InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A car is purchased for Rs. 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years? |
|
Answer» Present value of car, P = Rs.348000 Rate of depreciates for 1st year, p = 10% Rate of depreciates for 2nd year, q = 20% Time, n = 2 years Now, Value = P × (1 - p/100) × (1 - q/100) = 348000 × (1 - 10/100) × (1 - 20/100) = 348000 × (1 - 1/10) × (1 - 1/5) = 348000 × 9/10 × 4/5 = 34800 × 9 × 4/5 = 6960 × 9 × 4 = 25056 ∴ Value of the car after 2 years is Rs.25056. |
|
| 2. |
Arun took a loan of Rs. 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year? |
|
Answer» Present value, P = Rs.390625 Interest rate, R = 16% per annum Time, n = 1 year ∵ Compounded quarterly, ∴ Amount (A) = P [1 + (R/4)/100]4n [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 390625 [1 + (16/4)/100]4 ⇒ A = 390625 [1 + 4/100]4 ⇒ A = 390625 [1 + 1/25]4 ⇒ A = 390625 [26/25]4 ⇒ A = 390625 × 26/25 × 26/25 × 26/25 × 26/25 ⇒ A = 390625 × 456976/390625 ⇒ A = 1 × 456976 ⇒ A = 456976 ∴ Amount = Rs.456976 ∴ Arun has to pay Rs.45976 after 1 year. |
|
| 3. |
A scooter is bought at Rs. 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years? |
|
Answer» Present value of scooter, P = Rs.56000 Time, n = 3 years Rate of depreciates, R = 10% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ Value = P (1 - R/100)n [∵ Rate decreases] = 56000 (1 - 10/100)3 = 56000 (1 - 1/10)3 = 56000 (9/10)3 = 56000 × 729/1000 = 56 × 729 = 40824 ∴ Value of scooter after 3 years will be Rs.40824. |
|
| 4. |
A machine is purchased for Rs. 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years? |
|
Answer» Present value of machine, P = Rs.625000 Time, n = 2 years Rate of depreciates, R = 8% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ Value = P (1 - R/100)n [∵ Rate decreases] = 625000 (1 - 8/100)2 = 625000 (1 - 2/25)2 = 625000 (23/25)2 = 625000 × 729/625 = 1000 × 529 = 529000 ∴ Value of machine after 2 years will be Rs.529000. |
|
| 5. |
Rs. 62500 for 2 years 6 months at 12% per annum compounded annually. |
|
Answer» Present value = Rs.62500 Interest rate = 12 % per annum Time = 2 years 6 month = (2 + 1/2) years = (5/2) years Amount (A) = P (1 + R/100)n [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 62500 (1 + 12/100)2 × [1 + (1/2 × 12)/100] ⇒ A = 62500 (1 + 3/25)2 × [1 + 6/100] ⇒ A = 62500 (28/25)2 × [106/100] ⇒ A = 62500 × 28/25 × 28/25 × 106/100 ⇒ A = 625 × 784/625 × 106 ⇒ A = 1 × 784 × 106 ⇒ A = 83104 ∴ Amount = Rs.83104 ∴ Compound interest = Rs.(83104 – 62500) = Rs.20604 |
|
| 6. |
Rs. 16000 was borrowed by a man from a bank where compound interest is computed quarterly. Annual interest rate is 10%. How much he has to pay after 9 months to clear his liabilities? |
|
Answer» Amount borrowed = Rs 16000 Interest rate = 10% 9 Months = 3 quarterly years Amount to be repaired after 9 months = 16000 × (1 + \(\frac{2.5}{100}\))3 = 16000 × (\(\frac{102.5}{100}\))3 = 16000 × \(\frac{102.5}{100}\) × \(\frac{102.5}{100}\) × \(\frac{102.5}{100}\) = 17230.25 |
|
| 7. |
Rekha deposited Rs. 16000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year. |
|
