If amount on the principal of Rs 6,000

Question: If amount on the principal of Rs 6,000 is written as 6000 [1+5/100]3and compound interest payable half yearly, then rate of interest p.a. is ____ and time in years is ______. Solution: Rate 10% and 1.5 years...

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In the first year on an investment of Rs. 6,00,000

Question: In the first year on an investment of Rs. 6,00,000 the loss is 5% and in the second year the gain is 10%, the net result is 627000. Solution: 627000 Explanation: Investment amount = 600000 Loss in first year = 5%. So, investment in first year = 600000 (5/100) x 600000 = 600000 30000 = 570000 In second year, the gain is 10%. So, net result = 570000 + (10/100) x 570000 = 570000 + 57000 = 627000...

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Find the sum of the GP :

Question: Find the sum of the GP : $\mathbf{x}^{3}+\mathbf{x}^{5}+\mathbf{x}^{7}+\ldots$ To $\mathbf{n}$ terms Solution: Sum of a G.P. series is represented by the formula, $S_{n}=a \frac{r^{n}-1}{r-1}$ when r1. Sn represents the sum of the G.P. series upto nth terms, a represents the first term, r represents the common ratio and n represents the number of terms. Here $a=x^{3}$ $r=($ ratio between the $n$ term and $n-1$ term $) x^{5} \div x^{3}=x^{2}$ n terms $\therefore S_{n}=x^{3} \times \frac...

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The compound interest on Rs 8,000

Question: The compound interest on Rs 8,000 for one year at 16% p.a. compounded half yearly is ______, given that (1.08)2= 1.1664. Solution: Rs. 9331.2 Explanation: Principal = Rs.8000 Time period = 1 year Rate = 16% = 16/100 = 0.16 Amount = P ( 1+r/n)nt n = 2 (compounded half yearly in a year) A = 8000(1+0.16/2)21= 8000 (1+0.08)2= 8000 (1.08)2 A = 8000 1.1664 A = 9331.2...

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The marked price of an article

Question: The marked price of an article when it is sold for Rs. 880 after a discount of 12% is ______ Solution: Rs.1000 Explanation: selling price = Rs.880 Discount percentage = 12% Let x be the marked price. Since, discount is calculated on marked price, thus; x x (12/100) = 880 88x /100 = 880 x = 10 100 = 1000...

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Percentages are equal to fractions

Question: Percentages are equal to fractions with _________ equal to 100. Solution: Denominator...

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When principal P is compounded semi-annually

Question: When principal P is compounded semi-annually at r % per annum for t years, then Amount = ______ Solution: A = P(1+R/200)2t...

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The discount on an item for

Question: The discount on an item for sale is calculated on the _________ Solution: Marked price...

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Find the sum of the GP :

Question: Find the sum of the GP : $1-a+a^{2}-a^{3}+\ldots$ to $n$ terms $(a \neq 1)$ Solution: Sum of a G.P. series is represented by the formula $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$ when r1. Sn represents the sum of the G.P. series upto nth terms, a represents the first term, r represents the common ratio and n represents the number of terms. Here a = 1 $r=$ (ratio between the $n$ term and $n-1$ term) $-a \div 1=-a$ n terms $\therefore \mathrm{S}_...

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________ expenses are the additional expenses

Question: ________ expenses are the additional expenses incurred by a buyer for an item over and above its cost of purchase. Solution: Overhead...

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The time period after which the interest

Question: The time period after which the interest is added each time to form a new principal is called the __________ Solution: Conversion period...

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Sales tax = tax % of _______

Question: Sales tax = tax % of _______ Solution: Bill amount...

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Amount when interest is compounded

Question: Amount when interest is compounded annually is given by the formula_____ Solution: A = P(1+R/100)T[P = Principal, R = Rate, T = time]...

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_______ is charged on the sale of an item

Question: _______ is charged on the sale of an item by the government and is added to the bill amount. Solution: Sales tax...

