model 6 comparing sum in different years Practice Questions Answers Test with Solutions & More Shortcuts

Question : 1 [SSC CGL Prelim 2002]

A sum of money invested at compound interest amounts in 3 years to Rs.2,400 and in 4 years to Rs.2,520. The interest rate per annum is :

a) 12%

b) 5%

c) 10%

d) 6%

Answer: (b)

S.I. on Rs.2400 for 1 year

= Rs.(2, 520 - 2, 400) = Rs.120

Rate = $\text"S.I. × 100"/ \text"Principal × Time"$%

= ${120 × 100}/{2400 × 1}$ = 5%

Using Rule 7(i),

Here, b - a = 4 - 3 = 1

B = Rs.2520, A = Rs.2400

R% = $(B/A - 1)$ × 100%

= $[2520/2400 - 1] × 100%$

= $[{2520 - 2400}/2400] × 100%$

= $120/2400 × 100%$ = 5%

Question : 2 [SSC CGL 2005]

A sum becomes Rs.4500 after two years and Rs.6750 after four years at compound interest. The sum is

a) Rs.3050

b) Rs.4000

c) Rs.3000

d) Rs.2500

Answer: (c)

P$(1 + r/100)^2$ = 4500 ...(i)

P$(1 + r/100)^4$ = 6750 .....(ii)

On dividing equation (ii) by equation (i), we get

$(1 + r/100)^2 = 6750/4500$

From equation (i),

P × $6750/4500$ = 4500

P = ${4500 × 4500}/6750$ = Rs.3,000

Using Rule 7(ii),

Here, b - a = 4 - 2 = 2

B = Rs.6750, A = Rs.4500

R% = $[(B/A)^{1/2} - 1]$ × 100%

= $[(6750/4500)^{1/2} - 1] × 100%$

= $[(3/2)^{1/2} - 1] ×$ 100%

$(3/2)^{1/2} = 1 + R/100$

$3/2 = (1 + R/100)^2$

A = P$(1 + R/100)^2$

4500 = P × $3/2$ ⇒ P = Rs.3000

Question : 3 [SSC CAPFs 2016]

On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?

a) Rs.35

b) Rs.3.50

c) Rs.14

d) Rs.7

Answer: (d)

Principal = $\text"S.I. × 100"/ \text"Time × Rate"$

= ${350 × 100}/{2 × 4}$ = Rs.4375

C.I. = P$[(1 + R/100)^T - 1]$

= 4375$[(1 + 4/100)^2 - 1]$

= 4375$[(1 + 1/25)^T - 1]$

= 4375$[(26/25)^2 - 1]$

= 4375$(676/625 - 1)$

= ${4375 × 51}/625$ = Rs.357

Required difference

= Rs.(357 - 350) = Rs.7

Question : 4 [SSC CGL Tier-II 2016]

A sum of money is invested at 20% compound interest (compounded annually). It would fetch Rs. 723 more in 2 years if interest is compounded half yearly. The sum is

a) Rs.7,500

b) Rs.15,000

c) Rs.20,000

d) Rs.30,000

Answer: (d)

Let the principal be Rs. x.

When the interest is compounded annually,

C.I. = P$[(1 + R/100)^T - 1]$

= P$[(1 + 20/100)^2 - 1]$

= P$[(6/5)^2 - 1]$

= P$(36/25 - 1)$ = Rs.${11P}/25$

When the interest is compounded half–yearly,

C.I. = P$[(1 + 10/100)^4 - 1]$

= P$[(11/10)^4 - 1]$

= P$(14641/10000 - 1)$

= Rs.${4641P}/10000$

${4641P}/10000 - {11P}/25$ = 723

${4641P - 4400P}/10000$ = 723

${241P}/10000$ = 723

P = ${723 × 10000}/241$ = Rs.30000

Question : 5 [SSC CGL Tier-II 2012]

An amount of money at compound interest grows up to Rs.3,840 in 4 years and up to Rs.3,936 in 5 years. Find the rate of interest.

a) 2.05%

b) 2.5%

c) 3.5%

d) 2%

Answer: (b)

A = P$(1 + R/100)^T$

3840 = P$(1 + R/100)^4$ ....(i)

3936 = P$(1 + R/100)^5$ ...(ii)

Dividing equation (ii) by equation (i),

$3936/3840 = 1 + R/100$

$R/100 = 3936/3840$ - 1

= ${3936 - 3840}/3840 = 96/3840$

R = $96/3840 × 100$ = 2.5%

Using Rule 7(i),

Here, b - a = 5 - 4 = 1

B = Rs.3,936, A = Rs.3,840

R%= $(B/A - 1)$ × 100%

= $(3936/3840 - 1)$ × 100%

= $({3936 - 3840}/3840) × 100%$

= $96/3840 × 100% = 10/4%$ = 2.5%

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