Summary
Simple machines are those in which effort is applied at a more convenient point in a more convenient direction. This note contains information about various types of simple machine with its example, principal of simple machine.
Subjective Questions
Q1:
Compute compound interest on Rs 5000 for 2 years at 12% per annum.
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal(P) = Rs 5000,<br>Rate (R) = 12%<br>Time (T) = 2 years<br>Compound interest (C.I) = ?<br>Now,</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= 5000 \left[ \left( 1 + \frac{12}{100}\right)^2 - 1\right]\\ &= 5000[(1.12)^2 -1]\\ &= 5000[1.2544 - 1] \\ &= 5000 \times 0.2544\\ &= Rs \: 1272\: \: _{Ans.}\end{align*}</p>
Q2:
At what rate percent compound interest will Rs 847 in 2 years?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal (P) = Rs 700<br>Compound interest (A) = Rs 847<br>Time (T) = 2 years<br>Rate of interest (R) = ?</p> <p>\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ or, 847 &= 700\left(1 + \frac{R}{100}\right)^2 \\ or, \frac{847}{700} &= \left(1 + \frac{R}{100}\right)^2 \\ or, 1.21 &= \left(1 + \frac{R}{100}\right)^2\\ or,(1.1)^2 &= \left(1 + \frac{R}{100}\right)^2 \\ or, 1.1 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.1 - 1\\ or, R &= 0.1 \times 100 \\ \therefore R &= 10\% \:\:\:_{Ans.}\end{align*}</p>
Q3:
What is the compound interest on Rs 50,000 for 2 years at 10% per annum payable half yearly.
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal(P) = Rs 50,000<br>Time (T) = 2 years<br>Rate (R) = 10%<br>Compound interest (C.I) = ?</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^2T - 1\right]\\ &= 50,000\left[ \left( 1 + \frac{10}{100}\right)^{2 \times 2} - 1\right] \\ &= 50,000 [(1 + 0.05)^4 - 1]\\ &= 50,000 [1.2155 -1] \\ &= 50,000 \times 0.2155 \\ &= Rs \: 10775 \:\: \: _{Ans.} \end{align*}</p>
Q4:
Find the compound Interest on Rs 12000 for 1 year and 6 months at rate 10% per annum?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principle (P) = Rs 12,000<br>Time (T) = 1 year<br>Month(M)=6 month<br>Rate (R) = 10%<br>Compound interest (C.I) = ?</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T \left( 1 + \frac{MR}{1200}\right)- 1\right]\\ &= 12,000 \left[ \left( 1 + \frac{10}{100}\right)^1 \left( 1 + \frac{6 \times 10}{1200}\right) - 1 \right] \\ &= 12,000 [(1.1) (1.05) - 1]\\ &= 12,000\: [1.155 - 1] \\ &= 12,000\times 0.155 \\&= Rs 930\\ \end{align*} \(\therefore Compound \: interest = Rs \: 930 \:\: _{Ans.}\)</p>
Q5:
How much will Rs 25000 amount to 2 years compounded yearly. If the rate of the successive years be 4 % and 5 % respectively.
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principle (P) = Rs 25,000<br>Time(T) = 2 years<br>Rate (R<sub>1</sub>) = 4%<br>Rate (R<sub>2</sub>) = 5%<br>Compound Amount (CA) = ?</p> <p>\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right)<br>\\ &= 25000\left(1 + \frac{4}{100}\right) \left(1 + \frac{5}{100}\right) \\ &= 2500 \times 1.04 \times 1.05 \\ &= Rs \: 27300 \:\:\: _{Ans.} \end{align*}</p>
Q6:
Find the sum of money on which the compound amount is Rs 5159.68 for 1 year at 8% per annum payable half yearly?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Compound Amount (CA) = Rs 5191.68<br>Time (T) = 1 year<br>Rate (R) = 8%<br>Principal (P) = ?</p> <p>\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{200}\right)^{2T} \\ or, 5191..68 &= P \left(1 + \frac{8}{200}\right)^2\\ or, 5191.68 &= P(1.04)^2\\ or, P &= \frac{5191.68}{1.0816}\\ &= \: Rs \: 4800 \end{align*}</p> <p>\(\therefore\) The principal (P) = Rs \: 4800 \(_{Ans.}\)</p> <p></p>
