type 3 money multiples in ‘n’ years Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Let principal be Rs. x.

Amount = Rs.2x

Interest = Rs.(2x - x) = Rs.x

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

= ${x × 100}/{x × 8} = 25/2$

= 12$1/2%$ per annum

Answer: (d)

Let the principal be x.

Case-I

$2x = {x × R × 15}/100$

R = ${2 × 100}/15 = 40/3%$

Case-II

SI = 4x

$4x = {x × 40 × T}/300$

T = ${4 × 300}/40 = 30$ years

Using Rule 3,

R = ${(3 - 1)}/15 × 100%$

= $2/15 × 100%$

= $2/3 × 20% = 40/3%$

T = ${(n - 1)}/R$ Years

= ${(5 - 1)}/{40/3} × 100$ = 30 years.

Answer: (b)

Principal = P

Amount = ${7p}/6$

S.I. = ${7p}/6 - P = P/6$

R = ${S.I × 100}/{P × T} = {P × 100}/{6 × p × 3}$

= $50/9 = 5{5}/9%$

R% = ${(7/6 - 1) × 100%}/3$

= $1/18 × 100% = 50/9%$

= 5$5/9$%

Answer: (b)

According to the question,

If principal be Rs. x, then

S.I. = Rs. x

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

= ${x × 100}/{x × 25/4} = 400/25$ = 16 years

Using Rule 3,

T = ${(n - 1)}/R × 100%$

= ${2 - 1}/{25/4} × 100$

= $400/25$ = 16 years.

Answer: (d)

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

= ${x × 100}/{x × 25/4}$ = 16 years

Using Rule 3,

T = ${(n - 1)}/{R%} years$

= ${(2 - 1)}/{25/4} × 100$ years = 16 years.

Answer: (c)

Let P be the principal and R% rate of interest.

S.I. = ${\text"PR" × 10}/100 = \text"PR"/10$

According to the question,

${PR}/10 = (P + {PR}/10) × 2/5$

$R/10 = (1 + R/10) × 2/5$

$R/10 = 2/5 + R/25$

$R/10 - R/25 = 2/5$

${5R - 2R}/50 = 2/5$

${3R}/50 = 2/5$

R = ${50 × 2}/{3 × 5} = 20/3 = 6{2}/3$%