model 6 comparing sum in different years Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Principal = Rs.P (let)

Rate = R% per annum

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

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

$650/P = (1 + R/100)$ ...(i)

Again, 676 = P$(1 + R/100)^2$

676 = P$(650/P)^2$

= ${P × 650 × 650}/P^2$

P = ${650 × 650}/676$ = Rs.625

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)

Difference = 238.50 - 225 = Rs.13.50

= S.I. on Rs.225 for 1 year

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

= ${13.50 × 100}/{225 × 1}$ = 6% per annum

Using Rule 7(i),

Here, b - a = 1

B = Rs.238.50, A = Rs.225

R% = $(B/A - 1)$ × 100%

= $({238.50}/225 - 1) × 100%$

= $({238.50 - 225}/225) × 100%$

= $({13.5}/225) × 100%$ = 6%

Answer: (b)

Let the rate of interest be r% per annum,

According to the question,

4840 = P$(1 + r/100)^2$ ..... (i)

and 5324 = P$(1 + r/100)^3$....(ii)

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

$1 + r/100 = 5324/4840 = 1 + 484/4840$

$r/100 = 484/4840$ ⇒ r = 10%

Here, b - a = 3 - 2 = 1, B = Rs.5,324, A = Rs.4,840

R% = $(B/A - 1)$ × 100%

= $(5324/4840 -1)$ × 100%

= $({5324 - 4840}/4840) × 100%$

= $484/4840 × $100% = 10%

Answer: (b)

Rate of interest = 12% p.a.

= 1% per month

Time = 12y months

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

64 = 1$(1 + 1/100)^{12y}$

64 = $1(1.01)^{12y}$