Avinash bought an electric iron for Rs 900
Question: Avinash bought an electric iron for Rs 900 and sold it at a gain of 10%. He sold another electric iron at 5% loss which was bought Rs 1200. On the transaction he has a: (a) Profit of Rs 75 (b) Loss of Rs 75 (c) Profit of Rs 30 (d) Loss of Rs 30 Solution: (c) Profit of Rs 30 Explanation: Price of electric iron = Rs. 900 Sold at 10% profit Now, selling price of the electric iron =(10/1000) x 900 + 900 = 90+ 900 = Rs.990 Another electric iron sold at 5% loss. Cost price of another electri...
Read More →Find the sum of the GP :
Question: Find the sum of the GP : $\frac{2}{9}-\frac{1}{3}-\frac{1}{2}-\frac{3}{4}+\ldots .$ To 6 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, $a=\frac{2}{9}$ $r=($ ratio between the $n$ term and $n-1$ term $)-\frac{1}{3} \d...
Read More →The original price of a washing machine
Question: The original price of a washing machine which was bought for Rs. 13500 including of 8% VAT, is (a) Rs. 12420 (b) Rs. 14580 (c) Rs.12500 (d) Rs. 13492 Solution: (a) Rs 12,420 Explanation: The original price of the washing machine = Rs.13500 VAT = 8%. The original price of the washing machine including of 8% VAT = 13500-13500 x8/100 = 13500-135 x 8 = 13500-1080 = Rs.12420...
Read More →If the solve the problem
Question: If $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ prove that $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\cos \theta-\theta \sin \theta), \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\sin \theta+\theta \cos \theta)$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{3} \theta}{\mathrm{a} \theta}$ Solution: Basic idea: Second order derivative is nothing but derivative of derivative i.e. $\frac{\ma...
Read More →A sum is taken for two years at 16% p.a.
Question: A sum is taken for two years at 16% p.a. If interest is compounded after every three months, the number of times for which interest is charged in 2 years is: (a) 8 (b) 4 (c) 6 (d) 9 Solution: (a) 8 Explanation: Rate of interest is compounded after every three months. Thus, the time period for amount in a year will be 4 times. If amount is taken for 2 year, then 42 = 8 times charged in 2 year....
Read More →Find the sum of the GP :
Question: Find the sum of the GP : $\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2 \sqrt{2}}+\ldots \ldots$ to 8 terms Solution: Sum of a G.P. series is represented by the formula $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}$ hen |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, $a=\sqrt{2}$ $r=($ ratio between the $n$ term and $n-1$ term $) \frac{1}{\...
Read More →A jacket was sold for Rs 1,120
Question: A jacket was sold for Rs 1,120 after allowing a discount of 20%. The marked price of the jacket is: (a) Rs 1440 (b) Rs 1400 (c) Rs 960 (d) Rs 866.66 Solution: (b) Rs. 1400 Explanation: Let marked price = x Discount = 20% Selling price = 1120 Hence, 1120 = x x 20/100 1120 = x x/5 1120 = 4x/5 x = (11205)/4 = 1400...
Read More →Find the sum of the GP :
Question: Find the sum of the GP : $1-\frac{1}{2}+\frac{1}{4}=\frac{1}{8}+\ldots$ to 9 terms Solution: Sum of a G.P. series is represented by the formula $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}$ 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 $)-\frac{1}{2} \div 1=-\frac{1}{2}$ n...
Read More →Latika bought a teapot for Rs 120
Question: Latika bought a teapot for Rs 120 and a set of cups for Rs 400. She sold teapot at a profit of 5% and cups at a loss of 5%. The amount received by her is: (a) Rs 494 (b) Rs 546 (c) Rs 506 (d) Rs 534 Solution: (c) Rs 506 Explanation: Price of teapot = Rs. 120 Price of set of cups = Rs. 400 Latika sold teapot at a profit of 5% Selling price of teapot = 5/100 x 120 + 120 =120/20 +120 = 6 + 120 = Rs.126 Also, cups were sold at a loss of 5%. Now, selling price of cups = 400 5/100 x 400 = 40...