Answer» Given details are, Principal (p) = Rs 16000 Rate (r) = 20% per annum = 20/4 = 5% (quarterly) Time (t) = 1 year = 4 quarters of a year By using the formula, A = P (1 + R/100) n = 16000 (1 + 5/100)4 = 16000 (105/100)4 = Rs 19448.1 ∴ CI = Rs 19448.1 – 16000 = Rs 3448.1 |
|
| 8. |
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs. 482 more. Find the sum. |
|
Answer» Given details are, Rate (r) = 20% per annum = 20/2 = 10% (half yearly) Time (t) = 2 years = 2 × 2 = 4 half years Principal be = Rs P P (1 + R/100) n – P (1 + R/100) n = 482 P (1 + 10/100)4 – P (1 + 20/100)2 = 482 P (110/100)4 – P (120/100)2 = 482 P (1.4641) – P (1.44) = 482 0.0241P = 482 P = 482/0.0241 = 20000 ∴ Amount is Rs 20000. |
|
| 9. |
Meera borrowed a sum of Rs. 1000 from Sita for two years. If the rate of interest is 10% compounded annually, find the amount that Meera has to pay back. |
|
Answer» Given, Principal = Rs.1000 Rate of interest = 10% p.a Time = 2 years Hence, Compound interes \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={1000}[({1}+\frac{10}{100})^2-{1}]\) \(={1000}\times\frac{21}{100}\) =Rs. 210 So, Amount that Meera has to pay back = Rs.(1000+210) = Rs.1210 |
|
| 10. |
Rahman lent Rs. 16000 to Rasheed at the rate of 12 1/2%per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years. |
|
Answer» Given, Principal = Rs.16000 Rate \(={12}\frac{1}{2}\text%\) per annum \(=\frac{25}{2}\text%\) Time = 3 years Hence, Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={16000}[({1}+\frac{25}{(2\times100)})^3-{1}]\) \(={16000}[(\frac{9}{8})^3-{1}]\) \(={16000}\times\frac{217}{512}\) =Rs. 6781.25 So, Amount payable by Rasheed to Rahman after 3 years = Rs(16000+6781.25) = Rs.22781.25 |
|
| 11. |
Two friends A and B jointly lent out Rs 81600 at 4% compound interest. After 2 years A gets the same amount as B gets after 3 years. The investment made by B was (a) Rs 40000 (b) Rs 30000 (c) Rs 45000 (d) Rs 38000 |
|
Answer» (a) Rs 40000 Let the investment made by A be Rs x. Then, investment made by B = Rs (81600 – x) Given, x\(\big(1+\frac{4}{100}\big)^2\) = (81600-x)\(\big(1+\frac{4}{100}\big)^3\) \(\Rightarrow\)\(\frac{X}{(81600-X)}\) = \(\frac{\big(1+\frac{4}{100}\big)^3}{\big(1+\frac{4}{100}\big)^2}\) \(\Rightarrow\) \(\frac{X}{(81600-X)}\) - \(\big(1+\frac{4}{100}\big)\) - \(\frac{26}{25}\) \(\Rightarrow\) 25x = (81600 – x) × 26 \(\Rightarrow\) 25x + 26x = 81600 × 26 \(\Rightarrow\) 51x = 81600 × 26 \(\Rightarrow\) x = \(\frac{81600\times26}{51}\) = 41600 \(\therefore\) Investment made by B = Rs (81600 – Rs 41600) = Rs 40000. |
|
| 12. |
A man deposited Rs 6000 in a bank at 5% simple interest. Another man deposited Rs 5000 at 8% compound interest. After 2 years, the difference of their interests will be (a) Rs 230 (b) Rs 232 (c) Rs 600 (d) Rs 832 |
|
Answer» (b) Rs 232 First man's S.I = \(\frac{Rs\,6000\times5\times2}{100}\) = Rs 600 Second man's C.I = Rs 5000\(\Big[\big(1+\frac{8}{100}\big)^2-1\Big]\) = Rs 5000\(\Big[\big(\frac{27}{25}\big)^2-1\Big]\) =Rs 5000\(\Big[\frac{729-625}{625}\Big]\) = Rs \(\frac{5000\times104}{625}\) = Rs 832 \(\therefore \) Difference in interest = Rs 832 – Rs 600 = Rs 232. |
|
| 13. |
