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1.

An amount compounded at the rate of interest of 20% per annum matures to be Rs. 14400 after 2 years. Find the simple interest on the same amount at the end of 2 years.1). Rs. 20002). Rs. 25003). Rs. 40004). Rs. 6000

Answer»

<P>GIVEN data -

RATE of INTEREST = 20% per annum

Let the principle amount be P Rs.

When compounded annually -

As we know,

Matured amount = $(Principle\;amount \times {\left( {1 + \frac{{rate\;of\;interest}}{{100}}} \right)^{time\;of\;investment}})$

$(\RIGHTARROW 14400 = {\rm{P}} \times {\left( {1 + \frac{{20\;}}{{100}}} \right)^2})$

14400 = P × (1.2)2

∴ 14400 = 1.44P

∴ P = Rs. 10000

Simple interest on same amount for 2 years at 20% rate of interest-

Simple interest = $(\frac{{principle\;amount\; \times \;rate\;of\;interest\; \times \;time\;of\;investment}}{{100}})$

Simple interest = $(\frac{{10000\; \times \;20\; \times \;2}}{{100}})$

∴ Simple interest = Rs. 4000
Discussion
2.

1). 5%2). 8%3). 10%4). 15%

Answer»

On a sum of RS. P at a rate of R% per ANNUM,

Compound interest after n years = P[(1 + R/100)n - 1]

On a sum of Rs. 62400,

CI after 2 years = 62400[(1 + R/100)2 - 1]

⇒ 20124 = 62400[(1 + R/100)2 - 1]

⇒ 0.3225 = (1 + R/100)2 - 1

⇒ 1.3225 = (1 + R/100)2

⇒ 1.15 = 1 + R/100

⇒ R/100 = 1.15 - 1 = 0.15

⇒ R = 100 × 0.15 = 15%
Discussion
3.

Veer Invested X Rs. in SBI at the rate of 8% for 2 year and obtained a total simple interest of Rs. 6400. If he invested Rs. 10000 more for same period at the rate of 5% for first year and at 10% for 2nd year compound annually, then find the total compound interest obtained by Veer after 2 year.1). Rs. 60002). Rs. 50003). Rs. 77504). Rs. 7000

Answer»

PRINCIPAL = Rs. X

Rate of interest = 8%

Time = 2 years

According to PROBLEM,

⇒ X × 8 × 2/100 = 6400

⇒ 0.16X = 6400

⇒ X = 40000

In case of compound interest in 1st YEAR,

Principal = Rs. 40000 + 10000 = Rs. 50000

Rate of interest = 5%

∴ Amount = 50000(1 + 5/100) = Rs. 52500

In case of 2nd year,

Principal = Rs. 52500

Rate of interest = 10%

∴ Amount = 52500(1 + 10/100) = Rs. 57750

∴ Total compound interest obtained = Rs. (57750 - 5000) = Rs. 7750
Discussion
4.

Vijay invests a principal amount of Rs. 300000 in a fixed deposit for 2 years.The rate at which he invests is 20% per annum. What will be the simple interest earned by him at the end of his investment tenture?1). Rs. 220002). Rs. 130003). Rs. 1200004). Rs. 15600

Answer»

<P>PRINCIPAL = 300000

Rate = 20%

Time = 3 years

SI = (P × R × T ) / 100 = (300000 × 20 × 2 ) / 100 = 120000
Discussion
5.

The simple interest on a sum of money at 4% per annum for 2 years is Rs. 80. The compound interest in the same sum for the same period is1). Rs. 82.602). Rs. 82.203). Rs. 84.804). Rs. 81.60

Answer»
Discussion
6.

