Quantitative Aptitude > Interest
COMPOUND INTEREST MCQs
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Answer: Option B. -> 600
Answer: (b)Using Rule 6,Difference = ${PR}^2/(100)^2$1.50 = ${P × 5 × 5}/(100)^2$P = 400 × 1.5 = Rs.600
Answer: (b)Using Rule 6,Difference = ${PR}^2/(100)^2$1.50 = ${P × 5 × 5}/(100)^2$P = 400 × 1.5 = Rs.600
Answer: Option B. -> 4000
Answer: (b)Let the principal be x.Compound interest= P$[(1 + R/100)^t - 1]$= $x[(1 + 10/100)^2 - 1]$= $x[(1.1)^2 - 1]$= x (1.21 - 1) = 0.21xSI = ${x × 2 × 10}/100 = x/5 = 0.2x$According to the question,0.21x - 0.2x = 40$0.01x = 40 ⇒ x = 40/{0.01}$ = Rs.4000Using Rule 6,Here, C.I. - S.I. = Rs.40, R = 10%, T = 2 years, P = ?C.I. - S.I. = P$(R/100)^2$40 = P$(10/100)^2$ ⇒ P = Rs.4000
Answer: (b)Let the principal be x.Compound interest= P$[(1 + R/100)^t - 1]$= $x[(1 + 10/100)^2 - 1]$= $x[(1.1)^2 - 1]$= x (1.21 - 1) = 0.21xSI = ${x × 2 × 10}/100 = x/5 = 0.2x$According to the question,0.21x - 0.2x = 40$0.01x = 40 ⇒ x = 40/{0.01}$ = Rs.4000Using Rule 6,Here, C.I. - S.I. = Rs.40, R = 10%, T = 2 years, P = ?C.I. - S.I. = P$(R/100)^2$40 = P$(10/100)^2$ ⇒ P = Rs.4000
Answer: Option A. -> 20%
Answer: (a)A = P$(1 + R/100)^T$1.44P = P$(1 + R/100)^2$$(1.2)^2 = (1 + R/100)^2$$1 + R/100$ = 1.2R = 0.2 × 100 = 20%Using Rule 8,Here, n = 1.44, t = 2 yearsR% = $(n^{1/6} - 1) × 100%$= $[(1.44)^{1/2} - 1] × 100%$= [(1.2) - 1] × 100%= 0.2 × 100% ⇒ R% = 20%
Answer: (a)A = P$(1 + R/100)^T$1.44P = P$(1 + R/100)^2$$(1.2)^2 = (1 + R/100)^2$$1 + R/100$ = 1.2R = 0.2 × 100 = 20%Using Rule 8,Here, n = 1.44, t = 2 yearsR% = $(n^{1/6} - 1) × 100%$= $[(1.44)^{1/2} - 1] × 100%$= [(1.2) - 1] × 100%= 0.2 × 100% ⇒ R% = 20%
Answer: Option D. -> 15 years
Answer: (d)Let the Principal be P and rate of interest be r%.2 P = P$(1 + r/100)^2$2 = $(1 + r/100)^5$ ...(i)On cubing both sides,8 = $(1 + r/100)^15$Time = 15 yearsUsing Rule 5,Here, m = 2, t = 5 yearsIt becomes 8 times = $2^3$ timesin t × n = 5 × 3 = 15 years
Answer: (d)Let the Principal be P and rate of interest be r%.2 P = P$(1 + r/100)^2$2 = $(1 + r/100)^5$ ...(i)On cubing both sides,8 = $(1 + r/100)^15$Time = 15 yearsUsing Rule 5,Here, m = 2, t = 5 yearsIt becomes 8 times = $2^3$ timesin t × n = 5 × 3 = 15 years
Answer: Option C. -> 8 years
Answer: (c)A = P$(1 + R/100)^T$Let P be Rs.1, then A = Rs.22 = 1$(1 + R/100)^4$$2^2 = (1 + R/100)^8$Time = 8 yearsUsing Rule 11,Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(2)^{1/4} = (4)^{1/n_2}$$(2)^{1/4} = (2^2)^{1/n_2}$$2^{1/4} = 2^{1/n_2}$$1/4 = 2/n_2$$n_2$ = 8 years
Answer: (c)A = P$(1 + R/100)^T$Let P be Rs.1, then A = Rs.22 = 1$(1 + R/100)^4$$2^2 = (1 + R/100)^8$Time = 8 yearsUsing Rule 11,Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(2)^{1/4} = (4)^{1/n_2}$$(2)^{1/4} = (2^2)^{1/n_2}$$2^{1/4} = 2^{1/n_2}$$1/4 = 2/n_2$$n_2$ = 8 years
Answer: Option D. -> 4 years