The difference between compound interest and simple interest on a sum for 2 years at 8% p.a. is Rs 768. The sum is (a) Rs 100000 (b) Rs 110000 (c) Rs 120000 (d) Rs 170000 |
|
Answer» (c) Rs 120000 Let the sum be Rs 100. Then, S.I = Rs\(\frac{100\times2\times8}{100}\) = Rs 16 C.I = Rs 100\(\Big[\big(1+\frac{8}{100}\big)^2-1\Big]\) = Rs 100\(\Big[\big(\frac{27}{25}\big)^2-1\Big]\) = Rs 100 x \(\Big[\frac{729-625}{625}\Big]\) = Rs\(\big(\frac{100\times104}{625}\big)\) = Rs 16.64 \(\therefore\) Difference = Re 0.64 If the difference between C.I. and S.I. is Re 0.64, Sum = Rs 100 If the difference is Rs 768, sum =Rs \(\big(\frac{100}{0.64}\times768\big)\) = Rs 120000. |
|
| 14. |
The compound interest on a sum for two years is Rs 832 and the simple interest on the same sum at the same rate for the same period is Rs 800. What is the rate of interest ? (a) 6% (b) 8%(c) 10% (d) 12% |
|
Answer» (b) 8% Let the sum be Rs P and rate of interest per annum be R% Then, P\(\Big[\big(1+\frac{R}{100}\big)^2-1\Big]\) - \(\frac{2PR}{100}\) = Rs 832 – Rs 800 = Rs 32 \(\Rightarrow\) P\(\Big[1+\frac{2R}{100}+\frac{R^2}{10000}-1\Big]\) - \(\frac{2PR}{100}\) = 32 \(\Rightarrow\) \(\frac{PR^2}{10000}\) = 32 \(\Rightarrow\) PR x R = 320000.......(i) Also, \(\frac{2PR}{100}\) = 800 (S.I) \(\Rightarrow\) PR = 40000 ..........(ii) \(\therefore\) From (i) and (ii) 40000 × R = 320000 \(\Rightarrow\) R = 8% p.a |
|
| 15. |
A money lender borrows money at 4% p.a. simple interest and pays interest at the end of the year. He lends it at 6% p.a. compound interest compounded half-yearly and receives the interest at the end of the year. Thus he gains Rs 104.50 a year. The amount of money he borrows is (a) Rs 4500 (b) Rs 5000(c) Rs 5500(d) Rs 6000 |
|
Answer» (b) Rs 5000 Let the amount of money he borrows be Rs x. Then, Interest paid by the money lender= \(\frac{X\times4\times1}{100}\) = Rs \(\frac{4X}{100}\) Interest received by the money lender = x\(\Big[\big(1+\frac{3}{100}\big)^2-1\Big]\) = x\(\Big[\frac{(103)^2}{(100)^2}-1\Big]\) = x\(\Big[\frac{103^2-100^2}{10000}\Big]\) = x\(\Big[\frac{103+100)(103-100)}{10000}\Big]\) = \(\frac{609X}{10000}\) Given, \(\frac{609X}{10000}\)-\(\frac{4X}{100}\) = 104.50 \(\Rightarrow\) \(\frac{209X}{10000}\) = 104.50 \(\Rightarrow\) 209x = 1045000 \(\Rightarrow\) x = \(\frac{1045000}{209}\) = Rs 5000. |
|
| 16. |
If the compound interest on a certain sum for three years at 10% is Rs 993, what would be the simple interest on the same sum at the same rate for the same time ? (a) Rs 930 (b) Rs 920 (c) Rs 900 (d) Rs 890 |
|
Answer» (c) Rs 900 Given, 993= P\(\Big[\big(1+\frac{10}{100}\big)^3-1\Big]\) \(\Rightarrow\) 993 = P\(\Big[\big(\frac{11}{10}\big)^3-1\Big]\) \(\Rightarrow\) 993 = P\(\Big[\frac{1331-1000}{1000}\Big]\) \(\Rightarrow\) P = \(\frac{993\times1000}{331}\) \(\Rightarrow\) P = Rs 3000. \(\therefore\) S.I = Rs\(\frac{3000\times3\times10}{100}\)= Rs 900. |
|
| 17. |
A sum of money at compound interest amounts to three times of itself in three years. In how many years will it be nine times of itself ? (a) 6 years (b) 5 years (c) 9 years (d) 7 years |
|