Manisha borrowed a total amount of Rs. 40000, part of it on simple interest rate of 16% per annum and remaining on simple interest rate of 12% per annum. If at the end of 3 years she paid in all Rs. 56320 to settle the loan amount, what was the amount borrowed at 16% per annum?1). Rs. 160002). Rs. 180003). Rs. 175004). Rs. 12000

Answer»

We know that,

Interest = Amount - Principal

Interest = 56320 - 40000 = Rs. 16320

Let’s assume the amount borrowed at rate of 16% per annum be X

⇒ Remaining amount = 40000 - X

We know the formula for simple Interest,

SI = (P × R × t)/100

Where,

SI = Simple Interest

P = Principal

R = Rate of Interest

t = TIME period

Here Rate of interest is different for each part, so we can calculate SI for each part then we will add SI for each part to calculate total SI

SI for first part of amount X,

Rate of Interest = 16%

⇒ SI1 = (X × 16 × 3)/100 = 48X/100

SI for SECOND part of amount (40000 - X),

Rate of Interest = 12%

$(\Rightarrow {\rm{S}}{{\rm{I}}_2} = \frac{{\left( {40000 - {\rm{X}}} \right) \TIMES 12 \times 3}}{{100}} = \frac{{1440000}}{{100}} - \frac{{36X}}{{100}} = 14400 - \frac{{36X}}{{100}})$

Total Simple Interest = SI1 + SI2 = Rs. 16320 (as calculated above)

⇒ (48X/100) + 14400 - (36X/100) = 16320

⇒ 12X/100 = 16320 - 14400

⇒ X = 1920/0.12 = Rs. 16000

∴ The amount borrowed at rate of 16% per annum = Rs. 16000
Discussion
7.

The simple interest on a certain sum of money after x years is Rs. 247.50 and after 2 more years is Rs. 412.50. Find the simple interest on the same sum at the same rate of interest after 10 years.1). Rs. 4952). Rs. 5773). Rs. 6604). Rs. 825

Answer»

Let the sum of money be RS. ’P’ and the RATE of interest be ‘R%’ PER annum

As we know, Simple interest

⇒ SI = (P × R × T)/100

Given, SI = Rs. 247.50 after T = x YEARS

⇒ 247.50 = (PR × x)/100

⇒ PR = 24750/x----(1)

Also, SI = Rs.412.50 after T = (x + 2) years

⇒ 412.50 = (PR × (x + 2))/100

⇒ PR = 41250/(x + 2)----(2)

Equating (1) and (2), we get,

⇒ 41250/(x + 2) = 24750/x

⇒ 41250x = 24750x + 49500

⇒ 16500x = 49500

⇒ x = 49500/16500

⇒ x = 3

Substituting in (1),

PR = 24750/3 = 8250

Hence, SI after T = 10 years,

SI = (PR × 10)/100

⇒ SI = (8250 × 10)/100

⇒ SI = Rs. 825
Discussion
8.

Find the compound interest on Rs. 20000 for 3 year, if the rate of interest is 5% for first year, 10% for the second year and 15% for the third year?1). Rs. 65152). Rs. 62653). Rs. 65554). Rs. 6665

Answer»

In 1st year,

PRINCIPAL = Rs. 20000

Rate of interest = 5%

∴ AMOUNT,

⇒ 20000(1 + 5/100)

⇒ 21000

In 2nd year,

Principal = Rs. 21000

Rate of interest = 10%

∴ Amount,

⇒ 21000(1 + 10/100)

⇒ 23100

In 3rd year,

Principal = Rs. 23100

Rate of interest = 15%

∴ Amount,

⇒ 23100(1 + 15/100)

⇒ 26565

∴ Interest EARNED = (26565 - 20000) = Rs. 6565
Discussion
9.

The difference between CI and SI on an amount of Rs. 15625 for 2 years is Rs. 625. What is the rate of interest per annum?1). 12%2). 15%3). 16%4). 20%

Answer»

ACCORDING to the question

CI - SI = 625

⇒ CI - SI = P(R/100)2

⇒ 15625 (R/100)2 = 625

⇒ (R/100)2 = 625/15625

⇒ R = 20%

∴ the REQUIRED RATE of interest is 20%
Discussion
10.

