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

Answer: (c)

Interest on Rs.650 for 1 year

= 676 - 650 = Rs.26

So, r = $26/650 × 100$

r = 4% per annum

P = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$

= $650/{26/25} = 650 × 25/26$ = Rs.625

Using Rule 7(i),

Here, b - a = 1

B = Rs.676, A = Rs.650

R% = $(B/A - 1)$ × 100%

= $[676/650 - 1] × 100%$

= $[{676 - 650}/650] × 100%$

= $26/650 × 100% = 100/25$ = 4%

Amount= P$(1 + R/100)^1$

650 = P$(1 + 4/100)$

P = ${650 × 100}/104$ = Rs.625

Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, then

P = $A_1(A_1/A_2)^n$

Answer: (a)

Let the rate of interest = R% per annum.

We know that

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

2420 = P$(1 + R/100)^2$ ....(i)

2662 = P$(1 + R/100)^3$ ...(ii)

Dividing equation (ii) by (i),

$1 + R/100 = 2662/2420$

$R/100 = 2662/2420$ - 1

$R/100 = {2662 - 2420}/2420$

= $242/2420 = 1/10$

R = $1/10 × $100 = 10%

Using Rule 7(i),

Here, b - a = 3 - 2 = 1

B = Rs.2,662, A= Rs.2,420

R% = $(B/A - 1)$ × 100%

= $(2662/2420 -1)$ × 100%

= $[{2662 - 2420}/2420]$ × 100%

= $242/2420 × 100%$ = 10%

Answer: (a)

If the principal be Rs.P, then

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

1440 = P$(1 + R/100)^2$ ...(i)

and 1728 = P$(1 + R/100)^3$ ...(ii)

On dividing equation (ii) by (i),

$1728/1440 = 1 + r/100$

$r/100 = 1728/1440$ - 1

= ${1728 - 1440}/1440 = 288/1440$

r = ${288 × 100}/1440$

r = 20% per annum

Using Rule 7(i),

Here, b - a = 3 - 2 = 1

B = Rs.1728, A = Rs.1440

R% = $(B/A - 1)$ × 100%

= $(1728/1440 - 1) × 100%$

= $({1728 - 1440}/1440) × 100%$

= $[288/1440] × 100%$ = 20%

Answer: (d)

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

7000 = P$(1 + R/100)^4$....(i)

10000 = P$(1 + R/100)^8$.......(ii)

Dividing equation (ii) by (i)

$10000/7000 = (1 + R/100)^4$

$10/7 = (1 + R/100)^4$

From equation (i),

7000 = P × $10/7$ ⇒ P = Rs.4900

Using Rule 7(iii),

Here, b - a = 8 - 4 = 4

B = Rs.10,000, A = Rs.7000

R% = $((B/A)^{1/n} - 1)$ × 100%

R% = $[(10000/7000)^{1/4} - 1]$

= $[(10/7)^{1/4} - 1]$

$1 + R/100 = (10/7)^{1/4}$

$(1 + R/100)^4 = 10/7$

7000 = $P × 10/7$

Since, Amount = P$(1 + R/100)^4$

P = Rs.4900