## Sample Questions: Sales Tax and VAT Exercise 3(c)

In this Exercise, all the prices are excluding Vat unless specified

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Question 1: A person purchases an article for Rs.6200 and sells it to a customer for Rs.8500. if the sales-tax (under VAT) is 8%; find the VAT paid by the person.

Answer:

Cost Price of the article $= Rs. \ 6200$

Selling Price of the article $= Rs. \ 8500$

$VAT = Tax \ Charges - Tax \ Paid$

$= Tax \ on \ Selling \ Prices - Tax \ on \ Cost \ Prices$

$= \frac{8}{100} \times 8500- \frac{8}{100} \times 6200$

$= Rs. \ 184$

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Question 2: A purchases an article for Rs.3600 and sells it to B for Rs.4800. B in turn, sells the article to C for Rs.5500. If the Sales-Tax (under VAT) is 10%, find the VAT levied on A and B.

Answer:

Cost Price of the article for A $= Rs. \ 3600$

Selling Price of the article  A $= Rs. \ 4800$

VAT levied on A $= \frac{10}{100} \times (4800-3600) = Rs. \ 120$

Cost Price of the article for B $= Rs. \ 4800$

Selling Price of the article  B $= Rs. \ 5500$

VAT levied on B $= \frac{10}{100} \times (5500-4800) = Rs. \ 70$

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Question 3: A manufacturer buys raw material for Rs.60,000 and pays 4% tax. He sells the ready stock for Rs.92,000 and charged 12.5% tax. Find the VAT paid by the manufacturer.

Answer:

Cost price of the raw material  $= Rs. \ 60000$

Tax paid by the manufacturer  $= \frac{4}{100} \times 60000 = Rs. \ 2400$

Selling price of the stock  $= Rs. \ 92000$

Tax charged  $= \frac{12.5}{100} \times 92000 = Rs. \ 11500$

$VAT = Tax \ recovered \ on \ sale - Tax \ paid \ on \ purchase$

$= Rs. \ 11500 - Rs. \ 2400 = Rs. \ 9100$

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Question 4: The cost of an article is Rs.6000 to a distributor. He sells it to a trader for Rs.7500 and the trader sells it to a customer for Rs.8000. If the VAT rate is 12.5%; Find the VAT paid by the:
i. Distributor
ii. Trader

Answer:

Cost Price of the article for distributor $= Rs. \ 6000$

Selling Price of the article for distributor $= Rs. \ 7500$

VAT levied on distributor $= \frac{12.5}{100} \times (7500-6000) = Rs. \ 187.50$

Cost Price of the article for Trader $= Rs. \ 7500$

Selling Price of the article  for Trader $= Rs. \ 8000$

VAT levied on Trader $= \frac{12.5}{100} \times (8000-7500) = Rs. \ 62.50$

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Question 5: The printed price of an article is Rs.2500. A wholesaler sells it to a retailer at 20% discount and charges sales-tax at the rate of 10%. Now the retailer, in turn, sells the article to a customer at its list price and charges the sales-tax at the same rate. Find:
i. The amount that retailer pays to the wholesaler.
ii. The VAT paid by the retailer.

Answer:

Printed Price of the article for wholesaler $= Rs. \ 2500$

Cost Price of the article for retailer $= \frac{100-20}{100} \times 2500 =Rs. \ 2000$

Amount paid by the retailer $= \frac{10}{100} \times 2000 +2000 = Rs. \ 2200$

Selling Price of the article for retailer $= Rs. \ 2500$

VAT levied on distributor $= \frac{10}{100} \times (2500-2000) = Rs. \ 50$

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Question 6: A retailer buys an article for Rs.800 and pays the sale-tax at the rate of 8%. The retailer sells the same article to a customer for Rs.1000 and charges sales-tax at the same rate. Find:
i. The price paid by a customer to buy this article.
ii. The amount of VAT paid by the retailer.

Answer:

Cost price of the article  $= Rs. \ 800$

Tax paid by the retailer  $= \frac{8}{100} \times 800 = Rs. \ 64$

Selling price of the article  $= Rs. \ 1000$

Tax charged  $= \frac{8}{100} \times 1000 = Rs. \ 80$

Price paid by the customer $= Rs. \ 1000 + Rs. \ 80 = Rs. \ 1080$

$VAT = Tax \ recovered \ on \ sale - Tax \ paid \ on \ purchase$

$= Rs. \ 80 - Rs. \ 64 = Rs. \ 16$

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Question 7: A shopkeeper buys 15 identical articles for Rs.840 and pays sales-tax at the rate of 8%. He sells 6 of these articles at Rs.65 each and charges sales-tax at the same rate. Calculate the VAT paid by the shopkeeper against the sale of these six articles.

Answer:

Cost price of the article  $= Rs. \ \frac{840}{15} = Rs. \ 56$

Tax paid by the retailer  $= \frac{8}{100} \times 56 \ times 6 = Rs. \ 26.88$

Selling price of the article  $= Rs. \ 65$

Tax charged  $= \frac{8}{100} \times 65 \times 6 = Rs. \ 31.20$

$VAT = Tax \ recovered \ on \ sale - Tax \ paid \ on \ purchase$

$= Rs. \ 31.20 - Rs. \ 26.88 = Rs. \ 4.32$

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Question 8: The marked price of an article is Rs.900 and the rate of sales-tax on it is 6%. If on selling the article at its marked price, a retailer has to pay VAT=Rs.4.80; Find the money paid by him (including sales-tax) for purchasing this article.

Answer:

Let Cost price of the article  $= Rs. \ x$

Tax paid by the retailer  $= \frac{6}{100} \times x = Rs. \ 0.06x$

Selling price of the article  $= Rs. \ 900$

Tax charged  $= \frac{6}{100} \times 900 = Rs. \ 54$

$VAT = Tax \ recovered \ on \ sale - Tax \ paid \ on \ purchase$

$= Rs. \ 54 - Rs. \ 0.06x = Rs. \ 4.80 \Rightarrow x = 820$

Money paid $= Rs. \ 820 + Rs. \ 0.06 \times 820 = Rs. \ 869.20$

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Question 9: A manufacturer marks an article at Rs.5000. He sells this article to a wholesaler at a discount of 25% on the marked price and the wholesaler sells it to a retailer at a discount of 15% on its marked price. If the retailer sells the article without any discount and at each stage the sales-tax is 8%, calculate the amount of VAT paid by:
i. The Wholesaler
ii. The Retailer

Answer:

Market Price of the article for wholesaler $= Rs. \ 5000$

Cost Price of the article for wholesaler $= \frac{100-25}{100} \times 5000 =Rs. \ 3750$

Amount of tax paid by the wholesaler $= \frac{8}{100} \times 3750 = Rs. \ 300$

Cost Price of the article for retailer $= \frac{100-15}{100} \times 5000 =Rs. \ 4250$

Amount of tax paid by the retailer $= \frac{8}{100} \times 4250 = Rs. \ 340$

Selling price of the article for retailer $= Rs. \ 5000$

Amount of tax paid by the end customer $= \frac{8}{100} \times 5000 = Rs. \ 400$

Vat paid by retailer $= 400 - 340 = Rs. 60$

Vat paid by wholesaler $= 340 - 300 = Rs. 40$

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Question 10: A shopkeeper buys an article at a discount of 30% and pays sales-tax at the rate of 8%. The shopkeeper, in turn, sells the article to a customer at the printed price and charges sales tax at the same rate. If the printed price of the article is Rs.2500; Find:
i. The price paid by the shopkeeper.
ii. The price paid by the customer.
iii. The VAT paid by the shopkeeper.

Answer:

Printed Price of the article for shopkeeper $= Rs. \ 2500$

Cost Price of the article for shopkeeper $= \frac{100-30}{100} \times 2500 =Rs. \ 1750$

Tax paid by the shopkeeper $= \frac{8}{100} \times 1750 = Rs. \ 140$

Price paid by the shopkeeper $= Rs. \ 1750 + \ Rs. 140 = Rs. \ 1890$

Selling Price of the article for shopkeeper $= Rs. \ 2500$

Tax paid by the customer $= \frac{8}{100} \times 2500 = Rs. \ 200$

Price paid by the customer $= Rs. \ 2500 + \ Rs. 200 = Rs. \ 2700$

VAT paid by the shopkeeper $= \frac{8}{100} \times (2500-1750) = Rs. \ 60$

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Question 11: A shopkeeper sells an article at its list price Rs.3000 and charges sales-tax at the rate of 12%. If the VAT paid by the shopkeeper is Rs.72, at what price did the shopkeeper buy the article inclusive of sales-tax?

