MATHS M.C.Q. [1 Marks Each]

The car with $25$ litres of petrol can go $= 150km$
Then, with $1$ litre of petrol can go $=\frac{150}{25}=6$
With $30$ litres of petrol can go $= 30 \times 6 = 180km$

$R_1=12 \% $
$R_2=10 \% $
$P_1=\text { Rs. } 8000 $
$P_2=\text { Rs. } 9100$
Let their amounts be equal in T years.
$\text { Amount }_1= S _{ I } I _1+ P _1 $
$=\frac{ P _1 \times r _1 \times T }{100}+ P _1 $
$=960 T+8000 $
$\text { Amount }_2= S . I _2+ P _2 $
$=\frac{ P _2 \times r _2 \times T }{100}+ P _2 $
$=\frac{9100 \times 10 \times T }{100}+9100 $
$=910 T+9100$
Amount $_1=$ Amount $_2$
$\Rightarrow 960 T+8000=910 T+9100 $
$\Rightarrow 960 T-910 T=9100-8000 $
$\Rightarrow 50 T=1100 $
$\Rightarrow T =22$
Hence, after $22$ years their amounts will be equal.

$\frac{4}{7}\ \text{of}\ 49\%\ \text{of x}=21$
$\Rightarrow\frac{4}{7}\times\frac{49}{100}\times\text{x}=21$
$\Rightarrow\frac{\text{7x}}{25}=21$
$\Rightarrow\frac{7x\times25}{25}=21\times25$
$\Rightarrow\text{x}=\frac{21\times25}{7}$
$\Rightarrow\text{x}=75$
Hence, the correct option is $(d).$

$72\%$ of 25 $=\frac{72}{100}\times25=18$
Number of students not good in hindi $= 25 - 18 = 7$

Let the required number be $x.$
According to the question,
$35\%\ \text{of x}+39=\text{x}$
$\Rightarrow\frac{35}{100}\times\text{x}+39=\text{x}$
$\Rightarrow\frac{35\text{x}\times100}{100}=(\text{x}-39)\times100$
$\Rightarrow35\text{x}=100\text{x}-3900$
$\Rightarrow100\text{x}-35\text{x}=3900$
$\Rightarrow65\text{x}=3900$
$\Rightarrow\text{x}=\frac{3900}{65}$
$\Rightarrow\text{x}=60$
Hence, the correct option is $(a).$

Given, total parts $= 10 \times 10 = 100$
$\therefore$ Shaded parts $= 36$
$\therefore$ Per cent of shaded parts $=\frac{36}{100}\times100\%=36\%$
Then, per cent of unshaded parts $= 100 - 60 = 40\%$
Hence, the unshaded region is $36\%$

$0.1=\frac{1}{10}$
$=\frac{1}{10}\times100\%$
$=10\%$

Amount $= Rs. 420$
And ratio $= 3 : 4$
Sum of ratios $= 3 + 4 = 7$
A’s share $=\frac{420\times3}{7}=\text{Rs. }60\times3=\text{Rs. }180$

Let the sum (P) be Rs. $x$
Then, the simple interest (I) $=\text{Rs.}\frac{16}{25}\text{x}$
Also,
Rate $(R) = R\%$
Time (T) = R years ($\because$ the rate percent per annum and the time are numerically equal)
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{R}=\frac{100\ \times\ \text{I}}{\text{P}\ \times\ \text{T}}$
$\Rightarrow\text{R}=\frac{100\ \times\ \frac{16}{25}\text{x}}{\text{x}\ \times\ \text{R}}$
$\Rightarrow\text{R}\ \times \text{R}=\frac{{64\text{x}}}{\text{x}}$
$\Rightarrow\text{R}^2=8^2$
$\Rightarrow\text{R}=8\%$
Hence, the correct option is $(a).$

Given, $P = Rs. 3000,$
$T = 2$ year
$R = 11\%$
$\therefore\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}=\frac{3000\times11\times2}{100}=\text{Rs. }660$
Now, amount $(A) = P + I$
$= 3000 + 660$
$= Rs. 3660$

Given, total parts $= 10 \times 10 = 100$
$\therefore$ Shaded parts $= 60$
$\therefore$ Per cent of shaded parts $=\frac{60}{100}\times100\%=60\%$
Then, per cent of unshaded parts $= 100 - 60 = 40\%$
Hence, the unshaded region is $40\%$

Let the present age of Ravish and Shikha be $3x$ and $8x$, respectively,
After six years,
Age of Ravish $= (3x + 6)$ years and
Age of Shikha $= (8x + 6)$ years
Since, $(\text{3x}+6):(8\text{x}+6)=4:9$
$\Rightarrow\frac{\text{(3x}+6)}{\text{(8x}+6)}=\frac{4}{9}$
$\Rightarrow9(3\text{x}+6)=4(8\text{x}+6)$
$\Rightarrow27\text{x}+54=32\text{x}+24$
$\Rightarrow-5\text{x}=-30$
$\Rightarrow\text{x}=\frac{-30}{-5}$
$\Rightarrow\text{x}=6$
$\therefore\text{3x}=3\times6=18$
So, the present age of Ravish is $18$ years.
Hence, the correct alternative is option $(a).$

