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 $(d).$

$SP$ of stool $= Rs. 630$
Loss$\% = 10\%$
$\text{C.P.}=\frac{\text{S.P}\times100}{100-\text{loss%}}$
$=\frac{630\times100}{100-10}$
$=\text{Rs. }\frac{630\times100}{90}$
$=\text{Rs. }700$

Given number is multiplied by $100$ and then divided by $100.$

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

$25\%=\frac{25}{100}$
$=\frac{25\div25}{100\div25}$
$=\frac{1}{4}$
Hence, the correct option is $(a).$

Number of students who like to watch $T.V.$
$= 30\%$ of $100$
$=\frac{30}{100}\times100$
$=30$

$27$ days $- 3$ days $= 24$ days
Men $= 500 + 300 = 800$
For $500$ men, provision is sufficient $= 24$ days
For $1$ man, provision will be $= 24 \times 500$ (less man, more days)
and for $500 + 300 = 800$ men provision
will be sufficient $=\frac{24\times500}{800}=15\text{ days}$
(more men, less days)

Weight of $4.5\ m$ rod $= 17.1\ kg$
Wight of $1\ m$ rod $=\frac{17.1}{4.5}\text{kg}$
And wight of $12\ m$ rod
$=\frac{17.1\times12}{4.5}=\frac{171\times12\times100}{100\times45}\text{kg}$
$=\frac{19\times12}{5}=\frac{228}{5}=45.6\text{kg}$

Let the original salary of the man be $x$.
According to the question,
$\text{x}+25\%\ \text{of x}=20000$
$\Rightarrow\text{x}+\frac{25}{100}\times\text{x}=20000$
$\Rightarrow\text{x}+\frac{\text{x}}{4}=20000$
$\Rightarrow\frac{4\text{x}+4}{4}=20000$
$\Rightarrow\frac{5\text{x}}{4}\times4=20000\times4$
$\Rightarrow5\text{x}=80000$
$\Rightarrow\text{x}=\frac{80000}{5}$
$\Rightarrow\text{x}=16000$
Therefore, Original salary of the man is $Rs. 16000.$
Hence, the correct option is $(b).$

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 ($d).$

Times $= 5$ years
Simple interest $=\frac25\text{P}$
$\Rightarrow\frac{\text{P}\times\text{Rate}\times\text{time}}{100}=\frac{2}{5}\text{P}$
$\Rightarrow\frac{\text{Rate}\times5}{100}=\frac25$
$\Rightarrow\text{Rate}=\frac{2\times100}{5\times5}$
$\Rightarrow\text{Rate}=8\%$

Let the number that must be added to each term of the ratio $9 : 16$ be $x.$
Then,
$(9+\text{x}):(16+\text{x})=2:3$
$\Rightarrow\frac{(9+\text{x})}{(16+\text{x})}=\frac{2}{3}$
$\Rightarrow3(9+\text{x})=2(16+\text{x})$
$\Rightarrow27+\text{3x}=32+\text{x}$
$\Rightarrow3\text{x}-2\text{x}=32-27$
$\therefore\text{x}=5$
So, $5$ must be added to each term of the ratio $9 : 16$ to make the ratio $2 : 3.$
Hence, the correct alternative is option $(a).$

Let the $CP$ be $x.$
$SP = Rs. 924$
Gain $= 10\%$
Therfore, $SP$ is more than $CP.$
Now,
Gain$=10\%$ of $CP$ and $SP = CP +$ gain
So, $SP = CP + 10\%$ of $CP$
$\Rightarrow924=\text{x}+\frac{10}{100}\times\text{x}$
$\Rightarrow924=\Big(1+\frac{1}{10}\Big)\text{x}$
$\Rightarrow924=\Big(\frac{11}{10}\Big)\text{x}$
$\Rightarrow\text{x}=\Big(\frac{9240}{11}\Big)$
$\Rightarrow\text{x}=840$
Thus, $CP = Rs. 840$
Hence, the correct option is $(d).$