Answer:

Let Cost Price of the article for shopkeeper $= Rs. \ x$

Tax paid by the shopkeeper $= \frac{12}{100} \times x = Rs. \ 0.12x$

Selling Price of the article for shopkeeper $= Rs. \ 3000$

Tax paid by the customer $= \frac{12}{100} \times 3000 = Rs. \ 360$

$VAT = Tax \ recovered \ on \ sale - Tax \ paid \ on \ purchase$

$= Rs. \ 360 - Rs. \ 0.12x = Rs. \ 72 \Rightarrow x = Rs. \ 2400$

Tax paid by the shopkeeper $= 0.12 \times 2400 = Rs. 288$

Price paid by the shopkeeper $= 2400 + 288 = Rs. 2688$

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Question 12: A manufacturer marks an article for Rs.10,000. He sells it to a wholesaler at 40% discount. The wholesaler sells this article to a retailer at 20% discount on the marked price of the article. If retailer sells the article to a customer at 10% discount and the rate of sales-tax is 12% at each stage, find the amount of VAT paid by the:
i. Wholesaler
ii. Retailer

Answer:

Market Price of the article for wholesaler $= Rs. \ 10000$

Cost Price of the article for wholesaler $= \frac{100-40}{100} \times 10000 =Rs. \ 6000$

Amount of tax paid by the wholesaler $= \frac{12}{100} \times 6000 = Rs. \ 720$

Cost Price of the article for retailer $= \frac{100-20}{100} \times 10000 =Rs. \ 8000$

Amount of tax paid by the retailer $= \frac{12}{100} \times 8000 = Rs. \ 960$

Selling price of the article for retailer $= \frac{100-10}{100} \times 10000 =Rs. \ 9000$

Amount of tax paid by the end customer $= \frac{12}{100} \times 9000 = Rs. \ 1080$

Vat paid by retailer $= 1080 - 960 = Rs. 120$

Vat paid by wholesaler $= 960 - 720 = Rs. 240$

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## Sample Questions: Sales Tax and VAT Exercise 3(b)

Question 1: A person buys an unfinished article for Rs.1800 and spends Rs.600 on its finishing, pacing, transportation, etc. He marks the article at such a price that will give him 20% profit. How much will a customer pay for the article including 12% Sales Tax.

Answer:

Price paid for the unfinished article $= Rs. \ 1800$

Overheads   $= Rs. \ 600$

Therefore Cost of the article $= Rs. \ 1800 + Rs. \ 600 = Rs. \ 2400$

Desired profit   $= 20\%$

Sale Price   $= (\frac{100+20}{100}) \ of \ Rs. \ 2400 = (\frac{100+20}{100}) \times 2400 = Rs. \ 2880$

Therefore Money paid by the customer = Sale Price  of the article+Sales Tax on the article

$= Rs. \ 2880 + \frac{12}{100} \times 2880 = Rs. \ 3225.60$

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Question 2: A person buys an article for Rs.800 and spends Rs.100 on its transportation, etc. He marks the article at a certain price and then sells it for Rs.1287 including 10% Sales Tax. Find this profit as per cent.

Answer:

Price paid for the article $= Rs. \ 800$

Overheads   $= Rs. \ 100$

Therefore Cost of the article $= Rs. \ 800 + Rs. \ 100 = Rs. \ 900$

Let the Sale Prices $= x$

Sales Tax on the article $= \frac{10}{100} \times x = 0.1x$

Cost to the customer $= x + 0.1x = 1.1x = 1287 \Rightarrow x = Rs.\ 1170$

Therefore Profit $= 1170-900 = Rs. \ 270$

Therefore profit % $= \frac{270}{900} \times 100 = 30\%$

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Question 3: A person announces a discount of 15% on his goods. If the marked price of an article, in his shop, is Rs.6000; How much a customer has to pay for it, if the rate of Sales Tax is 10%

Answer:

Marker Price $= Rs. \ 6000$

Discount $= 15\%$ of the marked price

Discounted Price $= \frac{100-15}{100} \times 6000 = Rs. \ 5100$

Sales Tax $= \frac{10}{100} \times 5100 = Rs. \ 510$

Price paid by the customer $= 5100 + 510 = Rs. \ 5610$

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Question 4: The catalog price of a colored T.V. is Rs.24000. The shopkeeper gives a discount of 8% on the list price. He gives a further off season discount of 5% on the balance, But Sales Tax at 10% is charged on the remaining amount. Find:

1. The sales Tax a customer has to pay.
2. The final price he has to pay for the T.V. [2001]

Answer:

Catalog price of a colored T.V. $= Rs. \ 24000$

Price after 8% discount $= \frac{100-8}{100} \times 24000 = 22080$

Price after 5% off season discount $= \frac{100-5}{100} \times 22080 = Rs. \ 20976$

Sales Tax $= \frac{10}{100} \times 20976 = Rs. \ 2097.60$

Price paid by the customer $= Rs. \ 20976 + Rs. \ 2097.60 = Rs. \ 23073.60$

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Question 5: A person marks his goods 40% above the cost price and then allows discount of 20%. Find how much will be a customer pay for an article which costs the person Rs.200 and a Sales Tax of 10% is levied on the sales price of the article.

Answer:

The cost price $= Rs. \ 200$

Mark Price of goods  $= 200+0.4 \times 200= Rs. \ 280$

Discount given  $=20\%$

Discounted Price  $= \frac{100-20}{100} \times 280 = Rs. 224$

Amount paid by the customer  $224 + \frac{10}{100} \times 224 = Rs. \ 246.40$

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Question 6: A toy is purchased for Rs.591.36 which includes 12% rebate on the printed price and 12% Sales Tax on the sales price of the toy. Find the printed price of the toy.

Answer:

Let the printed Sale Prices $= x$

After discount the Sale Price $= \frac{100-12}{100} \times x = 0.88x$

Price paid by the customer $= 0.88x+ \frac{12}{100} \times 0.88x = 0.9856x = 591.36 \Rightarrow x = Rs. \ 600.27$

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Question 7: The catalog price on an article is Rs.20,000. The person allows two successive discounts 15% and 10%. If Sales Tax at the rate of 10% is charged on the remaining amount. Find:

1. The sales Tax amount a customer has to pay.
2. The final total price that customer has to pay for the article.

Answer:

Catalog price of a article $= Rs. \ 20000$

Price after 8% discount $= \frac{100-15}{100} \times 20000 = 17000$

Price after 5%  discount $= \frac{100-10}{100} \times 17000 = Rs. \ 15300$

Sales Tax $= \frac{10}{100} \times 15300 = Rs. \ 1530$

Price paid by the customer $= Rs. \ 15300 + Rs. \ 1530 = Rs. \ 16830$

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Question 8: A person buys an article of Rs.1700 at a discount of 15% on its printed price. He raises the printed price of the article by 20% and then sells it for Rs.2688 including sales tax on the new marked price. Find:

1. The rate of sales tax
2. The trader’s profit as per cent.

Answer:

Let the printed price $= Rs. \ x$

Price the trader paid $= \frac{100-15}{100} \times x = 0.85x$

Given $0.85x = 1700 \Rightarrow x= Rs. \ 2000$

New Printed Prices $= \frac{100+20}{100} \times 2000 = Rs. \ 2400$

Let the sales tax % $= x$

Final Selling prices $2688 = 2400 (1+\frac{x}{100}) \Rightarrow x = 12\%$

Profit $= 2400-1700 = Rs. \ 700$

Profit % $= \frac{700}{1700} \times 100 = 41\frac{3}{17}\%$

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Question 9: A person buys an article at a rebate of 20% on its marked price and then spends Rs.300  on its transportation, etc. If he sells the article for Rs.4160 (including sales tax at the rate of 4% of the marked price), find the person’s profit as per cent.

