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

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

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%

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

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

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%