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

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: (c)

Principal = Rs.P and time = T years

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

According to the question,

P + ${PT × 5}/100$ = 2200

P + ${PT}/20$ = 2200 ...(i)

Again, ${PT × 8}/100 - {PT × 5}/100$

= 2320 - 2200

${3PT}/100$ = 120

PT = ${120 × 100}/3$ = 4000 ...(ii)

From equation (i),

P + $4000/20$ = 2200

P = 2200 - 200 = Rs. 2000

From equation (ii),

PT = 4000

T = $4000/2000$ = 2 years

Alternative Method

Difference in rates

= 8 - 5 = 3%

Since,3% ≡ 2320 - 2200 = 120

5% ≡ $120/3$ × 5 = 200

Principal = Rs.(2200 - 200) = Rs.2000

Time = ${200 × 100}/{2000 × 5}$ = 2 years

Answer: (d)

According to the question,

Principal = Rs.x.

Interest = Rs.x.

Time = $50/3$ years

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

= ${x × 100}/{x × {50/3}}$

= ${100 × 3}/50$ = 6% per annum

Answer: (a)

Using Rule 1,

SI = Rs.(7200–6000) = Rs.1200

SI = $\text"PRT"/100$

1200 = ${6000 × R × 4}/100$

R = ${1200 × 100}/{6000 × 4}$ = 5%

New rate of R = 5×1.5 = 7.5%

Then, SI = ${6000 × 7.5 × 5}/100$ = Rs.2250

Amount = Rs.(6000 + 2250) = Rs.8250

Answer: (b)

Let the principal be Re.1

S.I. = $41/40 - 1 = 1/40$

Now, rate = ${\text"Interest" ×100}/{\text"Principal × Time"}$

= ${1/40 × 100}/{1 × 1/4} = {100 × 4}/40$ = 10%

Using Rule 3,

R = ${(41/40 - 1) × 100%}/{1/4}$

= $1/40 × 4 × 100%$ = 10%