Find the difference between CI and SI Rs. 2000 at a rate of 6% for 3 years compounded annually?1). 22.032). 243). 114). 21.75

Answer»

For THREE years, difference, by formula

⇒ A = P (1 + R/100)3

⇒ A = 2000(1 + 6/ 100)3

⇒ A = 2000(106/100) 3

⇒ A = 2382.032

⇒ CI = 2382.032 - 2000 = 382.032

⇒ SI = (P × R × t)/(100)

⇒ SI = (2000 × 6 × 3)/100 = 360

∴ CI - SI = 382.032 - 360 = 22.032
Discussion
11.

A person lent a sum of Rs. 9000 on simple interest at 10% for 3 years. Find the difference in interest received by him, if he had lent the same sum on compound interest at 20% for 2 years.1). Rs. 19602). Rs. 11603). Rs. 12604). Rs. 2700

Answer»

Given,

Principal = 9000, Rate% = 10% and Time = 3 years

We KNOW that,

SIMPLE interest = (P×R×T)/100

⇒ SI = (9000×10×3)/100 = 2700

At compound interest,

$(Amount = P{\left[ {1 + \frac{{r\% }}{{100}}} \right]^{time}})$

$(Amount = 9000 \times {\left[ {1 + \frac{{20}}{{100}}} \right]^2})$

⇒ Amount = 12960

⇒ CI = Amount - Principal

⇒ CI = 12960 - 9000 = 3960

⇒ Difference of interest = CI - SI = 3960 - 2700 = 1260

Hence, 1260 will be the answer.

 
Discussion
12.

1). PQR = 12). P2 = QR3). Q2 = PR4). R = P2Q

Answer»

Let the rate be R and time be t in both cases.

As given, Q is the SI on P, we can write,

⇒ Q = P ? r ? t/100----- (1)

Also given that, R is SI on Q, we can write,

⇒ R = Q ? r ? t/100----- (2)

Dividing equation (2) by equation (1), we get,

⇒ Q/R = P/Q

∴ Q2 = PR
Discussion
13.

Quantity B: The simple interest on a certain amount for 5 years at 4% per annum is 1/4th of compound interest on Rs. 5500 for a year at 10%per annum. The difference between sum placed in Compound interest and sum placed in Simple interest is –1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

QUANTITY A:

PRINCIPAL = 6000

Principal amount BECOMES double after 4 YEARS then,

Using Compound interest formula,

Let P = principal, R = rate of interest and N = time period

Amount = P(1 + R/100)N

12000 = 6000(1 + R/100)4

2 = (1 + R/100)4

After 16 years it will becomes,

(1 + R/100)16 = ?

[(1 + R/100)4]4 = 24

24 = 16

In 16 years it will becomes 16 times = 6000 x 16 = 96000

Quantity B:

Let Principal amount for SI be Rs.a.

Simple interest = (a × 5 × 4)/100 = a/5

Compound interest calculated annually,

Compound interest = P(1 + R/100)N – P

Compound interest = 5500(1 + 10/100) – 5500

Compound interest = 5500 × 1.1 – 5500 = 550

Given,

Simple interest = (1/4) × Compound interest

Simple interest = 550/4

Simple interest = 137.5

Simple interest = a/5 = 137.5

a = 687.5

Difference between the sum placed in Compound interest and simple interest =

= 5500 – 687.5

= 4812.5

From above SOLUTION,

Quantity A > Quantity B
Discussion
14.

