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

As, $\text{a}:\text{b}=5:7$ and $\text{b}:\text{c}=6:11$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac57$ and $\frac{\text{b}}{\text{c}}=\frac{6}{11}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{5\times6}{7\times6}$ and $\frac{\text{b}}{\text{c}}=\frac{6\times7}{11\times7}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{30}{42}$ and $\frac{\text{b}}{\text{c}}=\frac{42}{77}$
$\Rightarrow\text{a}:\text{b}=30:42$ and $\text{b}:\text{c}=42:77$
$\therefore\text{a}:\text{b}:\text{c}=30:42:77$
Hence, the correct alternative is option $(b).$

$\frac{2}{5}\ \text{of}\ 50\%\ \text{of x}=10$
$\Rightarrow\frac{2}{5}\times\frac{50}{100}\times\text{x}=10$
$\Rightarrow\frac{\text{x}}{5}=10$
$\Rightarrow\frac{x\times5}{5}=10\times5$
$\Rightarrow\text{x}=50$
Hence, the correct option is $(b).$

Given that $60\%$ of $x$ is $1200$
$\frac{60}{100}\times\text{x}=1200$
$\text{x}=1200\times\frac{100}{60}$
$\text{x}=2000$

$45$ years $: 10$ years
$\Rightarrow 9 : 2$

As, $\text{a}:\text{b}=2:5$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{2}{5}$
Let $\text{a}=\text{2x}$ and $\text{b}=5\text{5x}.$
Then,
$\frac{3\text{a}+\text{2b}}{4\text{a}+\text{b}}$
$=\frac{3\times2\text{x}+2\times\text{5x}}{4\times2\text{x}+5\text{x}}$
$=\frac{6\text{x}+10\text{x}}{8\text{x}+5\text{x}}$
$=\frac{16\text{x}}{13\text{x}}$
$=\frac{16}{13}$
Hence, the correct alternative is option $(a).$

$\frac{1}{4}=\frac{1}{4}\times100\%$
$=25\%$

It is given that,
Sum $(P) = Rs. 2000$
Rate $(R) = 2\%$
Time $(T) = 10$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$=\frac{2000\ \times\ 2\ \times\ 10}{100}$
$=\text{Rs. }400$
Amount $= I + P = Rs. 400 + Rs. 2000 = Rs. 2400$
Hence, the correct option is $(b).$

The increase in the number of wins (or amount of change) $= 6 - 4 = 2$
Percentage increase $=\frac{\text{Amount of change}}{\text{Origanal amount or base}}\times100$
$=\frac{2}{4}\times100$
$=50\%$

Given number is multiplied by $100$ then solved.

Let the required number be $x.$
$\text{x}\%\ \text{of}\ 75=12$
$\Rightarrow\frac{\text{x}}{100}\times75=12$
$\Rightarrow\frac{3\text{x}}{4}=12$
$\Rightarrow\frac{3\text{x}\times4}{4}=12\times4$
$\Rightarrow3\text{x}=48$
$\Rightarrow\text{x}=\frac{48}{3}$
$\Rightarrow\text{x}=16$
Hence, the correct option is $(d).$

$15$ is multiplied by $100$ and product is divided by $25.$

Rs. $10 : 50$ paise
$= 10 \times 100$ paise: $50$ paise
$= 1000 : 50$
$= 20 : 1$

Amount $=\frac{49}{40}$ times the sum $=\frac{49}{40}\text{P}$
Simple interest $(I) =$ Amount - Sum $=\frac{49}{40}\text{P}-\text{P}=\frac{9}{40}\text{P}$
Let the sum $(P)$ be $x.$
Then, simple interest $(I) =\frac{9}{40}\text{x}$
Rate $(R) = R%$
Time (T) $=2\frac{1}{2}\text{ years}=\frac{5}{2}\text{ years}$
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{R}=\frac{100\ \times\ \text{I}}{\text{P }\times\text{ T}}$
$=\frac{100\ \times\frac{9}{40}\text{x}}{\text{x } \times\ \frac{5}{2}}$
$=\frac{45}{5}$
$=9\%$
Hence, the correct option is $(d).$

$10$ pipe can fill tank in $= 24$ minutes
$1$ pipe will fill it in $= 24 \times 10$ minutes (less pipe, more time)
and $10 - 2 = 8$ pipes will fill the tank in
$=\frac{24\times10}{8}=30$ minutes.

