Vinod sold a watch to Arun at a gain of 12% and Arun had to sell it to Manoj at a loss of 5%.
Question: If manoj paid ₹ 3990 for it, how much did vinod pay for the watch? Solution: Let the CP of the watch for Vinod be ₹x. SP = Gain + CP $=12 \%$ of $\mathrm{CP}+x$ $=\frac{12}{100} x+x$ $=₹ \frac{112}{100} x$ Now, SP of the water for Vinod will be the CP of the watch for Arun. SP of the watch for Arun $=\mathrm{CP}-$ Loss $=\frac{112}{100} x-5 \%$ of $\frac{112}{100} x$ $=\frac{112}{100} x-\frac{5}{100}\left(\frac{112}{100} x\right)$ $=\frac{112}{100} x\left(1-\frac{5}{100}\right)$ $=₹ \f...
Read More →What would happen if green plants
Question: What would happen if green plants disappear from earth ? Solution: Herbivores will die of starvation followed by carnivores and then decomposers...
Read More →Is nutrition a necessity for an
Question: Is nutrition a necessity for an organism ? Discuss. Solution: Yes, nutrition is a must for an organism because of the following reasons : Importance of Nutrition/Food Food provides energy: Energy is required by the body all the time, whether asleep, taking rest or doing work. When the body is not doing any apparent work, energy is still being consumed in maintaining order. Further, biosynthetic activities continue for replacing materials being consumed or degraded. A number of other ac...
Read More →Consider the function
Question: Consider the function $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$, defined by Write its domain and range. Also, draw the graph of f(x). Solution: Given: f(x) To Find: Domain and Range of $\mathrm{f}(\mathrm{x})$ When $f(x)=1-x \mid x0$ In this case there is no value of $x(x0)$ which makes the above expression undefined. Therefore Domain(f) $=(-\infty, 0) \ldots(1)$ When f(x) = x | x = 0 In this case there is no value other than 0 which makes the above expression undefined. Therefor...
Read More →Differentiate between an autotroph
Question: Differentiate between an autotroph and a heterotroph. Solution:...
Read More →Solve this
Question: If $\sin ^{2} y+\cos x y=k$, find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$ Solution: We are given with an equation $\sin ^{2} y+\cos (x y)=k$, we have to find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$ by using the given equation, so by differentiating the equation on both sides with respect to $x$, we get, $2 \sin y \cos y \frac{d y}{d x}-\sin (x y)\left[(1) y+x \frac{d y}{d x}\right]=0$ $\frac{d y}{d x}[2 \sin y \cos y-x \sin (x y)]=y \sin (x y)$ $\frac{d y}{d x}=\frac{y \sin (x ...
Read More →A shopkeeper sold two fans for ₹ 1140 each.
Question: A shopkeeper sold two fans for ₹ 1140 each. On one he gains 14%, while on the other he loses 5%. Calculate his gain or loss per cent in the whole transaction. Solution: SP of first fan $=₹ 1,140$ Gain $=14 \%$ Let the CP of first fan be ₹ $x$. CP $=$ SP - Gain $\Rightarrow x=1140-\frac{14}{100} x$ $\Rightarrow x+\frac{14}{100} x=1140$ $\Rightarrow \frac{114}{100} x=1140$ $\Rightarrow x=1000$ $\therefore$ CP of first fan $=$ ₹ 1,000 SP of second fan $=$ ₹ 1,140 Loss $=5 \%$ Let the CP ...
Read More →Why do fishes die when taken out of water ?
Question: Why do fishes die when taken out of water ? Solution: Fish taken out of water die due to Inability to obtain oxygen from air Collapsing of gill lamellae so that no space is left for gaseous exchange....
Read More →If a plant is releasing carbon dioxide
Question: If a plant is releasing carbon dioxide and taking in oxygen during the day, does it mean that there is no photosynthesis occurring ? Justify your answer. Solution: A plant releases carbon dioxide and takes in oxygen only when photosynthesis is either absent or too small as not to compensate for respiration. (In photosynthesis, plants absorb CO2and release O2. The normal rate of photosynthesis is many times the rate of respiration. As a result, CO2produced during respiration is consumed...
Read More →Luxmi sold two sarees for ₹ 1980 each.
