Question

Shanti sweets stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of the dimensions 25cm\[ \times \]20cm\[ \times \]5cm and the smaller of dimensions 15cm\[ \times \]12cm\[ \times \]5cm. For all the overlaps, \[5\% \]of the total surface area is required extra. If the cost of the cardboard is Rs.4 for 1000\[c{m^2}\], find the cost of the cardboard required for supplying 250 boxes of each kind.

Answer 1

Hint: We use the formula of total surface area of cuboid for a bigger box. Calculate 5% of that area and add it to the obtained area to calculate the total area with extra 5%. Use a unitary method to find the cost of cardboard required for supplying 250 boxes. Repeat the same procedure for the small box.
* Total surface area of cuboid having dimensions: length ‘l’, breadth ‘b’ and height ‘h’ is given by the formula \[2(lb + bh + hl)\]
* We can write \[m\% \] of \[x\] as \[\dfrac{m}{{100}}x\]
* Unitary methods help us to find the value of multiple objects if we are given the value of one object by just multiplying the value of a single object to the number of objects.

Complete step-by-step solution:
We know Shanti sweets stall was placing order for making packaging boxes in two sizes, smaller and bigger.
Big Boxes:
Length of big box, \[l = 25\]cm
Breadth of big box, \[b = 20\]cm
Height of the big box, \[h = 5\]cm
\[\because \]Total surface area of cuboid\[ = 2(lb + bh + hl)\]
Substitute the values of ‘l’, ‘b’ and ‘h’ in the formula
\[ \Rightarrow \] Total surface area of cuboid\[ = 2(25 \times 20 + 20 \times 5 + 5 \times 25)\]\[c{m^2}\]
\[ \Rightarrow \] Total surface area of cuboid\[ = 2(500 + 100 + 125)\]\[c{m^2}\]
\[ \Rightarrow \] Total surface area of cuboid\[ = 2(725)\]\[c{m^2}\]
\[ \Rightarrow \] Total surface area of cuboid\[ = 1450\]\[c{m^2}\].....................… (1)
Now we calculate 5% of the Total surface area from equation (1)
\[ \Rightarrow 5\% \]of \[1450 = \dfrac{5}{{100}} \times 1450\]
Cancel same terms from numerator and denominator in RHS
\[ \Rightarrow 5\% \]of \[1450 = \dfrac{5}{{10}} \times 145\]
\[ \Rightarrow 5\% \]of \[1450 = 72.5\]
So, the extra surface area is 72.5\[c{m^2}\].................… (2)
The total Surface area after adding 5% extra is given by adding surface area from equations (1) and (2)
\[ \Rightarrow \]Total surface area after 5% extra\[ = (1450 + 72.5)\]\[c{m^2}\]
\[ \Rightarrow \]Total surface area after 5% extra\[ = 1522.5\]\[c{m^2}\]
Now we have a surface area of 1 big box as 1522.5\[c{m^2}\].
We use a unitary method to calculate the total surface area of cardboard required to make 250 boxes.
\[\because \]Surface area of 1 cardboard box \[ = 1522.5\]\[c{m^2}\]
\[ \Rightarrow \] Surface area of 250 cardboard boxes \[ = \left( {1522.5 \times 250} \right)\]\[c{m^2}\]
\[ \Rightarrow \] Surface area of 250 cardboard boxes \[ = 380625\]\[c{m^2}\]
Again we use a unitary method to find the cost of cardboard required for supplying 250 big boxes.
We are given cost of the cardboard is Rs.4 for 1000\[c{m^2}\]
\[ \Rightarrow \]Cost of 1000\[c{m^2}\]of cardboard\[ = \]Rs.4
\[ \Rightarrow \]Cost of 1\[c{m^2}\]of cardboard\[ = \]Rs\[\dfrac{4}{{1000}}\]
\[ \Rightarrow \]Cost of 380625\[c{m^2}\]of cardboard\[ = \]Rs\[\dfrac{4}{{1000}} \times 380625\]
\[ \Rightarrow \]Cost of 380625\[c{m^2}\]of cardboard\[ = \]Rs1522.5………………..… (3)
Small Boxes:
Length of small box, \[l = 15\]cm
Breadth of small box, \[b = 12\]cm
Height of the small box, \[h = 5\]cm
\[\because \]Total surface area of cuboid\[ = 2(lb + bh + hl)\]
Substitute the values of ‘l’, ‘b’ and ‘h’ in the formula
\[ \Rightarrow \] Total surface area of cuboid\[ = 2(15 \times 12 + 12 \times 5 + 5 \times 15)\]\[c{m^2}\]
\[ \Rightarrow \] Total surface area of cuboid\[ = 2(180 + 60 + 75)\]\[c{m^2}\]
\[ \Rightarrow \] Total surface area of cuboid\[ = 2(315)\]\[c{m^2}\]
\[ \Rightarrow \] Total surface area of cuboid\[ = 630\]\[c{m^2}\].................… (4)
Now we calculate 5% of the Total surface area from equation (1)
\[ \Rightarrow 5\% \] of \[630 = \dfrac{5}{{100}} \times 630\]
Cancel same terms from numerator and denominator in RHS
\[ \Rightarrow 5\% \] of \[630 = \dfrac{5}{{10}} \times 63\]
\[ \Rightarrow 5\% \] of \[630 = 31.5\]
So, the extra surface area is 72.5\[c{m^2}\]................… (5)
The total Surface area after adding 5% extra is given by adding surface area from equations (4) and (5)
\[ \Rightarrow \]Total surface area after 5% extra\[ = (630 + 31.5)\]\[c{m^2}\]
\[ \Rightarrow \]Total surface area after 5% extra\[ = 661.5\]\[c{m^2}\]
Now we have a surface area of 1 small box as 661.5\[c{m^2}\].
We use a unitary method to calculate the total surface area of cardboard required to make 250 boxes.
\[\because \]Surface area of 1 cardboard box \[ = 661.5\]\[c{m^2}\]
\[ \Rightarrow \] Surface area of 250 cardboard boxes \[ = \left( {661.5 \times 250} \right)\]\[c{m^2}\]
\[ \Rightarrow \] Surface area of 250 cardboard boxes \[ = 165375\]\[c{m^2}\]
Again we use a unitary method to find the cost of cardboard required for supplying 250 small boxes.
We are given cost of the cardboard is Rs.4 for 1000\[c{m^2}\]
\[ \Rightarrow \]Cost of 1000\[c{m^2}\]of cardboard\[ = \]Rs.4
\[ \Rightarrow \]Cost of 1\[c{m^2}\]of cardboard\[ = \]Rs\[\dfrac{4}{{1000}}\]
\[ \Rightarrow \]Cost of 165375\[c{m^2}\]of cardboard\[ = \]Rs\[\dfrac{4}{{1000}} \times 165375\]
\[ \Rightarrow \]Cost of 165375\[c{m^2}\]of cardboard\[ = \]Rs661.5………………..… (6)
The total cost of making 250 boxes of each kind is given by adding costs from equations (3) and (6)
\[ \Rightarrow \] The total cost of making 250 boxes of each kind\[ = \]Rs\[\left( {1522.5 + 661.5} \right)\]
\[ \Rightarrow \] The total cost of making 250 boxes of each kind\[ = \]Rs 2184

\[\therefore \] The total cost of making 250 boxes of each kind is Rs 2184.

Note: Students many times make mistakes while calculating the cost using unitary method as they leave the cost value in fractional terms. Keep in mind the value of cost should always be simplified and should never be in fraction form. Also, we add a value of 5% as it is given extra, remember that percentage is always with reference to the given number, many students just write the value of 5% and not 5% of the number given.