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Question: If $(x-y) e^{\frac{x}{x-y}}=a$, prove that: $\frac{d y}{d x}=\frac{2 y-3 x}{2 x-1}$ Solution: Given: $(x-y) e^{\frac{x}{x-y}}=a$ Taking log on both sides we get, $\log (x-y)+\frac{x}{x-y} \log (e)=\log a$ (Using $\log a^{b}=b \log a$ and $\log (e)=1$ ) Differentiating both sides we get, $\frac{1}{x-y}\left[1-\frac{d y}{d x}\right]+\frac{(x-y) \frac{d}{d x}(x)+x\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}=0$ Taking L.C.M and solving the equation we get, $(x-y)\left[1-\frac{d y}{d x}\right...
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Question: Mark (✓) against the correct answer: 6 pipes can fill a tank in 24 minutes. One pipe can fill it in (a) 4 minutes (b) 30 minutes (c) 72 minutes (d) 144 minutes Solution: (d) 144 minutes Let one pipe takexmin to fill the tank. Clearly, one pipe will take more time to fill the tank. So, it is a case of inverse proportion. Now, $6 \times 24=1 \times x$ $\Rightarrow x=6 \times 24$ $\Rightarrow x=144$ One pipe can fill the tank in 144 minutes....
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Question: If $\cos y=x \cos (a+y)$, where $\cos a \neq \pm 1$, prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$ Solution: Here, $\cos y=x \cos (a+y)$, where $\cos a \neq \pm 1$ Differentiating both sides with respect to $x$, we get $-\sin y \frac{d y}{d x}=x\left(-\sin (a+y) \frac{d y}{d x}\right)+\cos (a+y)$ $\frac{d y}{d x}[x \sin (a+y)-\sin y]=\cos (a+y)$ $\frac{d y}{d x}=\frac{\cos (a+y)}{x \sin (a+y)-\sin y}$ Multiplying the numerator and the denominator by $\cos (a+y)$ on th RHS ...
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Question: Prove that $\left(1+\mathrm{i}^{2}+\mathrm{i}^{4}+\mathrm{i}^{6}+\mathrm{i}^{8}+\ldots .+\mathrm{i}^{20}\right)=1$ Solution: L.H.S $=\left(1+i^{2}+i^{4}+i^{6}+i^{8}+\ldots .+i^{20}\right)$ $\sum_{n=0}^{n=20} i^{n}$ $=1+-1+1+-1+\ldots \ldots \ldots . .+1$ As there are 11 times 1 and 6 times it is with positive sign as ${ }^{i}{ }^{0}=1$ as this is the extra term and there are 5 times 1 with negative sign. So, these 5 cancel out the positive one leaving one positive value i.e. 1 $\sum_{n...
Read More →A garrison of 200 men had provisions for 45 days.
Question: A garrison of 200 men had provisions for 45 days. After 15 days, 40 more men join the garrison. Find the number of days for which the remaining food will last. Solution: Clearly, the remaining food is sufficient for 200 men for (45 15), i.e., 30 days. Total number of men = 200 + 40 = 240 Let the remaining food last forxdays. Clearly, more men will take less number of days to finish the food. So, it is a case of inverse proportion. Now, $200 \times 30=240 \times x$ $\Rightarrow x=\frac{...
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Question: If $(\cos x)^{y}=(\cos y)^{x}$ find $\frac{d y}{d x}$. Solution: Here, $(\cos x)^{y}=(\cos y)^{x}$ Taking log on both sides, $\log (\cos x)^{y}=\log (\cos y)^{x}$ $y \log (\cos x)=x \log (\cos y)$ Differentiating it with respect to $x$ using the chain rule and product rule, $\frac{d}{d x}(y \log \cos x)=\frac{d}{d x}(x \log \cos y)$ $y \frac{d}{d x} \log \cos x+\log \cos x \frac{d y}{d x}=x \frac{d}{d x} \log \cos y+\log \cos y \frac{d x}{d x}$ $y \frac{1}{\cos x}(-\sin x)+\log \cos x ...
