AAO Exam-CT 12: Quant (Interest)

AAO Exam-CT 12-Quant (Interest)

0  1542
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

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

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

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

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

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

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

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

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+r100)n$�=�{\left(1+\frac{\mathrm{r}}{100}\right)}^{�}$

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+20100)n$�{\left(1+\frac{20}{100}\right)}^{�}$

Amount after n years in second fund = P(1+10100)n$�{\left(1+\frac{10}{100}\right)}^{�}$

ATQ,

P(1+20100)n/P(1+10100)n=1728/1331$�{\left(1+\frac{20}{100}\right)}^{�}\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}\mathrm{P}{\left(1+\frac{10}{100}\right)}^{�}=1728\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}1331$

12n11n=1728/1331$\frac{{12}^{�}}{{11}^{�}}=1728\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}1331$

12n11n=123113$\frac{{12}^{�}}{{11}^{�}}=\frac{{12}^{3}}{{11}^{3}}$

⇒ 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

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.
• 449%$4\frac{4}{9}\mathrm{%}$
• 559%$5\frac{5}{9}\mathrm{%}$
• 21617%$2\frac{16}{17}\mathrm{%}$
• 3%
• 249%$2\frac{4}{9}\mathrm{%}$

Given:

Ratio of principal and amount = 9 : 17

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

Formula used:

I=P×t×r100$\mathrm{I}=\frac{\mathrm{P}\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}}{100}$

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$4=\phantom{\rule{thickmathspace}{0ex}}\frac{9\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}8\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}}{100}$

⇒ r = 50/9%

∴ Rate of Interest is 559%$5\frac{5}{9}\phantom{\rule{thickmathspace}{0ex}}\mathrm{%}$

Interest always calculated on principal in case of simple interest.

⇒ 4=17×8×r100=5017=21617$4=\phantom{\rule{thickmathspace}{0ex}}\frac{17×\phantom{\rule{thickmathspace}{0ex}}8\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}}{100}=\frac{50}{17}=2\frac{16}{17}$ [Wrong]