type 1 basic simple interest using formula Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Simple interest for 2 years

= Rs.(568 - 520) = Rs.48

Interest for 5 years

= Rs.$48/2 × 5$ = Rs.120

Principal = Rs.(520 - 120) = Rs.400

Using Rule 12,

P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$

=$({568 × 5 - 520 × 7}/{5 - 7})$

= $({2840 - 3640}/{-2})$

=${- 800}/{- 2}$ = Rs.400

Answer: (b)

Using Rule 1,

Time from 11 May to 10 September, 1987

= 21 + 30 + 31 + 31 + 10

= 123 days

Time = 123 days = $123/365$ year

S.I. = ${7300 × 123 × 5}/{365 × 100}$ = Rs.123

Answer: (b)

Let each instalment be x Then,

$(x + {x × 5 × 1}/100) + (x + {x × 5 × 2}/100)$

+ $(x + {x × 5 × 3}/100) + x = 6450$

$(x + x/20) + (x + x/10)$

+ $(x + {3x}/20)$ + x= 6450

${21x}/20 + {11x}/10 + {23x}/20 + x = 6450$

${21x + 22x + 23x + 20x}/20$

= 6450

${86x}/20 = 6450$

$x = {6450 × 20}/86$ = Rs.1500

Equal instalment = ${6450 × 200}/{4[200 + (4 - 1) × 5]}$

= ${6450 × 200}/{4(215)}$

= ${6450 × 50}/215$ = Rs.1500

Answer: (b)

Using Rule 1,

Let the rate of interest be R per cent per annum.

${400 × 2 × R}/100 + {550 × 4 × R}/100$

+ ${1200 × 6 × R}/100 = 1020$

8R + 22 R +72 R = 1020

102 R= 1020

R = $1020/102$ = 10%

Answer: (d)

Using Rule 1,

Let the sum be x.

Using formula, I = $\text"PRT"/100$ we have

${x × 15/12 × 15/2}/{100 - x} - {x × 8/12 × 25/2}/100$

= 32.50

${25x}/2400$ = 32.50

$x = {32.50 × 2400}/25$ = 3120

Required sum = Rs.3120

Answer: (d)

Using Rule 1,

P = ${A × 100}/{100 + r × t}$

=${7000 × 100}/{100 + 10/3 × 5}$

= ${7000 × 100 × 3}/350$ = Rs.6000