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 $(a).$

$5\%\text{ of }200=\frac{5}{100}\times200=10$

$4m$ long shadow is of a tree of height $= 6m$
$1m$ long shadow of flagpole will of height $=\frac{6}{4}\text{m}$
$50m$ long shadow, the height of pole 6 will be $=\frac{6}{4}\times50=75\text{m}$

Amount = Rs. $3605$
Time $=\frac{219}{365}$ days $=\frac{219}{365}$ days
Rate = 5% per annum
Amount $=\text{Sum}+\frac{\text{Sum}\times\text{Rate}\times\text{Time}}{100}$
Amount $=\text{Sum}\Big(1+\frac{\text{Rate}\times\text{Time}}{100}\Big)$
$\text{Sum}=\frac{3605}{1+\frac{5}{100}\times\frac{219}{365}}$
$=\frac{3605\times36500}{37595}$
$\text{Sum}=\text{Rs. }3500$

As, $\text{a}:\text{b}=4:5$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{4}{5}$
Also, $\text{b}:\text{c}=2:\text{3}$
$\Rightarrow\frac{\text{b}}{\text{c}}=\frac{2}{3}$
So, $\text{a}:\text{c}=\frac{\text{a}}{\text{c}}$
$=\frac{\text{ab}}{\text{bc}}$
$=\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{c}}$
$=\frac45\times\frac23$
$=\frac{8}{15}$
$=8:15$
Hence, the correct alternative is option $(b).$

Profit $= S.P. - C.P.$

Let the total number of students be $x.$
There are $60$ girls out of $100$ students be $x.$
$60\%$ of $x$ is girls. Therefore, $60\%$ of $x = 18$
So, $18$ girls are out of how many students?
$\frac{60}{100}\times\text{x}=18$
$\text{x}=\frac{18\times100}{60}=30$
The total number of students are: $30$
Number of boys $= 30 - 18 = 12$
The ratio is $=\frac{18}{12}=3.2$

Six is multiplied by $100$ and the product is divided by $5.$

Required percentage $=\frac{50}{100}\times100\%=50\%$

$12\%=\frac{12}{100}$
$=\frac{12\div4}{100\div4}$
$=\frac{3}{25}$
Hence, the correct option is $(a).$

Total number of radios $= 25.$
Number of radios out of order $= 16$
$\%$ of the radios of order $=\frac{16}{25}\times100=64\%$

$C.P. = Rs. 80$
$S.P. = Rs. 100$
Gain $= S.P. - C.P. = Rs. 100 - 80 = Rs. 20$
$\therefore\text{Gain%}=\frac{\text{Gain}\times100}{\text{C.P}}$
$=\frac{20\times100}{80}$
$=25\%$

Given, $90%$ of $x = 315\ km.$
$\Rightarrow\frac{90}{100}\times\text{x}=315\Rightarrow\text{x}=\frac{315\times100}{90}$
$\therefore\text{x}=350\text{km.}$

Suppose that the number of boys in the school is $x$
Then,$ x : 320 = 9 : 5$
$\Rightarrow 5x = 2880$
$\Rightarrow x = 576$
Hence, total strength of the school $= 576 + 320 = 896$

Required amount of protein
$=\frac{25}{100}\times4$
$=1\text{Kg}$

$12.5\%=\frac{12.5}{100}$
$=\frac{125}{1000}$
$=\frac{125\div25}{1000\div25}$
$=\frac{5}{40}$
$=\frac{5\div5}{40\div5}$
$=\frac{1}{8}$
Hence, the correct option is $(a).$

$CP$ of $6$ lemons = Re $1$
$CP$ of $1$ lemon $=\text{Rs. }\frac{1}{6}$
$CP$ of $4$ lemons = Re $1$
$SP$ of $1$ lemon $=\text{Rs. }\frac{1}{4}$
$\therefore$ Gain $= SP - CP$
$=\frac{1}{4}-\frac{1}{6}$
$=\frac{3-2}{12}$
$=\text{Rs. }\frac{1}{12}$
Gain $= 10\%$
$\text{Gain%}=\frac{\text{Gain}\times100}{\text{C.P}}$
$=\frac{\frac{1}{12}\times100}{\frac{1}{6}}$
$=\frac{100\times6}{12}$
$=50\%$

$\frac{1}{5}=\frac{1}{5}\times100\%=20\%$

Let the $CP$ of an article be $x.$
$SP$ of the article = Rs. $144$
Loss $= 10\%$
Therefore, $CP$ is more than $SP.$
Now, Loss = $CP - SP$ and Loss = Loss percent $\times CP$
Thus, $CP - SP $= Loss percent $\times CP$
$\Rightarrow\text{x}-144=\frac{10}{100}\times\text{x}$
$\Rightarrow\text{x}-144=\frac{1}{10}\times\text{x}$
$\Rightarrow10\text{x}-1440=\text{x}$
$\Rightarrow10\text{x}-\text{x}=1440$
$\Rightarrow9\text{x}=1440$
$\Rightarrow\text{x}=\frac{1440}{9}$
$\Rightarrow\text{x}=160$
Therefore, $CP$ of the article = Rs. $160$
Now, in order to gain $10\%$, let the new $SP$ be $y.$
Gain = Gain percent $\times CP$
$=\frac{10}{100}\times160$
$= Rs. 16$
$SP = CP +$ Gain
$= Rs. 160 + Rs. 16$
$= Rs. 176$
Hence, the correct option is $(d).$