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

Answer: (c)

Let the annual instalment be x

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

$x = P_1(1 + 25/200)$

$x = P_1 × 9/8$

$P_1 = 8/9x$

Similarly, $P_2 = 64/81x$

$P_1 + P_2$ = 6800

$8/9x + 64/81x$ = 6800

${72x + 64x}/81 = 6800$

${136x}/81 = 6800$

$x = {6800 × 81}/136$ = Rs.4050

Using Rule 9(i),

Here, P = Rs.6800, R = $25/2$%, n = 2

Each instalment

= $p/{(100/{100 + r}) + (100/{100 + r})^2$

= $6800/{(100/{100 +{25/2}}) + (100/{100 + {25/2}})^2$

= $6800/{200/225 + (200/225)^2}$

= $6800/{200/225(1 + {200/225})}$

= ${6800 × 225 × 225}/{200 × 425}$ = Rs.4050

Answer: (d)

Using Rule 9(i),

Let the value of each instalment be Rs.x

Principal = Present worth of Rs.x due 1 year hence, present worth of Rs. x due 2 years hence

210 = $x/(1 + R/100) + x/(1 + R/100)^2$

210 = $x/(1 + 10/100) + x/(1 + 10/100)^2$

210 = $x/{1 + 1/10} + x/(1 + 1/10)^2$

210 = $x/{11/10} + x/(11/10)^2$

210 = ${10x}/11 + {100x}/121$

210 = ${110x + 100x}/121$

210 × 121 = 210 x

$x = {210 × 121}/210$ = Rs.121

Answer: (c)

Using Rule 1,

Let the amount deposited in Post Office be Rs.x lakhs.

Amount deposited in bank = Rs.(3 - x) lakhs

According to the question,

${x × 10 × 1}/{100 × 12} + {(3 - x) × 6 × 1}/{100 × 12}$

= $2000/100000 = 1/50$

10x + 18 - 6x

= $1/50$ × 1200 = 24

4x = 24 - 18 = 6

x = $6/4$ = Rs.$3/2$ lakhs

∴ Required difference = 0