type 3 money multiples in ‘n’ years Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

Case-I

Let the principal be x

Amount = 3x

Interest = 2x

Time = 20 years

I = $\text"PRT"/100$

2x = ${x × R × 20}/100$ ⇒ R = 10%

Case-II

I = x, P = x, R = 10, T = ?

I = $\text"PRT"/100$

x = ${x × 10 × T}/100$ ⇒ T = 10 years.

Using Rule 3,

R% = ${(3 - 1)}/20$ × 100% = 10%

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

T = ${2 - 1}/10 × 100$ = 10 years

Answer: (a)

Case I,

Principal = Rs.x

Interest = Rs.x

Time = 6 years

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

= ${x × 100}/{x × 16} = 50/3%$ per annum

Case II,

Interest = ${x × 12 × 50}/{100 × 3}$ = Rs.2x

i.e., Amount is thrice the principal.

Answer: (c)

If the principal be x, the amount = 2x

SI = x

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

= ${x × 100}/{x × 15} = 20/3 = 6{2}/3$ years

Using Rule 3,

T = ${(n - 1)}/R × 100%$

= $({2 - 1}/15) × 100$

= $100/15 = 20/3$ Years

= 6$2/3$ years

Answer: (d)

If the principal be Rs.100 then S.I. = Rs.100.

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

= ${100 × 100}/{100 × 12} = 25/3$ years

= 8 years 4 months

Using Rule 3,

T = ${(n - 1)}/R × 100%$

= ${(2 - 1)}/12 × 100%$

= $100/12 = 25/3$ years.

= 8$1/3$ years

= 8 years, 4 months.

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