
A certain concrete mix needs cement, fine aggregate, and coarse aggregate in the ratio $1:2:4$ . If the contractor has 25 kg of fine aggregate. How much quantity of the cement and that of the coarse aggregate is required to prepare the mix?
Answer
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Hint: In this question, we have to find the quantity of the cement and the coarse aggregate. Thus, we will solve this problem using ratios. As we know, the ratio is the quantitative relation between the amounts showing the number of times one value contains within the other. Thus, we start solving this problem by letting the common ratio be x, then we will get the ratio in terms of x. Then, we will solve for x through the amount of fine aggregate. In the last, we will find the quantity of cement and coarse aggregate, which is the required solution.
Complete step by step solution:
According to the problem, we have to find the value of cement and coarse aggregate.
Thus, we will use the ratios to get the solution.
First, we will let the value of the common ratio be equal to x.
Thus, the ratio of cement, fine aggregate, and coarse aggregate is $1x:2x:4x$ ------- (1)
Now, as per the problem, the contractor has 25kg of fine aggregate, thus, from equation (1), we get
$2x=25$
Now, we will divide 2 on both sides in the above equation, we get
$\dfrac{2}{2}x=\dfrac{25}{2}$
On further simplification, we get
$x=\dfrac{25}{2}$ ------- (2)
Thus, the value of cement from equation (1) and (2) is
$1x=\dfrac{25}{2}$
Thus, the amount of cement is $\dfrac{25}{2}$ .
Thus, the value of coarse aggregate from equation (1) and (2) is
$4x=\dfrac{25}{2}$
Now, we will divide 4 on both sides in the above equation, we get
$\dfrac{4}{4}x=\dfrac{25}{2.(4)}$
Therefore, we get
$x=\dfrac{25}{8}$
Thus, the amount of coarse aggregate is $\dfrac{25}{8}$ .
Therefore, for the mix, the amount of cement be $\dfrac{25}{2}kg$ and the amount of coarse aggregate is $\dfrac{25}{8}kg$ .
Note: While solving this problem, do mention all the steps properly to avoid an error. Do not forget to put x in the ratio, because it is the common multiple of the amount of the mix. At the end of your solution, do mention the unit of the cement and coarse aggregate that is kg to get an accurate answer.
Complete step by step solution:
According to the problem, we have to find the value of cement and coarse aggregate.
Thus, we will use the ratios to get the solution.
First, we will let the value of the common ratio be equal to x.
Thus, the ratio of cement, fine aggregate, and coarse aggregate is $1x:2x:4x$ ------- (1)
Now, as per the problem, the contractor has 25kg of fine aggregate, thus, from equation (1), we get
$2x=25$
Now, we will divide 2 on both sides in the above equation, we get
$\dfrac{2}{2}x=\dfrac{25}{2}$
On further simplification, we get
$x=\dfrac{25}{2}$ ------- (2)
Thus, the value of cement from equation (1) and (2) is
$1x=\dfrac{25}{2}$
Thus, the amount of cement is $\dfrac{25}{2}$ .
Thus, the value of coarse aggregate from equation (1) and (2) is
$4x=\dfrac{25}{2}$
Now, we will divide 4 on both sides in the above equation, we get
$\dfrac{4}{4}x=\dfrac{25}{2.(4)}$
Therefore, we get
$x=\dfrac{25}{8}$
Thus, the amount of coarse aggregate is $\dfrac{25}{8}$ .
Therefore, for the mix, the amount of cement be $\dfrac{25}{2}kg$ and the amount of coarse aggregate is $\dfrac{25}{8}kg$ .
Note: While solving this problem, do mention all the steps properly to avoid an error. Do not forget to put x in the ratio, because it is the common multiple of the amount of the mix. At the end of your solution, do mention the unit of the cement and coarse aggregate that is kg to get an accurate answer.
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