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

Answer: (d)

Using Rule 1,

Case I :

S.I. = ${5000 × 2 × 4}/100$ = Rs.400

Case II :

S.I. = ${5000 × 25 × 2}/{100 × 4}$ = Rs.625

Gain = Rs.(625 - 400) = Rs.225

Answer: (b)

Using Rule 1,

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

= ${1080 × 100}/{3000 × 12} = 3$ years

Answer: (d)

Interest for 1 year

= Rs.(925 - 850) = Rs.75

If a sum becomes $a_1$ in $t_1$ years and $a_2$ in $t_2$ years then rate of interest

= ${100(a_2 - a_1)}/({a_1t_2 - a_2t_1})%$

= ${100(925 - 850)}/{850 × 4 - 3 × 925} = 7500/625$ = 12%

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

= ${75 × 100}/{1 × 12}$ = Rs.625

Using Rule 12,

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

= ${925 × 3 - 850 × 4}/{3 - 4}$

= ${2775 - 3400}/{-1}$

= ${- 625}/{- 1}$ = Rs.625

Answer: (c)

Principal + interest for 8 years= Rs.2900... (i)

Principal + interest for 10 years = Rs.3000 ... (ii)

Subtracting equation (i) from (ii)

Interest for 2 years = Rs.100

Interest for 8 years

= $100/2 × 8$ = Rs.400

From equation (i),

Principal = Rs.(2900 - 400) = Rs.2500

Rate = ${S.I × 100}/{\text"Time × Principal"}$

= ${400 × 100}/{8 × 2500} = 2%$

Using Rule 12,

R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100

= $({2900 - 3000}/{3000 × 8 - 2900 × 10})$ × 100

=$({- 100}/{24000 - 29000}) × 100$

= ${-100}/{-5000}$ × 100 = 2%

Answer: (c)

Using Rule 1,

Let the sum lent to C be x

According to the question,

${2500 × 7 × 4}/100 + {x × 7 × 4}/100 = 1120$

or 2500 × 28 + 28x = 112000

or 2500 + x = 4000

or x = 4000 - 2500 = 1500