A sum was put at simple interest at a certain rate for 3 years. Had it been put at 3% higher rate, it would have fetched Rs. 27 more. The sum is?1). Rs. 4002). Rs. 2503). Rs. 3004). Rs. 500

Answer»

Formula for Simple Interest -

$(SI = \FRAC{{P \times R \times T}}{{100}})$

Where,

P = Principal

R = RATE of interest

T = Time period

Let the sum be Rs. x and the original rate be R%,

Then, $({\RM{SI}} = \frac{{{\rm{X}} \times {\rm{\;R}} \times {\rm{}}3}}{{100}})$

When rate is increased by 3%

Then, R = R + 3

And $(SI = \frac{{{\rm{X}} \times {\rm{}}\left( {{\rm{R}} + 3} \RIGHT){\rm{}} \times {\rm{}}3}}{{100}})$

According to the question, the difference between the two EQUATIONS is Rs. 27

$(\therefore \frac{{{\rm{X}} \times {\rm{}}\left( {{\rm{R}} + 3} \right){\rm{}} \times {\rm{\;}}3}}{{100}}{\rm{}} - {\rm{}}\frac{{{\rm{X}} \times {\rm{\;R}} \times {\rm{}}3}}{{100}} = 27)$

$(\Rightarrow \frac{{3{\rm{XR}} + 9{\rm{X}} - 3{\rm{XR}}}}{{100}} = 27)$

$(\Rightarrow \frac{{9{\rm{X}}}}{{100}} = 27)$

⇒ 9X = 2700

⇒ X = 2700/9

⇒ X = 300

∴ The sum is of Rs. 300.
Discussion
15.

If a sum of money becomes 3 times of itself in 18 years at simple interest, then calculate the rate percent per annum charged by the financer?1). 16.66%2). 18.5%3). 11.11%4). 22.22%

Answer»

Let principal amount be Rs. P

Thus, SIMPLE INTEREST charged = 3P – P = Rs. 2P

Simple interest = (Principal × Rate × Time)/100

⇒ 2P = (P × Rate × 18)/100

⇒ Rate = (200/18) = 11.11%

∴ Rate% = 11.11%
Discussion
16.

Raju makes a deposit of Rs. 50000 in the SBI for a period of 5/2 years. If the rate of interest is 20% per annum compounded half-yearly. Find the maturity value of the money deposited by him?1). 81525.502). 1244163). 80525.504). 79565.75

Answer»

rate of interest is 20% COMPOUNDED half-yearly

Rate of interest compounded annually = 10% and t will be 5 years

MATURITY value of MONEY at the end of 5/2 years

⇒ 5000(1 + 10/100)5 = 50000 (11/10)5

∴ Maturity value = 80525.50
Discussion
17.

A sum of Rs. 2600 is lent out in two parts in such a way that the simple interest on one part at 10% per annum for 5 years is equal to that on another part at 9% per annum for 6 years. The sum lent out at 9% is1). Rs. 5002). Rs. 10003). Rs. 12504). Rs. 1350

Answer»

Total AMOUNT is Rs. 2600

Let the part given on 10% per ANNUM is = P

Then sum lent at 9% per annum is = 2600 – P

S.I. on amount P for 5 years at 10% = $(\frac{{P \TIMES 10 \times 5}}{{100}})$

S.I. on amount (2600 – p) for 6 yr at 9 % $(= \frac{{\left( {2600\;-\;p} \RIGHT) \times 6 \times 9}}{{100}})$

Given, Both S.I. are equal:

$(\frac{{\left( {P \times 10 \times 5} \right)}}{{100}} = \frac{{\left( {2600\;-\;p} \right) \times 6 \times 9}}{{100}})$

⇒ 50P = 140,400 - 54P

⇒ 104P = 140,400

⇒ P = 140,400/104 = 1350

∴ Sum lent at 9% per annum is = 2600 – 1350 = Rs. 1250
Discussion
18.

1). Quantity B > Quantity A2). Quantity B < Quantity A3). Quantity B ≥ Quantity A4). Quantity B ≤ Quantity A

Answer»

<P>FIRST we will find Quantity A,

Quantity A:

As per the GIVEN INFORMATION,

the average weight of 12 men is 68 kg and the average weight of 38 women is 53 kg,

∴ total weight of 12 men = 12 × 68 = 816

∴ total weight of 38 women = 38 × 53 = 2014

∴ total weight of all 50 people = 816 + 2014 = 2830

∴ average weight of 50 people = 2830/50 = 56.6

Quantity A = 56.6

Now,

Quantity B:

The simple interest on Rs. 5000 after 3 months at 4.2% per annum.