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Find the sum of the GP :

Question: Find the sum of the GP : $1-\frac{1}{3}+\frac{1}{3^{2}}-\frac{1}{3^{3}}+$ to n terms Solution: Sum of a G.P. series is represented by the formula $S_{n}=a \frac{1-r^{n}}{1-r}$ when |r|1. Sn represents the sum of the G.P. series upto nth terms, a represents the first term, r represents the common ratio and n represents the number of terms. Here a = 1 r = (ratio between the n term and n-1 term) $-\frac{1}{3} \div 1=-\frac{1}{3}$ n terms $\therefore S_{\mathrm{n}}=1 \times \frac{1-\frac{-...

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If the solve the problem

Question: If $y=e^{x} \cos x$, prove that $\frac{d^{2} y}{d x^{2}}=2 e^{x} \cos \left(x+\frac{\pi}{2}\right)$ Solution: Basic idea: Second order derivative is nothing but derivative of derivative i.e. $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$ $\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v$, i.e. $f=v(u(x))$. For the sake of simplicity just assume $t=u(x)$ Then $f=v(t)...

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Discount = Discount % of ________

Question: Discount = Discount % of ________ Solution: Discount = Discount % of Marked Price....

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Find the sum of the GP :

Question: Find the sum of the GP : $\sqrt{7}+\sqrt{21}+3 \sqrt{7}+\ldots$ to $\mathrm{n}$ terms Solution: Sum of a G.P. series is represented by the formula $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$ when r1. Sn represents the sum of the G.P. series upto nth terms, a represents the first term, r represents the common ratio and n represents the number of terms. Here, $a=\sqrt{7}$ $r=($ ratio between the $n$ term and $n-1$ term $) \sqrt{7} \div \sqrt{21}=\s...

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Discount = _______-_______.

Question: Discount = _______-_______. Solution: Discount = Marked Price Selling Price....

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15% increase in price of an article,

Question: 15% increase in price of an article, which is Rs.1,620, is the increase of ____. Solution: Rs.212 Explanation: Let x is the price of the article. Thus, as per given question; 1620 = x + x (15/100) 1620 = 115x/100 115x = 1620 100 x = (1620100)/115 x = 1408 Hence, increase in price = 1620 1408 = 212....

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Increase of a number from 150

Question: Increase of a number from 150 to 162 is equal to increase of ____ per cent. Solution: 8% Explanation: Increase of a number from 150 to 162 = 162-150 = 12 Percentage of increased number = 12/150 100 = 120/15 = 8%...

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________ is a reduction on the marked

Question: ________ is a reduction on the marked price of the article. Solution: Discount...

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Radhika bought a car for Rs 2,50,000.

Question: Radhika bought a car for Rs 2,50,000. Next year its price decreased by 10% and further next year it decreased by 12%. In the two years overall decrease per cent in the price of the car is (a) 3.2% (b) 22% (c) 20.8% (d) 8% Solution: (c) 20.8% Explanation: Radhika bought a car for Rs. 250000. Cost price = Rs.250000 Its price decreased next year for 10%. Thus, new price = 250000 (10/100) 250000 = 250000 25000 = 225000 Again, the price of car decreased by 12% next year. So the price will b...

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Prove the following

Question: 40% of [100 20% of 300] is equal to: (a) 20 (b) 16 (c) 140 (d) 64 Solution: (b) 16 Explanation: 40% of [100 20% of 300] = 40% [100 (20/100300)] = 40% [100 60] = 40/100 40 = 16...

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A TV set was bought for Rs 26,250 including 5% VAT.

Question: A TV set was bought for Rs 26,250 including 5% VAT. The original price of the TV set is (a) Rs 27,562.50 (b) Rs 25,000 (c) Rs 24,937.50 (d) Rs 26,245 Solution: (c) Rs 24,937.50 Explanation: Cost price of TV set = Rs. 26250. VAT including = 5% Original price = Cost price of article including VAT = 26250 (5/100)x 26250 = 26250-1312.5 =24,937.50 Therefore, original price of TV set is = Rs. 24,937.50...

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