model 5 ci with instalments Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

Using Rule 1,

Let the income of company in 2010 be Rs.P

According to the question,

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

2664000 = P$(1 + 20/100)^2$

2664000 = P$(1 + 1/5)^2$

2664000 = P × $(6/5)^2$

P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000

Answer: (b)

Using Rule 9(i),

Sum borrowed = Present worth of Rs.17640 due 1 year hence + Present worth of Rs.17640 due 2 years hence

= Rs.$(17640/{(1 + 5/100)} + 17640/{(1 + 5/100)^2})$

= Rs.$(17640 × 20/21 + 17640 × 20/21 × 20/21)$

= Rs.(16800 + 16000) = Rs.32800

Answer: (d)

Let principal (present worth) for first year be $P_1$ and that for two years be $P_2$.

16224 = $P_1(1 + 4/100)$

16224 = $P_1(1 + 1/25) = {26P_1}/25$

$P_1 = {16224 × 25}/26$ = Rs.15600

Again,

$16224 = P_2(1 + 4/100)^2$

16224 = $P_2(26/25)^2 = {676P_2}/625$

$P_2 = {16224 × 625}/676$ = 15000

Cash value of the scooter

= Rs.(16224 + 15600 + 15000) = Rs.46824

Answer: (d)

Using Rule 9(i),

Let each instalment be x.

$x/(1 + 5/100) + x/(1 + 5/100)^2 = 12300$

${20x}/21 + (20/21)^2x = 12300$

${20x}/21(1 + 20/21)$ = 12300

${20x}/21 × 41/21 × x = 12300$

$x = {12300 × 21 × 21}/{20 × 41}$ ⇒ x = 6615