Answer:

Let the printed price $= Rs. \ x$

Price the trader paid $= \frac{100-20}{100} \times x = 0.80x$

Overheads $= Rs. \ 300$

Sales Tax $= \frac{4}{100} \times x = 0.04x$

Total Price paid by the customer $= 4160$

Therefore $4160 = x+0.04x \Rightarrow x = Rs. \ 4000$

Total cost paid by the shopkeeper $= 0.8 \times 4000+ 300 = Rs. \ 3500$

Profit % $= \frac{4160-3500}{3500} \times 100 = 14\frac{2}{7}\%$

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Question 10: A person buys an article for Rs.2400 from a wholesaler at 20% rebate on its list price. He marks up the list price of the article bought by 10% and then sells it for Rs.3498 including sales Tax on the marked-up price. Find:

1. The rate of sales tax
2. The person’s profit as per cent.

Answer:

Let the printed price $= Rs. \ x$

Price the trader paid $= \frac{100-20}{100} \times x = 0.80x$

Given $0.80x = 2400 \Rightarrow x= Rs. \ 3000$

New Printed Prices $= \frac{100+10}{100} \times 3000 = Rs. \ 3300$

Let the sales tax % $= x$

Final Selling prices $3498 = 3300 (1+\frac{x}{100}) \Rightarrow x = 6\%$

Profit $= 3300-2400 = Rs. \ 300$

Profit % $= \frac{900}{2400} \times 100 = 37.5\%$

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## Sample Questions: Sales Tax and VAT Exercise 3(a)

Question 1: A person purchases a wrist-watch costing Rs. 540. The rate of sale Tax is 8%. Find the total amount paid by the person for the watch.

Answer:

Sales price of the wrist watch $= Rs. 540$

Sales Tax $= 8\% \ of \ Rs. \ 540 = \frac{8}{100} \times 540 = Rs. \ 43.2$

Total amount to be paid   $= Rs. \ 540+ Rs. 43.2 = Rs. \ 583.20$

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Question 2: A person paid Rs.345.60 as Sales Tax on a purchase of Rs.3840. Find the rate of Sales Tax.

Answer:

Sales price of the wrist watch $= Rs. 3840$

Sales Tax $= x\% \ of \ Rs. \ 3840 \Rightarrow \frac{x}{100} \times 3840 = 345.60 \Rightarrow x = 9\%$

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Question 3: The price of a washing machine, inclusive of Sales tax, is Rs.13530/-. If the Sales Tax is 10% find its basic cost price.

Answer:

Let Sales Price of the washing machin be  $= Rs. x$

Total amount paid   $= Rs. \ 13530$

Sales Tax $= 10\% \ of \ Rs. \ x$

Therefore $x+10\% \ of \ Rs. \ x = 13530 \Rightarrow x = Rs. \ 12300$

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Question 4: A person purchases biscuits costing Rs.158 on which the rate of Sales Tax is 6%. That person also purchases some cosmetic good costing Rs.354 on which the rate of Sales Tax is 9%. Find the total amount to be paid by the person.

Answer:

Sales price of the biscuit $= Rs. 158$

Sales Tax $= 6\% \ of \ Rs. \ 158 = \frac{6}{100} \times 158 = Rs. \ 9.48$

Sales price of the cosmetic good $= Rs. 354$

Sales Tax $= 9\% \ of \ Rs. \ 354 = \frac{9}{100} \times 354 = Rs. \ 31.86$

Total amount to be paid   $= Rs. \ (158+9.48)+ Rs. (354+31.86) = Rs. \ 553.34$

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Question 5: A person purchases some article costing Rs.5460. The shopkeeper charged Sales Tax at 8% As the person wanted to take the articles purchased outside the State, the shopkeeper charged 3% as Central Sales Tax on the actual price of the articles. Find the total mount the person had to pay for the articles.

Answer:

Sales price of the article $= Rs. 5460$

Sales Tax $= (8+3)\% \ of \ Rs. \ 5460 = \frac{11}{100} \times 5460 = Rs. \ 600.60$

Total amount to be paid   $= Rs. \ 5460+ Rs. 600.60 = Rs. \ 6060.60$

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Question 6: The marked price of two articles A and B together is Rs.6000. The sale tax on article A is 8% and that on article B is 10%. If on selling both the articles, the total sales tax collected is Rs.552, find the marked price of each of the articles A and B.

Answer:

Sales price of the article A $= Rs. x$

Sales price of the article B $= Rs. (6000-x)$

Sales Tax  on A $=8\% \ of \ Rs. \ x = \frac{8}{100} \times x$

Sales Tax  on B $=10\% \ of \ Rs. \ (6000-x) = \frac{10}{100} \times (6000-x)$

Total Sales Tax paid  $= Rs. \ 552$

$\Rightarrow \frac{11}{100} \times x + \frac{10}{100} \times (6000-x) = 552$

Sales price of the article A $= x = Rs. \ 2400$

Sales price of the article B $= (6000-x) = Rs. \ 3600$

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Question 7: A person buys a leather coat costing Rs.3600, the rate of sales Tax being 10%. She asks the shopkeeper to reduce the price of the coat to such an extent that she does not have to pay anything more than Rs.3366 including Sales Tax. Calculate:

1. The reduction needed in the cost price of the coat.
2. The reduction as percent.

Answer:

Sales price of the Leather coat $= Rs. 3600$

Let discounted Sales price of the Leather cost $= Rs. x$

Sales Tax  on the coat $=10\% \ of \ Rs. \ x = 0.1 x$

Total amount paid $x+0.1x = 1.1x = 3366 \Rightarrow x = Rs. \ 3060$

Therefore the discount $= 3600-3060 = Rs. \ 540$

$\% \ reduction = \frac{540}{3600} \times 100 = 15\%$

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Question 8: The price of a T.V. set inclusive of Sales Tax of 9% is Rs.13407. Find its marked price. If Sales Tax is increased to 13%, how much more does the customer has to pay for the T.V. [2002]

Answer:

Let Sales price of the T.V. $= Rs. x$

Sales Tax  on the T.V. $=9\% \ of \ Rs. \ x = 0.09 x$

Total amount paid $x+0.09x = 1.09x = 13407 \Rightarrow x = Rs. \ 12300$

If Sales Tax  on the T.V.  $=13 \% \ of \ Rs. \ 12300$

The customer will have to pay  $(13-9)\% \times 12300 = Rs. \ 492$  more.

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Question 9: The price of an article is Rs.8250 which includes Sales Tax at 10%. Find how much more or less does a customer pay for the article, if the Sales Tax on the article:

1. Increases to 15%
2. Decreases to 6%
3. Increases by 2%
4. Decreases by 3%

Answer:

Total amount of the article $= Rs. 8250$

Let the Sales Price of the article $= Rs. x$

Sales Tax  on the article $=10\% \ of \ Rs. \ x = 0.1 x$

Therefore $x+0.1x = 8250 \Rightarrow x = Rs. \ 7500$

Sales Tax paid  on the article $=10\% \ of \ Rs. \ 7500 = Rs. \ 750$

If you Increases Sales Tax to 15%, then the person will have to pay $\frac{(15-10)}{100} \times 7500 = Rs. \ 375$ more.

If you Decrease Sales Tax to 6%, then the person will have to pay $\frac{4}{100} \times 7500 = Rs. \ 300$ less.

If Sales Tax increases by 2%, then the person will have to pay $\frac{2}{100} \times 7500 = Rs. \ 150$ more.

If Sales Tax decreases by 3%, then the person will have to pay $\frac{3}{100} \times 7500 = Rs. \ 225$ less.

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Question 10: A bicycle is available for Rs.1664 including Sales Tax. If the list price of the bicycle is Rs.1600, Find:

1. The rate of Sales Tax.
2. The price, a customer will pay for the bicycle if the Sales Tax is increased by 6%.

Answer:

Total price of the bicycle $= Rs. 1664$

Sales Price of the bicycle $= Rs. 1600$

Sales tax paid $= (1664-1600) = Rs. \ 64$

Rate of Sales Tax \$latex = \frac{64}{1600} \times 100 = 4\% &s=1%

Sales Tax  on the bicycle $=10\% \ of \ Rs. \ 1600 = 1600$

Total amount paid $1600+160 = Rs. \ 1760$

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Question 11: When the rate of Sale-tax is decreased from 9% to 6% for a colored T.V. a person will save Rs.780 in buying this T.V. Find the list price of the T.V.

Answer:

Let Sales Price of the T.V. $= Rs. x$

Saving on Sales Tax $= \frac{3}{100} \times x = 780 \Rightarrow x = Rs. \ 26000$

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Question 12: A shopkeeper sells an article for Rs.21384 including 10% sales-Tax. However, the actual rate of sales-tax is 8%. Find the extra profit made by the dealer.

