Question

➡️please answer the last(16th) question.
☯️And also please help me by reporting my all questions except this one.
${} \bf \red{ \gamma \: do \: not \: spam \: \gamma }$
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Answer 1

water required each day = 1200 X 150

water available in the tank = 30 X 25 X10 X 1000

no. of days = amount of water

------------------------------

Daily need

30 X 25 X 10 X 1000

--------------------------------

1200 X 150

= 7,500,000

--------------------

180,000

= 41.666

Please mark as brainiest :-)

Answer 2

Given Information:

• Total population of locality = 1,200

• Water required by per head, per day = 150 l

• Dimensions of the tank = 30 m × 25 m × 10 m

Solution:

Here,as the water required by per head daily is 150 l. So, water required by total population would be :

$\sf{ \longrightarrow \: (total \: population \times water \: required \: by \: per \: head) \: litres} \\ \\ \sf{ \longrightarrow \:(1200 \times 150) \: litres } \\ \\ \sf{ \longrightarrow \:180000 \: litres } \: \: \: \green{ \bigstar}$

Now, let us calculate the volume of the tank in order to calculate the days the water of this tank will last.

$\sf{ \longrightarrow \: Volume_{(Tank)} = l \times b \times h \: cubic \: units} \\ \\ \sf{ \longrightarrow \: Volume_{(Tank)} = (30 \times 25 \times 10) \: {m}^{3} } \\ \\ \sf{Volume_{(Tank)} = 7500 \: {m}^{3} }$

Here, we got the volume in metre cube. But, we'll change it into litres as the water required by the population is also in litres.

$\: \: \: \: \: \: \: \underline{ \sf{Conversion \: of \: {m}^{3} \: into \: litres}} \\ \\ \sf{ \longrightarrow \: {1 \: m}^{3} = 1000 \: l} \\ \\ \sf{ \longrightarrow \: {7500 \: m}^{3} =( 7500 \times 1000) \: l} \\ \\ \sf{ \longrightarrow \: {7500 \: m}^{3} = 7500000 \: l} \: \: \: \green{ \bigstar}$

Now, let us assume the number of the day the water of the tanks will last be "x". So,

→ Water required by total population × the number of days = Volume of tank (in litres)

$\sf{ \longrightarrow \: 180000 \: litres \times x = 7500000 \: litres} \\ \\ \sf{ \longrightarrow \: x = \dfrac{7500000 \:}{180000 \: } \: days} \\ \\ \sf{ \longrightarrow \: x = \dfrac{750 \:}{18 \: } \: days} \\ \\ \longrightarrow \: \underline{ \boxed{ \sf{x = 41.66 \: or \: 41 \: days} }} \: \red{ \bigstar}$

Therefore, for 41 days , the water of the tank will last