Answer» (a) 6 years Given, 3P = P\(\big(1+\frac{r}{100}\big)^3\) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^3\) = 3 Let t be the time in years in which the sum will be nine times of itself. Then, 9P = P\(\big(1+\frac{r}{100}\big)^t\) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = 9 = 32 \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = \(\Big[\big(1+\frac{r}{100}\big)^3\Big]^2\) (From (i)) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = \(\big(1+\frac{r}{100}\big)^6\) \(\Rightarrow\) t = 6 years. |
|
| 18. |
On what sum will the compound interest at 5% per annum for 2 years compounded annually be Rs 164? |
|
Answer» Given, Rate of interest= 5% p.a Time = 2 years Compound interest = Rs.164 Let principal = P By applying formula , Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={P}[({1}+\frac{5}{100})^2-{1}]\) = 164 \(={P}[(\frac{21}{20})^2-{1}]\) = 164 \(={P}\times\frac{41}{400}\) = 164 \(={P}=\frac{164\times400}{41}\) =Rs. 1600 Hence, Principal = Rs. 1600. |
|
| 19. |
A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded annually. Find the sum. |
|
Answer» Given details are, Rate = 10 % per annum Amount = Rs 756.25 Time (t) = 2 years By using the formula, A = P (1 + R/100) n 756.25 = P (1 + 10/100)2 P = 756.25 / (1 + 10/100)2 = 756.25/1.21 = 625 ∴ The principal amount is Rs 625. |
|
| 20. |
In how many years will the compound interest on Rs 1000 at the rate of 10% p.a. be Rs 210 ? |
|
Answer» P = Rs 1000, C.I. = Rs 210 A = P + C.I. = Rs 1000 + Rs 210 = Rs 1210 r = 10% p.a., n = ? \(\therefore\) 1210 = 1000\(\big(1+\frac{10}{100}\big)^n\) \(\Rightarrow\) \(\frac{1210}{1000}\) = \(\big(1+\frac{1}{10}\big)^n\) \(\Rightarrow\) \(\frac{121}{100}\) = \(\big(\frac{11}{10}\big)^n\) = \(\big(\frac{11}{10}\big)^2\) = \(\big(\frac{11}{10}\big)^n\) \(\Rightarrow\) n = 2years. |
|
| 21. |
Find the principal if the interest compounded annually at the rate of 10% for two years is Rs. 210. |
|
Answer» Given details are, Rate = 10 % per annum Compound Interest (CI) = Rs 210 Time (t) = 2 years By using the formula, Let P be ‘x’ CI = A – P 210 = P (1 + R/100) n – P = P [(1 + R/100)n – 1] = x [(1 + 10/100)2 – 1] = x [(110/100)2 – 1] 210 = x ((1.1)2 – 1) x = 210 / ((1.1)2 – 1) = 210/0.21 = Rs 1000 ∴ The required sum is Rs 1000. |
|
| 22. |
A bank offers, continuously increasing principal at the rate of 8% per year. Rohit thought to invest 16000Rs then how much long will it take for him to double the value invested by him? |
|
Answer» R = 8%, P = Rs. 16000, A = Rs. 32000 Let after n years Rohit will get money to double the value invested by him. \(\because A=P(1+\frac R{100})^n\) ⇒ 32000 = 16000(1 + \(\frac{8}{100}\))n ⇒ (1 + \(\frac{2}{25}\))n = \(\frac{32000}{16000}=2\) ⇒ \((\frac{27}{25})^n=2\) ⇒ n log \(\frac{27}{25}\) = log 2 ⇒ n = \(\frac{log2}{log1.08}\) = \(\frac{0.3010}{0.33423}\approx9\) Almost, n = 9 years Rohit will double money as he invested. |
|
| 23. |
Manu deposited Rs 15000 in a financial institution. Interest is calculated in every 3 months and added to the amount. Rate of interest in 8 %. How much he gets after 1 year. |
|
Answer» Amount deposited = Rs 15000 Rate of interest = 8% Amount he gets after 1 year |
|
| 24. |