Question: Luxmi sold two sarees for ₹ 1980 each. On one, she lost 10%, while on the other she gained 10%. Find her gain or loss per cent in the whole transaction. Solution: SP of first saree $=₹ 1,980$ Loss $=10 \%$ Let the CP of first saree be ₹ $x .$ CP $=$ Loss $+$ SP $\Rightarrow \frac{10}{100} \times x+1980=x$ $\Rightarrow x-\frac{10}{100} x=1980$ $\Rightarrow \frac{90}{100} x=1980$ $\Rightarrow x=2200$ $\therefore$ CP of first saree $=$ ₹ 2,200 SP of second saree $=$ ₹ 1,980 Gain $=10 \%$ ...
Read More →Two green plants are kept separately in oxygen free containers,
Question: Two green plants are kept separately in oxygen free containers, one in dark and the other in continuous light. Which one will live longer ? Give reasons. (CCE 2010) Solution: Plant kept in continuous light will live longer due to Manufacture of food and hence its availability to the plant for maintenance and growth, Production of oxygen in photosynthesis and its availability for respiration of the plant. Plant kept in oxygen free container kept in dark will die within a few days due to...
Read More →A radio is sold for Rs 3120 at a loss of 4%.
Question: A radio is sold for Rs 3120 at a loss of 4%. What will be the gain or loss per cent if it is sold for Rs 3445? Solution: Let the original price be $x$. SP = Rs 3120 Now, SP = CP - loss $\Rightarrow 3120=x-\frac{4}{\frac{100}{x}}$ $\Rightarrow 3120=x-\frac{x}{25}$ $\Rightarrow 3120=\frac{24 x}{25}$ $\Rightarrow \frac{3120 \times 25}{24}=x$ $\Rightarrow x=3250$ So, the cost price is Rs 3250 . If it is sold for Rs 3445 , then its a gain because $S PC P$. Now, gain = SP - CP $=\operatornam...
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Question: If $\sin ^{2} y+\cos x y=k$, find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$ Solution: We are given with an equation $\sin ^{2} y+\cos (x y)=k$, we have to find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$ by using the given equation, so by differentiating the equation on both sides with respect to $x$, we get, $2 \sin y \cos y \frac{d y}{d x}-\sin (x y)\left[(1) y+x \frac{d y}{d x}\right]=0$ $\frac{d y}{d x}[2 \sin y \cos y-x \sin (x y)]=y \sin (x y)$ $\frac{d y}{d x}=\frac{y \sin (x ...
Read More →How do the guard cells regulate
Question: How do the guard cells regulate opening and closing of stomatal pores ? (CCE 2010, 2012) Solution: Opening and closing of stomata is regulated by gain or loss of turgidity of their guard cells. During opening of stomata, guard cells withdraw K+ions from surrounding epidermal cells, followed by absorption of water from them. As a result, guard cells swell up and become turgid. Their outer thin and elastic walls bend outwardly followed by outward movement of thicker inner walls. The latt...
Read More →“All plants give out oxygen during day
Question: All plants give out oxygen during day and carbon dioxide during night. Do you agree with the statement ? Give reason.(CCE 2010) Solution: Yes. Respiration is going on throughout day and night. Photosynthesis occurs only during the day. Rate of photosynthesis is several times the rate of respiration. All the CO2produced in respiration is also consumed in photosynthesis during the day time. Therefore, during day time, plants give out oxygen, which is a product of photosynthesis. However,...
Read More →By selling an umbrella for ₹ 336, a shopkeeper loses 4%.
Question: By selling an umbrella for ₹ 336, a shopkeeper loses 4%. At what price must he sell it to gain 4%? Solution: Let the CP of the umbrella be ₹ $x$. SP of the umbrella = ₹336 Loss $=4 \%$ of $₹ x=₹ \frac{4}{100} x$ CP $-$ Loss $=$ SP $\Rightarrow x-\frac{4}{100} x=336$ $\Rightarrow \frac{96}{100} x=336$ $\Rightarrow x=₹ 350$ $\therefore$ CP of the umbrella $=₹ 350$ Now, for gain of $4 \%$, SP $=\mathrm{CP}+$ Gain $\Rightarrow \mathrm{SP}=350+\frac{4}{100} \times 350$ $\Rightarrow \mathrm{...
Read More →By selling a bouquet for Rs 322, a florist gains 15%.
