Question

Two types of boxes $ A,B $ are to be placed in a truck having the capacity of $ 10 $ tons. When $ 150 $ boxes of type $ A $ and $ 100 $ boxes of type $ B $ are loaded in the truck weighs $ 10 $ tons. But when $ 260 $ boxes of type $ A $ are loaded in the truck, it can still accommodate $ 40 $ boxes of type $ B $ , so that it is fully loaded, Find the sum of the weight of each type of box.

Answer 1

Hint: As we know that the above given question is a word problem. A problem is a mathematical question written as one sentence or more describing a real life scenario where that problem needs to be solved by the way of mathematical calculation. We can solve the given problem by applying the method of mathematical equations and then simplify it.

Complete step by step solution:
We need to first understand the requirement of the question which is the weight of each type of box. Now we will form the equation according to the parts of the question.
Let us assume the weight of box $ A $ be $ x $ and the weight of box $ B $ be $ y $ . It is given that when $ 150 $ boxes of type $ A $ and $ 100 $ boxes of type $ B $ are loaded in the truck, it weighs $ 10 $ tons.
We know that $ 1 $ ton $ = 1000kg $ . So $ 10 $ tons is equal to $ 10000kg $ .
Now we have the first equation: $ 150x + 100y = 10000 $ . By simplifying it in the simpler form, we divide the whole equation by $ 50 $ . It gives us
 $ \dfrac{{150x}}{{50}} + \dfrac{{100y}}{{50}} = \dfrac{{10000}}{{50}} $ .
So the new equation is $ 3x + 2y = 200 $ .
We have another statement which says that when $ 260 $ boxes of type $ A $ are loaded in the truck, it can still accommodate $ 40 $ boxes of type $ B $ , so that it is fully loaded. So the second equation is $ 260x + 40y = 10000 $ . We will simplify the equation again by dividing it with $ 10 $ . It gives $ \dfrac{{260x}}{{20}} + \dfrac{{40y}}{{20}} = \dfrac{{10000}}{{20}} $ . So the new equation is: $ 3x + 2y = 500 $ .
We will now subtract the first equation from the second: $ 13x - 3x + 2y - 2y = 500 - 200 $
It gives us
 $ 10x = 300 \\
\Rightarrow x = \dfrac{{300}}{{10}}
 $ .
So the value of $ x = 30 $ . Now by substituting the value of $ x $ in the first equation:
 $ 3 \times 30 + 2y = 200 $ . On further solving we have
 $ 90 + 2y = 200 \\
\Rightarrow 2y = 200 - 90
 $ .
 $ \Rightarrow 2y = 110 \\
\Rightarrow y = \dfrac{{110}}{0} = 55
 $ .
Hence the weight of box $ A $ is $ 30kg $ and the weight of box $ B $ is $ 55kg $ .
So, the correct answer is “The weight of box $ A $ is $ 30kg $ and the weight of box $ B $ is $ 55kg $ ”.

Note: We should always be careful what the question is asking i.e. weight of each type of box which includes two different types of boxes. Based on the requirement and by observing all the necessary information that is already available in the question we gather the information and then create an equation or by unitary method whichever is applicable, then we solve the problem and then verify the answer by putting the value in the problem and see whether we get the same answer or not.