We know that SI = (P × R × T)/100

Where, P = Principal, R = % rate of interest, T = time in years

Now, as per the given information,

SI = (5000 × 3 × 4.2)/(100 × 12)

∴ SI = 52.5

Quantity B = 52.5

Clearly, Quantity B < Quantity A
Discussion
19.

Pankaj borrowed a sum of Rs. ___ from Kamal at simple interest. The rate of interest is 10% per annum for the first 2 years and 12% per annum for the next 3 years and 15% per annum thereafter. He paid Rs. 5332 as interest after 7 years.1). Rs. 62202). Rs. 60203). Rs. 60004). Rs. 6200

Answer»

LET the SUM be Rs. p

Then according to the question,

(p × 2 × 10)/100 + (p × 3 × 12)/100 + (p × 2 × 15)/100 = 5332

⇒ (20 × p)/100 + (36 × p)/100 + (30 × p)/100 = 5332 

⇒ (86 × p)/100 = 5332 

⇒ p = 5332 × 100/86 = Rs. 6200
Discussion
20.

1). Rs. 10002). Rs. 7503). Rs. 1204). Rs. 500

Answer»

SIMPLE interest = (P × R × T) /100

⇒ Simple interest on RS 6000 for 5 years at 15% RATE = (6000 × 15 × 5) /100 = Rs. 4500

⇒ Simple interest on Rs 5000 for 4 years at 20% rate = (5000 × 20 × 4) /100 = Rs. 4000

∴ Difference = 4500 – 4000 = Rs. 500
Discussion
21.

1). Rs. 6802). Rs. 20003). Rs. 12504). Rs. 3500

Answer»

LET the amount LENT at 5% be Rs. x

∴ MONEY lent at 3% = Rs. (4000 – x)

SIMPLE interest on Rs. x = (Principal × Time × RATE)/100

⇒ Simple interest = (x × 4 × 5)/100 = Rs. x/5

Total interest = Rs. 640

∴ Interest on Rs. (4000 – x) = Rs. (640 – x/5)

⇒ [(4000 – x) × 4 × 3)]/100 = 640 – x/5

⇒ 3(4000 – x)/25 = 640 – x/5

⇒ 12000 – 3x = 16000 – 5x

⇒ 2x = 4000

⇒ x = 2000

∴ Money lent at 5% is Rs. 2000
Discussion
22.

Ajay took a loan from ABC bank at a simple interest rate of 8% in the first year with an increase of 0.5% in each subsequent year. He paid interest of Rs. 4860 after Five years. How much loan did he take?1). Rs. 108002). Rs. 331403). Rs. 155004). Cannot be determined

Answer»

Let’s assume Ajay took a LOAN of Rs. X.

We know the formula for simple Interest,

SI = (P × R × t)/100

Where,

SI = Simple Interest

P = Principal

R = Rate of Interest

t = Time period

Here Rate of interest is varying each year, so we can calculate SI at each year then we will add SI at each year to calculate total SI

SI for First Year,

Rate of Interest = 8%

⇒ SI1 = (X × 8 × 1)/100 = 8X/100

SI for Second Year,

Rate of Interest = 8.5%

⇒ SI2 = (X × 8.5 × 1)/100 = 8.5X/100

Similarly,

SI for third, fourth and fifth years are as follows

⇒ SI3 = 9X/100

⇒ SI4 = 9.5X/100

⇒ SI5 = 10X/100

Total Simple Interest = SI1 + SI2 + SI3 + SI4 + SI5 = 4860 (GIVEN)

⇒ (8X/100) + (8.5X/100) + (9X/100) + (9.5X/100) + (10X/100) = 4860

⇒ 45X/100 = 4860

⇒ X = 486000/45 = Rs. 10800

∴ AMOUNT of loan Ajay took = Rs. 10800
Discussion
23.

