Answer: Option D. -> 18$3/4$%
Answer: (d)Let the principal be xPrincipal + SI = ${7x}/4$SI = ${7x}/4 - x = {3x}/4$Rate = ${SI × 100}/\text"Principal × Time"$= ${3x × 100}/{4 × x × 4} = 18{3}/4%$Using Rule 3,R = ${(7/4 - 1)}/4 × 100%$= $3/16$ × 100%= $75/4% = 18{3}/4%$

Answer: Option D. -> 20 years
Answer: (d)If principal be x, interest = xand rate = r% p.a. thenRate = ${SI × 100}/\text"Principal × Time"$= ${x × 100}/{x × 10}$ = 10%Now, p = x, interest = 2xThen, time = ${SI × 100}/\text"Principal × Rate"$= ${2x × 100}/{x × 10}$ = 20 yearsUsing Rule 3,R = ${(2 - 1)}/10$ × 100%R = 10%T = ${(n - 1)}/R × 100$ years= ${3 - 1}/10 × 100$ = 20 years.

Answer: Option D. -> 8$1/3$%
Answer: (d)Let the principal be x.Amount = 2xInterest = (2x - x) = xRate = ${S.I. × 100}/\text"Principal × Time"$= ${x × 100}/{x × 12} = 25/3 = 8{1}/3%$Using Rule 3,R = ${(2 - 1)}/12 × 100%$R = $25/3% = 8{1}/3$%

Answer: Option A. -> 32 years
Answer: (a)Case IPrincipal = Rs. xInterest = Rs. xRate = ${SI × 100}/\text"Principal × Time"$= ${x × 100}/{x × 16} = 25/4%$ per annumCase IIInterest = Rs. 2xTime = ${SI × 100}/\text"Principal × Rate"$= ${2x × 100 × 4}/{x × 25}$ = 32 yearsUsing Rule 3,R = ${(n - 1)}/T × 100%$= ${(2 - 1)}/16 × 100%$= $25/4 % = 6{1}/4%$Now, T = ${(n - 1)}/R × 100%$= ${(3 - 1)}/{25/4} × 100$= $800/25$ = 32 years.

Answer: Option B. -> 6%
Answer: (b)Rate = ${\text"SI" × 100}/\text"Principal × Time"$= $9/25 × 100/6$ = 6% per annumUsing Rule 5,Here, n = $9/25$, T = 6 years.R = ${n × 100}/T$R = $9/25 × 100/6$R = 6%

Answer: Option D. -> 8%
Answer: (d)Rate of interest = r % per annumS.I. = ${\text"Principal × Rate × Time"/100$According to the question,${3200 × 5 × r}/{100 × 2} - {3000 × 5 × r}/200$ = 4080r - 75r = 405r = 40 ⇒ r = $40/5$= 8% per annum Using Rule 13The difference between the S.I. for a certain sum $P_1$ deposited for time $T_1$ at $R_1$ rate of interest and another sum $P_2$ deposited for time $T_2$ at $R_2$ rate of interest isS.I. = ${P_2R_2T_2 - P_1R_1T_1}/100$

Answer: Option D. -> 5 years
Answer: (d)Let the principal be xInterest = $2/5$ xRate = 8% per annumTime = ${\text"Interest" × 100}/\text"Principal × Rate"$=${{2/5}x × 100}/{x × 8} = 40/8$ = 5 years Using Rule 5If Simple Interest (S.I.) becomes 'n' times of principal i.e.S.I. = P × n then.RT = n × 100

Answer: Option A. -> 6$2/3$%
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${5R - 2R}/20$ = 13R = 20R = $20/3 = 6{3}2%$Using Rule 5,Here, S.I. = $2/5$ amountS.I. = $2/5$ (P + S.I.)S.I. = $2/5$ S.I. + $2/5$ P$3/5$ S.I. = $2/5$ PS.I. = $2/3$PNow, n = $2/3$, T = 10 years.R= ${n × 100}/T$= $2/3 × 100/10$= $20/3 = 6{2}/3%$

Answer: Option A. -> 5%
Answer: (a)Let Prinicpal = Rs.100S.I. = $100 × 1/5$ = Rs.20Rate = ${20 × 100}/{100 × 4}$ = 5%Using Rule 5,Here, n = $1/5$, T = 4 years.R = ${n × 100}/T$R = $1/5 × 100/4$R = 5%

Answer: Option C. -> 12%
Answer: (c)Rate = ${\text"SI" × 100}/\text"Principal × Time"$= $12/25 × 100/4$ = 12% per annumUsing Rule 5,Here, n = $12/25$, T = 4 years.R = ${n × 100}/T$R = $12/25 × 100/4$R = 12%