Answer: (d)Let the principal be Rs.1.A = P$(1 + R/100)^T$8 = 1$(1 + R/100)^3$$2^3 = 1(1 + R/100)^3$2 = 1$(1 + R/100)^1$$2^4 = (1 + R/100)^4$Time = 4 yearsUsing Rule 11,Here, $x = 8, n_1 = 3, y = 16, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(8)^{1/3} = (16)^{1/n_2}$$(2^3)^{1/3} = (2^4)^{1/n_2}$$2^1 = 2^{4/n_2}$1= $4/n_2$$n_2$ = 4 years
Answer: (d)Let the principal be Rs.1.A = P$(1 + R/100)^T$8 = 1$(1 + R/100)^3$$2^3 = 1(1 + R/100)^3$2 = 1$(1 + R/100)^1$$2^4 = (1 + R/100)^4$Time = 4 yearsUsing Rule 11,Here, $x = 8, n_1 = 3, y = 16, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(8)^{1/3} = (16)^{1/n_2}$$(2^3)^{1/3} = (2^4)^{1/n_2}$$2^1 = 2^{4/n_2}$1= $4/n_2$$n_2$ = 4 years
Answer: Option C. -> 18 years
Answer: (c)Let the sum be x. Then,$2x = x(1 + r/100)^6$2 = $(1 + r/100)^6$Cubing both sides,8 =$((1 + r/100)^6)^3$8 = $(1 + r/100)^18$$8x = x(1 + r/100)^18$The sum will be 8 times in 18 years.i.e., Time = 18 yearsUsing Rule 5,Here, m = 2, t = 6 yearsIt will becomes 8 times of itself= $2^3$ times of it selfin t × n years = 6 × 3 = 18 years
Answer: (c)Let the sum be x. Then,$2x = x(1 + r/100)^6$2 = $(1 + r/100)^6$Cubing both sides,8 =$((1 + r/100)^6)^3$8 = $(1 + r/100)^18$$8x = x(1 + r/100)^18$The sum will be 8 times in 18 years.i.e., Time = 18 yearsUsing Rule 5,Here, m = 2, t = 6 yearsIt will becomes 8 times of itself= $2^3$ times of it selfin t × n years = 6 × 3 = 18 years
Answer: Option A. -> 50%
Answer: (a)Suppose P = Rs.100and amount A = Rs.225A = P$(1 + r/100)^t$or 225 = $100(1 + r/100)^2$or $225/100 = [1 + r/100]^2$or 1 + $r/100 = 15/10$or ${100 + r}/100 = 15/10$or 100 + r = 150or r = 50% Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$
Answer: (a)Suppose P = Rs.100and amount A = Rs.225A = P$(1 + r/100)^t$or 225 = $100(1 + r/100)^2$or $225/100 = [1 + r/100]^2$or 1 + $r/100 = 15/10$or ${100 + r}/100 = 15/10$or 100 + r = 150or r = 50% Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$
Answer: Option D. -> 20 years
Answer: (d)Let the sum be x which becomes 2x in 10 years.Hence, 4x in 20 yearsMethod 2 : Unitary Method can also be used.Using Rule 5,Here, m = 2, t = 10Time taken to become 4 times = $2^2$ times= t × n = 10 × 2 = 20 years
Answer: (d)Let the sum be x which becomes 2x in 10 years.Hence, 4x in 20 yearsMethod 2 : Unitary Method can also be used.Using Rule 5,Here, m = 2, t = 10Time taken to become 4 times = $2^2$ times= t × n = 10 × 2 = 20 years
Answer: Option A. -> Rs.1,92,000
Answer: (a)A = P$(1 + R/100)^T$24000 = 12000$(1 + R/100)^5$2 = $(1 + R/100)^5$$2^4 = (1 + R/100)^20$i.e. The sum amounts to Rs.192000 after 20 years. Using Rule 11,A certain sum at C.I. becomes x times in $n_1$ year and y times in $n_2$ years then $x^{1/n_1} = y^{1/n_2}$
Answer: (a)A = P$(1 + R/100)^T$24000 = 12000$(1 + R/100)^5$2 = $(1 + R/100)^5$$2^4 = (1 + R/100)^20$i.e. The sum amounts to Rs.192000 after 20 years. Using Rule 11,A certain sum at C.I. becomes x times in $n_1$ year and y times in $n_2$ years then $x^{1/n_1} = y^{1/n_2}$
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