Read More →A bought a tape recorder for Rs 8,000
Question: A bought a tape recorder for Rs 8,000 and sold it to B. B in turn sold it to C, each earning a profit of 20%. Which of the following is true: (a) A and B earn the same profit. (b) A earns more profit than B. (c) A earns less profit than B. (d) Cannot be decided. Solution: (c) A earns less profit than B Explanation: Cost price of tape recorder bought by A = Rs.8000 Cost price of tape recorder for B =20% profit on cost price for A =20/100 x 8000 + 8000 =20 x 80 + 8000 =1600 + 8000 =Rs.96...
Read More →Find the sum of the GP :
Question: Find the sum of the GP : $0.15+0.015+0.0015+\ldots .$ To 6 terms Solution: Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}$ 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 = 0.15 r = (ratio between the n term and n-1 term) 0.015 0.15 = 0.1 n = 6 terms $\Rightarrow \mathrm{S}_{\m...
Read More →The marked price of an article
Question: The marked price of an article is Rs 80 and it is sold at Rs 76, then the discount rate is: (a) 5% (b) 95% (c) 10% (d) appx. 11% Solution: (a) 5% Explanation: Marked price = Rs. 80 Sold price = Rs.76 We know that, Selling price = Marked price Discount Discount = Marked price Selling price Discount = Rs.80-Rs.76 = Rs.4 Discount % = 4/80 x 100 = 5%...
Read More →Shyama purchases a scooter costing
Question: Shyama purchases a scooter costing Rs 36,450 and the rate of sales tax is 9%, then the total amount paid by her is: (a) Rs 36,490.50 (b) Rs 39,730.50 (c) Rs 36,454.50 (d) Rs 33,169.50 Solution: (b) Rs 39,730.50 Explanation: Scooter cost Rs.36450 at the rate of sales tax = 9%. Total cost of scooter paid by Shyama = 9% of 36450 + 36450 = (9/100 36450) + 36450 = 3280.5 + 36450 = 39730.5...
Read More →Find the sum of the GP :
Question: Find the sum of the GP : $1+\sqrt{3}+3+3 \sqrt{3}+\ldots . .$ to 10 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 = 1 $r=($ ratio between the $n$ term and $n-1$ term $) \sqrt{3} \div 1=\sqrt{3}=1.732$ n = 10 terms...
Read More →For calculation of interest compounded half yearly,
Question: For calculation of interest compounded half yearly, keeping the principal same, which one of the following is true. (a) Double the given annual rate and half the given number of years. (b) Double the given annual rate as well as the given number of years. (c) Half the given annual rate as well as the given number of years. (d) Half the given annual rate and double the given number of years. Solution: (d) Half the given annual rate and double the given number of years....
Read More →Ashima took a loan of Rs 1,00,000 at 12% p.a.
Question: Ashima took a loan of Rs 1,00,000 at 12% p.a. compounded half-yearly. She paid Rs.1,12,360. If (1.06)2is equal to 1.1236, then the period for which she took the loan is: (a) 2 years (b) 1 year (c) 6 months (d) 1(1/2) years Solution: (b) 1 year Explanation: P = Rs.100000, R = 12% per annum compounded half-yearly. Amount = Rs.112360 Since we know, A = P (1+R/100)T 112360 = 100000(1+12/100)T 112360/100000 = (1+12/100)T (1.1236)1= (1.12)T If we compare the base terms, 1.1236 is approximate...
Read More →Find the sum of the GP :
Question: Find the sum of the GP : 1 + 3 + 9 + 27 + . To 7 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 = 1 r = (ratio between the n term and n-1 term) 3 1 = 3 n = 7 terms $\therefore \mathrm{S}_{\mathrm{n}}=1 \frac{3^{7}-1...