Read More →30 men can finish a piece of work in 28 days.
Question: 30 men can finish a piece of work in 28 days. How many days will be taken by 21 men to finish it? Solution: Letxbe the number of days taken by 21 men to finish the piece of work. More men will take less time to complete the work. So, this is a case of inverse proportion. Now, $30 \times 28=21 \times x$ $\Rightarrow x=\frac{30 \times 28}{21}$ $\Rightarrow x=40$ 21 men will take 40 days to finish the piece of work....
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Question: Prove that $\sqrt{-16}+3 \sqrt{-25}+\sqrt{-36}-\sqrt{-625}=0$. Solution: L.H.S $=\sqrt{-16}+3 \sqrt{-25}+\sqrt{-36}-\sqrt{-625}$ Since we know that $\mathrm{i}=\sqrt{-1}$. So, $=\sqrt{16} i+3 \sqrt{25} i+\sqrt{36} i-\sqrt{625} i$ $=4 \mathrm{i}+15 \mathrm{i}+6 \mathrm{i}-25 \mathrm{i}$ $=0$ L.H.S = R.H.S Hence proved....
Read More →10 people can dig a trench in 6 days.
Question: 10 people can dig a trench in 6 days. How many people can dig it in 4 days? Solution: Letxpeople dig the trench in 4 days. More people will take less number of days to dig the trench. Hence, this is a case of inverse proportion. Now,$10 \times 6=x \times 4$ $\Rightarrow x=\frac{60}{4}$ $\Rightarrow x=15$ 15 people can dig the trench in 4 days....
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Question: If $y^{x}+x^{y}+x^{x}=a^{b}$, find $\frac{d y}{d x}$. Solution: Given that, $y^{x}+x^{y}+x^{x}=a^{b}$ Putting, $u=y^{x}, v=x^{y}, w=x^{x}$, we get $u+v+w=a^{b}$ Therefore, $\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0$ .......(i) Now, $u=y^{x}$, Taking log on both sides, we have $\log u=x \log y$b Differentiating both sides with respect to $x$, we have $=x \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot 1$ So, $\frac{d u}{d x}=u\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)$ $=y^{...
Read More →10 people can dig a trench in 6 days.
Question: 10 people can dig a trench in 6 days. How many people can dig it in 4 days? Solution: Letxpeople dig the trench in 4 days. More people will take less number of days to dig the trench. Hence, this is a case of inverse proportion. Now,$10 \times 6=x \times 4$ $\Rightarrow x=\frac{60}{4}$ $\Rightarrow x=15$ 15 people can dig the trench in 4 days....
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Question: Prove that $(1-\mathrm{i})^{\mathrm{n}}\left(1-\frac{1}{\mathrm{i}}\right)^{\mathrm{n}}=2^{\mathrm{n}}$ for all values of $\mathrm{n} \mathrm{N}$ Solution: L.H.S $=(1-\mathrm{i})^{n}\left(1-\frac{1}{i}\right)^{\mathrm{n}}$ $=(1-\mathrm{i})^{\mathrm{n}}\left(1-\mathrm{i}^{-4^{*} 1+3}\right)^{\mathrm{n}}$ $=(1-i)^{n}\left(1-i^{3}\right)^{n}$ Since, $i^{4 n+3}=-1$ $=(1-i)^{n}(1+i)^{n}$ Applying $a^{n} b^{n}=(a b)^{n}$ $=((1-i)(1+i))^{n}$ $=\left(1-\mathrm{i}^{2}\right)^{\mathrm{n}}$ $=2^{...
Read More →The railway fare for 61 km is Rs 183.
Question: The railway fare for 61 km is Rs 183. Find the fare for 53 km. Solution: Let Rsxbe the railway fare for a journey of distance 53 km. The lesser the distance, the lesser will be the fare. So, it is a case of direct proportion . Now, $\frac{61}{183}=\frac{53}{x}$ $\Rightarrow x=\frac{53 \times 183}{61}$ $\Rightarrow x=159$ The railway fare for a journey of distance 53 km is Rs 159....
