Fill in the blanks.

Question 11 Mark

If the capacity of a cylindrical tank is $1848 m^3$ and the diameter of its base is 14 m , then the depth of the tank is _________________.

Answer

12 m , because
Let the depth of the tank be $h$ metres. Then,
$
\begin{aligned}
\pi \times(7)^2 \times h & =1848 \\

\end{aligned}
$
$\Rightarrow h =1848 \times \frac{7}{22} \times \frac{1}{7 \times 7}=12 m$

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Question 21 Mark

Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new cube, then the surface area of the cube so formed is _________________ .

Answer

$486 cm^2$, because
Volume of cube $=$ volume of cube with side $1 cm+$ volume of cube with side 6 cm + volume of cube with side 8 cm
$=1^3+6^3+8^3=729 cm^2$
Edge of new cube $=\sqrt[3]{729}=9 cm$
$\therefore$ Surface area of the new cube $=6 \times(9)^2=486 cm^2$
$\left[\because\right.$ Surface area of cube $\left.=6 \times(\text { side })^2\right]$

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Question 31 Mark

If the volume of a wall 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu . metres, then the breadth of the wall is _________________ .

Answer

40 cm , because
Let the breadth of the wall be $x$ metres
then, $\quad$ Height $=5 x$ metres
and Length $=8 \times(5 x)=40 x$ metres
$
\therefore \quad x \times 5 x \times 40 x=12.8
$
$
\Rightarrow \quad x^3=\frac{12.8}{200}=\frac{128}{2000}=\frac{64}{1000}
$
So,$
x=\frac{4}{10} m=\left(\frac{4}{10} \times 100\right) cm=40 cm
$

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Question 41 Mark

The length of the longest pole that can be placed in a room 12 m long 8 m broad and 9 m high is _________________ .

Answer

17 m , because
Length of longest pole $=$ length of the diagonal of the room
$\begin{array}{ll} & =\sqrt{l^2+b^2+h^2} \\ \text { Here, } \quad l & =12 m, b=8 m, h=9 m\end{array}$
$\begin{aligned} \therefore \text { Length of longest pole } & =\sqrt{(12)^2+(8)^2+(9)^2} \\ & =\sqrt{289}=17 m .\end{aligned}$

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Question 51 Mark

The time taken by a boy to run around a square field of sides 35 metres, if he runs at the rate of 9 $km / hr$ is _________________ sec.

Answer

56 seconds, because
$
\begin{aligned}
\text { speed } & =9 km / hr \\
& =\left(9 \times \frac{5}{18}\right) m / sec \\
& =\frac{5}{2} m / sec
\end{aligned}
$
Distance travelled by a boy
$
\begin{array}{l}
\qquad=35 \times 4=140 m \\
{[\text { Perimeter of square }=4 \times \text { side }=4 \times 35=140 m]}
\end{array}
$
$\begin{aligned} \therefore \quad \text { Time taken } & =\frac{\text { Distance }}{\text { speed }}=\frac{140}{5 / 2} \\ & =56 \text { seconds } .\end{aligned}$

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Question 61 Mark

A man is walking at the rate of $5 km / hr$ crosses a bridge in 15 minutes. Then, length of the bridge in metres is _________________ .

Answer

1250 metres, because
$
\begin{aligned}
\text { Speed } & =\left(5 \times \frac{5}{18}\right) m / sec \\
& =\frac{25}{18} m / sec
\end{aligned}
$
Distance covered in 15 minutes
$
\begin{aligned}
= & \left(\frac{25}{18} \times 15 \times 60\right) m \\
= & 1250 m \\
& {\left[\text { Using, Speed }=\frac{\text { Distance }}{\text { Time }}\right] }
\end{aligned}
$

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Question 71 Mark

A tyre has two punctures. The first puncture alone would have made the tyre falt in 9 minutes and second alone would have done it in 6 minutes. If air leaks out at a constant rate, then after _________________ minutes both the punctures together make the tyre falt?

Answer

$3 \frac{3}{5}$ minutes, because
1 minute's work of both the punctures
$
=\frac{1}{9}+\frac{1}{6}=\frac{5}{18}
$
So, both the punctures will make the tyre falt in
$
\frac{18}{5}=3 \frac{3}{5} min
$

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Question 81 Mark

'A' can finish a work in 18 days ' $B$ ' can do the same work in half the time taken by 'A' then, working together, _________________ part of the same work they can finish in a day.