Surabhi borrowed a sum of Rs. 12000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years. |
|
Answer» Given details are, Principal (p) = Rs 12000 Rate (r) = 5 % Time = 3years By using the formula, A = P (1 + R/100) n = 12000 (1 + 5/100)3 = 12000 (105/100)3 = Rs 13891.5 ∴ Compound Interest = A – P = Rs 13891.5 – Rs 12000 = Rs 1891.5 |
|
| 25. |
Anil borrowed a sum of Rs. 9600 to install a hand pump in his dairy. If the rate of interest is 5 ½ % per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years. |
|
Answer» Given details are, Principal (p) = Rs 9600 Rate (r) = 5 ½ % = 11/2 % Time = 3years By using the formula, A = P (1 + R/100) n = 9600 (1 + 11/2×100)3 = 9600 (211/200)3 = Rs 11272.71 ∴ Compound Interest = A – P = Rs 11272.71 – Rs 9600 = Rs 1672.71 |
|
| 26. |
Anil borrowed a sum of Rs. 9600 to install a handpump in his dairy. If the rate of interest is 5 1/2 % per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years. |
|
Answer» Given, Rate of interest \(={5}\frac{1}{2}\text%\)\(=\frac{11}{2}\text%\) annually Time = 3 years Hence, \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={9600}[({1}+\frac{11}{2\times100})^3-{1}]\) = 9600× 0.174 = Rs.1672.72 So, |
|
| 27. |
Swati took a loan of Rs. 16000 against her insurance policy at the rate of 12 % per annum. Calculate the total compound interest payable by Swati after 3 years. |
|
Answer» Given, Principal = Rs.16000 Rate \(={12}\frac{1}{2}\text%\) Time = 3 years Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={16000}[({1}+\frac{25}{2\times100})^3-{1}]\) \(={16000}\times\frac{217}{512}\) =Rs. 6781.25 |
|
| 28. |
Roma borrowed Rs. 64000 from a bank for 1 1/2 years at the rate of 10% per annum. Compare the total compound interest payable by Roma after 1 1/2 compound interest payable by Roma after |
|
Answer» Given, Principal = Rs. 64000 Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\times{2}\) = 3 half years Rate \(={10}\text%\) \(=\frac{10}{2} \) \(={5}\text%\) half yearly Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={64000}[({1}+\frac{5}{100})^3-{1}]\) \(={64000}\times\frac{1261}{8000}\) =Rs. 10088 |
|
| 29. |
Swati took a loan of Rs. 16000 against her insurance policy at the rate of 12 1/2 % per annum. Calculate the total compound interest payable by Swati after 3 years. |
|
Answer» Given details are, Principal (p) = Rs 16000 Rate (r) = 12 ½ % = 12.5% Time = 3years By using the formula, A = P (1 + R/100) n = 16000 (1 + 12.5/100)3 = 16000 (112.5/100)3 = Rs 22781.25 ∴ Compound Interest = A – P = Rs 22781.25 – Rs 16000 = Rs 6781.25 |
|
| 30. |
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs. 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half. |
|
Answer» Given, Principal = Rs.64000 Rate of interest = 5% per annum \(=\frac{5}{2}\text%\) half yearly Time \({1}\frac{1}{2}\) years \(=\frac{2}{2}\times{2}\) = 3 half years Hence, Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={64000}[({1}+\frac{5}{2\times100})^3-{1}]\) \(={64000}[(\frac{41}{40})^3-{1}]\) \(={64000}\times\frac{4921}{64000}\) Rs. 4921 So, interest paid by Abha = Rs.4921 |
|
| 31. |