It is given that,
Sum ($P_1$) $= Rs. 1000$
Rate ($R_1$) $= 12%$
Time ($T_1$) $= 3 years$
$\text{I}_1=\frac{\text{P}_1\ \times\ \text{R}_1\ \times\ \text{T}_1}{100}$
$=\frac{1000\ \times\ 12\ \times\ 3}{100}$
$=\text{Rs. }360\ ...(\text{i})$
Sum ($P_2$​​​​​​​) $= Rs. 1500$
Rate ($R_2$​​​​​​​) $= 8%$
Time ($T_2$) $=1\frac{1}{2}\ \text{year}=\frac{3}{2}\ \text{year}$
$\text{I}_2=\frac{\text{P}_2\ \times\ \text{R}_2\ \times\ \text{T}_2}{100}$
$=\frac{1500\ \times\ 8\ \times\ 3}{100\ \times\ 2}$
$=\text{Rs. }180\ ...(\text{ii})$
Subtracting $(ii)$ from $(i)$, we get
$I_2 - I_1 = Rs. 360 - Rs. 180$
$= Rs. 180$
Hence, the correct option is $(c).$

Let $P$'s share be $Rs. x.$
Then $Q$'s share $= Rs. (840 - x)$
As, $P'$s share : $Q$'s share $= 3 : 4$
$\Rightarrow\frac{\text{P's share}}{\text{Q's share}}=\frac34$
$\Rightarrow\frac{\text{x}}{(840-\text{x})}=\frac{3}{4}$
$\Rightarrow\text{4x}=3(840-\text{x})$
$\Rightarrow\text{4x}=3\times840-\text{3x}$
$\Rightarrow\text{4x}+\text{3x}=3\times840$
$\Rightarrow7\text{x}=3\times840$
$\text{x}=\frac{3\times840}{7}$
$\Rightarrow\text{x}=3\times120$
$\therefore\text{x}=360$
So, $P$'s share is Rs. $360.$
Hence, the correct alternative is option $(c).$

First decimal is removed and then multiplied by $100.$

Let the rate of popukation be $p, q, r$ respectively.
Therefore, population after $3$ years is
$=\text{P}(1+\frac{\text{P}}{100})(1+\frac{\text{q}}{100})(1+\frac{\text{r}}{100})$
$=2500\times(1+\frac{4}{100})(1+\frac{5}{100})(1+\frac{8}{100})$
$=25000\times\frac{26}{25}\times\frac{21}{20}\times\frac{27}{25}$
$=29,484$

$\text{A}=\frac{1}{3}\text{B, B}=\frac{1}{2}\text{C}$
$\frac{\text{A}}{\text{B}}=\frac{1}{3},\frac{\text{B}}{\text{C}}=\frac{1}{2}$
$\Rightarrow\text{A : B}=1:3$
$\text{B : C}=1:2=3:6$
(Multiplying by $3)$
$\therefore\text{A : B : C}=1 : 3 : 6$

Amount that Meenu gets
$= (100\% - 60\%) × 100$
$=\frac{40}{100}\times100=\text{Rs. }40$

$50\%\ \text{of}\ 150=\frac{50}{100}\times150$
$=75$
$70\%\ \text{of}\ 300=\frac{70}{100}\times300$
$=210$
Therefore, $50\%$ of $150 + 70\%$ of $300 = 75 + 210 = 285.$
Hence, the correct option is $(b).$

$3 : 5 = 15 : x$
$\frac{3}{5}=\frac{15}{\text{x}}$
$x = 25$

$100\% - 40\% = 60\%.$

Let the rate be $R \%$
$S.I. = A - P$
$= 720 - 600$
$= Rs. 120$
Time $= 4$ years
$\text{R}=\frac{10\times\text{S.I.}}{\text{P}\times\text{T}}$
$\text{R}=\frac{100\times120}{600\times4}$
$=5$
Rate of interest $= 5\%$
Now,
$\text{R}=(5+2)\%=7\%$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{600\times7\times4}{100}$
$\text{Rs. }168$
Amount $= S.I. + P$
$= 600 + 168$
$= Rs. 768$

