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

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: (c)

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

525 = P$[(1 + 10/100)^2 - 1]$

525 = P$(121/100 - 1)$

525 = ${P × 21}/100$

P = ${525 × 100}/21$ = Rs.2500

Again, new rate = 5% per annum

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

= ${2500 × 5 × 4}/100$ = Rs.500

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: (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: (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: (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