Maths MCQ

Let the cost price of one pen be $₹1.$
Then, $CP$ of $20$ pens $= Rs. 20$
and $SP$ of $20$ pens $= Rs. 15 (\because SP$ of $20$ pens $= CP$ of $15$ pens$)$
Therfore, $CP$ is more than $SP.$
So, Loss $= CP - SP$
$= Rs. 20 - Rs. 15$
$= Rs. 5$
Loss $% =\frac{\text{Loss}}{\text{CP}}\times100$
$=\frac{5}{20}\times100$
$=25\%$
Hence, the correct option is $(a).$

Let the $CP$ of a pen be $x.$
$SP$ of a pen $= Rs. 48$
Loss $= 20\%$
Therefore, $CP$ is more than $SP.$
Now, Loss $= CP - SP$ and Loss $=$ Loss percent $\times CP$
Thus, $CP - SP =$ Loss percent $\times CP$
$\Rightarrow\text{x}-48=\frac{20}{100}\times\text{x}$
$\Rightarrow100\text{x}-4800=20\text{x}$
$\Rightarrow100\text{x}-20\text{x}=4800$
$\Rightarrow80\text{x}=4800$
$\Rightarrow\text{x}=\frac{4800}{80}$
$\Rightarrow\text{x}=60$
Therefore, $CP$ of the pen $= Rs. 60$
Now, in order to gain $20\%$, let the new $SP$ be $y.$
Gain $=$ Gain percent $\times CP$
$=\frac{20}{100}\times60$
$= Rs.12$
$SP = CP +$ Gain
$= Rs. 60 + Rs. 12$
$= Rs. 72$
Hence, the correct option is $(d).$

Let the $SP$ be $x.$
$CP = Rs. 3300$
Loss $= 10\%$
Therfore, $CP$ is more than $SP.$
Now,
Loss $= 10\%$ of $CP$ and $SP = CP -$ loss
So, $SP = CP - 10\%$ of $CP$
$\Rightarrow\text{x}=3300-\frac{10}{100}\times3300$
$\Rightarrow\text{x}=\Big(1-\frac{1}{10}\Big)3300$
$\Rightarrow\text{x}=\Big(\frac{9}{10}\Big)3300$
$\Rightarrow\text{x}=2970$
Thus, $SP = Rs. 2970$
Hence, the correct option is $(d).$

Let the total lemons be $12.$
$CP$ of $6$ lemons $= Rs. 1$
then, $CP$ of $12$ lemons $= Rs. 2$
Also, $SP$ of $4$ lemons $= Rs. 1$
then, $SP$ of $12$ lemons $= Rs. 3$
Therefore, $SP$ is more than $CP.$
So, Gain $= SP - CP$
$= Rs. 3 - Rs. 2$
$= Rs. 1$
Gain % $=\frac{\text{Profit}}{\text{CP}}\times100$
$=\frac{1}{2}\times100$
$=50\%$
Hence, the correct option is $(a).$

Let the $CP$ be $x.$
$SP = Rs. 630$
Loss $= 10\%$
Therfore, $CP$ is more than $SP.$
Now,
Loss $= 10\%$ of $CP$ and $SP = CP -$ loss
So, $SP = CP - 10\%$ of $CP$
$\Rightarrow630=\text{x}-\frac{10}{100}\times\text{x}$
$\Rightarrow630=\Big(1-\frac{1}{10}\Big)\text{x}$
$\Rightarrow630=\Big(\frac{9}{10}\Big)\text{x}$
$\Rightarrow\text{x}=\frac{6300}{9}$
$\Rightarrow\text{x}=700 $
Thus, $CP = Rs. 700$
Hence, the correct option is $(c).$

Let the cost price of one pencil be $Rs. 1.$
Then, $CP$ of $5$ pencils $= Rs. 5$
and $SP$ of $5$ pencils $= Rs. 6 (\because \ SP$ of $5$ pencils $= CP$ of $6$ pencils$)$
Therfore, $SP$ is more than $CP.$
So, Profit $= SP - CP$
$= Rs. 6 - Rs. 5$
$= Rs. 1$
Gain $% =\frac{\text{Gain}}{\text{CP}}\times100$
$=\frac{1}{5}\times100$
$=20\%$
Hence, the correct option is $(a).$

Since, $CP$ is more than $SP.$
Therefore, loss $= CP − SP$
$= Rs. 120 - Rs. 80$
$= Rs. 40$
Hence, the correct option is $(a).$