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

$55\%\ \text{of}\ 1000=\frac{55}{100}\times1000$
$=550$
$60\%\ \text{of}\ 2000=\frac{60}{100}\times2000$
$=1200$
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$\% = 20\%$
$\therefore\text{C.P.}=\frac{\text{S.P}\times100}{100+\text{gain%}}$
$=\text{Rs. }\frac{120\times100}{100+20}$
$=\text{Rs. }\frac{720\times100}{120}$
$=\text{Rs. }600$

Let x be the third proportional to $9$ and $12$ then
$9 : 12 :: x : 12$
$\Rightarrow 9 \times x = 12 \times 12$
$\Rightarrow\text{x}=\frac{12\times12}{9}=\frac{144}{9}=16$
Third proportional $= 16$

$\text{a : b}=3:4,\text{b : c}=8:9$
Or $\frac{\text{a}}{\text{b}}=\frac{3}{4}$ and $\frac{\text{b}}{\text{c}}=\frac{8}{9}$
Multiplying,
$\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{c}}=\frac{3}{4}\times\frac{8}{9}$
$\Rightarrow\frac{\text{a}}{\text{c}}=\frac{2}{3}$
$\therefore\text{a : c}=2:3$

Let the cost price of the jug be $Rs. x$
$\text{Loss}=\text{Rs. }\frac{\text{x}}{7}$
SP of the jug $= CP$ - Loss
$=\text{Rs. }\Big(\text{x}-\frac{\text{x}}{7}\Big) $
$=\text{Rs. }\frac{6}{7}\text{x}$
SP of the jug = Rs. 168 (given)
$\therefore\frac{6}{7}\text{x}=168$
$\Rightarrow\frac{168\times7}{6}$
$\Rightarrow\text{x}=\frac{1176}{6}$
$\Rightarrow\text{x}=196$
$CP$ of the jug,$= Rs. 196$

No. of men in the beginning $= 500$
More men arrived $= 300$
No. of total men $= 500 + 300 = 800$
For $500$ men, provision are for $= 24$ days
For $1$ man, provision will be $= 24 \times 500$ days (less men, more days)
and for $800$ men, provision will be $=\frac{24}{800}\times500\text{ days}$
(more men less days)
$= 15$ days

$\%$ of students absent $=\frac{8}{32}\times100=25\%$
$\%$ of students present $=100-25=75\%$

Let $x$ be the mean proportional of $6$ and $24.$
Then,
$\frac{\text{6}}{\text{x}}=\frac{\text{x}}{24}$
$\Rightarrow\text{x}^2=6\times24$ (By cross multiplication)
$\Rightarrow\text{x}^2=144$
$\therefore\text{x}=12$
So, the mean proportional of $6$ and $24$ is $12.$
Hence, the correct alternative is option $(b).$

Let sum $= P$ and Interest rate $= R$
Now Simple Interest $= 2P$
Simple Interest $=\frac{\big(\text{sum}\times\text{interest rate}\times\text{time}\big)}{100}$
$\Rightarrow2\text{P}=\frac{(\text{P}\times16\times\text{R})}{100}$
$\Rightarrow2=\frac{16\text{R}}{100}$
$\Rightarrow16\text{R}\times2$
$\Rightarrow16\text{R}=200$
$\Rightarrow\text{R}=\frac{200}{16}$
$\Rightarrow\text{R}=\frac{25}{2}$ $(200$ and $16$ is divided by $16)$
$\Rightarrow\text{R}=12.5\%$

$16$ men can reap a field in $= 30$ days
$1$ man will reap the field in $= 30 \times 16$ days
and $20$ men will reap the field in $=\frac{30\times16}{20}=24\text{ days}$

$16$ men can reap the field in $30$ days.
$1$ man can reap the field in $30 \times 16$ days. [Less men, more days]
$\therefore 20$ men can reap the field in $\frac{30\times16}{20}=24\text{ days}$ [More men, less days]

Given, ratio $= 3 : 8$
In percentage, $\frac{3}{8}\times100\%=37.5\%$

The boys and girls in a school are in the ratio of $8 : 5$.
If the number of girls is $160$, what is the total strength of the school?
Given ratio of boys and girls is $8 : 5$
Let number of boys $= 8x$
and number of girls $= 5x$
Now given number of girls $= 160$
$\Rightarrow 5x = 160$
$\Rightarrow\text{x}=\frac{160}{5}$
$\Rightarrow\text{x}=32$
Number of boys $= 8x = 8 \times 32 = 256$
So total strength $= 256 + 160 = 416$

