Question

Three pieces of cake of weights $4\dfrac{1}{2}lbs,6\dfrac{3}{4}lbs\text{ and }7\dfrac{1}{5}lbs$ respectively are to be divided into parts of equal weights. Further, each must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained?
(a) 54
(b) 72
(c) 20
(d) 41

Answer 1

We have given the three pieces of the cakes weighing $4\dfrac{1}{2}lbs,6\dfrac{3}{4}lbs\text{ and }7\dfrac{1}{5}lbs$. First of all convert them into proper fractions. Then, we are going to find the heaviest piece of cake which is equal to the H.C.F of these three weights. The weights are in proper fraction and the formula for the H.C.F of the fractions is equal to $\dfrac{H.C.F\text{ of numerators}}{L.C.M\text{ of denominators}}$. After that, to find the maximum number of guests that could be entertained which will be calculated by dividing the heaviest weight with each of the three given weights separately and then adding all the three divisions.

Complete step-by-step answer:
The weight of the three pieces of cakes is equal to:
$4\dfrac{1}{2}lbs,6\dfrac{3}{4}lbs\text{ and }7\dfrac{1}{5}lbs$
As you can see the above three weights are in improper fraction so we are going to convert them into proper fraction.
$\left( 4+\dfrac{1}{2} \right)lbs,\left( 6+\dfrac{3}{4} \right)lbs,\left( 7+\dfrac{1}{5} \right)lbs$
Taking 2, 4, 7 as L.C.M in the first, second and third bracket we get,
$\begin{aligned} & \dfrac{8+1}{2}lbs,\dfrac{24+3}{4}lbs,\dfrac{35+1}{5}lbs \\\ & =\dfrac{9}{2}lbs,\dfrac{27}{4}lbs,\dfrac{36}{5}lbs \\\ \end{aligned}$
Now, we are going to find the heaviest piece of cake by finding the H.C.F of the above three dfractions.
We know that the formula for H.C.F of fractions is equal to:
$\dfrac{H.C.F\text{ of numerators}}{L.C.M\text{ of denominators}}$
The numerators of the three fractions are:
9, 27 and 36
We are going to find H.C.F of the above three numbers:
Now, we are going to find the prime factorization of 9, 27 and 36.
$\begin{aligned} & 9=\underline{3\times 3\times 1} \\\ & 27=3\times \underline{3\times 3\times 1} \\\ & 36=2\times 2\times \underline{3\times 3\times 1} \\\ \end{aligned}$
We know that H.C.F is the common factors which are present in all the above three numbers so in the above prime factorization, the underlying factors are common in all the three numbers so the H.C.F of the above three numbers is:
$\begin{aligned} & 3\times 3\times 1 \\\ & =9 \\\ \end{aligned}$
Now, the denominators of the three fractions are as follows:
2, 4 and 5
We are going to find the L.C.M of the above three numbers. First of all, we are going to find the prime factorization of 2, 4 and 5 are:
$\begin{aligned} & 2=2\times 1 \\\ & 4=2\times 2\times 1 \\\ & 5=5\times 1 \\\ \end{aligned}$
The L.C.M of the above three numbers is calculated by multiplying the common factors which are present in all the three numbers with the factors which are not common. The point to be noted is that if in any two numbers one factor is common but that factor is not there in the third number then also we will multiply that factor only once. Just because that factor is not common in all the three numbers we cannot multiply that factor two times. For e.g. in the above three numbers, 2 is common in 2 and 4 but not common in 5 but we are multiplying 2 only once with the other factors.
In the above three numbers, 1 is common in all the three numbers and the uncommon factors are $2\times 2\times 5$ so multiplying 1 with $2\times 2\times 5$ we get,
$\begin{aligned} & 2\times 2\times 5\times 1 \\\ & =20 \\\ \end{aligned}$
Now, we got the H.C.F of numerators as 9 and L.C.M of the denominators as 20 so dividing 9 by 20 we will get the H.C.F of the three fractions.
$\dfrac{9}{20}lbs$
The above weight is the heaviest piece of the cake. Now, the maximum guest that could be entertained by this heavy piece of cake is found by dividing this heaviest piece by each of the 3 weights given above and then we will add all the three divisions.
$\dfrac{\dfrac{9}{2}}{\dfrac{9}{20}}+\dfrac{\dfrac{27}{4}}{\dfrac{9}{20}}+\dfrac{\dfrac{36}{5}}{\dfrac{9}{20}}$
Simplifying above expression we get,
$\begin{aligned} & \dfrac{9}{2}\times \dfrac{20}{9}+\dfrac{27}{4}\times \dfrac{20}{9}+\dfrac{36}{5}\times \dfrac{20}{9} \\\ & =10+15+16 \\\ & =41 \\\ \end{aligned}$
Hence, the maximum number of guests that could be entertained by a piece of cake is 41.

So, the correct answer is “Option (d)”.

Note: The mistake that could be happened in the above problem is that while calculating the H.C.F of fractions you might have take the L.C.M of the three denominators and then add the three numerators in a way that we do when we have given three fractions with addition sign. In the below, I am going to show that:
$\dfrac{9}{2}lbs,\dfrac{27}{4}lbs,\dfrac{36}{5}lbs$
Taking L.C.M of the denominator which will be 20 then we are going to find the heaviest piece of cake as follows:
$\begin{aligned} & \dfrac{90+135+144}{20} \\\ & =\dfrac{369}{20} \\\ \end{aligned}$
This is completely wrong so make sure you won’t make this mistake.