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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.
Home NCERT Solutions Class 9 Maths Question 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. 🚀 Checkout eSaral Courses 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 inver...
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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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