$1\ m = 100\ cm$
$\therefore 1.5\ m = 150\ cm$
$\text{x}=\frac{9}{150}=\frac{3}{5}$
The ratio is $3 : 5$

$\text{Interest}=\frac{2500\times4\times4}{100}=\text{Rs. }400$

Number of cows that eat as much as $15$ buffaloes $= 21$
Number of cows that eat as much as $1$ buffalo $=\frac{21}{15}$
$\therefore$ Number of cows that eat as much as $35$ buffaloes $=\frac{21}{15}\times35=49$

$SP$ of a football $= Rs. 100$
Gain $= Rs. 15$
Gain $= SP - CP$
$\Rightarrow 15 = 100 - CP$
$\Rightarrow CP = Rs. (100 - 15)$
$\Rightarrow CP = Rs. 85$

It is given that,
Sum $(P) = Rs. 1500$
Rate $(R) = 8\%$
Time $(T) = 3$ years
Simple interest $=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$=\frac{1500\ \times\ 8\ \times\ 3}{100}$
$=\text{Rs. }360$
Therefore, simple interest $(I) = Rs. 360$
Hence, the correct option is $(b).$

No. of rounds of rope $= 140$
Radius of base of cylinder $= 14\ cm$
Radius of second cylinder of cylinder $= 20\ cm$
If radius is $14\ cm$, then rounds of rope are $= 140$
If radius is $1\ cm$, then round $= 140 \times 14$ (less radius more rounds)
and if radius is $20\ cm$, then rounds will
be $=\frac{140\times14}{20}=98$ (more radius less rounds)

$3m : 60cm = 3 \times 100cm : 60cm$
$= 300 : 60 = 5 : 1.$

Let x be the maximum marks Pass marks $= (160 + 20) = 180$
$36\%$ of $x = 180$
$\Rightarrow\Big(\frac{36}{100}\times\text{x}\Big)=180$
$\Rightarrow\Big(\frac{180\times100}{36}\Big)$
$=(5\times100)=500$
Hence, maximum marks $= 500$

$15$ buffaloes can eat as much as $= 21$ cows
$1$ buffalo will eat as much as $=\frac{21}{15}\text{ cows}$
$35$ buffaloes will eat as much as
$=\frac{21\times35}{15}\text{cm}=49\text{ cows}$

Since, $SP$ is more than $CP.$
Therefore, profit $= SP - CP$
$= Rs 250 - Rs. 200$
$= Rs. 50$
Hence, the correct option is $(b).$

Given, cost price of bicycle $= ? 1800$ and profit $=12\%$
As we know,
$\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times100$
$\Rightarrow12=\frac{\text{Profit}}{1800}\times100$
$\Rightarrow\text{Profit}=12\times18=\text{Rs.}216$
$\therefore SP = CP +$ Profit $= 1800 + 216 = 2016.$
Hence, the seling price for bicycle is $Rs. 2016.$

$\frac{\text{x}}{\text{y}}=\frac{3}{4}$
$\text{x}=\frac{3\text{y}}{4}$
$\therefore\frac{7\text{x}+3\text{y}}{7\text{x}-3\text{y}}=\frac{7\frac{3\text{y}}{4}+3\text{y}}{7\frac{3\text{y}}{4}-3\text{y}}$
$=\frac{21\text{y}+12\text{y}}{21\text{y}-12\text{y}}=\frac{33\text{y}}{9\text{y}}=\frac{11}{3}$
Hence, $(7x + 3y) : (7x - 3y) = 11 : 3$
The correct option is $(C).$

A worker makes toy in $\frac{2}{3}\text{hr}=1$
He will make toys in $1 hr =1\times\frac{3}{2}$
and will make toys in $\frac{22}{3}\text{ hrs}=1\times\frac{3}{2}\times\frac{22}{3}$
$= 11$ (more time more toys)

$1$ rupee $= 100$ paise
$5$ rupees $= 500$ paise
$\text{Ratio}:\frac{50}{500}=1:10$

We have $\text{x}=\frac{3}{4}\text{y}$
Now, $\frac{7\text{x}+3\text{y}}{7\text{x}-3\text{y}}$
$=\frac{7\times\frac{3}{4}\text{y}+3\text{y}}{7\times\frac{3}{4}\text{y}-3\text{y}}=\frac{21\text{y}+12\text{y}}{21\text{y}-12\text{y}}$
$\frac{33\text{y}}{9\text{y}}=\frac{11}{3}$

Number of children who like playing football $=\frac{25}{100}\times40=10$

Let the cost price of one pen be Rs. $1.$
Then, $CP$ of $20$ pens $= Rs. 20$
and $SP$ of $20$ pens $= Rs. 15$ ($\because SP$ of $20$ pens $= CP$ of $15$ pens)
Therefore, $CP$ 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 $SP$ be $x.$
$CP = Rs. 950$
Gain $= 6\%$
Therfore, $SP$ is more than $CP.$
Now,
Gain $= 6\%$ of $CP$
$=\frac{6}{100}\times950$
$= 3 \times 19$
$= 57$
Thus, $SP = CP +$ gain
$= Rs. 950 + Rs. 57$
$= Rs. 1007$
Hence, the correct option is $(c).$

$C.P. = Rs. 120$
$S.P. = Rs. 105$
Loss $= C.P. - S.P. = Rs. 120 - 105 = Rs. 15$
$\therefore\text{Loss}\%=\frac{15\times100}{120}$
$=\frac{25}{2}\%$
$=12\frac12\%$