Question

From a sack of potatoes weighing 120 kg, a merchant sells portions weighing \[6\text{kg, }5\dfrac{1}{4}\text{kg, 9}\dfrac{1}{2}\text{kg and 9}\dfrac{3}{4}\text{kg}\] respectively.
(i) How many kgs did he sell?
(ii) How many kgs are still left in the sack?

Answer 1

Hint: We have to find the weights of the potatoes the merchant sold. For this, we will add the weights \[6\text{kg, }5\dfrac{1}{4}\text{kg, 9}\dfrac{1}{2}\text{kg and 9}\dfrac{3}{4}\text{kg}\] after converting mixed fractions into simple fractions. This is done by considering the mixed fraction form as $Q\dfrac{R}{d}$ and writing the simple fractions using $\dfrac{Q\times d+R}{d}$. We have to find the weight of potatoes still left in the sack. We will do this by subtracting the weights of the potatoes the merchant sold from the given total weight, that is. 120 kg.

Complete step-by-step solution
It is given that a sack of potatoes weighs 120 kg. From this \[6\text{kg, }5\dfrac{1}{4}\text{kg, 9}\dfrac{1}{2}\text{kg and 9}\dfrac{3}{4}\text{kg}\] were sold. We need to find how many kgs did the merchant sell and also the weight of the remaining potatoes.
(i) Let us find the weights of the potatoes the merchant sold. For this, we have to add \[6\text{kg, }5\dfrac{1}{4}\text{kg, 9}\dfrac{1}{2}\text{kg and 9}\dfrac{3}{4}\text{kg}\]
Weights of potatoes sold by the merchant \[=6\text{kg}+5\dfrac{1}{4}\text{kg+ 9}\dfrac{1}{2}\text{kg + 9}\dfrac{3}{4}\text{kg}...\text{(a)}\]
Let us see how to convert a mixed fraction into a fraction.
For a mixed fraction of the form $Q\dfrac{R}{d}$, the fraction form can be obtained by multiplying Q and d and adding R to its result, and dividing it with d. This is denoted below:
$Q\dfrac{R}{d}=\dfrac{Q\times d+R}{d}$ .
Hence, we can write (a) as
Weights of potatoes sold by the merchant \[=6+\dfrac{5\times 4+1}{4}\text{+}\dfrac{9\times 2+1}{2}\text{+}\dfrac{9\times 4+3}{4}\]
Now, let us solve the numerators of each term. We will get
Weights of potatoes sold by the merchant \[=6+\dfrac{20+1}{4}\text{+}\dfrac{18+1}{2}\text{+}\dfrac{36+3}{4}\]
Now we can add the numerators of each term. The result is shown below.
Weights of potatoes sold by the merchant \[=6+\dfrac{21}{4}\text{+}\dfrac{19}{2}\text{+}\dfrac{39}{4}\]
To add the fractional terms, we must take the LCM of the denominators. That is
\[\begin{align}
  & 2\left| \!{\underline {\,
  4,2,4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2,1,2 \,}} \right. \\
 & \text{ }1,1,1 \\
\end{align}\]
Hence, the LCM is $2\times 2=4$ .
Now, we will multiply the numerator and denominator of the fractions with suitable numbers so that their denominator becomes 4 and we will not do this on those fractions whose denominators are 4.
Weights of potatoes sold by the merchant \[=\dfrac{6\times 4}{1\times 4}+\dfrac{21}{4}\text{+}\dfrac{19\times 2}{2\times 2}\text{+}\dfrac{39}{4}\]
Now, we solve the numerators. We will get
Weights of potatoes sold by the merchant \[=\dfrac{24}{4}+\dfrac{21}{4}\text{+}\dfrac{38}{4}\text{+}\dfrac{39}{4}\]
Now we can add all the numerators. That is
Weights of potatoes sold by the merchant \[=\dfrac{24+21+38+39}{4}\]
\[=\dfrac{122}{4}\]
We can further simplify this by canceling the common factors from numerator and denominator. We can see that 2 is the common factor of the numerator and denominator. Hence, we get
Weights of potatoes sold by the merchant \[=\dfrac{61}{2}\text{ kg}\]
Now, let us write this in a mixed fraction. We will divide 61 by 2. This is shown below.
\[2\overset{30}{\overline{\left){\begin{align}
  & 61 \\
 & -6 \\
 & \_\_\_ \\
 & \text{ 1} \\
 & \text{ -0} \\
 & \_\_\_\_ \\
 & \text{ 1} \\
\end{align}}\right.}}\]
We can write the mixed fraction in the following form.
$\text{Quotient}\dfrac{\text{Remainder}}{\text{Divisor}}$
Hence,
Weights of potatoes sold by the merchant \[=30\dfrac{1}{2}\text{ kg}\]
(ii) We need to find the weight of potatoes still left in the sack. We can do this by subtracting the weight of potatoes sold from the total weight of potatoes. That is
Weight of potatoes still left in the sack $=120-\dfrac{61}{2}$
The LCM of 1 and 2 $=1\times 2=2$ . Hence,
Weight of potatoes still left in the sack $=\dfrac{120\times 2}{1\times 2}-\dfrac{61}{2}$
Let us solve the numerator. We will get
Weight of potatoes still left in the sack $=\dfrac{240}{2}-\dfrac{61}{2}$
We can write this as
Weight of potatoes still left in the sack $=\dfrac{240-61}{2}=\dfrac{179}{2}\text{ kg}$
Now, we have to find the mixed fraction of $\dfrac{179}{2}$ .
\[2\overset{89}{\overline{\left){\begin{align}
  & 179 \\
 & -16 \\
 & \_\_\_\_ \\
 & \text{ 19} \\
 & \text{ -18} \\
 & \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \text{ 1} \\
\end{align}}\right.}}\]
Let us write this in the form $\text{Quotient}\dfrac{\text{Remainder}}{\text{Divisor}}$ . We will get
Weight of potatoes still left in the sack $=89\dfrac{1}{2}\text{ kg}$

Note: Be careful with the units given. If the all given units are not similar, convert it into a single unit, that is, if some units are in kg and others in g, convert those in g to kg. Do not forget to write the units in the results. When doing $Q\dfrac{R}{d}=\dfrac{Q\times d+R}{d}$, you may add d and R first and then multiply the result with Q. This is wrong. First, multiply Q with d and add the result with R.