Q7:
A certain sum of money amount to Rs 1323 in 2 years and to Rs 1389.15 in 3 years. Find the compound rate of interest and the sum.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>First compound amount (CA<sub>1</sub>) = Rs 1323<br>Time (T<sub>1</sub>) = 2 years<br>Second compound amount (CA<sub>2</sub>) = Rs 1389.15<br>Time (T<sub>2</sub>) = 3 years<br>We know that,</p> <p>\begin{align*} CA &= P \left(1 + \frac{R}{100}\right)^T \\ 1323 &= P \left(1 + \frac{R}{100}\right)^2 \: \: \: \: \: ........ (1)\\ or, 1389.15 &= P \left(1 + \frac{R}{100}\right)^3 \:\:\:\:\: ........(2) \\ \end{align*}</p> <p>\(\text{The equation (2 )is divided by equation (1) } \)</p> <p>\begin{align*} \frac{1389.15}{1323} &= \frac{P \left(1 + \frac{R}{100}\right)^3 }{P \left(1 + \frac{R}{100}\right)^2 }\\ or, 1.05 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.05 - 1\\ or, R &= 0.05 \times 100\%\\ R &= 5\% \end{align*}</p> <p>\(\text{Putting Value of R in}\:\: eq^n (1)\)</p> <p>\begin{align*}P \left(1 + \frac{R}{100}\right)^2 &= 1323 \\ or, P \left(1 + \frac{5}{100}\right)^2 &=1323 \\ or, P \times 1.1025 &= 1323\\ P &= \frac{1323}{1.1025}\\ \therefore P &= 1200\\ \end{align*} \(\therefore Principle = Rs \: 1200 \:\: and \: rate = 5\% \)</p>
Q8:
According to the system of compound interest, a sum of money in 2 years amount to Rs 7260 and is 3 years amount to Rs 7986. Find the rate of interest.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>First compound interest (CA<sub>1</sub>) = Rs 7260<br>Time (T<sub>1</sub>) = 2 years<br>Second compound interest (CA<sub>2</sub>) = Rs 7986<br>Time (T<sub>2</sub>) = 3 years<br>We know that,</p> <p>\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ 7260 &= P\left( 1 + \frac{R}{100}\right)^2 \: \: \: \: ....... (1) \\ 7986 &= P\left( 1 + \frac{R}{100}\right)^3 \:\:\:\: ....... (2)\\ \end{align*}</p> <p>\( \text{The equation (2) is divide d by equation (1)}\)</p> <p>\begin{align*} \frac{7986}{7260} &= \frac{P\left( 1 + \frac{R}{100}\right)^3 }{P\left( 1 + \frac{R}{100}\right)^2}\\ or, 1.1 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.1 - 1 \\ or, \frac{R}{100} &= 1.1 -1 \\ or, R &= 0.1 \times 100 \\ R &= 10\% \end{align*}</p> <p></p>
Q9:
The sum of simple interest and compound interest after 2 years is Rs 202.50 and the rate of interest is 10% per annum. Find the principle.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Let, Principle (P) = Rs x<br>Time (T) = 2 years<br>Rate (R) = 5%</p> <p>\begin{align*} Simple \: Interest \: (S.I) &= \frac{P\:T\:R}{100} \\ &= \frac{x \times 2 \times 5}{100} \\ &= x \times 0.1 \end{align*}</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= x \left[ \left( 1 + \frac{5}{100}\right)^2 - 1\right] \\ &= x [(1.05)^2 - 1] \\ &= x[1.1025 - 1]\\ &= x \times 0.1025 \end{align*}</p> <p>From question,</p> <p>\begin{align*} S.I + C.I &= Rs \: 202.50 \\ x \times 0.1 + x \times 0.1025 &= Rs \: 202.50\\ or, x(0.1 + 0.1025) &= Rs \: 202.50\\ x &= \frac{202.50}{0.2025} \\ \therefore x &= Rs \: 10000 \end{align*}</p> <p>\(\therefore\) Principle (P) = Rs 10000 \(_{Ans.}\)</p>
Q10:
The compound interest of a sum of money in 1 year and 2 years are Rs 450 and 945 respectively. Find the rate of interest compounded yearly and sum.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>First compound interest (CI<sub>1</sub>) = Rs 450<br>Time (T<sub>1</sub>) = 1 year<br>Second compound interest (C<sub>2</sub>) = Rs 945<br>Time (T<sub>2</sub>) = 2 years<br>Let Principal = P and rate = R<br>We know that,</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ C.I_1&= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right] \\ 450 &= P \left[ \left( 1 + \frac{R}{100}\right)^1 - 1\right] \: \: ....... (1) \\ C.I_2&= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right] \\ 945 &= P \left[ \left( 1 + \frac{R}{100}\right)^2 - 1\right] \:\:\: ....... (2)\end{align*}</p> <p>Equation (2) is divided by equation (1)</p> <p>\begin{align*} \frac{ P \left[ \left( 1 + \frac{R}{100}\right)^2 - 1\right]}{ P \left[ \left( 1 + \frac{R}{100}\right)^1 - 1\right]} &= \frac{945}{450}\\ or, \frac{\left[ \left( 1 + \frac{R}{100}\right) + 1\right] \left[ \left( 1 + \frac{R}{100}\right) - 1\right]}{\left[ \left( 1 + \frac{R}{100}\right)^1 - 1\right]} &= 2.1 \\ or, 2 + \frac{10}{100} &= 2.1\\ or, \frac{R}{100} &= 2.1 - 2 \\ or, R &= 0.1 \times 100\% \\ \therefore R &= 10\%\end{align*}</p> <p>Putting the value of R in equation (1)</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right) - 1\right] \\ or, P \times \frac{10}{100} &= 450\\ or, P &= 450 \times 10\\ P &= 4500 \\ \\ \therefore The \: sum = 4500, rate (R) = 10\% \end{align*}</p> <p></p>
Q11:
A person took a loan of Rs 46,875. If the rate of the compound interest is 4 paisa per rupee per year, in how many years will the compound interest be Rs 5,853.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal (P) = Rs. 46875<br>Compound interest (C.I.) = Rs. 5853<br>In 1-year compound interest of Rs 1 is 4 paisa.<br>In 1-year compound interest of Rs 100 is <strong>4 × 100</strong> paisa. <br>In 1-year compound interest of Rs 100 is Rs. \( \frac{4 \times 100}{100} \) = Rs 4</p> <p>Interest rate (R) = 4%<br>Time (T) =?<br>We know that,</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ 5853 &= 46875 \left[ \left( 1 + \frac{4}{100}\right)^T - 1\right] \\ \frac{5853}{46875} &=\left( 1 + \frac{4}{100}\right)^T - 1\\ or, 0.124864 + 1 &=(1.04)^T\\ or, (1.04)^3 &= (1.04)^T \\ \therefore Time &= 3 \: years \end{align*}</p>
Q12:
The compound amount of a sum of money is 2 years is Rs 8820 and in 3 years is Rs 9261. Find the sum and rate of interest.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>First compound amount (CA<sub>1</sub>) = Rs 8820<br>Time (T<sub>1</sub>) = 2 years<br>Second compounded amount (CA<sub>2</sub>) = Rs 9261<br>Time (T<sub>2</sub>) = 3 years<br>Let principal = P , Rate = R</p> <p>\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ 8820 &= P \left(1 + \frac{R}{100}\right)^2\:\:\: ...... (1)\\ 9261 &= P \left(1 + \frac{R}{100}\right)^3 \:\:\: ..... (2) \end{align*}</p> <p>Equation (2) is divided by equation (1)</p> <p>\begin{align*} \frac{P \left(1 + \frac{R}{100}\right)^3}{ P \left(1 + \frac{R}{100}\right)^2} &= \frac{9261}{8820} \\ or, 1 + \frac{R}{100} &= 1.05\\ or, \frac{R}{100} &= 1.05 - 1 \\ or, R &= 0.05 \times 100\% \\ R &= 5\% \end{align*}</p> <p>Putting value of R in equation (1)</p> <p>\begin{align*} P \left(1 + \frac{R}{100}\right)^T &= 8820 \\ or, P \left(1 + \frac{5}{100}\right) &= 8820\\ or, P \times (1.05)^2 &= 8820 \\ or, P \times 1.1025 &= 8820\\or,P &= \frac{8820}{1.1025}\\ P &= Rs \: 8000 \end{align*}</p> <p>\(\therefore \) The principal = Rs 8000 and Rate = 5% \(_{Ans.} \)</p>
Q13:
The compound interest on a sum of money in 3 years at 20% per annum will Rs 384 more than simple interest. Find the sum.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Time (T) = 3 years<br>Rate (R) = 20%<br>Principal (P) = ?<br>Compound Interest (C.I) - Simple Interest (S.I) = Rs. 384<br>We know that,</p> <p>\begin{align*} S.I &= \frac{P \: T \: R}{100}\\ &= \frac{P \times 3 \times 20}{100}\\ &= P \times 0.6 \end{align*}</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= P \left[ \left( 1 + \frac{20}{100}\right)^3 - 1\right]\\ &=P\: [(1.2)^3 - 1]\\ &= P[1.728 - 1]\\ &= P \times 0.728 \end{align*}</p> <p>From question,</p> <p>\begin{align*} C.I - S.I &= Rs\: 384\\ P \times 0.728 - P \times 0.6 &= Rs\: 384 \\P\:(0.728 - 0.6)&= 384\\ or, P \times 0.128 &= 384\\ or, P &= \frac{384}{0.128}\\ \therefore P &= Rs\:3000 \:\:\:\: _{Ans.} \end{align*}</p>
Q14:
Prakash lend altogether Rs 6000 to Anima and Eline for 2 years. Anima agree to pay simple interest at 10% p.a. and Eline agree to pay compound interest at the rate of 8%. If Eline paid Rs 50 more than Amina as the interest. Find how much did he lent to each.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>For Anima,<br>Let, Principal (P) = Rs x,<br>Time (T) = 2 years<br>Rate (R) = 10%</p> <p>\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{x \times2 \times10 }{100}\\ &= x \times 0.2 \end{align*}</p> <p>For Eline,<br>Principal (P) = Rs 6000 - x<br>Time (T) = 2 years<br>Rate (R) = 5%</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= (6000 - x)\left[ \left( 1 + \frac{5}{100}\right)^2 - 1\right]\\ &= (6000 - x ) [(1.08)^2 - 1] \\ &=Rs \: 998.4 \times Rs \: 0.1664x \end{align*}</p> <p>From question,</p> <p>\begin{align*} C.I. - S.I &= Rs \: 50\\ or, 998.4 - x \times 0.1664 - x \times 0.2 &=50\\ or, 998.4 - x \times 0.3664&=50\\ or, -x \times 0.3664&=50 - 998.4\\ or, x &= \frac{-948.4}{-0.3664}\\ &= Rs\: 2588.43\end{align*}</p> <p>For Anima sum = Rs 2588.43<br>For Eline sum = (6000 - 2588.43) = Rs 3411.57</p>
Q15:
Sita borrowed Rs 170000 from Radha at the rate of 21% per annum, at the end of 1 year 6 months.
(i) How much simple interest will she have to pay?
(ii) How much compound interest will she have to pay?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal (P) = Rs 170000<br>Rate (R) = 21%<br>Time (T) = 1 year 6 month = \( 1 + \frac{6}{12} = \frac{3}{2} \: years \)<br><br></p> <p>\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{170000\times3 \times21 }{100\times 2}\\ &= Rs\: 53550 \:\:\:\:_{Ans.} \end{align*}</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ C.I &= 170000 \left[ \left( 1 + \frac{21}{100}\right)^{\frac{3}{2}} - 1\right] \\&= 170000 \left[ \left( \frac{121}{100}\right)^{\frac{3}{2}} - 1\right]\\ &= 170000\left[ \left( \frac{11}{10}\right)^{\frac{2 \times3}{2}} - 1\right] \\&= 170000[(1.1)^3 - 1]\\ &=170000[1.331 - 1]\\ &= Rs\: 170000 \times 0.331\\ &= Rs\: 56270 \:\:\:_{Ans.} \end{align*}</p> <p></p>
Q16:
What is difference between the compound interest and simple interest on Rs 5120 for 3 years at 12.5% per annum?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principle (P) = Rs 5120<br>Time (T) = 3 years<br>Rate (R) = 12.5%</p> <p><br>\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{5120 \times3 \times12.5 }{100} \\ &= Rs\:1920\end{align*}</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= 5120 \left[ \left( 1 + \frac{12.5}{100}\right)^3 - 1\right]\\ &=5120 [1.4238281 - 1] \\ &= 5120 \times 0.4238281 \\&= Rs\: 2170 \end{align*}</p> <p>\begin{align*}Difference &= C.I - S.I\\ &= 2170-1920\\ &= Rs\: 250 \end{align*}</p>
Q17:
Ram borrowed Rs 4250 from Shyam at the rate of 12% per annum at the and of one year.
(i) How much simple interest will Ram have to pay?