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 $(d).$

$2\text{A}=3\text{B}=4\text{C}=\text{x}$
$\therefore\text{A}=\frac{\text{x}}{2},\text{B}=\frac{\text{x}}{3},\text{C}=\frac{\text{x}}{4}$
$\therefore\text{A : B : C}=\frac{\text{x}}{2}:\frac{\text{x}}{3}:\frac{\text{x}}{4}=\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$
$\frac{6\ :\ 4\ :\ 3}{12}=6:4:3$

As, $\frac{1}{\text{a}}:\frac{1}{\text{b}}:\frac{1}{\text{c}}=3:4:5$
$\Rightarrow\frac{1}{\text{a}}:\frac{1}{\text{b}}=3:4$ and $\frac{1}{\text{b}}:\frac{1}{\text{c}}=4:5$
$\Rightarrow\frac{1}{\text{a}}\div\frac{1}{\text{b}}=\frac{3}{4}$ and $\frac{1}{\text{b}}\div\frac{1}{\text{c}}=\frac45$
$\Rightarrow\frac{1}{\text{a}}\times\frac{\text{b}}{1}=\frac{3}{4}$ and $\frac{1}{\text{b}}\times\frac{\text{c}}{1}=\frac{4}{5}$
$\Rightarrow\frac{\text{b}}{\text{a}}=\frac{3}{4}$ and $\frac{\text{c}}{\text{b}}=\frac{4}{5}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{4}{3}$ and $\frac{\text{b}}{\text{c}}=\frac{5}{4}$ (Reciprocal of both sides)
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{4\times5}{3\times5}$ and $\frac{\text{b}}{\text{c}}=\frac{5\times3}{4\times3}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{20}{15}$ and $\frac{\text{b}}{c}=\frac{15}{12}$
$\Rightarrow\text{a}:\text{b}=20:15$ and $\text{b}:\text{c}=15:12$
$\therefore\text{a}:\text{b}:\text{c}=20:15:12$
Hence, the correct alternative is option $(b).$

For bufalo, $CP = Rs. = 44000$
Loss$\% = 5\%$
$\therefore\text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times100\%$
$\Rightarrow5=\frac{\text{Loss}}{44000}\times100$
$\Rightarrow\text{Loss}=5\times440=\text{Rs.}2200$
So, $SP = CP -$ Loss $= 44000 - 2200 = Rs. 41800$
For cow, $CP = Rs. 18000$
Profit$\% = 10\%$
$\therefore\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times100\%$
$\Rightarrow10=\frac{\text{Profit}}{18000}\times100$
Profit $= Rs. 1800$
So, $SP = CP +$ Profit $= 18000 + 1800 = Rs. 19800$
Total $CP$ of buffalo and cow $= 44000 + 18000 = Rs. 62000$
Total $SP$ of buffalo and cow $= 41800 + 19800 = Rs. 61600$
Net loss $= CP - SP = 62000 - 61600 = Rs. 400$

$120 : 600 = 3 : x$
$\Rightarrow\frac{120}{600}=\frac{3}{\text{x}}$
$\Rightarrow\text{x}$

$C.P$. of $12$ pencils
$= S.P$. of $15$ pencils
$=$ Re $1$ (suppose)
$\therefore$ $C.P$ Of $1$ pencil $= \text{Rs. }\frac{1}{12}$
And $S.P$ of $1$ pencil $=\text{Rs. }\frac{1}{15}$
$\therefore$ Loss $= C.P - S.P$
$=\frac{1}{12}-\frac{1}{15}$
$=\frac{5-4}{60}$
$=\text{Rs. }\frac{1}{60}$
$\therefore\text{Loss%}=\frac{\text{Loss}\times100}{\text{C.P.}}$
$=\frac{\frac{1}{60}\times100}{\frac{1}{12}}$
$=\frac{1\times100\times12}{60\times1}$
$=20\%$