1). Rs. 13162). Rs. 1232.53). Rs. 1234.54). Rs. 1332.5

Answer»

In CASE of SIMPLE interest,

Principal = P

Time = 4 years

RATE = 5%

Interest accrued = Rs. 2600

Simple interest = (P × R × T)/100, where P = principal, R = rate of interest and T = time period

According to problem,

⇒ 2600 = (P × 5 × 4)/100

⇒ P = (2600 × 100)/20

⇒ P = 13000

∴ In case of compound interest,

Principal = 13000

Rate = 5%

Number of years = 2

∴ Amount, A = 13000(1 + 5/100)2

⇒ Rs. 14332.5

∴ Compound interest accrued,

⇒ 14332.5 - 13000

⇒ Rs. 1332.5
Discussion
24.

1). 6666.672). 3333.343). 5412.74). 4587.3

Answer»

LET assume part of 10,000 put under the interest rate of 8% = x

Part of 10,000 put under the interest rate of 12% = y,

x + y = 10,000

For part x,

Interest = PRT/100,

Where, P = Rs. x,

R = 8%,

T = 3 year

Interest = (x × 8 × 3)/100 = 0.24x---- (1)

For part y,

Interest = PRT/100,

Where, P = Rs. y

R = 12%,

T = 1 year

Interest = (y × 12 × 1)/100 = 0.12y---- (2)

COMPARING both equation,

⇒ 0.24x = 0.12y

⇒ x/y = 0.12/0.24

⇒ x/y = 1/2

⇒ 2x = y---- (3)

Substitute the value of Eq(3),

⇒ x + 2x = 10000

⇒ 3x = 10000

⇒ x = Rs.3333.33

Put the value of x in Eq(3);

⇒ 2(3333.33) = y

∴ y = Rs. 6666.67
Discussion
25.

Quantity B: If Rs. 9000 amounts to Rs. 11979 in 3 years compounded annually. Find the rate of interest per annum.1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

Quantity A:

Let the principal = P

Amount = (9/8)P

Simple interest = (9/8)P – P

Simple interest = (1/8)P

N = 5

Using simple interest FORMULA,

(1/8)P = (P × R × 5)/100

R = 2.5%

The rate of interest is 2.5% per annum.

Quantity B:

Let P = principal, R = rate of interest and N = time period

Amount = P(1 + R/100)N

11979 = 9000(1 + R/100)3

11979 = (9/1000) × (R + 100)3

(R + 100)3 = 1331000

Taking cube root,

R + 100 = 110

R = 10

Rate of interest is 10% per annum.

From above SOLUTION,

Relation between Quantity A < Quantity B is established.
Discussion
26.

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 yeaRs. The sum is:1). 6502). 6903). 6584). 698

Answer»

As SUM of money amounts to Rs. 815 in 3 years and Rs. 854 in 4 years

So,

S.I. for 1 year = Rs. 854 – 815 = Rs. 39

S.I. for 3 years = Rs. 39 x 3 = Rs. 117

Principal = Amount – S.I. (we are considering the CASE for 3 years)

Principal = Rs. (815 - 117) = Rs. 698
Discussion
27.

At what rate per annum on simple interest will a sum of money triple itself in 10 years?1). 20%2). 17%3). 22%4). 15%

Answer»

Formula for Simple Interest :

$(SI = \frac{{P \times R \times T}}{{100}})$

Where,

P = Principal

R = Rate of interest

T = Time period

Let the REQUIRED rate of interest be X.

According to the QUESTION, SI must be equal to 2 × P in order to make the final sum three times the original principal amount after 10 YEAR.

