type 1 basic simple interest using formula Practice Questions Answers Test with Solutions & More Shortcuts

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: (c)

Using Rule 1,

After 10 years,

SI = ${1000 × 5 × 10}/100$ = Rs.500

Principal for 11th year

= 1000 + 500 = Rs.1500

SI = Rs.(2000 - 1500) = Rs.500

T = ${SI × 100}/{P × R} = {500 × 100}/{1500 × 5}$

= $20/3$ years = 6$2/3$ years

Total time = 10 + 6$2/3$

= 16$2/3$ years

Answer: (c)

If the principal be x and rate of interest be r% per annum, then

SI after 1 year = 920 - 880 = Rs.40

SI after 2 years = Rs.80

880 = x + 80

x = Rs.(880 - 80) = Rs.800

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

= $({920 × 2 - 880 × 3}/{2 - 3})$

= $({1840 - 2640}/{-1})$

= ${- 800}/{- 1}$ = Rs.800

Answer: (b)

Using Rule 1,

If rate of interest be R% p.a. then,

SI = ${\text"Principal × Rate × Time"/100$

${6000 × 2 × R}/100 + {1500 × 4 × R}/100$

= 900

120 R + 60R = 900

180R = 900

R = $900/180$ = 5%