$\text{x}\%\ \text{of}\ 2000=600$
$\Rightarrow\frac{\text{x}}{100}\times2000=600$
$\Rightarrow20\text{x}=600$
$\Rightarrow\text{x}=\frac{600}{20}$
$\Rightarrow\text{x}=30$
Hence, the correct option is $(b).$

$\frac{2}{5}=\frac{2}{5}\times100\%$
$=40\%$

Sum of the ratio terms $= 4 + 3 = 7$
B’s share $840\times\frac{3}{7}=360$

One hour $= 60$ minutes
and $1$ minute $= 60$ seconds
So, one hour $= 60 \times 60$ seconds
Now, the ratio of one hour and $60$ seconds $=\frac{\big(60\times60\big)}{60}=\frac{60}{1}=60:1$

We have, $5.2$
In percentage, $\frac{52}{10}\times100\%=520\%$

$2\text{A}=3\text{B}$
$\Rightarrow\frac{\text{A}}{\text{B}}=\frac{3}{2}$
$4\text{B}=5\text{C}$
$\Rightarrow\frac{\text{B}}{\text{C}}=\frac{5}{4}$
Multiplying, We get
$\frac{\text{A}}{\text{B}}\times\frac{\text{B}}{\text{C}}=\frac{3}{2}\times\frac{5}{4}$
$\Rightarrow\frac{\text{A}}{\text{C}}=\frac{15}{8}$
$\therefore\text{A : C}=15:8$

$\%$ of sdefective parts $=\frac{5}{500}\times100=1$
$\%$ of non-defective parts $= 100 - 1 = 99$

Required percentage
$= 100\% – (50\% + 30\%) = 20\%.$

Simple interest = Rs. $x$
Rate $= x \%$ per annum
Time $= x$ years
Simple interest $=\frac{\text{Princepal}\times\text{Rate}\times\text{Time}}{100}$
$\Rightarrow\text{x}=\frac{\text{Princepal}\times\text{x}\times\text{x}}{100}$
$\Rightarrow\text{Principal}=\text{Rs. }\frac{100}{\text{x}}$

By unit method cost of $10$ bowls is calculated.

It is given that,
Sum $(P) = Rs. 3000$
Rate $(R) = 3%$
Time $(T) = 20$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$=\frac{3000\ \times\ 3\ \times\ 20}{100}$
$=\text{Rs. }1800$
Amount $= I + P = Rs. 1800 + Rs. 3000 = Rs. 4800$
Hence, the correct option is $(c).$

$1\ km = 1000\ m$
Now, ratio of 1km and 500m $=\frac{1000}{500}=\frac{2}{1}=2:1$

Amount $= 2$ times the sum $= 2P$
Simple interest $(I) =$ Amount - Sum $= 2P - P = P$
Let the sum $(P)$ be $x.$
Then, simple interest $(I) = x$
Rate $(R) = R\%$
Time $(T) = 10$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{R}=\frac{100\ \times\ \text{I}}{\text{P }\times\text{ T}}$
$=\frac{100\ \times\text{ x}}{\text{x } \times\ 10}$
$=10\%$
Hence, the correct option is $(b).$

Let the Raman’s income be x. it is given that Mohan’s income is $25\%$ more than Raman’s income.
The, Mohan's income $=\text{x}+20\%\text{ of } \text{x}+\frac{25}{100}\text{x}=\text{x}\Big(1+\frac{25}{100}\Big)$
$=\text{x}\Big(\frac{100+25}{100}\Big)=\frac{125}{100}\text{x}$
$\therefore$ Raman's income as compared to Mohan's income.
$=\frac{\text{Mohan's income - Raman's income}}{\text{Mohan's income}}\times100\%$
$\frac{\frac{125}{100}\text{x-x}}{\frac{125}{100}\text{x}}\times100\%=\frac{\frac{125\text{x}-100\text{x}}{100}}{\frac{125}{100}\text{x}}\times100\%$
$=\frac{25\text{x}}{125\text{x}}\times100\%=20\%$