Read More →The cost of 140 tennis balls is Rs 4900.
Question: The cost of 140 tennis balls is Rs 4900. Find the cost of 2 dozen such balls. Solution: Let Rsxbe the cost of 24 tennis balls. More tennis balls will cost more. Now, $\frac{140}{4900}=\frac{24}{x}$ $\Rightarrow x=\frac{24 \times 4900}{140}$ $\Rightarrow x=840$ The cost of 2 dozen tennis balls is Rs 840....
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Question: $\left\{i^{18}+\frac{1}{i^{25}}\right\}^{3}=2(1-i)$ Solution: L.H.S $=\left\{i^{18}+\frac{1}{i^{25}}\right\}^{3}$ $\Rightarrow\left\{i^{4 \times 4+2}+i^{-4 \times 7+3}\right\}^{3}$ Since $i^{4 n}=1$ $i^{4 n+1}=i$ $i^{4 n+2}=-1$ $i^{4 n+3}=-1$ $=\left\{i^{2}+i^{3}\right\}^{3} .$ $=(-1-i)^{3}$ Applying the formula $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ We have, $\left.+3 \mathrm{i}^{2}+3 \mathrm{i}+1\right)$ $i+3-3 i-1$ $=2(1-i)$ L.H.S = R.H.S Hence proved....
Read More →350 boxes can be placed in 25 cartons.
Question: 350 boxes can be placed in 25 cartons. How many boxes can be placed in 16 cartons? Solution: Letxbe the required number of boxes. Less number of boxes will require less number of cartons. So, it is a case of direct proportion. Now, $\frac{350}{25}=\frac{x}{16}$ $\Rightarrow x=\frac{350 \times 16}{25}$ $\Rightarrow x=224$ 224 boxes can be placed in 16 cartoons....
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Question: Tick (✓) the correct answer: A car takes 2 hours to reach a destination by travelling at 60 km/hr. How long will it take while travelling at 80 km/hr? (a) 1 hr 30 min (b) 1 hr 40 min (c) 2 hrs 40 min (d) none of these Solution: (a) 1 h 30 min Letxh be the time taken by the car travelling at 80 km/hr. The greater the speed, the lesser will be the time taken. So, it is a case of inverse proportion. Now, $60 \times 2=80 \times x$ $\Rightarrow x=\frac{120}{80}$ $\Rightarrow x=1.5$ Therefor...
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Question: Prove that $\left\{\mathrm{i}^{21}-\left(\frac{1}{\mathrm{i}}\right)^{46}\right\}^{2}=2 \mathrm{i}$. Solution: L.H.S. $=\left\{i^{21}-\left(\frac{1}{i}\right)^{46}\right\}^{2}$ $=\left\{i^{4 \times 5+1}-i^{-4 \times 12+2}\right\}^{2}$ Since $i^{4 n}=1$ $i^{4 n+1}=i$ $i^{4 n+2}=i^{2}=-1$ $i^{4 n+3}=i^{3}=-1$ $=\left\{i^{1}-i^{2}\right\}^{2}$ $=\{i+1\}^{2}$ Now, applying the formula $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $=i^{2}+1+2 i .$ $=-1+1+2 i$ $=2 i$ L.H.S = R.H.S Hence proved....
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Question: If $x y=e^{x-y}$, find $\frac{d y}{d x}$ Solution: The given function is $x y=e^{x-y}$ Taking log on both sides, we obtain $\log (x y)=\log \left(e^{x-y}\right)$ $\log x+\log y=(x-y) \log e$ $\log x+\log y=(x-y) \times 1$ $\log x+\log y=x-y$ Differentiating both sides with respect to $x$, we obtain $\frac{\mathrm{d}}{\mathrm{dx}}(\log x)+\frac{\mathrm{d}}{\mathrm{dx}}(\log y)=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\frac{\mathrm{dy}}{\mathrm{dx}}$ $\frac{1}{x}+\frac{1}{y} \frac{d y}...
