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2) If three metallic spheres of radii $6$ cms, $8$ cms and $10$ cms are melted to from a single sphere, the diameter of the new sphere will be:

3) Sixty men can build a wall in $40$ days, but though they begin the work together, $5$ men quit every ten days. The time needed to build the wall is:

4) Shyam is travelling on his cycle and has calculated to reach point 'A' at $2$ PM If he travels at $10$ kmph. he will reach there at $12$ noon if he travels at $15$ kmph. At what speed must he travel to reach point 'A' at $1$ PM:

5) An aeroplane travelling at $700$ kmph in level flight drops a bomb from a height of $1000$ metres on a target. The time taken from releasing the bomb to hitting the target is nearest to the figure:

6) In a History examination, the average for the entire class was $80$ marks. If $10\%$ of the students scored $95$ marks and $20\%$ scored $90$ marks. What was the average marks of the remaining students of the class

$10×95=950\\

20×90=1800$

$950+1800+70x=80×100\\

x=75$

$d=(t-2)×15$

$t=6$ hrs

$d=60$

$60=5×s$

speed $=12$ km/hr

$60×10+55×10+50×10+45×10=2100$ man days

$\dfrac{300}{40}=7.5$ days

Total days $=10+10+10+10+7.5=47.5$ days

$r×r×r=1728$

$r=12$ cm

$12×12\left(\dfrac{4}{3}×12×\dfrac{1}{2}+h\right)=3312\\

\dfrac{4}{3}×12×\dfrac{1}{2}+h=23\\

8+h=23\\

h=23-8=15\text{ m}$

$s=ut+0.5at^2$

here $s=1000$

$u=0$ (initial speed)

$a=g=9.8$ m/s

^{2}

So time can be calculated as $14.14$ sec

__Answer for Q2:__

If three metallic spheres melted down and formed a new sphere, volume of all three sphere will be equivalent to the volume of new sphere and final equation will be

$(r_1)^3+(r_2)^3+(r_3)^3=R^3$

here,

$6^3+8^3+10^3=R^3\\

\Rightarrow 216+512+1000=R^3\\

\Rightarrow R=1728^{1/3}\\

\Rightarrow R=12\text{ cm}$

Hence, diameter of the new sphere will be $(2R)$

i.e $24$ cm.

$s_1×t_1=s_2×t_2\quad$($s_1=10$ kmph, $s_2=15$ kmph)

$10(14-t)=15(12-t)\quad$ ($2$ o'clock on $24$ hrs clock is $14,$ while $12$ is $12$)

$140-10t=180-15t\quad$ ($t$ is the start time)

$5t=40$

$t=8$ hrs, means he had started journey at $8$ am

$d=s_1(14-t)\\

d=10(14-8)\\

d=60\text{ km}$

Time to reach destination $=13-8=5$ hrs

Speed $=\dfrac{d}{t}=\dfrac{60}{5}=12$ kmph

$x=\dfrac{d}{10}~~\cdots(1)$

Now, $(x-2)=\dfrac{d}{15}$

$\Rightarrow 15x-30=d\\

\Rightarrow 15x=30+d\\

\Rightarrow x=\dfrac{d+30}{15}~~\cdots(2)$

From $(1)$ and $(2)$, we have

$\dfrac{d+30}{15}=\dfrac{d}{10}\\

\Rightarrow 10d+300=15d\\

\Rightarrow 5d=300\\

\Rightarrow d=60\text{ km}$

Now,

$x-1=\dfrac{d}{10}-1\\

x-1=\dfrac{60}{10}-1=5$

[We are talking about $1$ p.m and we have chosen the time duration of up to to $2$ p.m to be $x$, so up to $1$ p.m,it is $(x-1)$ hours]

So,the time taken up to $1$ p.m is $5$ hours.

Therefore, speed = distance/time $=\dfrac{60}{5}$ km/hr

$=12$ km/hr

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