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

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)

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: (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