Question: By selling a bouquet for Rs 322, a florist gains 15%. At what price should he sell it to gain 25%? Solution: SPof the bouquet = Rs 322 Gain percentage = 15% $\mathrm{CP}$ of the bouquet $=\left(\frac{100}{100+\text { gain } \%}\right) \times \mathrm{SP}$ $=\operatorname{Rs}\left(\frac{100}{100+15}\right) \times 322$ $=\mathrm{Rs} \frac{100}{115} \times 322$ $=$ Rs 280 Now, CP = Rs 128 and desired gain percentage $=25 \%$ $\therefore$ Desired SP $=\left(\frac{100+\text { gain } \%}{100}...
Read More →Name the following :
Question: Name the following : The process in plants that links light energy with chemical energy. Organisms that can prepare their own food. Cells that surround a stomatal pore. The cell organelle where photosynthesis occurs. Organisms that cannot prepare their own food. An enzyme secreted by gastric glands in stomach that acts on proteins. Solution: Photosynthesis Autotrophs Guard cells Chloroplast Heterotrophs Pepsin....
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Question: If $\cos y=x \cos (a+y)$, with $\cos a \neq \pm 1$, prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$ Solution: We are given with an equation $\cos y=x \cos (a+y)$, we have to prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$ by using the given equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get, $-\sin y \frac{d y}{d x}=\cos (a+y)...
Read More →If the selling price of a flower vase is
Question: If the selling price of a flower vase is $\frac{5}{6}$ of its cost price, find the loss per cent. Solution: Let the $\mathrm{CP}$ of the flower vase set be ₹ $x$. SP of the flower vase $=\frac{5}{6} \mathrm{CP}=₹ \frac{5}{6} x$ Loss $=\mathrm{CP}-\mathrm{SP}=x-\frac{5}{6} x=₹ \frac{x}{6}$ Loss $\%=\frac{\text { Loss }}{\text { CP }} \times 100 \%=\frac{\frac{x}{6}}{x} \times 100 \%=\frac{100}{6} \%=\frac{50}{3} \%=16 \frac{2}{3} \%$...
Read More →If the selling price of a TV set is equal to
Question: If the selling price of a TV set is equal to $\frac{6}{5}$ of its cost price, find the gain per cent. Solution: Let the CP of the TV set be ₹ $x$. SP of the TV set $=\frac{6}{5} \mathrm{CP}=₹ \frac{6}{5} x$ Gain $=$ SP of the TV set $-$ CP of the TV set $=\frac{6}{5} x-x=$ ₹ $\frac{x}{5}$ Gain $\%=\frac{\text { Gain }}{\text { CP }} \times 100 \%=\frac{\frac{x}{5}}{x} \times 100 \%=\frac{100}{5} \%=20 \%$...
Read More →Give an account of the process adopted
Question: Give an account of the process adopted by Mendeleev for the classification of elements. How did he arrive at Periodic Law ? Solution: The basis of classification of elements adopted was atomic masses of the elements. The elements were arranged in order of increasing atomic masses. Elements with similar properties were kept in a particular group in order of increasing atomic masses. Elements placed in a particular group (or sub group) were having same valency. Note :Please note that whe...
Read More →Which group of elements could be placed
Question: Which group of elements could be placed in Mendeleevs Periodic Table without disturbing the original order ? Give reason. Solution: Group of noble gases (called zero group) could be placed in the Mendeleevs Periodic Table with out disturbing the original order. Reasons : Elements present in a group have same valency in the Mendeleevs Periodic Table. Based on their electronic configuration, the members of noble gas family have zero valency. That is why they are called inert gases. They ...
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Question: If $e^{x}+e^{y}=e^{x+y}$, prove that $\frac{d y}{d x}=-\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ or, $\frac{d y}{d x}, e^{y-x}=0$ Solution: We are given with an equation $\mathrm{e}^{x}+\mathrm{e}^{y}=\mathrm{e}^{x+y}$, we have to prove that $\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ by using the given equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentia...
Read More →Consider the real function
Question: Consider the real function $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}: \mathrm{f}(\mathrm{x})=\mathrm{x}+5$ for all $x \in R$. Find its domain and range. Draw the graph of this function. Solution: Given: $f(x)=x+5 \forall x \in R$ To Find: Domain and Range of f(x). The domain of the given function is all real numbers expect where the expression is undefined. In this case, there is no real number which makes the expression undefined. As f(x) is a polynomial function, we can have any...
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