Question

How do you solve $\dfrac{350}{p}=\dfrac{1050}{60}$ ?

Answer 1

Hint: We have to solve the value of p given in the above equation. To solve the value of p, we are going to rearrange the given equation in such a way so that on one side of the equation we get p and on the other side we have constant terms.

Complete step-by-step answer:
The equation given in the above problem is as follows:
$\dfrac{350}{p}=\dfrac{1050}{60}$
To solve the above equation means we are asked to find the value of “p” in the above equation which we are going to find by rearranging the equation in such a manner so that “p” must be on one side of the equation and the constant term on the other side of the equation.
Cross multiplying the above equation we get,
$\Rightarrow 350\times 60=1050p$
Multiplying the left hand side (L.H.S) of the above equation we get,
$\Rightarrow 21000=1050p$
Dividing 1050 on both the sides of the above equation we get,
$\dfrac{21000}{1050}=\dfrac{1050}{1050}p$
In the above equation, on the right hand side (R.H.S) of the above equation, 1050 will be cancelled from the numerator and the denominator. Also, one zero will be cancelled out from the numerator and the denominator of the L.H.S of the above equation and we get,
$\Rightarrow \dfrac{2100}{105}=p$
Now, the numerator and the denominator of the L.H.S of the above equation will be divided by 5 and we get,
$\Rightarrow \dfrac{420}{21}=p$
Now, the numerator and the denominator of the L.H.S of the above equation will be cancelled out by 21 and we get,
$\Rightarrow 20=p$
Hence, we have found the value of p from the given equation is 20.

Note: You can check the value of “p” found in the above solution by substituting this value of “p” in the given equation and then see whether this value of “p” is satisfying the given equation or not.
The value of “p” which we have found out in the above solution is 20 so substituting this value of “p” in given equation we get,
$\dfrac{350}{20}=\dfrac{1050}{60}$
In the above equation, one 0 will be cancelled out from the numerator and the denominator in both the sides of the above equation and we get,
$\Rightarrow \dfrac{35}{2}=\dfrac{105}{6}$
In the R.H.S of the above equation, the numerator and the denominator will be cancelled out by 3 and we get,
$\Rightarrow \dfrac{35}{2}=\dfrac{35}{2}$
As L.H.S = R.H.S in the above equation so the value of “p” which we have calculated above is correct.