Read More →If a % is the discount per cent
Question: If a % is the discount per cent on a marked price x, then discount is (a) (x/a) 100 (b) (a/x) 100 (c) x (a/100) (d) 100/(x a) Solution: (c) x (a/100) (Discount = Discount% on Marked Price)...
Read More →If the solve the problem
Question: If $x=a \sec \theta, y=b \tan \theta$, prove that $\frac{d^{2} y}{d x^{2}}=-\frac{b^{4}}{a^{2} y^{3}} .$ Solution: Idea of parametric form of differentiation: If $y=f(\theta)$ and $x=g(\theta)$ i.e. $y$ is a function of $\theta$ and $x$ is also some other function of $\theta$. Then $d y / d \theta=f^{\prime}(\theta)$ and $d x / d \theta=g^{\prime}(\theta)$ We can write : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$ Given, $x=a \sec \theta \ldots \...
Read More →To gain 25% after allowing a discount
Question: To gain 25% after allowing a discount of 10%, the shopkeeper must mark the price of the article which costs him Rs 360 as (a) Rs 500 (b) Rs 450 (c) Rs 460 (d) Rs 486 Solution: (a) Rs 500 Explanation: Say, marked price = x Cost price = Rs.360 As per the question; x [x(10/100)] [(25360)/100] = 360 x x/10 90 = 360 9x/10 = 360 + 90 9x = 4500 x = 500...
Read More →If 90% of x is 315 km,
Question: If 90% of x is 315 km, then the value of x is (a) 325 km (b) 350 km (c) 350 m (d) 325 m Solution: (b) 350 km Explanation: 90% of x is 315 km 90/100 x = 315 X = 315 100/90 = 315 10/9 = 350...
Read More →If a and b are the roots of
Question: If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c$ and $d$ are the roots of $x^{2}-$ $12 x+q=0$, where $a, b, c, d$ from a GP, prove that $(q+p):(q-p)=17: 15$ Solution: Given data is, $x^{2}-3 x+p=0 \rightarrow(1)$ a and b are roots of (1) So, $(x+a)(x+b)=0$ $x^{2}-(a+b) x+a b=0$ So, $a+b=3$ and $a b=p \rightarrow(2)$ Given data is, $x^{2}-12 x+q=0 \rightarrow(3)$ c and d are roots of (1) So, $(x+c)(x+d)=0$ $x^{2}-(c+d) x+c d=0$ So, $c+d=12$ and $c d=q \rightarrow(4)$ a, b, c, d a...
Read More →If marked price of an article is Rs 1,200
Question: If marked price of an article is Rs 1,200 and the discount is 12% then the selling price of the article is (a) Rs 1,056 (b) Rs 1,344 (c) Rs 1,212 (d) Rs 1,188 Solution: (a) Given, marked price of an article = Rs. 1200 Discount % = 12% Discount = Discount % on marked price = A x 1200=12 x 12 = Rs.144 Selling price = Marked price Discount Selling price = Rs.11200 Rs. 144 = Rs.1056 Hence, option (a) is correct....
Read More →The compound interest on Rs 50,000 at
Question: The compound interest on Rs 50,000 at 4% per annum for 2 years compounded annually is (a) Rs 4,000 (b) Rs 4,080 (c) Rs 4,280 (d) Rs 4,050 Solution: (b) Rs 4,080 Explanation: P = Rs.50000, R = 4%, T = 2 years A = P(1+R/100)T= 50000(1+4/100)2= 50000(1+1/25)2 A = 50000(26/25)2= 54080 Compound interest = A P = 54080 50000 = Rs. 4080...
Read More →Suppose a certain sum doubles in 2 yr at r% rate
Question: Suppose a certain sum doubles in 2 yr at r% rate of simple interest per annum and R/o rate of interest per annum compounded annually. Then, (a) r R (b) Rr (c)R = r (d) Cannot be determined Solution: (b) If the total amount received after 2 yr is same for both simple interest and compound interest on same principal, then the rate of simple interest is greater than the rate of compound interest. i.e.Rr Hence, option (b) is correct....
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