Answer:

Total amount of the article $= Rs. 21384$

Let Sales Price of the artice $= Rs. x$

Sales Tax  on the article $=10\% \ of \ Rs. \ x = 0.1 x$

Therefore $x+0.1x = 21384 \Rightarrow x = Rs. \ 19400$

But the acutal rate of sales tax should be 8%. That means the dealer is making 2% extra gain $= \frac{2}{100} \times 19400 = Rs. \ 388.80$

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## ICSE Board Problems Solved (Class 10): Compound Interest

Question 1: A person invests Rs. 10000 for 3 years at a certain rate of interest compounded annually. At the end of one year this sum amounts to Rs. 11200. Calculate:

1. the rate of interest
2. the amount at the end of second year
3. the amount at the end of third year [2006]

Answer:

1. The rate of interest:

$P=10000 \ Rs.; A= 11200 \ Rs. ; r=x\%$

$A=P(1+\frac{r}{100})^n \Rightarrow 11200=10000(1+\frac{x}{100})^1 \Rightarrow x=12\%$

2. Amount at the end of the second year

$A=P(1+\frac{r}{100})^n \Rightarrow A = 10000(1+\frac{12}{100})^2 = 12544 \ Rs.$

3. Amount at the end of third year

$A=P(1+\frac{r}{100})^n \Rightarrow A = 10000(1+\frac{12}{100})^3 = 14049.28 \ Rs.$

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Question 2: A sum of Rs. 9600 is invested for 3 years at 10% per annum at compound interest.

1. What is the sum due at the end of the first year?
2. What is the sum due at the end of the second year?
3. Find the difference between the answers of the 2) and 1) and find the interest on this sum (difference) for one year
4. Hence, write down the compound interest for the third year [1996]

Answer:

1. sum due at the end of the first year?

$P=9600 \ Rs.; A= ? \ Rs. ; r=10\%; n=1$

$A=P(1+\frac{r}{100})^n \Rightarrow A = 9600(1+\frac{10}{100})^1 = 10560 \ Rs.$

2. sum due at the end of the second year?

$P=9600 \ Rs.; A= ? \ Rs. ; r=10\%; n=2$

$A=P(1+\frac{r}{100})^n \Rightarrow A = 9600(1+\frac{10}{100})^2 = 11616 \ Rs.$

3. difference between the answers of the 2) and 1)

$11616-10560=1056 \ Rs.$

$Interest = 1056 \times \frac{10}{100} \times 1 = 105.6 \ Rs.$

4. Compound Interest for third year

$=1056+105.6 =1161.60 \ Rs.$

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Question 3: What sum of money will amount to Rs. 9261 in 3 years at 5% per annum compound interest? [2009]

Answer:

Given

$P=x \ Rs.; A= 9261 \ Rs. ; r=5\%; n=3$

$A=P(1+\frac{r}{100})^n \Rightarrow 9261 = x(1+\frac{5}{100})^3 \Rightarrow x = 8000 \ Rs.$

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Question 4: In what period of time will Rs. 12000 yield Rs. 3972 as compound interest at 10%, if compounded on a yearly basis. [2011]

Answer:

Given $P=12000 \ Rs.; A= (12000+3972)= 15972 \ Rs. ; r=10\%; n=n$

$A=P(1+\frac{r}{100})^n \Rightarrow 15972 = 12000(1+\frac{10}{100})^n \Rightarrow n = 3 \ years$

$\\$

Question 5: On what sum of money will the difference between the compound interest and the simple interest for two years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a.? [2012]

Answer

Let the sum be $x \ Rs.$

Simple Interest for 2 years $= x \times \frac{5}{100} \times 2 = 0.1x$

Amount Compound Interest

$A=P(1+\frac{r}{100})^n \Rightarrow A = x(1+\frac{5}{100})^2 \Rightarrow A = 1.1025x \ Rs.$

Given difference = 25 Rs.

Therefore $C.I. - S.I. = 25 \Rightarrow (1.1025x-x)-(0.1x) = 25 \Rightarrow x = 10000 \ Rs.$

$\\$

Question 6: The simple interest on a sum of money for 2 years at 4% p.a. is Rs. 340. Find

1. the sum of the money
2. the compound interest on this sum for one year payable half-yearly at the same rate [2008]

Answer:

Simple Interest

1. Given: $I = 340 \ Rs.; n=2 years; r=4\%; P=x \ Rs.;$

$340 = x \times \frac{4}{100} \times 2 \Rightarrow x = 4250 \ Rs.$

2. Compound Interest

$P= 4250 \ Rs.; n=1 years; r=4\% compounded half \ yearly;$

$A=P(1+\frac{r}{100})^n \Rightarrow A = 4250(1+\frac{4}{2 \times 100})^2 \Rightarrow A = 4421.7 \ Rs.$

$C.I. = 4421.7-4250 = 171.70 \ Rs.$

$\\$

Question 7: Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest on the same sum and at the same rate for  1 year, if the interest on the same sum and at the same rate for  year, if the interest  is reckoned half-yearly. [1997]

Answer:

Simple Interest

$P=x \ Rs.; T=2 \ Years; r=4\%$

$S.I=x \times \frac{4}{100} \times 2 = \frac{2}{25}x$

$450 = \frac{2}{25}x \Rightarrow x= 5625 \ Rs.$

Compounded Half Yearly

$P=5625\ Rs.; \ r=4\%; Compounded \ half \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{2 \times 100})^{1 \times 2} = 5625(1+\frac{4}{2 \times 100})^{1 \times 2} = 5852.25 \ Rs.$

$Compound \ Interest =5852.25-5625= 227.25 \ Rs.$

$\\$

Question 8: Rohit borrows Rs. 86000 from Arun for 2 years at 5% per annum simple interest. He immediately lends his money to Akshay at 5% compounded interest annually for the same period. Calculate Rohits profit at the end of two years. [2010]

Answer:

Simple Interest for 2 years

$S.I. = P \times \frac{r}{100} \times 2 \ Rs.$

$S.I. = 86000 \times \frac{5}{100} \times 2 = 8600 \ Rs.$

Compound Interest for 2 years

$P=8600\ Rs.; \ r=5\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} \Rightarrow A= 86000(1+\frac{5}{100})^{2} \Rightarrow A = 9481.50 \ Rs.$

Gain $=(94815-86000)-8600 = 215 \ Rs.$

$\\$

Question 9: A person invests Rs.6000 for two years at a certain rate of interest compounded annually, At the end of first year it amounts to Rs.6720. Calculate;

1. The rate of interest;
2. The amount at the end of the second year. [2010]

Answer:

Compound Interest for 1 year

$P=6000\ Rs.; \ r=x\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{x}{100})^{1}$

Given $6000(1+\frac{x}{100})^{1}=6720 \Rightarrow x= 12\%$

Amount at the end of second year

$A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{12}{100})^{2} = 7526.40 \ Rs.$

$\\$

Question 10: If the interest is compounded half-yearly, calculate the amount when principal is Rs.7400; the rate of interest is 5% per annum and the duration is one year. [2005]

Answer:

$P=7400\ Rs.; \ r=5\%; Compounded \ half \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{2 \times 100})^{n \times 2} = 12000(1+\frac{5}{2 \times 100})^{1 \times 2} = 7774.63 \ Rs.$

$\\$

Question 11: At what rate per cent will a sum of Rs.4000 yield Rs.1324 as compound interest in 3 years? [2013]

Answer:

$P=4000\ Rs.; \ r=x\%; \ n=3 \ year; \ Interest=1324 \ Rs.$

$A=P(1+\frac{r}{100})^n = 4000(1+\frac{x}{100})^3.$

Given $Interest = 1324 \ Rs.$

$\Rightarrow 4000(1+\frac{x}{100})^3 - 4000 = 1324 \Rightarrow x= 10\%$

$\\$

Question 12: The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at $10\%$ for a year is $Rs.15$ . Find the sum of money lent out. [1998]

Answer:

Simple Interest

$P=x \ Rs.; T=1 \ Years; r=10\%$

$S.I=x \times \frac{10}{100} \times 1 = 0.1x$

Compound Interest

$P=x; A= A; \ r=10\%; \ n=2 \ half \ years$

$A= x \times (1+\frac{10}{200})^2 = x \times (\frac{21}{20})^2$

$C.I. = x \times (\frac{21}{20})^2 - x$

Given $C.I. - S.I. = x \times (\frac{21}{20})^2 - x - 0.1x = 15$

$\Rightarrow x= 6000 \ Rs.$

$\\$

Question 13: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 1320$  and for the third year is $Rs. \ 1452$ . Calculate the rate of interest and the original sum of money. [2014]