To convert a fraction into percent, we multiply it by $100$
Ex: to convert $\frac{2}{5}$ in percent, we have
$\big(\frac{2}{5}\big)\times100=\frac{\big(2\times100\big)}{5}=\frac{200}{5}=40\%$

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

$1\ km = 1000\ m$
$\therefore 3\ km = 3000\ m$
$\frac{3000}{300}=10:1$
So, it is $10 : 1$

$25\%\ \text{of x}=100$
$\Rightarrow\frac{25}{100}\times\text{x}=100$
$\Rightarrow\frac{\text{x}}{4}=100$
$\Rightarrow\frac{4\times\text{x}}{4}=4\times100$
$\Rightarrow\text{x}=400$
Hence, the correct option is $(d).$

$3$ computers for every $6$ students.
That means $1$ computer for $2$ students.
Then, for $24$ students need computers $=\frac{24}{2}=12$

$8. 4 : 9 = 40 : x$
$\Rightarrow\frac{4}{9}=\frac{40}{\text{x}}$
$\Rightarrow\text{x}=90$

Let the sum be Rs. $P$
Times $= x$ years
Rate of interest
$r = x \%$ per annum
$S.I. = Rs. x$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\text{P}=\frac{\text{x}\times\text{100}}{\text{x}\times\text{x}}$
$\text{P}=\text{Rs. }\frac{100}{\text{x}}$

Money paid as house rent $= 15\%$ of $160000=\frac{15}{100}\times160000=\text{Rs. }24000$
Remaining amount $= 160000 - 24000 = Rs 136000$
Money spent on child’s education $= 10\%$ of $136000=\frac{10}{100}\times136000=\text{Rs. }13600$
Money left with her $= 136000 - 13600= Rs 122400$

Let the present ages of Renu and Ravi be $5x$ and $6x.$
As, the sum of their present ages $= 44$ years
$\Rightarrow5\text{x}+6\text{x}=44$
$\Rightarrow11\text{x}=44$
$\Rightarrow\text{x}=\frac{44}{11}$
$\therefore\text{x}=4$
Now, the present age of Renu $= 5 \times 4 = 20$ years and
the present ages of Ravi $= 6 \times 4 = 24$ years
So, the difference of their ages $= 24 - 20 = 4$ years.
Hence, the correct alternative is option $(a).$

$12$ workers take to complete the work $= 4$ hrs.
$1$ worker will take $= 4 \times 12$ hrs. (less worker, more time)
$15$ workers will take $=\frac{4\times12}{15}$ hrs. (more workers, less time)
$=\frac{16}{5}\text{ hr}=3\text{ hrs. }12\text{ min} $

$40\%\ \text{of}\ 150=\frac{40}{100}\times150$
$=60$
$60\%\ \text{of}\ 150=\frac{60}{100}\times150$
$=90$
Therefore, $40\%$ of $150 + 60\%$ of $150 = 60 + 90 = 150.$
Hence, the correct option is $(b).$

$\frac{2}{2}=\frac{2}{2}\times100\%=100\%$

As, $\text{p}:\text{q}=2:5$
$\Rightarrow\frac{\text{p}}{\text{q}}=\frac{2}{5}$
Let $\text{p}=\text{2x}$ and $\text{q}=5\text{x}$
Now, $\frac{25\text{p}+14\text{q}}{5\text{p}+\text{7q}}$
$=\frac{25\times2\text{x}+14\times5\text{x}}{5\times2\text{x}+7\times5\text{x}}$
$=\frac{50\text{x}+70\text{x}}{10\text{x}+35\text{x}}$
$=\frac{120\text{x}}{45\text{x}}$
$=\frac83$
$=8:3$
Hence, the correct alternative is option $(c).$

Let age of $A = 3x$
And age of $B = 8x$
$6$ years hence, their ages will be $3x + 6$ and $8x + 6$
$\frac{3\text{x}+6}{8\text{x}+6}=\frac{4}{9}$
$\Rightarrow 9(3x + 6) = 4(8x + 6)$
$\Rightarrow 27x + 54 = 32x + 24$
$\Rightarrow 32x - 27x = 54 - 24$
$\Rightarrow 5x = 30$
$\Rightarrow x = 6$
A’s age $= 3x = 3 \times 6 = 18$ years

$0.05=\frac{5}{100}$
$=\big(\frac{5}{100}\big)\times100$
$=\big(\frac{5\times100}{100}\big)$
$=5\%$