type 5 si on ‘n’ years & ‘x/y ‘of sum Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

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

= $9/25 × 100/6$ = 6% per annum

Using Rule 5,

Here, n = $9/25$, T = 6 years.

R = ${n × 100}/T$

R = $9/25 × 100/6$

R = 6%

Answer: (d)

Rate of interest = r % per annum

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

According to the question,

${3200 × 5 × r}/{100 × 2} - {3000 × 5 × r}/200$ = 40

80r - 75r = 40

5r = 40 ⇒ r = $40/5$

= 8% per annum

Here, $P_1 = Rs.3000, R_1 = R, T_1 = 5/2$ years

$P_2 = Rs.3200, R_2 = R, T_2 = 5/2$ years

Difference S.I. = Rs.40

40 = ${3200 × R × 5/2 - 3000 × R × 5/2}/100$

4000 = 8000 R - 7500 R

R = 8%

Answer: (d)

Let the principal be x

Interest = $2/5$ x

Rate = 8% per annum

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

=${{2/5}x × 100}/{x × 8} = 40/8$ = 5 years

Here, n = $2/5$ and R = 8%

RT = (n × 100)

T = ${n × 100}/R$

T = $2/5 × 100/8$ = 5 years

Answer: (a)

Amount after 10 years

= P$(1 + {RT}/100)$ = P$(1 + {R × 10}/100)$

= Rs. P$(1 + R/10)$

Interest = Rs.P$(1 + R/10) × 2/5$

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

R = ${P(1 + R/10) × 2/5 × 100}/{P × 10}$

R = 4$(1 + R/10)$

$R/4 = 1 + R/10$

$R/4 - R/10$ = 1

${5R - 2R}/20$ = 1

3R = 20

R = $20/3 = 6{3}2%$

Using Rule 5,

Here, S.I. = $2/5$ amount

S.I. = $2/5$ (P + S.I.)

S.I. = $2/5$ S.I. + $2/5$ P

$3/5$ S.I. = $2/5$ P

S.I. = $2/3$P

Now, n = $2/3$, T = 10 years.

R= ${n × 100}/T$

= $2/3 × 100/10$

= $20/3 = 6{2}/3%$

Answer: (a)

Let Prinicpal = Rs.100

S.I. = $100 × 1/5$ = Rs.20

Rate = ${20 × 100}/{100 × 4}$ = 5%

Using Rule 5,

Here, n = $1/5$, T = 4 years.

R = ${n × 100}/T$

R = $1/5 × 100/4$

R = 5%