As, $8 : x :: 16 : 35$
$\Rightarrow\frac{\text{8}}{\text{x}}=\frac{16}{35}$
$\Rightarrow\text{16x}=8\times35$ (By cross multiplication)
$\Rightarrow\text{x}=\frac{8\times35}{16}$ (Transposing $16$ to $R.H.S.)$
$\therefore\text{x}=\frac{35}{2}$
Hence, the correct alternative is option $(c).$

Let the sum be Rs. $x$
Rate of interest $= r\%$
Time $=2\frac12$ years $=\frac52$ years
Amount $=\frac65\times\text{Sum}$
Rate $= ?$
Amount $=\frac65\times\text{Sum}$
Principal $+ S.I. = $Amount
$\text{Principal}+\frac{\text{Principal}\times\text{Rate}\times\text{Time}}{100}=\frac65\times\text{Principal}$
$\Rightarrow\text{x}+\frac{\text{x}\times\text{r}\times5}{100\times2}=\frac65\text{x}$
$\Rightarrow\text{x}\Big(1+\frac{5\text{r}}{100\times2}\Big)=\frac65\text{x}$
$\Rightarrow1+\frac{\text{r}}{40}=\frac65$
$\Rightarrow\text{r}=40\times\frac15$
$\Rightarrow\text{r}=8$
So, the rate of interest is $8\%$

Given, $P = Rs. 30000, T = 3$ year, $R = 15%$
We know that, $\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}=\frac{30000\times15\times3}{100}=\text{Rs. 13500}$

$6kg : 400g = 6 \times 1000g : 400g$
$= 6000 : 400 = 15 : 1$

Given cost of $6$ bowls $= ₹ 90$
Cost of $1$ bowl $=\frac{90}{6}=₹\ 15$
Therefore, cost of $10$ such bowls $= 15 \times 10 = ₹ 150$

Given, $I = Rs. 126, R = 14\%$ and $T = 2$year
$\therefore\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}\Rightarrow126=\frac{\text{P}\times14\times2}{100}$
$\Rightarrow126\times100=\text{P}\times14\times2$
$\Rightarrow\text{P}=\frac{12600}{14\times2}\Rightarrow\text{P}=\text{Rs. }450$

$\frac{5}{4}=\frac{5}{4}\times\frac{100}{100}$
$=\frac{5\times25}{100}$
$=\frac{125}{100}$
$=125\%$
Hence, the correct option is $(b).$

As, $a : b = 3 : 4$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac34$
So, $\text{4a}:\text{3b}=\frac{4\text{a}}{3\text{b}}$
$=\frac43\times\frac{\text{a}}{\text{b}}$
$=\frac43\times\frac34$
$=\frac{12}{12}$
$=\frac11$
$=1:1$
Hence, the correct alternative is option $(c).$

$\frac{\text{A}}{3}=\frac{\text{B}}{4}=\frac{\text{C}}{5}=1$ (Suppose)
$\text{A}=3,\text{B}=4,\text{C} =5$
$\text{A : B : C}=3:4:5$

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 $(c).$

Since,
$\frac{1}{12}:\frac{1}{60}$
$=\frac{1}{12}\div\frac{1}{60}$
$=\frac{1}{12}\times\frac{60}{1}$
$=\frac{60}{12}$
$=\frac{5}{1}$
$=5:1$
Hence, the correct alternative is option $(c).$

Number of Girls $= 50 - 40 = 10$
Required ratio $= 40 : 10$
$= 4 : 1$

Percentage of boys $=\frac{10}{40}\times100\%$
$=25\%$

Let the required number be $x.$
$\text{x}\%\ \text{of}\ \frac{2}{7}=\frac{1}{35}$
$\Rightarrow\frac{\text{x}}{100}\times\frac{2}{7}=\frac{1}{35}$
$\Rightarrow\frac{\text{x}}{350}=\frac{1}{35}$
$\Rightarrow\frac{350\times\text{x}}{350}=\frac{1\times350}{35}$
$\Rightarrow\text{x}=\frac{350}{35}$
$\Rightarrow\text{x}=10$
Hence, the correct option is $(d).$