Answer:

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 1452-1320 = Rs. \ 132$

$\Rightarrow Rs. \ 132$ is the interest on $Rs. \ 1320$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 132}{1320 \times 1} \% = 10\%$

Let the sum of money $= Rs. \ 100$

Therefore Interest on it for 1st Year $= 10\% \ of \ Rs. \ 100 = Rs. \ 10$

$\Rightarrow Amount \ in \ one \ year = 100 + 10 = Rs. \ 110$

$\therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 10\% \ of \ 110 = Rs. \ 11$

$\Rightarrow Amount \ in \ 2nd \ year = 110 + 11 = Rs. \ 121$

When interest of 2nd year $= Rs. \ 11, sum \ is \ Rs. \ 100$

$\Rightarrow \ When \ interest \ of \ 2nd \ year= Rs. \ 1320$, then

$sum= \frac{100}{11} \times 1320 = Rs. \ 12000$

$\\$

Question 14: Ramesh invests $Rs. \ 12800$ for three years at the rate of $10\%$ per annum compound interest. Find;

1. The sum due to that person at the end of the first year.
2. The interest he earns for the second year.
3. The total amount due to him at the end of the third year. [2007]

Answer:

For 1st year: $P = Rs. \ 12800; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{12800 \times 10 \times 1}{100} = Rs. \ 1280$

and, $Amount = 12800+1280 = Rs. \ 14080$

For 2nd year: $P = Rs. \ 14080; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{14080 \times 10 \times 1}{100} = Rs. \ 1408$

and, $Amount = 14080+1408 = Rs. \ 15488$

For 3rd year: $P = Rs. \ 15488; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{15488 \times 10 \times 1}{100} = Rs. \ 1548.80$

and, $Amount = 15488+1548.80 = Rs. \ 17036.80$

$\\$

Question 15: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 880$ and for the third year is $Rs. \ 968$. Calculate the rate of interest and the sum of money. [1995]

Answer:

Difference between the Compound interest of two successive years

$= 968-880 = Rs. \ 88$

$\Rightarrow Rs. \ 88$ is the interest on $Rs. \ 880$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 88}{880 \times 1} \% = 10\%$

For 1st year: $P = Rs. \ x; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{x \times 10 \times 1}{100} = Rs. \ 0.1x$

and, $Amount = x+ 0.1x = Rs. \ 1.1x$

For 2nd year: $P = Rs. \ 1.1x; \ R=10 \% \ and \ T=1 \ year$

Therefore $Interest = \frac{1.1x \times 10 \times 1}{100} = Rs. \ 0.11x$

Given $0.11x = 880 \Rightarrow x=Rs. \ 8000$

$\\$

## Sample Questions: Compound Interest Exercise 2(d)

Question 1: Find the effective rate per cent per annum equivalent to a nominal rate of 10% per annum interest payable half-yearly?

Answer:

Compounded Half Yearly

$P=100\ Rs.; \ r=10\%; Compounded \ half \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{2 \times 100})^{1 \times 2} = 100(1+\frac{10}{2 \times 100})^{1 \times 2} = 110.25 \ Rs.$

$\therefore Equivalent \ Nominal \ rate = 10.25\%$

$\\$

Question 2: A property decreases, in value, every year at the rate of 6% at the beginning of that year. If its value at the end of 2 years is Rs. 2,25,000, what was its worth at the beginning of these 2 years?

Answer:

At the start of $1^{st} \ year$

$P=x\ Rs.; \ r=6\%; \ n=1 \ Year$

At the end of $1^{st} \ year$

$A=P(1-\frac{r}{100})^n = x(1-\frac{6}{100})^1 = 0.94x \ Rs.$

$P=0.94x\ Rs.$

At the end of $2^{nd} year$

$A=P(1+\frac{r}{100})^n = 0.94x(1-\frac{6}{100})^1 = 0.8836x \ Rs.$

Given $0.8836x=225000 \Rightarrow x = 254640.10 \ Rs.$

$\\$

Question 3: A man wishes to accumulate capital worth Rs. 50,440 at the begining of 3 years from now. If he can invest his savings at 5% per annum compound interest, what equal sum must be put aside each year, beginning at the end of the first, to obtain the required amount.

Answer:

Start of 1st Year. Let us say that x Rs. are set aside each year.

$P=x\ Rs.; \ r=5\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = x(1+\frac{5}{100})^1 = 1.05x \ Rs.$

Start of 2nd Year

$P=(1.05x+x)=2.05x\ Rs.; \ r=5\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 2.05x(1+\frac{5}{100})^1 = 2.1525x \ Rs.$

Given $2.125x+x=50440 \Rightarrow = x= 16000 \ Rs.$

$\\$

Question 4: Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest on the same sum and at the same rate for  1 year, if the interest on the same sum and at the same rate for  year, if the interest  is reckoned half-yearly. [1997]

Answer:

Simple Interest

$P=x \ Rs.; T=2 \ Years; r=4\%$

$S.I=x \times \frac{4}{100} \times 2 = \frac{2}{25}x$

$450 = \frac{2}{25}x \Rightarrow x= 5625 \ Rs.$

Compounded Half Yearly

$P=5625\ Rs.; \ r=4\%; Compounded \ half \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{2 \times 100})^{1 \times 2} = 5625(1+\frac{4}{2 \times 100})^{1 \times 2} = 5852.25 \ Rs.$

$Compound \ Interest =5852.25-5625= 227.25 \ Rs.$

$\\$

Question 5: Find the compound interest to the nearest rupee on Rs.10,800 for 2 ½ years at 10% per annum.

Answer:

Compounded Yearly

$P=10800\ Rs.; \ r=10\%; Compounded \ yearly \ n=\frac{5}{2} \ year$

$A=P(1+\frac{r}{2 \times 100})^{\frac{5}{2} \times 2}$

$A=10800(1+\frac{10}{2 \times 100})^{\frac{5}{2} \times 2} = 13783.84 \ Rs.$

$C.I. = 13783.84-10800 = 2983.85 \ Rs.$

$\\$

Question 6: A bought a plot of land for Rs.70,000 and a car for Rs.32,000 on the same day. The value of the plot appreciates uniformly at the rate of 10% every year while the value of the car depreciates by 20% for the first year and by 10% for the second year. If A sells the plot of land as well as the car after 2 years, what will be the profit or loss on the whole?

Answer:

Value of Plot of Land at the end of 2 years

$P=70000\ Rs.; \ r=10\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} = 7000(1+\frac{10}{100})^{2} = 84700 \ Rs.$

Value of Car at the end of 2 years

$P=32000\ Rs.; \ r=20\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^{1} = 32000(1-\frac{20}{100})^{1} = 25600 \ Rs.$

After 2 years

$A=P(1+\frac{r}{100})^{1}=25600(1-\frac{10}{100})^{1}=23040 \ Rs.$

Total investment $= 70000+32000 = 102000 \ Rs.$

Total Return $= 84700+23040 = 107740 \ Rs.$

$\Rightarrow gain = 107740-102000 = 5740 \ Rs.$

$\\$

Question 7: The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is Rs. 97200, find;

1. Its value after 2 years.
2. Its value when it was purchased.

Answer:

Value after 2 years

$P=97200\ Rs.; \ r=10\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{1} = 97200(1-\frac{10}{100})^{2} = 78732 \ Rs.$

Value of Machine at the end of 1st year

$P=x\ Rs.; \ r=10\%; Compounded \ yearly \ n=1 \ year$

$A=x(1+\frac{r}{100})^{1} = x(1-\frac{10}{100})^{1} = 0.9x \ Rs.$

After 2 years

$A=P(1+\frac{r}{100})^{1} = 0.9x(1-\frac{10}{100})^{1} = 0.81x \ Rs.$

Given $0.81x = 97200 \Rightarrow x = 120000 \ Rs.$

$\\$

Question 8: A man borrowed a sum of money and agrees to pay it off by paying Rs.9450 at the end of the first year and Rs. 13,230 at the end of the second year. If the rate of compound interest is 5% per annum, find the sum borrowed.