(ii) How much interest compounded half-yearly will Ram has to pay?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal (P) = Rs 4250<br>Rate (R) = 12%<br>Time (T) = 1 year</p> <p>\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{4250 \times 1 \times12 }{100}\\ &=Rs\: 510 \end{align*}</p> <p>For half yearly,</p> <p>\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{200}\right)^{2T} - 1\right]\\ &= 4250\left[ \left( 1 + \frac{12}{200}\right)^{2\times 1} - 1\right] \\ &= 4250[(1.06)^2 -1]\\ &= 4250 [1.1236-1]\\ &=4250 \times 0.1236\\ &= Rs \:525.30 \:\:_{Ans}\end{align*}</p>
Q18:
What will be the compound interest on Rs 5000 for 2 years at the rate of 20% compound per annum? At what time the same interest on the same sum at the same rate of simple interest?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal (P) = Rs 5000<br>Time (T) = 2 years<br>Rate (R) = 20%</p> <p>\begin{align*} Compound\:Interest\:(C.I) &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= 5000\left[ \left( 1 + \frac{20}{100}\right)^2 - 1\right]\\ &= 5000[(1.2)^2 - 1]\\ &= 5000 \times 0.44 \\ &= Rs\:2200 \end{align*}</p> <p>Let, Simple interest (S.I) = Rs 2200<br>Principle (P) = Rs 5000<br>Rate (R) = 20%<br>Time (T) = ?</p> <p>\begin{align*} T &= \frac{I \times 100}{PR}\\ &= \frac{2200 \times 100}{5000 \times 20}\\ &= 2.2 \: years \end{align*}</p> <p></p>
Q19:
Mohan deposited Rs 5000 at 8% p.a compound interest in the bank. Find the difference between the amounts compounded yearly and half yearly in two years.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Principal (P) = Rs 5000<br>Rate (R) = 8%<br>Time (T) = 2 years<br>Compound amount (CA) = ?</p> <p><br>\begin{align*} For\: yearly,\\ CA &= P \left(1 + \frac{R}{100}\right)^T \\ &= 5000 \left(1 + \frac{8}{100}\right)^2\\ &=5000 \times \frac{108}{100} \times \frac{108}{100} \\ &= Rs \: 5832 \:\:\: _{Ans.} \end{align*}</p> <p>\begin{align*} For\: half \:yearly \: \\CA &= P \left(1 + \frac{R}{200}\right)^{2T} \\ &= 5000 \left(1 + \frac{8}{200}\right)^{2\times 2}\\ &= 5000 (1.04)^4\\ &= 5849.29 \end{align*}</p> <p>\begin{align*} Difference \: between &= 5849.29 - 5832\\ &= Rs \:17.29 \:\:\: _{Ans.} \end{align*}</p> <p></p> <p></p>
Q20:
Divide Rs 25200 into two parts such that the amount of are part in 2 years is the same as the amount of second part in 3 years the rate of compound interest being 10% per annum in both the cases.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>For first part<br>First part (P<sub>1</sub>) = Rs 25200 - x<br>Time (T<sub>1</sub>) = 2 years<br>Rate (R<sub>1</sub>) = 10%</p> <p>For second part<br>Second principal (P<sub>2</sub>) = Rs x<br>Time (T<sub>2</sub>) = 3 years<br>Rate (R<sub>3</sub>) = `10%</p> <p>\begin{align*} Compound \: amount \: (CA_1) &= P \left(1 + \frac{R}{100}\right)^T \\ &=(25200 - x) \left(1 + \frac{10}{100}\right)^2\\ &= (25200 - x) (1.1)^2\\ &= (25200-x) \times 1.21 \end{align*}</p> <p>\begin{align*} Compound \: amount \: (CA_2) &= P \left(1 + \frac{R}{100}\right)^T \\ &= x \left(1 + \frac{10}{100}\right)^3\\ &=x (1.1)^3 \\ &= x \times 1.331 \end{align*}</p> <p>From Question,</p> <p>\begin{align*} CA_1 &=CA_2 \\ (25200-x) \times 1.21 &= x\times 1.331 \\ or, 30492 - 1.21x &= x \times 1.331\\ or, 1.331x+1.21x&=30492\\ or, 2.541x &= 30492\\ x&= \frac{30492}{2.541}\\ \therefore x&=12000 \end{align*}</p> <p>\begin{align*} 1^{st} \: principle &= 25200 -x\\&= 25200-12000\\&= Rs\:13200 \:\:\:_{Ans.} \end{align*}</p> <p>\(2^{nd }\: principal = x= Rs 12000 \:\:\:_{Ans.}\)</p> <p></p> <p></p>