$\%$ of voters who voted yes $=\frac{90}{120}\times100=75\%$

$\frac{5}{2}=\frac{5}{2}\times100\%=250\%$

Let x to be added to each term of $3 : 5$
Then $\frac{3+\text{x}}{5+\text{x}}=\frac{5}{6}$
By cross multiplication
$18 + 6x = 25 + 5x$
$6x - 5x = 25 - 18$
$x = 7$
$7$ is to be added

In $20$ minutes, Raghu covers $= 5\ km$
in $1$ minutes, he will cover $=\frac{5}{10}\text{km}$
and in $50$ minutes, he will cover
$=\frac{5}{10}\times50$
$=\frac{25}{2}=12.5\text{km}$

$A = Rs. 3626$
$R = 6\%$
$T = 219$ days $=\frac{219}{365}\text{ years}$
Let the required sum be Rs. $x$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{\text{x}\times6\times219}{100\times365}$
$=\frac{1314\text{x}}{36500}$
$\text{A}=\text{P}+\text{S.I.}$
$=\text{x}+\frac{1314\text{x}}{36500}$
$=\frac{36500\text{x}+1314\text{x}}{36500}$
$=\frac{1314\text{x}}{36500}$
But the amount is Rs. 3626
$\therefore\frac{37814\text{x}}{36500}=3626$
$\text{x}=\frac{3626\times36500}{37814}=3500$
The required sum is $Rs. 3500$

Total cistern $= 1$
Filled in $1$ minute $=\frac{4}{5}$
Unfilled $=1-\frac{4}{5}=\frac{1}{5}$
$\frac{4}{5}$ cistern is filled in $= 1$ minutes $= 60$ seconds
$1$ full cistern can be filled in $=\frac{60\times5}{4}=75$ seconds
More time $= 75 - 60 = 15$ seconds

Since, $2:5=\frac25=\frac{2\times3}{5\times3}=\frac{6}{15}=6:15$
So, the ratio equivalent to $2 : 5$ is $6 : 15.$
Hence, the correct alternative is option $(a).$

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

$1\ kg$ is converted into grams then multiplied by $75$ and then divided by $100.$

Let the least number that is to be subtracted from each term of the ratio $15 : 19$ be $x.$
Then,
$(15-\text{x}):(19-\text{x})=3:4$
$\Rightarrow4(15-\text{x})=3(19-\text{x})$
$\Rightarrow60-\text{4x}=57-3\text{x}$
$\Rightarrow3\text{x}-4\text{x}=57-3\text{x}$
$\Rightarrow-\text{x}=-3$
$\therefore\text{x}=3$
So, $3$ is the least number to be subtracted from each term of the ratio $15 : 19$ to make the ratio $3 : 4.$
Hence, the correct alternative is option $(a).$

Out of Rs. $100$, money spent $= Rs. 80$
$\therefore$ Out of $Rs. 200,$ money spent
$=\frac{80}{100}\times200$
$= Rs. 160$
or
Money spent $= 80\%$ of $Rs. 200$
$=200\times\frac{80}{100}$
$= Rs. 60$
$\therefore$ Money left $= Rs. 200 - Rs. 160$
$= Rs. 40$

$10\% VAT$ on $Rs. 100$ will make it $Rs. 110$
So, for price including $VAT Rs. 110$, the original price is $Rs. 100$
Then, Price including $VAT Rs. 3300$, the original price
$=\text{Rs.} \big(\frac{100}{110}\times3300\big)$
$=\text{Rs. }3000$

$\text{Principal }=\Big(\frac{\text{I}\times100}{\text{R}\times\text{T}}\Big)$

It is given that,
Simple interest $(I) = Rs. n$
Rate $(R) = R\%$
Time $(T) = n$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{P}=\frac{100\ \times\ \text{I}}{\text{R}\ \times\ \text{T}}$
$\Rightarrow\text{P}=\frac{100\ \times\ \text{n}}{\text{R}\ \times\ \text{n}}$
$\Rightarrow\text{P}=\frac{100}{\text{R}}$
Hence, the correct option is $(c).$