Answer

$\frac{1}{6}$ th part, because
$
\begin{aligned}
\text { A's } 1 \text { day's work } & =\frac{1}{18} \\
B^{\prime} \text { 's } 1 \text { day's work } & =\frac{1}{9} \\
\therefore \quad(A+B) \text { 's } 1 \text { work } & =\left(\frac{1}{18}+\frac{1}{9}\right)=\frac{1}{6}
\end{aligned}
$
Hence, $A$ and $B$ together finish $\frac{1}{6}$ th part of the work in a day.

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Question 91 Mark

If there are $x$ weeks $x$ days, then total number of days are _________________ .

Answer

$8 x$ days, because
$x$ weeks $x$ days $=(7 x+x)$ days $=8 x$ days.

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Question 101 Mark

The hands of the clock are straight, _________________ times in a day.

Answer

44 times, because
In 12 hours, the hands coincide or are in opposite direction 22 times.
$\therefore$ In 24 hours, the hands coincide or are in opposite direction 44 times a day.

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Question 111 Mark

The hands of a clock coincide _________________ times in a day.

Answer

22 times, because
The hands of a clock coincides 11 times in every 12 hours (since between 11 and 1 , they coincide only once, i.e., at 12 o'clock).
$\therefore$ The hands coincide 22 times in a day.

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Question 121 Mark

The reflex angle between the hands of a clock at 10.25 is _________________ .

Answer

$197 \frac{1^{\circ}}{2}$, because.
$10.25 min=10+\frac{25}{60}=\frac{125}{12}$ hours$\therefore$
Angle traced by hour hand in$\frac{125}{12} hrs$
$
\begin{array}{l}
=\left(\frac{360}{12} \times \frac{125}{12}\right)^{\circ} \\
=312 \frac{1^{\circ}}{2}
\end{array}
$
Angle traced by minute hand in 25 min .
$
\begin{array}{l}
=\left(\frac{360}{60} \times 25^{\circ}\right) \\
=150^{\circ}
\end{array}
$
$\therefore \quad$ Reflex angle $=360^{\circ}-\left(312 \frac{1}{2}-150^{\circ}\right)$
$\begin{array}{l}=360^{\circ}-162 \frac{1^{\circ}}{2} \\ =197 \frac{1^{\circ}}{2} .\end{array}$

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Question 131 Mark

A batsman makes a score of 87 runs in the 17 th inning and thus increases his average by 3 . His average after the 17 th inning is _________________.

Answer

39 , because
Let the average after 17th inning $=x$
Then, average after 16th inning $=(x-3)$
$
\begin{aligned}
\therefore 16(x-3)+87 =17 x \\
\Rightarrow x =87-48=39 .
\end{aligned}
$

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Question 141 Mark

The average of four consecutive even numbers is 27, then the largest number is _________________ .

Answer

30, because
Let the numbers be $x, x+2, x+4$, and $x+6$. Then,
$\frac{x+(x+2)+(x+4)+(x+6)}{4}=27$(given)
$\begin{array}{lc}\Rightarrow \frac{4 x+12}{4}=27 \\ \Rightarrow x+3=27 \\ \Rightarrow x=24 \\ \therefore \text { Largest number }=(x+6) & =24+6=30 .\end{array}$

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Question 151 Mark

The average of first 40 natural numbers is _________________.

Answer

20.5, because
$
\begin{array}{l}
\text { Sum of first } n \text { natural numbers }=\frac{n(n+1)}{2} \\
\text { So, sum of first } 40 \text { natural numbers }=\frac{40 \times 41}{2}=820 \\
\therefore \quad \text { Required average }=\frac{820}{40}=20.5
\end{array}
$

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Question 161 Mark

The average of first nine prime numbers is _________________.

Answer

$11 \frac{1}{9}$, because
First nine prime numbers are :
$
\begin{array}{l}
2,3,5,7,11,13,17,19,23 \\
\therefore \text { Average }=\frac{2+3+5+7+11+13+17+19+23}{9} \\
=\frac{100}{9} \\
=11 \frac{1}{9} .
\end{array}
$

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