An empty tank was $\dfrac{2}{5}$ full after 1.5 liters of water was poured into it. The base area of the tank was $250\;c{m^2}$. What was the height of the tank?
Answer
649.5k+ views
Hint: This is a problem of mensuration and requires the concept of volumes of 3-D objects. We will also require the conversion of liters into square centimeters. The general form of the volume of any 3-D object is given by-
${\text{V}} = {\text{A}} \times {\text{h}}, \text{where A is the area of base and h is the height} $...........(1)
Complete step-by-step solution -
We have been given that the tank is two-fifth full when 1.5 litres of water is poured. This means that 1.5 litres is the two-fifths of the total volume of the tank. Let the volume of the tank be V. Mathematically-
$\dfrac{2}{5} \times {\text{V}} = 1.5$
${\text{V}} = 1.5 \times \dfrac{5}{2} = 3.75\;litres$
We have been given the base area of the tank. We also know that-
$1\;litre\; = \;1\;d{m^3}\;so,\;$
$1\;litre\; = \;\left( {10 \times 10 \times 10} \right)c{m^3}$
$1\;litre\; = \;1000\;c{m^3}$
So, we can write the value of V as-
${\text{V}} = 3.75\;litres$
${\text{V}} = 3.75 \times 1000\;c{m^3}$
${\text{V}} = 3750\;c{m^3}$
Now, to find the height of the tank, we will apply formula (1) as-
${\text{V}} = {\text{A}} \times {\text{h}}$
$3750 = 250{\text{h}}$
${\text{h}} = \dfrac{{3750}}{{250}} = 15\;cm$
This is the required height of the tank.
Note: The students often become confused in such questions because the type of the container is not given, that is, cylindrical, cuboidal, cubic, and so on. But here, we do not require the type of container because we can apply the general formula for volume which is the area of base times the height. Also, we should remember the formula of conversion of units of volume, and never forget to write the units in the final answer.
${\text{V}} = {\text{A}} \times {\text{h}}, \text{where A is the area of base and h is the height} $...........(1)
Complete step-by-step solution -
We have been given that the tank is two-fifth full when 1.5 litres of water is poured. This means that 1.5 litres is the two-fifths of the total volume of the tank. Let the volume of the tank be V. Mathematically-
$\dfrac{2}{5} \times {\text{V}} = 1.5$
${\text{V}} = 1.5 \times \dfrac{5}{2} = 3.75\;litres$
We have been given the base area of the tank. We also know that-
$1\;litre\; = \;1\;d{m^3}\;so,\;$
$1\;litre\; = \;\left( {10 \times 10 \times 10} \right)c{m^3}$
$1\;litre\; = \;1000\;c{m^3}$
So, we can write the value of V as-
${\text{V}} = 3.75\;litres$
${\text{V}} = 3.75 \times 1000\;c{m^3}$
${\text{V}} = 3750\;c{m^3}$
Now, to find the height of the tank, we will apply formula (1) as-
${\text{V}} = {\text{A}} \times {\text{h}}$
$3750 = 250{\text{h}}$
${\text{h}} = \dfrac{{3750}}{{250}} = 15\;cm$
This is the required height of the tank.
Note: The students often become confused in such questions because the type of the container is not given, that is, cylindrical, cuboidal, cubic, and so on. But here, we do not require the type of container because we can apply the general formula for volume which is the area of base times the height. Also, we should remember the formula of conversion of units of volume, and never forget to write the units in the final answer.
Recently Updated Pages
Vineet deposited Rs 15600 in a fixed deposit at simple class 10 maths CBSE

Puneet prepared two posters on National Integration class 10 maths CBSE

Acetyleneethyne burns in oxygen to give carbon dioxide class 10 chemistry CBSE

Sita sells a dining set to Neeta for Rs 6000 and gains class 10 maths CBSE

Match columnI with columnII and choose the correct class 12 biology NEET_UG

Match columnI with columnII and choose the correct class 12 biology NEET_UG

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

10 examples of evaporation in daily life with explanations

Cricket: What's a batter not out at innings end called?

What is the full form of POSCO class 10 social science CBSE

Define Potential, Developed, Stock and Reserved resources