Rate $=3\frac34\%$ per annum
$=\frac{15}{4}\%$ per annum
Time $=2\frac13$ years
$=\frac73$ years
$\text{S.I.}=\frac{\text{P}\times\frac{15}{4}\times\frac{7}{3}}{100}$
$\Rightarrow\text{P}=\frac{210\times100}{\Big(\frac{15}{4}\times\frac{7}{3}\Big)}$
$\Rightarrow\text{P}=600\times4$
$\Rightarrow\text{P}=\text{Rs. }2400$

Let x be subtracted from each of the term
$\frac{15-\text{x}}{19-\text{x}}=\frac{3}{4}$
$\Rightarrow 4 (15 - x) = 3 (19 - x)$
$\Rightarrow 60 - 4x = 57 - 3x$
$\Rightarrow -4x + 3x = 57 - 60$
$\Rightarrow -x = -3$
$x = 3$
Required number $= 3$

As, $24:36=\frac{24}{36}=\frac23=2:3$
So, the simplest from of $24 : 36 = 2 : 3$
Hence, the correct alternative is option $(d).$

Let the fourth term be $x.$
As, $3:5: :21:\text{x}$
$\Rightarrow\frac35=\frac{21}{\text{x}}$
$\Rightarrow3\text{x}=21\times5$
$\Rightarrow\text{x}=\frac{21\times5}{3}$
$\Rightarrow\text{x}=7\times5$
$\therefore\text{x}=35$
So, the fourth term is 35.
Hence, the correct alternative is option $(b).$

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

$644\ km$
$1\ cm$ represents $8 \ km.$
$\therefore$ $80.5\ cm$ represents $8 \times 80.5 = 644\ km.$

Amount $= 3$ times the sum $= 3P$
Simple interest $(I) =$ Amount - Sum $= 3P - P = 2P$
Let the sum $(P)$ be $x.$
Then, simple interest $(I) = 2x$
Rate $(R) = R\%$
Time $(T) = 16$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{R}=\frac{100\ \times\ \text{I}}{\text{P}\ \times\ \text{T}}$
$\Rightarrow\text{R}=\frac{100\ \times\ 2\text{x}}{\text{x}\ \times\ 16}$
$\Rightarrow\text{R}=12.5\%$
Hence, the correct option is $(d).$

We have, $225\%$
In percentage, $225\times\frac{1}{100}=\frac{225}{100}=\frac{9}{4}$
$\therefore$ Required ratio $= 9 : 4$
Note: The sign of percentage $(\%)$ is removed dividing the number by $100.$

None of these $0.8\ cm$ represent the map $= 8.8\ km$
$1\ cm$ will represent on the map $=\frac{8.8}{0.8}\text{km}$
$\therefore$ $80.5\ cm$ will represent on the map
$=\frac{8.8\times80.5}{0.8}=\frac{88\times805\times10}{10\times8\times10}$
$=\frac{8855}{10}=885.5\text{km}$

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

We have, $20\%$ of $700\text{m}=\frac{20}{100}\times700=20\times7=140\text{m}$
Hence, $20\%$ of $700 m$ is $140m.$

Let the sum be $Rs. x$
Let the rate be $R \%$ per annum
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{\text{x}\times\text{R}\times5}{100}$
$=\frac{\text{Rx}}{20}$
$\therefore\frac{3\text{x}}{5}=\frac{\text{Rx}}{20}$
$\Rightarrow\text{R}=\frac{3\text{x}\times20}{5\text{x}}=12\%$

Given, $P = Rs. 12000,$
$R = 10\ %$
$T = 1$ month $=\frac{1}{12}\text{yr}$
$\therefore\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}=\frac{12000\times10\times1}{100\times12}=\text{Rs. }100$

Given, cost of $6$ books $= 90$
So, the cost of $1$ book $=\frac{90}{6}=15$
So, the cost of $8$ books $= 15 \times 8 = Rs. 120$