Answer:

1st Year. Let $x$ be the sum borrowed.

$P=x\ Rs.; \ r=5\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^{1} = x(1+\frac{5}{100})^{1} = 1.05x \ Rs.$

2nd Year

$P=(1.05x-9450)\ Rs.; \ r=5\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^{1} = (1.05x-9450)(1+\frac{5}{100})^{1} = 1.05(1.05x-9450) \ Rs.$

Given $1.05(1.05x-9450) = 13230 \Rightarrow x = 21000 \ Rs.$

$\\$

Question 9: A and B each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. B received Rs. 64 more than A. Find the money lent by each and interest received.

Answer:

Let A’s S.I. investment be $x$

$S.I. = x \times \frac{8}{100} \times 2 = 0.16x$

Let B’s C.I. investment be $x$

$P=x\ Rs.; \ r=8\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} = x(1+\frac{8}{100})^{2} = 1.1664x \ Rs.$

$C.I. = 1.1664x-x=0.1664x$

Given $0.1664x-0.16x = 64 \Rightarrow x = 10000 \ Rs.$

$\\$

Question 10: Calculate the sum of money on which the compound interest (Payable annually) for 2 years be four times the simple interest on Rs. 4715 for 5 years, both at the rate of 5% per annum.

Answer:

Simple Interest

$S.I. = 4715 \times \frac{5}{100} \times 5 = 1178.75 \ Rs.$

$Compound \ Interest = 4 \ times 1178.75 = 4715 \ Rs.$

$P=x\ Rs.; \ r=5\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} \Rightarrow 4715+x= x(1+\frac{5}{100})^{2} \Rightarrow x = 46000 \ Rs.$

$\\$

Question 11: A sum of money was invested for 3 years. Interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to Rs. 4950, find the sum invested.

Answer:

$P=x \ Rs. ; r_1=10\%; r_2=15\%; r_3=18\%; n=3 \ years$

$A = P(1+\frac{r_1}{100})^1 \times (1+\frac{r_2}{100})^1$

$A = x(1+\frac{10}{100})^1 \times (1+\frac{15}{100})^1 = 1.265x$

Given $1.265x-1.1x=4950 \Rightarrow x= 30000 \ Rs.$

$\\$

Question 12: A sum of money is invested at 10% per annum compounded half-yearly. If the difference of amounts at the end of 6 months and 12 months is Rs.189, find the sum of money invested.

Answer:

Compounded Half Yearly

$P=x\ Rs.; \ r=10\%; Compounded \ half \ yearly \ n=1 \ year$

After first period

$A=P(1+\frac{r}{2 \times 100})^{1 \times 1} = x(1+\frac{10}{2 \times 100})^{1 \times 1} = 1.05x \ Rs.$

After second period

$A=P(1+\frac{r}{2 \times 100})^{1 \times 1} = 1.05x(1+\frac{10}{2 \times 100})^{1 \times 1} = 1.1025x \ Rs.$

Given $1.1025x-1.05x=189 \Rightarrow x=3600 \ Rs.$

$\\$

Question 13: Rohit borrows Rs. 86000 from Arun for 2 years at 5% per annum simple interest. He immediately lends his money to Akshay at 5% compounded interest annually for the same period. Calculate Rohits profit at the end of two years. [2010]

Answer:

Simple Interest for 2 years

$S.I. = P \times \frac{r}{100} \times 2 \ Rs.$

$S.I. = 86000 \times \frac{5}{100} \times 2 = 8600 \ Rs.$

Compound Interest for 2 years

$P=8600\ Rs.; \ r=5\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} \Rightarrow A= 86000(1+\frac{5}{100})^{2} \Rightarrow A = 9481.50 \ Rs.$

Gain $=(94815-86000)-8600 = 215 \ Rs.$

$\\$

Question 14: A person borrowed Rs.60,000 from a money lender at 5% simple interest for two year. The money lender deducts the interest that would be due at the end of the period and handed over the balance to the person. This person then  deposits this amount in a bank which pays 5% compound interest per annum. How much will the person will have to add to pay the money back to the money lender.

Answer:

Simple Interest for 2 years

$S.I. = P \times \frac{r}{100} \times 2 \ Rs.$

$S.I. = 60000 \times \frac{5}{100} \times 2 = 6000 \ Rs.$

Money given $= 60000-6000 = 54000 \ Rs.$

Compound Interest for 2 years

$P=54000\ Rs.; \ r=5\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} \Rightarrow A= 54000(1+\frac{5}{100})^{2} \Rightarrow A = 59535 \ Rs.$

Therefore money to be paid to square off the load $= 60000-59535= 465 \ Rs.$

$\\$

Question 15: The simple interest on a certain sum of money for 3 years at 5% per annum is Rs.1200. Find the amount due and the compound interest on this sum of money at the same rate and after 2 years, interest is reckoned annually.

Answer:

Simple Interest for 2 years. Let P $= x \ Rs.$

$S.I. = P \times \frac{r}{100} \times 3 \ Rs.$

$S.I. = x \times \frac{5}{100} \times 3 = \frac{3}{20}x \ Rs.$

Given $1200 = \frac{3}{20}x \Rightarrow x = 8000 \ Rs.$

Compound Interest for 2 years

$P=8000\ Rs.; \ r=5\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} \Rightarrow A= 8000(1+\frac{5}{100})^{2} \Rightarrow A = 8820 \ Rs.$

Compound Interest $=(8820)-8000 = 820 \ Rs.$

$\\$

Question 16: A person invests Rs.6000 for two years at a certain rate of interest compounded annually, At the end of first year it amounts to Rs.6720. Calculate;

1. The rate of interest;
2. The amount at the end of the second year. [2010]

Answer:

Compound Interest for 1 year

$P=6000\ Rs.; \ r=x\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{x}{100})^{1}$

Given $6000(1+\frac{x}{100})^{1}=6720 \Rightarrow x= 12\%$

Amount at the end of second year

$A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{12}{100})^{2} = 7526.40 \ Rs.$

## Sample Questions: Compound Interest Exercise 2(b)

Question 1: If the interest is compounded half-yearly, calculate the amount when principal is Rs.7400; the rate of interest is 5% per annum and the duration is one year. [2005]

Answer:

$P=7400\ Rs.; \ r=5\%; Compounded \ half \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{2 \times 100})^{n \times 2} = 12000(1+\frac{5}{2 \times 100})^{1 \times 2} = 7774.63 \ Rs.$

$\\$

Question 2: Find the difference between the compound interest compounded yearly and half-yearly on Rs.10000 for 18 months at 10% per annum.

Answer:

Compounded Yearly

$P=10000\ Rs.; \ r=10\%; Compounded \ yearly \ n=\frac{3}{2} \ year$

$A=P(1+\frac{r}{1 \times 100})^{1}.(1+\frac{r}{2 \times 100})^{\frac{1}{2} \times 2}$

$A=10000(1+\frac{10}{1 \times 100})^{1}.(1+\frac{10}{2 \times 100})^{\frac{1}{2} \times 2} = 11550 \ Rs.$

Compounded Half Yearly

$P=10000\ Rs.; \ r=10\%; Compounded \ half \ yearly \ n=\frac{3}{2} \ year$

$A=P(1+\frac{r}{2 \times 100})^{\frac{3}{2} \times 2} = 10000(1+\frac{10}{2 \times 100})^{\frac{3}{2} \times 2} = 11576.25 \ Rs.$

Difference $11576.25-11550 = 26.50 \ Rs.$

$\\$

Question 3: A man borrowed Rs.16000 for 3 years under the following terms:

1. 20% simple interest for the first 2 years;
2. 20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded semi-annually. Find the total amount to be paid at the end of the three years.

Answer:

Simple interest for the first two years

$S.I. = 16000 \times \frac{20}{100} \times 2 = 6400 \ Rs.$

$Amount = 16000+6400 = 22400 \ Rs.$

Compound interest for the remainder of the term

$P=10000\ Rs.; \ r=20\%; Compounded \ half \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{2 \times 100})^{1 \times 2} = 22400(1+\frac{20}{2 \times 100})^{1 \times 2} = 27104 \ Rs.$

$\\$

Question 4: What sum of money will amount to Rs.27783 in one and half years at 10% per annum compounded half yearly?