Kamal borrowed Rs. 57600 from LIC against her policy at 12 1/2 % per annum to build a house. Find the amount that she pays to the LIC after 1 1/2% years if the interest is calculated half-yearly. |
|
Answer» Given, Principal = Rs.57600 Rate \(={12}\frac{1}{2}\text%\) per annum \(=\frac{25}{4}\text%\) = 3 half years Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\times{2}\) = 3 half years Hence, Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={57600}{({1}+\frac{25}{4\times100})^3-{1}}]\) \(={57600}{(\frac{17}{16})^3-{1}}]\) \(={57600}\times\frac{817}{4096}\) =Rs. 11489.06 So, Amount that Kamal pays to LIC after \({1}\frac{1}{2}\) years = Rs.(57600 + 11489.06) = Rs.69089.06 |
|
| 32. |
Find the amount of Rs. 10000 after 2 years compounded annually; the rate of interest being 10% per anum during the first year and 12% per annum during the second year. Also, find the compound interest. |
|
Answer» Present value = Rs.10000 Interest rate for 1st year, p = 10 % per annum Interest rate for 2nd year, q = 12 % per annum Time = 2 years Amount (A) = P × (1 + p/100) × (1 + q/100) A = 10000 × (1 + 10/100) × (1 + 12/100) = 10000 × (1 + 1/10) × (112/100) = 10000 × 11/10 × 112/100 = 10 × 11 × 112 = 12320 ∴ Amount = Rs.12320 ∴ Compound interest = Rs.(12320 – 10000) = Rs.2320 |
|
| 33. |
In how many years will Rs. 1800 amount to Rs. 2178 at 10% per annum when compounded annually? |
|
Answer» Let time = n years P = Rs.1800 A = Rs.2178 R = 10% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ A = P (1 + R/100)n ⇒ 2178 = 1800 (1 + 10/100)n ⇒ (1 + 1/10)n = 2178/1800 ⇒ (11/10)n = 121/100 ⇒ (11/10)n = (11/10)2 ⇒ n = 2 ∴ Time = 2 years. |
|
| 34. |
A building worth Rs 133100 is constructed on land worth Rs 72900. After how many years will the value of both be same if the land appreciates at 10% p.a. and building depreciates at 10% p.a. ?(a) \(1\frac{1}{2}\) years(b) 2 years(c) 2\(\frac{1}{2}\) years(d) 3 years |
|
Answer» (d) 3 years Let the required time be t years. Then, 72900\(\big(1+\frac{10}{100}\big)^t\) = 133100\(\big(1-\frac{10}{100}\big)^t\) \(\Rightarrow\) \(\big(\frac{11}{10}\big)^t\)x \(\big(\frac{10}{9}\big)^t\) = \(\big(\frac{11}{9}\big)^3\) \(\Rightarrow\)\(\big(\frac{11}{9}\big)^t\) = \(\big(\frac{11}{9}\big)^3\) \(\Rightarrow\) t = 3 years. |
|
| 35. |
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population. |
|
Answer» Given details are, Growth rate of population of town is = 50/1000×100 = 5% Population after 2 years is = 22050 So, let present population of town be = x By using the formula, A = P (1 + R/100) 22050 = x (1 + 5/100) (1 + 5/100) 22050 = x (105/100) (105/100) 22050 = x (1.05) (1.05) 22050 = 1.1025x x = 22050/1.1025 = 20000 ∴ Present population of the town is 20000. |
|
| 36. |
The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. If the original count of the bacteria in a sample is 10000, find the bacteria count at the end of 3 hours. (a) 13310 (b) 10890 (c) 10990 (d) 11000 |
|
Answer» (b) 10890 Bacteria count after 3 years = 10000\(\big(1+\frac{10}{100}\big)\)\(\big(1-\frac{10}{100}\big)\)\(\big(1+\frac{10}{100}\big)\) = 10000 x \(\frac{11}{10}\)x \(\frac{9}{10}\)x \(\frac{11}{10}\) = 10890. |
|
| 37. |
