20%

Time = 10 years

Amount = 22800

Concept used:

Annual installment = (100 × A)/[100 × T + {R × T( T – 1)}/2]

Where,

A → Amount

T → Time

R → Rate

Calculations:

Annual installment = (100 × A)/[100 × T + {R × T(T – 1)/2}]

⇒ (100 × 22800)/[100 × 10 + {20 × 10(10 – 1)/2}]

⇒ (100 × 22800)/{1000 + (20 × 10 × 9/2)}

⇒ 2280000/(1000 + 900)

⇒ 2280000/1900

⇒ Rs. 1200

∴ Annual installment is Rs. 1200

5. Keya borrows Rs 25000 from shree at 10% compound rate of interest. At the end of each year she pays Rs 8000 back to shree. At the end of year 3 how much amount would be left for her to pay?

Rs 6795
Rs 5845
Rs 7965
Rs 4795
Rs 7435

Solution

Formula used:

C.I= P ((1 + (r / 100))^N -1))

Where, P = Principal amount,

R = Rate of interest

N = Number of years

Calculation:

Amount to be paid at end of year 1 = 25000 × (1 + 0.10)

⇒ Amount = 27500

⇒ P for Year 2 = 27500 – 8000

⇒ P = 19500

⇒ A for year 2 = 19500 × 1.10

⇒ A = 21450

⇒ P for year 3 = 21450 – 8000

⇒ P = 13450

⇒ A for year 3 = 13450 × 1.10

⇒ A = 14795

⇒ P for year 4 = 14795 – 8000

⇒ P for year 4 = 6795

∴ at end of year 3 keya will have to pay Rs 6795 to shree

6. Find the amount when Rs. 5000 is put on compound interest for 3 years at 10%, 15%, and 20% for first, second, and third year respectively.

8000
7590
4320
6520
7500

Solution

Formula Used:

Amount = P (1 + R₁/100) (1 + R₂/100) (1+R₃/100)

Calculations:

⇒ Amount = 5000 (1+10/100) (1+15/100) (1+20/100)

⇒ 5000 × 11/10 × 23/20 × 6/5

⇒ Rs. 7590

∴ Amount is Rs. 7590

7. Ravi invests Rs. 50,000 at 3% per annum for 2 years at simple interest. After he had completed 2 years, he also invests all the amount at 15% per annum for 2 years on compound interest. What is the total interest made by Ravi?

20,092.5
25,007.5
17,092.5
20,095
17,095

Solution

Calculation:

SI = (50000 × 3 × 2)/100

⇒ 3000

Amount = 50,000 + 3000

⇒ 53,000

Principal for CI = 53,000

CI = 53,000 × (1 + 15/100)² – 53,000

⇒ 53,000 × (115/100)² – 53000

⇒ 53,000 × (115/100) × (115/100) – 53000

⇒ 17,092.5

∴ Total interest = 3000 + 17092.5

⇒ Rs. 20,092.5

The total interest made by Ravi is Rs. 20,092.5.

8. Two different mutual funds give compound interest of 20% and 10% per annum respectively and ratio of amount received by Anik by investing same amount in these two different mutual funds is 1728 : 1331. Find out the time for which he invested the money.

3 years
4 years
2 years
5 years
1 year

Solution

GIVEN :

Rate of interest for two funds are 20% and 10% respectively and after some years ratio of amount received is 1728 : 1331

FORMULA USED :

$A = P {(1 + \frac{r}{100})}^{n}$

Where, A = amount (principle + interest), P = principle, r = rate of interest, n = number of years

ASSUMPTION :

Let us assume the principle invested by Anik in each of the funds is P and the time of investment is n.

CALCULATION :

Amount after n years in first fund = $P {(1 + \frac{20}{100})}^{n}$

Amount after n years in second fund = $P {(1 + \frac{10}{100})}^{n}$

ATQ,

P(1+20100)n/P(1+10100)n=1728/1331

⇒ 12n11n=1728/1331

⇒ $\frac{12^{n}}{11^{n}} = \frac{12^{3}}{11^{3}}$

⇒ n = 3 Time of investment = 3 years

9. Three equal installments, each of Rs. 300, were paid at the end of the year on a sum borrowed at 20% compounded annually. Find the sum.

Rs. 530.945
Rs. 631.945
Rs. 603.945
Rs. 731.945
Rs. 562.455

Solution

Given:

The Rate of interest is 20 %

The Installment is Rs. 300

Number of installment is 3

Formula used:

P( 1+ R/100)ⁿ= X( 1+ R/100)^n-1 + X( 1+ R/100)^n-2 + X( 1+ R/100)^{n -3} + X( 1+ R/100)^n-4 + ----

Where,

P = principal

R = Rate

n = number of installments

X = amount of installment

Calculation:

P(1 + 20/100)³ = 300(1+20/100)² + 300(1 + 20/100)¹+ 300

⇒ P(1.2)³ = 300 × 1.44 + 300 × 1.2 + 300

⇒ P(1.2)³ = 432 + 360 + 300

⇒ P = 631.945

∴ The sum is Rs. 631.95.

10. At a certain period of time the ratio of principal and amount is 9 : 17 at a certain rate of simple interest, after 8 years this ratio will become 3 : 7. Find the rate of interest.

$4 \frac{4}{9} %$
$5 \frac{5}{9} %$
$2 \frac{16}{17} %$
3%
$2 \frac{4}{9} %$

Solution

Given:

Ratio of principal and amount = 9 : 17

Ratio of principal and amount after 8 years = 3 : 7

Formula used:

I=P×t×r100

Where P = Principal, t = time, r = rate of interest, I = interest

Calculation:

Let principal and amount be 9x and 17x

After 8 years it will become 3y and 7y

⇒ As we know principal will be same for the cases

9x = 3y

⇒ x/y = 3/9 = 1/3

So, amount will be 17x = 17

⇒ 7y = 21

⇒ Interest = (21 – 17) = 4

⇒ 4=9×8×r100

⇒ r = 50/9%

∴ Rate of Interest is 559%

Interest always calculated on principal in case of simple interest.

⇒ 4=17×8×r100=5017=21617 [Wrong]

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