Answer: Option A. -> 6$2/3$%
Answer: (a)Using Rule 1Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ orS.I. = ${\text"P × R × T"/100$P = ${\text"S.I." × 100}/\text"R × T"$, R = ${\text"S.I." × 100}/\text"P × T"$, T = ${\text"S.I." × 100}/\text"P × R"$ A = P + S.I. or S.I. = A - P

Answer: Option A. -> 5%
Answer: (a)Using Rule 1,Let 'r' be the rate of interest190 = ${500 × 4 × r}/100 + {600 × 3 × r}/100$20r + 18r = 19038r = 190r = $190/38$ = 5%

Answer: Option A. -> 2$1/2$%
Answer: (a)Let the annual rate of interest = r%Time = r yearsLet the principal be x .Interest = $x/16$According to the question,$x/16 = {x × r × r}/100$ [Since, r = t]16$r^2$ = 100$r^2 = 100/16 = 25/4$r = $√ {25/4} = 5/2 = 2{1}/2$%Using Rule 5,Here, n = $1/16$, R = TRT = n × 100$R^2 = 100/16$R = $√{100/16} = 10/4 = 2{1}/2%$

Answer: Option C. -> $y^2 = zx$
Answer: (c)Using Rule 1,S.I. = ${P × R × T}/100$$y = {x × T × R}/100$and $z = {y × T × R}/100$So, $y/z = x/y = ⇒ y^2 = zx$

Answer: Option B. -> Rs.8,000
Answer: (b)Using Rule 1,Let the larger part of the sum be xSmaller part = Rs.(12000 - x)According to the question,${x × 3 × 12}/100 = {(12000 - x) × 9 × 16}/{2 × 100}$36 x = (12000 - x ) 72x = (12000 - x ) × 2x + 2x = 240003x = 24000$x = 24000/3$ = Rs.8000

Answer: Option C. -> Rs.840
Answer: (c)According to question,Interest of one year = Rs.42Rate = 5% and Time = 1 yearPrincipal = $\text"Interest × 100"/\text"Rate × Time"$= ${42 × 100}/{5 × 1}$ = Rs.840 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 B. -> 10%
Answer: (b)Using Rule 1,S.I. = ${\text"Principal × Rate × Time"/100$${4000 × 3 × x}/100$= ${5000 × 2 × 12}/100$$x = {5 × 2 × 12}/{4 × 3}$= 10% per annum

Answer: Option C. -> Rs.3600
Answer: (c)Let the sum be x${x × 5 × 15}/{100 × 12} - {x × 4 × 8}/{100 × 12}$ = 129$x/{100 × 12}$(75 - 32) = 129$x = {129 × 1200}/43$ = Rs.3600Using Rule 13,$P_1 = P, R_1 = 4%$$T_1 = 8 months = 8/12$ years$P_2 = P, R_2$ = 5%$T_2 = 15 month = 15/12$ yearsS.I. = Rs.129129 = ${P × 5 × 15/12 - P × {4 × 8}/12}/100$12900 = ${75P - 32P}/12$12900 = ${43P}/12$P = Rs.3600

Answer: Option D. -> 0.3%
Answer: (d)Let $r_1$, and $r_2$ be the required rate of interestThen, ${13.50} = {1500 × 3 × r_1}/100$– ${1500 × 3 × r_2}/100 = 4500/100(r_1 - r_2)$$r_1 - r_2 = 135/450 = 27/90$= $3/10$ = 0.3%Using Rule 13,$P_1 = Rs.1500, R_1 , T_1$ = 3 years.$P_2 = Rs.1500, R_2 , T_2$ = 3 years.S.I. = Rs.13.5013.50 = ${1500 × R_2 × 3 - 1500 × R_1 × 3}/100$$1350/100 = {4500(R_2 - R_1)}/100$$R_2 - R_1 = 1350/4500 = 27/90$= $3/10$ = 0.3%

Answer: Option D. -> 3$1/3$%
Answer: (d)$\text"Simple interest"/ \text"Principal" = 1/9$If the annual rate of interest be r%, thenRate = $\text"S.I. × 100"/ \text"Principal × Time"$$r = 1/9 × 100/r$$r^2 = 100/9$r = $√{100/9} = 10/3 = 3{1}/3$%Using Rule 5,Here, n = $1/9$, R = TRT = n × 100$R^2 = 1/9 × 100 = 100/9$R = $√{100/9} = 10/3 = 3{1}/3$%