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

$\text{x}\%\ \text{of}\ 2000=600$
$\Rightarrow\frac{\text{x}}{100}\times2000=600$
$\Rightarrow20\text{x}=600$
$\Rightarrow\text{x}=\frac{600}{20}$
$\Rightarrow\text{x}=30$
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 $100.$