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours? |
|
Answer» Given details are, Count of bacteria in sample is = 13125000 The increase and decrease of growth rates are = 10%, -8%, 12% So, let the count of bacteria after 3 hours be = x By using the formula, A = P (1 + R/100) x = 13125000 (1 + 10/100) (1 – 8/100) (1 + 12/100) x = 13125000 (110/100) (92/100) (112/100) x = 13125000 (1.1) (0.92) (1.12) = 14876400 ∴ Count of bacteria after three hours will be 14876400. |
|
| 38. |
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago? |
|
Answer» Given details are, Annual growth rate of population of city is = 8% Present population of city is = 196830 Let population of city 3 years ago be = x By using the formula, A = P (1 + R/100) 196830 = x (1 + 8/100) (1 + 8/100) (1 + 8/100) 196830 = x (27/25) (27/25) (27/25) 196830 = x (1.08) (1.08) (1.08) 196830 = 1.259712x x = 196830/1.259712 = 156250 ∴ Population 3 years ago was 156250. |
|
| 39. |
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours? |
|
Answer» Given, Count of bacteria in sample = 13125000 According to increase and decrease of growth rates, Let count of bacteria after 3 hours = X So, = 13125000\(({1}+\frac{10}{100})({1}-\frac{8}{100})({1}+\frac{12}{100})\) = x = 13125000\(\times\frac{11}{10}\times\frac{23}{25}\times\frac{28}{25} = x\) \(= x=\frac{1312500 \times 10 \times 25 \times 25}{11 \times 23 \times 28}\) = 14876400 Hence, Count of bacteria after 3 hours will be = 14876400 |
|
| 40. |
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000? |
|
Answer» Given, Population of city on last day of year 1998 = 72000 Increasing rate in 1999 = 7% Decreasing rate in 2000 = 10 % So, Population at the end of 2000 \(= 72000\times({1}+\frac{7}{100})({1}-\frac{10}{100})\) \(= 72000\times\frac{107}{100}\times\frac{9}{10}\) = 69336. |
|
| 41. |
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year? |
|
Answer» Given, Intial number of workers = 6400 At the end of first year = 25% retrenched At the end of second year = 25% retrenched At the end of third year = 25% increased So, Number of workers working during fourth year \(= 6400({1}-\frac{25}{100})({1}-\frac{25}{100})({1}+\frac{25}{100})\) \(= 6400\times\frac{3}{4}\times\frac{3}{4}\times\frac{5}{4}\) = 4500 workers |
|
| 42. |
Find the compound interest on Rs. 1000 at the rate of 8% per annum for 1 1/2 years when interest is compounded half yearly. |
|
Answer» Given details are, Principal (p) = Rs 1000 Rate (r) = 8% Time = 1 ½ years = 3/2 × 2 = 3 half years By using the formula, A = P (1 + R/100) n = 1000 (1 + 8/100)3 = 1000 (208/100)3 = Rs 1124.86 ∴ Compound Interest = A – P = Rs 1124.86 – Rs 1000 = Rs 124.86 |
|
| 43. |
In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound interest? |
|
Answer» Given details are, Principal = Rs 1000 Amount = Rs 1331 Rate = 10% per annum Let time = T years By using the formula, A = P (1 + R/100)n 1331 = 1000 (1 + 10/100)T 1331 = 1000 (110/100)T (11/10)T = 1331/1000 (11/10)T = (11/10)3 So on comparing both the sides, n = T = 3 ∴ Time required is 3years. |
|
| 44. |
The difference in simple interest and compound interest on a certain sum of money at 6 2/3 % per annum for 3 years in Rs. 46. Determine the sum. |
