Maths MCQ

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 $=\frac{6}{4}\times50=75\text{m}$

$27$ 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 $4.5m$ rod $= 17.1kg$
Wight of $1m$ rod $=\frac{17.1}{4.5}\text{kg}$
And wight of $12m$ rod
$=\frac{17.1\times12}{4.5}=\frac{171\times12\times100}{100\times45}\text{kg}$
$=\frac{19\times12}{5}=\frac{228}{5}=45.6\text{kg}$

Cost of $72$ eggs $= Rs 108$
Cost of $1$ egg $=\text{Rs. }\frac{108}{72}$
Cost of $108$ eggs $=\text{Rs. }\frac{108\times108}{72}=\text{Rs. }162$

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

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

A wall is constructed in $8$ days by $= 10$ men
It will be constructed in $1$ day by $= 10 \times 8 men ($less time, more men$) 10 \times 8$
And it will be constructed in $\frac{1}{2}=\frac{10\times8}{\frac{1}{2}}$
$=\frac{10\times8\times2}{1}=160\text{ men}$
More men required $= 160 - 10 = 150$

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

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
$=\frac{8.8\times80.5}{0.8}=\frac{88\times805\times10}{10\times8\times10}$
$=\frac{8855}{10}=885.5\text{km}$

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 $=\frac{45\times12}{9}=60$

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

$16$ 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$]$

$8$ men can finish the work in $= 40$ days
$1$ man will finish it in $= 40 \times 8$ days $($less men, more days$)$
$8 + 2 = 10$ men will finish it in $=\frac{40\times8}{10}\text{ days}$
$($more men, less days$)$
$= 32$ days.

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 $=\frac{21}{15}\times35=49$

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$)$

$15$ 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
$=\frac{21\times35}{15}\text{cm}=49\text{ cows}$