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

$\frac{2}{5}=\frac{2}{5}\times\frac{100}{100}$
$=\frac{40}{100}$
$=40\%$
Hence, the correct option is $(c).$

Loss $= Rs. 316 - Rs. 300 = Rs. 16.$

Given, ratio of income to savings $= 4 : 1$
Let income $= 4x$ and savings $= x.$
$\therefore$ Percentage of money saved by her $=\frac{\text{Savings}}{(\text{Income+Savimgs)}}\times100\%$
$=-\frac{\text{x}}{4\text{x}+\text{x}}\times100=\frac{\text{x}}{5\text{x}}\times100\%$
$=\frac{1}{5}\times100\%=20\%$

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

$9$ metre in cms are $900\ cm$, ratio of $900\ cm$ to $27$ is $100 : 3.$

Let the time be t years.
Principal $= Rs. 8000$
Amount $= Rs. 8360$
Rate $= 6\%$ per annum
Amount $=\text{Princepal}\Big(1+\frac{\text{Rate}\times\text{Times}}{100}\Big)$
$\frac{8360}{8000}=1+\frac{6\times\text{t}}{100}$
$\Rightarrow\frac{8360}{8000}-1=\frac{\text{6t}}{100}$
$\Rightarrow\text{t}=\Big(\frac{8360-8000}{8000}\Big)\times\frac{100}{6}$
$=\frac{360}{8000}\times\frac{100}{6}$
$=\frac68\times12\text{ months}$
$=9\text{ months}$

Given that $2cm = 1000km$
$2cm \rightarrow 1000km$
$2.5cm \rightarrow xkm$
By unitary method,
$\text{x}=\frac{2.5\times1000}{2}=1250\text{km}$

Number of cows that graze the field in $12$ days $= 45$
Number of cows that graze the field in $1$ day $= 45 \times 12$
$\therefore$ Number of cows that graze the field in $9$ days $=\frac{45\times12}{9}=60$

$7. 7 : x = 42 : 96$
$\Rightarrow\frac{7}{\text{x}}=\frac{42}{96}$
$\Rightarrow x = 16$

As, $\text{2a}=\text{3b}=4\text{c}$
$\Rightarrow\text{2a}=\text{3b}$ and $3\text{b}=\text{4c}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac32$ and $\frac{\text{b}}{\text{c}}=\frac43$
$\Rightarrow\text{a}:\text{b}=6:4$ and $\text{b}:\text{c}=4:3$
$\therefore\text{a}:\text{b}:\text{c}=6:4:3$
Hence, the correct alternative is option $(d).$

Let x be the number of bosy.
Then, $8 : 5 = x : 160$
$\Rightarrow\frac{8}{5}=\frac{\text{x}}{160}$
$\Rightarrow\text{x}=\frac{8\times160}{5}=256$
$\therefore$ Total strength of the school $= 256 + 160 = 416$

$55\%\ \text{of}\ 1000=\frac{55}{100}\times1000$
$=550$
$60\%\ \text{of}\ 2000=\frac{60}{100}\times2000$
$=1200$
Therefore, $55\%$ of $1000 ÷ 60\%$ of $2000$
$=\frac{550}{1200}=\frac{550\div50}{1200\div20}=\frac{11}{24}$
Hence, the correct option is $(a).$

$60\%\ \text{of}\ 200=\frac{60}{100}\times200$
$=120$
$30\%\ \text{of}\ 60\%\ \text{of}\ 200=\frac{30}{100}\times200$
$=36$
Hence, the correct option is $(d).$

Given number is multiplied by $100$ and then divided by $100.$

$SP$ of chair $= Rs. 720$
Gain$\% = 20\%$
$\therefore\text{C.P.}=\frac{\text{S.P}\times100}{100+\text{gain%}}$
$=\text{Rs. }\frac{120\times100}{100+20}$
$=\text{Rs. }\frac{720\times100}{120}$
$=\text{Rs. }600$