Answer: Option C. -> Nil
Answer: (c)Using Rule 1,Let the amount deposited in Post Office be Rs.x lakhs.Amount deposited in bank = Rs.(3 - x) lakhsAccording to the question,${x × 10 × 1}/{100 × 12} + {(3 - x) × 6 × 1}/{100 × 12}$= $2000/100000 = 1/50$10x + 18 - 6x= $1/50$ × 1200 = 244x = 24 - 18 = 6x = $6/4$ = Rs.$3/2$ lakhs∴ Required difference = 0

Answer: Option D. -> Rs.18,50,000
Answer: (d)Using Rule 1,Let the income of company in 2010 be Rs.PAccording to the question,A = P$(1 + R/100)^T$2664000 = P$(1 + 20/100)^2$2664000 = P$(1 + 1/5)^2$2664000 = P × $(6/5)^2$P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000

Answer: Option B. -> 64 = $(1.01)^{12y}$
Answer: (b)Rate of interest = 12% p.a.= 1% per monthTime = 12y monthsA = P$(1 + R/100)^T$64 = 1$(1 + 1/100)^{12y}$64 = $1(1.01)^{12y}$

Answer: Option C. -> Rs.500
Answer: (c)C.I. = P$[(1 + R/100)^T - 1]$525 = P$[(1 + 10/100)^2 - 1]$525 = P$(121/100 - 1)$525 = ${P × 21}/100$P = ${525 × 100}/21$ = Rs.2500Again, new rate = 5% per annumS.I. = $\text"Principal × Time × Rate"/100$= ${2500 × 5 × 4}/100$ = Rs.500

Answer: Option A. -> Rs.625
Answer: (a)Principal = Rs.P (let)Rate = R% per annumA = P$(1 + R/100)^T$650 = P$(1 + R/100)$$650/P = (1 + R/100)$ ...(i)Again, 676 = P$(1 + R/100)^2$676 = P$(650/P)^2$= ${P × 650 × 650}/P^2$P = ${650 × 650}/676$ = Rs.625

Answer: Option A. -> 6%
Answer: (a)Difference = 238.50 - 225 = Rs.13.50= S.I. on Rs.225 for 1 yearRate = $\text"S.I. × 100"/\text"Principal × Time"$= ${13.50 × 100}/{225 × 1}$ = 6% per annumUsing Rule 7(i),Here, b - a = 1B = Rs.238.50, A = Rs.225R% = $(B/A - 1)$ × 100%= $({238.50}/225 - 1) × 100%$= $({238.50 - 225}/225) × 100%$= $({13.5}/225) × 100%$ = 6%

Answer: Option B. -> 10%
Answer: (b)Let the rate of interest be r% per annum,According to the question,4840 = P$(1 + r/100)^2$ ..... (i)and 5324 = P$(1 + r/100)^3$....(ii)On dividing equation (ii) by equation (i), we have,$1 + r/100 = 5324/4840 = 1 + 484/4840$$r/100 = 484/4840$ ⇒ r = 10%Using Rule 7,If on compound interest, a sum becomes Rs.A in 'a' years and Rs.B in 'b' years then,(i) If b - a = 1, then, R% = $(B/A - 1)$ × 100%(ii) If b - a = 2, then, R% = $(√{B/A} - 1)$ × 100%(iii) If b - a = n then, R% = $[(B/A)^{1/n} - 1]$ × 100%where n is a whole number.

Answer: Option C. -> Rs.625
Answer: (c)Interest on Rs.650 for 1 year= 676 - 650 = Rs.26So, r = $26/650 × 100$r = 4% per annumP = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$= $650/{26/25} = 650 × 25/26$ = Rs.625Using Rule 7(i),Here, b - a = 1B = Rs.676, A = Rs.650R% = $(B/A - 1)$ × 100%= $[676/650 - 1] × 100%$= $[{676 - 650}/650] × 100%$= $26/650 × 100% = 100/25$ = 4%Amount= P$(1 + R/100)^1$650 = P$(1 + 4/100)$P = ${650 × 100}/104$ = Rs.625Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, thenP = $A_1(A_1/A_2)^n$

Answer: Option A. -> 20
Answer: (a)If the principal be Rs.P, thenA = P$(1 + R/100)^T$1440 = P$(1 + R/100)^2$ ...(i)and 1728 = P$(1 + R/100)^3$ ...(ii)On dividing equation (ii) by (i),$1728/1440 = 1 + r/100$$r/100 = 1728/1440$ - 1= ${1728 - 1440}/1440 = 288/1440$r = ${288 × 100}/1440$r = 20% per annumUsing Rule 7(i),Here, b - a = 3 - 2 = 1B = Rs.1728, A = Rs.1440R% = $(B/A - 1)$ × 100%= $(1728/1440 - 1) × 100%$= $({1728 - 1440}/1440) × 100%$= $[288/1440] × 100%$ = 20%

Answer: Option A. -> 10%
Answer: (a)Let the rate of interest = R% per annum.We know thatA = P$(1 + R/100)^T$2420 = P$(1 + R/100)^2$ ....(i)2662 = P$(1 + R/100)^3$ ...(ii)Dividing equation (ii) by (i),$1 + R/100 = 2662/2420$$R/100 = 2662/2420$ - 1$R/100 = {2662 - 2420}/2420$= $242/2420 = 1/10$R = $1/10 × $100 = 10%Using Rule 7(i),Here, b - a = 3 - 2 = 1B = Rs.2,662, A= Rs.2,420R% = $(B/A - 1)$ × 100%= $(2662/2420 -1)$ × 100%= $[{2662 - 2420}/2420]$ × 100%= $242/2420 × 100%$ = 10%