Answer:

$P=x\ Rs.; \ r=10\%; Compounded \ half \ yearly \ n=\frac{3}{2} \ year; A=27783 \ Rs.$

$A=P(1+\frac{r}{2 \times 100})^{n \times 2}$

$27783=x(1+\frac{10}{2 \times 100})^{\frac{3}{2} \times 2} \Rightarrow 27783 = 1.157625x \Rightarrow x= 2400 \ Rs.$

$\\$

Question 5: A  invests a certain sum of money at 20% per annum, interest compounded yearly. B invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If B gets Rs.33 more than A in 18 months, calculate the money invested by each.

Answer:

A’s investment: Compounded Yearly

$P=x\ Rs.; \ r=20\%; Compounded \ yearly \ n=\frac{3}{2} \ year$

$A=P(1+\frac{r}{1 \times 100})^{1}.(1+\frac{r}{2 \times 100})^{\frac{1}{2} \times 2}$

$A=x(1+\frac{20}{1 \times 100})^{1}.(1+\frac{20}{2 \times 100})^{\frac{1}{2} \times 2} = 1.32x \ Rs.$

Compounded Half Yearly

$P=x\ Rs.; \ r=20\%; Compounded \ half \ yearly \ n=\frac{3}{2} \ year$

$A=P(1+\frac{r}{2 \times 100})^{\frac{3}{2} \times 2} = x(1+\frac{20}{2 \times 100})^{\frac{3}{2} \times 2} = 1.331x \ Rs.$

Difference $1.331x-1.32x=33 \Rightarrow x= 3000 \ Rs.$

$\\$

Question 6: At what rate of interest per annum will a sum of Rs.62500 earn a compound interest of Rs.5100 in one year? The interest is to be compounded half-yearly.

Answer:

Compounded Half Yearly

$P=62500\ Rs.; A=(62500 + 5100) = 67600 \ Rs.; \ r=x\%; Compounded \ half \ yearly \ n=\frac{2}{2} \ year$

$67600=62500(1+\frac{x}{2 \times 100})^{\frac{2}{2} \times 2} \Rightarrow 1.0816 = (1+\frac{x}{200})^2 \Rightarrow x=8\%$

$\\$

Question 7: In what time will rs.1500 yield Rs.496.50 as compound interest at 20% per year compounded semi-annually?

Answer:

Compounded Half Yearly

$P=1500\ Rs.; A=(1500 + 496.50) = 1996.50 \ Rs.; \ r=20\%; Compounded \ half \ yearly \ n=n \ year$

$1996.50=1500(1+\frac{20}{2 \times 100})^{n \times 2} \Rightarrow 1.331 = (1+\frac{20}{200})^{2n} \Rightarrow n=\frac{3}{2} years$

$\\$

Question 8: Calculate the C.I. on Rs.3500 at 6% per annum for 3 years, the interest being compounded half-yearly.

Answer:

Compounded Half Yearly

$P=3500\ Rs.; \ r=6\%; Compounded \ half \ yearly \ n=3 \ year$

$A=3500(1+\frac{6}{2 \times 100})^{3 \times 2} \Rightarrow A= 4179.18 \ Rs.$

$C.I. = 4179.18-3500 = 679.18 \ Rs.$

$\\$

Question 9: Find the difference between compound interest and simple interest on Rs.12,000 and in $1 \frac{1}{2}$ at 10% compounded yearly.

Answer:

Compounded Yearly

$P=12000\ Rs.; \ r=10\%; Compounded \ yearly \ n=\frac{3}{2} \ year$

$A=P(1+\frac{r}{1 \times 100})^{1}.(1+\frac{r}{2 \times 100})^{\frac{1}{2} \times 2}$

$A=12000(1+\frac{10}{1 \times 100})^{1}.(1+\frac{10}{2 \times 100})^{\frac{1}{2} \times 2} = 13860 \ Rs.$

Simple interest for 1.5 years

$S.I. = 12000 \times \frac{10}{100} \times \frac{3}{2} = 1800 \ Rs.$

$Amount = 16000+6400 = 22400 \ Rs.$

$Difference = (13860-13000)-1800 = 60 \ Rs.$

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Question 10: The simple interest on a sum of money for 3 years at 5% per annum is Rs.900. Find:

1. The sum of money and
2. The compound interest on this sum for 1.5 years payable half-yearly at double the rate per annum.

Answer:

Simple interest for 3 years

$900 = x \times \frac{5}{100} \times 3 \Rightarrow x= 6000 \ Rs.$

$Amount = 16000+6400 = 22400 \ Rs.$

Compounded Half Yearly

$P=6000\ Rs.; \ r=10\%; Compounded \ half \ yearly \ n=\frac{3}{2} \ year$

$A=6000(1+\frac{10}{2 \times 100})^{\frac{3}{2} \times 2} \Rightarrow A= 6945.75 \ Rs.$

$Compound \ interest = 6945.75-6000 = 945.75 \ Rs.$

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Question 11: The compound interest in one year on a certain sum of money at 10% per annum compounded half-yearly exceeds the simple interest on the same sum at the same rate and for the same period by Rs.30. Calculate the sum.

Answer:

$\\$

Simple interest for 1 years

$S.I. = x \times \frac{10}{100} \times 1 = 0.1x \ Rs.$

$Amount = 16000+6400 = 22400 \ Rs.$

$Difference = (13860-13000)-1800 = 60 \ Rs.$

Compounded Half Yearly

$P=x\ Rs.; \ r=10\%; Compounded \ half \ yearly \ n=1 \ year$

$A=x(1+\frac{10}{2 \times 100})^{1 \times 2} \Rightarrow A= 1.1025x \ Rs.$

$Compound \ interest = 1.1025x-x = 0.1025x \ Rs.$

Difference $0.1025x-0.1x=30 \Rightarrow x= 12000 \ Rs.$

## Sample Questions: Compound Interest Exercise 2(a)

Question 1: Find the amount and the compound interest on Rs.12000 in 3 years at 5%; interest being compounded annually.

Answer:

$P=12000\ Rs.; \ r=5\%; \ n=3 \ years$

$A=P(1+\frac{r}{100})^n = 12000(1+\frac{5}{100})^3 = 13891.50 \ Rs.$

$Compound \ Interest = 13891.50-12000 = 1891.50 \ Rs.$

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Question 2: Calculate the amount, if Rs.15000 is lent at compound interest for 2 years and the rates for the successive years are 8% p.a. and 10% p.a. respectively.

Answer:

$P=15000\ Rs.; \ r=8\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 15000(1+\frac{8}{100})^1 = 16200 \ Rs.$

$P=15000\ Rs.; \ r=10\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 15000(1+\frac{10}{100})^1 = 17820\ Rs.$

$\\$

Question 3: Calculate the compound interest accrued on Rs.6000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.

Answer:

$P=6000\ Rs.; \ r=5\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 6000(1+\frac{5}{100})^1 = 6300 \ Rs.$

$P=6300\ Rs.; \ r=8\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 6300(1+\frac{8}{100})^1 = 6804 \ Rs.$

$P=6804\ Rs.; \ r=10\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 6804(1+\frac{10}{100})^1 = 7484.40 \ Rs.$

$Compound \ Interest = 7484.40-6000 = 1484.40 \ Rs.$

$\\$

Question 4:
What sum of money will amount to Rs.5445 in 2 years at 10% per annum compound interest?

Answer:

$A= 5445 \ Rs. ;P=x\ Rs.; \ r=10\%; \ n=2 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow 5445= x(1+\frac{10}{100})^2 Rs.$

$\Rightarrow x= 4500 \ Rs.$

$\\$

Question 5: On what sum of money will be compound interest for 2 years at 5 per cent per annum amount to Rs,768.75?

Answer:

$A= A \ Rs. ;P=x\ Rs.; \ r=5\%; \ n=2 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow A= x(1+\frac{5}{100})^2 Rs.$

$\Rightarrow C.I.=A-P=768.75$

$\Rightarrow C.I.=x(1+\frac{5}{100})^2-x=768.75 \Rightarrow x=7500 \ Rs.$

$\\$

Question 6: Find the sum on which the compound interest for 3 years at 10% per annum amounts to Rs.1655.

Answer:

$A= A \ Rs. ;P=x\ Rs.; \ r=10\%; \ n=3 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow A= x(1+\frac{10}{100})^2 Rs.$

$\Rightarrow C.I.=A-P=1655$

$\Rightarrow C.I.=x(1+\frac{10}{100})^2-x=1655 \Rightarrow x=5000 \ Rs.$

$\\$

Question 7: What principal will amount to Rs.9856 in two years, if the rates of interest for Question successive years are 10% and 12% respectively?