|
Answer» Given, Time = 3 years Rate = 6 2/3 % per annum = 20/3% Let principal = Rs P Compound Interest (CI) – Simple Interest (SI) = Rs 46 C.I – S.I = Rs 46 By using the formula, P [(1 + R/100)n – 1] – (PTR)/100 = 46 P [(1 + 20/3×100)3 – 1] – (P(3)(20/3))/100 = 46 P[(1 + 20/300)3 – 1] – P/5 = 46 P[721/3375] – P/5 = 46 721/3375P – 1/5P = 46 (721P-675P)/3375 = 46 46P = 46 × 3375 46P = 46 × 3375/46 = 3375 ∴ The sum is Rs 3375. |
|
| 45. |
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs. 20. Find the sum. |
|
Answer» Given details are, Rate = 4 % per annum Time = 2years Compound Interest (CI) – Simple Interest (SI)= Rs 20 By using the formula, CI – SI = 20 P [(1 + R/100)n – 1] – (PTR)/100 = 20 P [(1 + 4/100)2 – 1] – (P(2)(4))/100 = 20 P[51/625] – (2P)/25 = 20 51/625P – 2/25P = 20 (51P-50P)/625 = 20 P = 20 × 625 P = 20/7.918 = 12500 ∴ The sum is Rs 12500. |
|
| 46. |
Find the compound interest on Rs. 1000 at the rate of 8% per annum for 1 years when interest is compounded half yearly. |
|
Answer» Given, Principal = Rs.1000 Rate = 8% p.a \(=\frac{8}{2}={4}\text% \) half yearly Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\times{2}\) = 3 half years Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={1000}[({1}+\frac{4}{100})^3-{1}]\) \(={1000}[(\frac{26}{25})^3-{1}]\) \(={1000}\times\frac{1951}{1525}\) =Rs. 124.86 |
|
| 47. |
The difference between the S.I. and C.I. on a certain sum of money for 2 years at Rs.4 per annum is Rs. 20. Find the sum. |
|
Answer» Given, Rate = 4% Time = 2 years C.I – S.I = Rs.20 Let principal = P So, \(={P}[({1}+\frac{4}{100})^2-{1}]-\frac{P \times 4 \times 2}{100}\) = 20 \(=\frac{51p}{625}-\frac{2p}{25}\) = 20 \(=\frac{51p-50p}{625}\) = 20 = p = \({20}\times{625}\) = Es. 12500 Hence, Principal = Rs. 12500 |
|
| 48. |
In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest compounded half-yearly? |
|
Answer» Given, Principal = Rs. 4400 Amount = Rs.4576 Rate = 8% p.a \(=\frac{8}{2} ={4}\text%\) half yearly Let time = T years = 2T half years So, A \(={P}[({1}+\frac{4}{100})^T]\) \(={4400}[({1}+\frac{4}{100})^{2T}]\) = 4576 \(=(\frac{26}{25})^{2T}\) \(=\frac{4576}{4400}\) \(=\frac{26}{25}\) = 2T = 1 \(={T}=\frac{1}{2}\) years Hence, Time \(=\frac{1}{2}\) years |
|
| 49. |
In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest compounded half-yearly? |
|
Answer» Given details are, Rate = 8% per annum = 8/2 = 4% (half yearly) A = Rs 4576 Principal (P) = Rs 4400 Let n be ‘2T’ By using the formula, A = P (1 + R/100)n 4576 = 4400 (1 + 4/100)2T 4576 = 4400 (104/100)2T (104/100)2T = 4576/4400 (104/100)2T = 26/25 (26/25)2T = (26/25)1 So on comparing both the sides, n = 2T = 1 ∴ Time required is ½ year. |
|
| 50. |
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly. |
|
Answer» Given details are, Time = 2 years = 2×2 = 4 half years Let rate = R % per annum = R/2% half years Let principal be = P So, Amount becomes = 4P By using the formula, A = P (1 + R/100)n 4P = P (1 + R/2×100)4 (1 + R/200)4 = 4 (1 + R/200) = 41/4 1 + R/200 = 1.4142 R/200 = 1.4142-1 = 0.4142 R = 0.4142 × 200 = 82.84% ∴ Required Rate is 82.84% per annum. |
|