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Question: Tick (✓) the correct answer 3 persons can build a wall in 4 days, then 4 persons can build it in (a) $5 \frac{1}{3}$ days (b) 3 days (c) $4 \frac{1}{3}$ days (d) none of these Solution: (b) 3 days Letxbe number of days taken by 4 persons to build the wall. More number of persons will take less time to build the wall. So, it is a case of inverse proportion. Now, $3 \times 4=4 \times x$ $\Rightarrow x=3$ Therefore, 4 persons can build the wall in 3 days....
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Question: Tick (✓) the correct answer 3 persons can build a wall in 4 days, then 4 persons can build it in (a) $5 \frac{1}{3}$ days (b) 3 days (c) $4 \frac{1}{3}$ days (d) none of these Solution: (b) 3 days Letxbe number of days taken by 4 persons to build the wall. No. of persons 3 4 No. of days 4 x More number of persons will take less time to build the wall. So, it is a case of inverse proportion. Now, $3 \times 4=4 \times x$ $\Rightarrow x=3$ Therefore, 4 persons can build the wall in 3 days...
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Question: If $y=(\sin x-\cos x)^{\sin x-\cos x}, \frac{\pi}{4}x\frac{3 \pi}{4}$, find $\frac{d y}{d x}$. Solution: Here, $y=(\sin x-\cos x)^{(\sin x-\cos x)}$ .......(1) Taking log on both sides, $\log y=\log (\sin x-\cos x)^{(\sin x-\cos x)}$ $\log y=(\sin x-\cos x) \log (\sin x-\cos x)$ Differentiating it with respect to $x$ using product rule, chain rule, $\frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \frac{d}{d x}(\sin x-\cos x)+(\sin x-\cos x) \frac{d}{d x} \log (\sin x-\cos x)$ $\frac{1...
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Question: Prove that $\left(1+i^{10}+i^{20}+i^{30}\right)$ is a real number. Solution: L.H.S $=\left(1+\mathrm{i}^{10}+\mathrm{i}^{20}+\mathrm{j}^{30}\right)$ $=\left(1+i^{4 \times 2+2}+i^{4 \times 5}+i^{4 \times 7+2}\right)$ Since $\Rightarrow i^{4 n}=1$ $\Rightarrow i^{4 n+1}=i$ $\Rightarrow i^{4 n+2}=-1$ $\Rightarrow i^{4 n+3}=-1$ $=1+i^{2}+1+i^{2}$ $=1+-1+1+-1$ = 0, which is a real no Hence, $\left(1+\mathrm{i}^{10}+\mathrm{i}^{20}+\mathrm{i}^{30}\right)$ is a real number....
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Question: Tick (✓) the correct answer: 6 pipes fill a tank in 120 minutes, then 5 pipes will fill it in (a) 100 min (b) 144 min (c) 140 min (d) 108 min Solution: (b) 144 min Letxmin be the time taken by 5 pipes to fill the tank. No. of pipes 6 5 Time (in min) 120 x Now, $6 \times 120=5 \times x$ $\Rightarrow x=144$ Therefore, 5 pipes will take 144 min to fill the tank....
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Question: Tick (✓) the correct answer: A photograph of a bacteria enlarged 50000 times attains a length of 5 cm. The actual length of bacteria is (a) 1000 cm (b) 103cm (c) 104cm (d) 102cm Solution: (c) $10^{-4} \mathrm{~cm}$ Let $x \mathrm{~cm}$ be the actual length of the bacteria. The larger the object, the larger its image will be. Now, $\frac{x}{1}=\frac{5}{50000}=10^{-4} \mathrm{~cm}$ Hence, the actual length of the bacteria is $10^{-4} \mathrm{~cm}$....
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