Answer:

$P=x\ Rs.; \ r=10\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = x(1+\frac{10}{100})^1 = 1.1x \ Rs.$

$P=1.1x\ Rs.; \ r=12\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 1.1x(1+\frac{12}{100})^1 = 1.232x \ Rs.$

$\Rightarrow 1.232x=9856 \Rightarrow x=8000 \ Rs.$

$\\$

Question 8: On a certain sum, the compound interest in 2 years amounts to Rs.4240. If the rates of interest for successive year are 10% and 15% respectively, find the sum.

Answer:

$P=x\ Rs.; \ r=10\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = x(1+\frac{10}{100})^1 = 1.1x \ Rs.$

$P=1.1x\ Rs.; \ r=15\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n = 1.1x(1+\frac{15}{100})^1 = 1.265x \ Rs.$

$\Rightarrow 1.232x - x=4240 \Rightarrow x=16000 \ Rs.$

$\\$

Question 9: At what rate per cent per annum will Rs.6000 amount to Rs.6615 in 2 years when interest is compounded annually?

Answer:

$P=6000\ Rs.; A=6615\ Rs.; \ r=x\%; \ n=2 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow 6615= 6000(1+\frac{x}{100})^2 \Rightarrow x= 5\%.$

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Question 10: At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2years?

Answer:

$P=x\ Rs.; A=1.44x\ Rs.; \ r=r\%; \ n=2 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow 1.44x= x(1+\frac{r}{100})^2 \Rightarrow r= 20\%.$

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Question 11:
At what rate per cent will a sum of Rs.4000 yield Rs.1324 as compound interest in 3 years? [2013]

Answer:

$P=4000\ Rs.; \ r=x\%; \ n=3 \ year; \ Interest=1324 \ Rs.$

$A=P(1+\frac{r}{100})^n = 4000(1+\frac{x}{100})^3.$

Given $Interest = 1324 \ Rs.$

$\Rightarrow 4000(1+\frac{x}{100})^3 - 4000 = 1324 \Rightarrow x= 10\%$

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Question 12: A person invests Rs.5000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to Rs.6272. Calculate;

1. The rate of interest per annum
2. The amount at the end of the third year.

Answer:

$P=5000\ Rs.; \ r=x\%; \ n=2 \ years; \ A=6272 \ Rs.$

$A=P(1+\frac{r}{100})^n \Rightarrow 6272= 5000(1+\frac{x}{100})^2 \Rightarrow x = 12\%$

At the end of third year

$A=P(1+\frac{r}{100})^n \Rightarrow A= 5000(1+\frac{12}{100})^3 = 7024.64 \ Rs.$

$\\$

Question 13: In how many years will Rs.7000 amount to Rs.9317 at 10% per cent per annum compound interest?

Answer:

$P=7000\ Rs.; \ r=10\%; \ n=n \ years; \ A=9217 \ Rs.$

$A=P(1+\frac{r}{100})^n \Rightarrow 9317= 7000(1+\frac{10}{100})^n \Rightarrow n = 3 \ years$

$\\$

Question 14: Find the time, in years, in which Rs.4000 will produce Rs.630.50 as compound interest at 5% p.a. interest being compounded annually.

Answer:

$P=4000\ Rs.; \ r=5\%; \ n=n \ years; \ A=4630.50 \ Rs.$

$A=P(1+\frac{r}{100})^n \Rightarrow 4630.50= 4000(1+\frac{5}{100})^n \Rightarrow n = 3 \ years$

$\\$

Question 15: Divide Rs.28,730 between A and B so that when their shares are lent out at 10% compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.

Answer:

Let the share of $A = x \ Rs$. Therefore share of $B = (28730-x) \ Rs$.

For A

$P=x\ Rs.; \ r=10\%; \ n=3 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= x(1+\frac{10}{100})^3 \$

For B

$P=(28730-x)\ Rs.; \ r=10\%; \ n=5 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= (28730-x)(1+\frac{10}{100})^5 \$

Given

$x(1+\frac{10}{100})^3 = (28730-x)(1+\frac{10}{100})^5$

$x = (28730-x)(1.1)^2$

$\Rightarrow x= 15730 \ Rs$

Therefore B’s share = $(28730-15730)= 13000 \ Rs.$

$\\$

Question 16:
A sum of Rs.34522 is divided between A and B, 18 years and 21 years old respectively in such a way that if their shares be invested at 5% per annum compound interest, both will receive equal money at the age of 30 years. Find the shares of each out of Rs.34522.

Answer:

Let the share of A = x Rs. Therefore share of B = (34522-x) Rs.

For A

$P=x\ Rs.; \ r=5\%; \ n=12 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= x(1+\frac{5}{100})^12 \$

For B

$P=(34522-x)\ Rs.; \ r=10\%; \ n=9 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= (34522-x)(1+\frac{5}{100})^9 \$

Given

$x(1+\frac{5}{100})^12 = (34522-x)(1+\frac{5}{100})^9$

$x(1.05)^3 = (34522-x)$

$\Rightarrow x= 16000 \ Rs$

Therefore B’s share = $(28730-15730)= 13000 \ Rs.$

$\\$

Question 17: A sum of Rs.44200 is divided between A and B, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10% per annum compound interest, they will receive equal amounts on reaching 16 years of age.

1. What is the share of each out of Rs.44200?
2. What will each receive, when 16 years old?

Answer:

Let the share of A = x Rs. Therefore share of B = (44200-x) Rs.

For A

$P=x\ Rs.; \ r=10\%; \ n=4 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= x(1+\frac{10}{100})^4 \$

For B

$P=(44200-x)\ Rs.; \ r=10\%; \ n=2 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= (44200-x)(1+\frac{10}{100})^2 \$

Given

$x(1+\frac{10}{100})^4 = (44200-x)(1+\frac{10}{100})^2$

$x(1.1)^2 = (44200-x)$

$\Rightarrow x= 20000 \ Rs$

Therefore B’s share = $(44200-20000)= 24200 \ Rs.$

$\\$

Question 18: At the beginning of year 2011, a man had Rs.22000 in his bank account. He saved some money by the end of this year and deposited it in the bank. The bank pays 10% per annum compound interest and at the end of year 2012 he had Rs.39820 in his bank account. Find, what amount of money at the end of year 2011.

Answer:

year 2011

$P=22000\ Rs.; \ r=10\%; \ n=1 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow A= 22000(1+\frac{10}{100})^1 \ = 24200\ Rs.$

Lets us say he saves and deposits $x \ Rs.$ at the end of year 2011.

$P=(24200+x)\ Rs.; A=39820\ Rs.; \ r=10\%; \ n=1 \ years;$

$A=P(1+\frac{r}{100})^n \Rightarrow 39820= (24200+x)(1+\frac{10}{100})^1 \Rightarrow x = 12000 \ Rs.$

$\\$

Question 19: If the amounts of two consecutive years on a sum of money are in the ratio 20:21, find the rate of interest.

Answer:

$P=x\ Rs.; \ r=r\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow A1= x(1+\frac{r}{100})$

$P=x(1+\frac{r}{100})\ Rs.; \ r=r\%; \ n=1 \ year$

$A=P(1+\frac{r}{100})^n \Rightarrow A2= x(1+\frac{r}{100})(1+\frac{r}{100})^1$

Given $A1:A2= 20:21$

$x(1+\frac{r}{100}) : x(1+\frac{r}{100})^2 = 20: 21$

$\frac{1}{x(1+\frac{r}{100})} = \frac{20}{21}$

$\Rightarrow r=5\%$

$\\$

Question 20: On what sum of money will the difference between the compound interest and simple interest for 3 years be equal to Rs.930, if the rate of interest charged for both is 10% p.a.?

Answer:

Let the $P = x \ Rs.$

Simple Interest

$S.I = x \times \frac{10}{100} \times 3 = 0.3x$

Compound Interest

$P=x\ Rs.; \ r=10\%; \ n=3 \ year$

$A=P(1+\frac{r}{100})^n = x(1+\frac{10}{100})^3 = 1.331x \ Rs.$

$C.I. = 1.331x-x = 0.331x$

Given

$0.331x-0.3x= 930 \Rightarrow x = 30000 \ Rs.$

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