$(\begin{array}{l} \RIGHTARROW 2{\RM{P}} = {\rm{\;}}\frac{{{\rm{P\;}} \times {\rm{X\;}} \times 10}}{{100}}\\ \Rightarrow 2 = {\rm{\;}}\frac{{\rm{X}}}{{10}} \end{array})$

⇒ X = 20%

∴ Required rate of interest is of 20%.
Discussion
28.

Find the difference between the simple interest at the rate of 10.50% and compound interest at the rate of 10% per annum compounded annually, after 3 years on a sum of Rs. 1000.1). Rs. 102). Rs. 123). Rs. 144). Rs. 16

Answer»

Compound interest after 3 YEARS = 1000 × [(1 + 10/100)3 - 1] 

= 1000 × (1331 – 1000)/1000 = 331 

Simple interest = (1000 × 3 × 10.5)/100 = 315 

∴ required DIFFERENCE = 331 – 315 = 16
Discussion
29.

How much payment will discharge a debt of Rs. 375 at the end ofthree years at a rate of 20% per annum at compound interest?1). Rs. 6482). Rs. 6503). Rs. 7204). Rs. 730

Answer»

Amount = $(Principal\LEFT\{ {{{\left( {1\; + \;Rate\% } \RIGHT)}^{TIME}}} \right\})$

⇒ 375(1 + 0.2)3 = 648
Discussion
30.

Shubham took a loan of Rs. 250000 with simple interest for as many years as the rate of interest. If he paid Rs. 10000 as interest at the end of the loan period, what was the rate of interest for him?1). 4.7%2). 3.5%3). 3%4). 5%

Answer»
Discussion
31.

Bill Gates borrows $7000 from Zuckerberg at the rate of r% per annum. After 3 years he again borrows $3000 and paid a total of $5325 as simple interest after 8 years from the first time he borrowed the money. Find the value of ‘r’?1). 5.5%2). 9.5%3). 7.5%4). 12.5%

Answer»

Total interest = interest on 7000 for 3 years + interest on 10000 for 5 years 

According to the question,

⇒ (7000 × R × 3)/100 + (10000 × r × 5)/100 = 5325

⇒ (210 × r) + (500 × r) = 5325

⇒ 710 × r = 5325

Required RATE of interest ‘r’ = 7.5%
Discussion
32.

Sunil invested Rs. 25000 in the scheme of simple interest. If the invested amount is four times the interest earned in 1st three years, then find the interest earned at the end of 4 years.1). Rs. 49002). Rs. 54403). Rs. 63304). Rs. 7223.33

Answer»

Time of investment is 3 years.

Amount INVESTED is FOUR times the INTEREST earned in 1st 3 years.

& principle amount = Rs. 25000

∴ Principle amount = 4 × (Interest earned in 1st 3 years)

SIMPLE interest = $(\frac{{\left( {principle\;amount\; \times \;rate\;of\;interest\; \times \;time} \right)}}{{100}})$

$(\therefore 25000 = \frac{{4 \times 25000 \times rate\;of\;interest \times 3}}{{100}})$

∴ 100 × 25000 = 12 × 25000 × Rate of interest

∴ Rate of interest = 100/12

∴ To find interest earned at the end of 4 years -

Simple INTERST = $(\frac{{principle\;amount\; \times \;rate\;of\;interst\; \times \;time\;of\;investment}}{{100}})$

∴ Simple interst = $(\frac{{25000\; \times \;100\; \times \;4}}{{100\; \times \;12}})$

∴ Simple interest earned in 4 years = Rs. 8333.33

∴ At the end of 4 years Sunil will earn Rs. 8333.33 interest.

Alternate:

As we know, Amount invested is four times the interest earned in 1st 3 years

Let ‘I’ be interest for 1 year

According to the question;

⇒ 25000 = 4 × 3 × I

⇒ I = 25000/12 = 2083.33

∴ Interest for 4 years = 4 × I = 4 × 2083.33 = 8333.33
Discussion