OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Exercise 6 (D)

 

Question 1. A man buys 8 dozen mangoes, 10 dozen apples and 4 dozen bananas. Mangoes cost Rs 18 per dozen, apples Rs 9 per dozen and bananas Rs 6 per dozen. Represent the quantities bought by a row matrix and the prices by a column matrix and hence obtain the total cost.
Answer: Let Q be the quantity matrix and P be the price matrix.
The quantity matrix Q is a \( 1 \times 3 \) matrix, and the price matrix P is a \( 3 \times 1 \) matrix.
Thus, the quantity matrix is:
\( Q = \begin{bmatrix} 8 & 10 & 4 \end{bmatrix} \)
The price matrix is:
\( P = \begin{bmatrix} 18 \\ 9 \\ 6 \end{bmatrix} \)
The total cost is found by multiplying the quantity matrix Q by the price matrix P.
Required total cost \( = QP \)
\( = \begin{bmatrix} 8 & 10 & 4 \end{bmatrix} \begin{bmatrix} 18 \\ 9 \\ 6 \end{bmatrix} \)
This multiplication means multiplying each quantity by its respective price and adding them up.
\( = \begin{bmatrix} 8 \times 18 + 10 \times 9 + 4 \times 6 \end{bmatrix} \)
\( = \begin{bmatrix} 144 + 90 + 24 \end{bmatrix} \)
\( = \begin{bmatrix} 258 \end{bmatrix} \)
So, the required total cost is Rs 258. Matrix multiplication is useful for calculating total values from quantity and price lists.
In simple words: We put the number of items bought in a row and their prices in a column. Then, we multiply these two matrices together to find the total money spent.

🎯 Exam Tip: Remember that for matrix multiplication to be possible, the number of columns in the first matrix must equal the number of rows in the second matrix. The resulting matrix will have the number of rows of the first and columns of the second.

 

Question 2. A store has in stock 20 dozen shirts, 15 dozen trousers and 25 dozen pairs of socks. If the selling prices are Rs 50 per shirt, Rs 90 per trousers and Rs 12 per pair of socks, then find the total amount the store owner will get after selling all the items in the stock.
Answer: First, let's find the total number of each item since prices are given per single item, not per dozen.
Number of shirts \( = 20 \times 12 = 240 \)
Number of trousers \( = 15 \times 12 = 180 \)
Number of pairs of socks \( = 25 \times 12 = 300 \)
Let S be the stock matrix and P be the price matrix.
The stock matrix S will be a \( 1 \times 3 \) matrix, and the price matrix P will be a \( 3 \times 1 \) matrix.
Stock matrix:
\( S = \begin{bmatrix} 240 & 180 & 300 \end{bmatrix} \)
Price matrix:
\( P = \begin{bmatrix} 50 \\ 90 \\ 12 \end{bmatrix} \)
To find the total amount, we multiply the stock matrix by the price matrix.
Total amount \( = SP \)
\( = \begin{bmatrix} 240 & 180 & 300 \end{bmatrix} \begin{bmatrix} 50 \\ 90 \\ 12 \end{bmatrix} \)
\( = \begin{bmatrix} (240 \times 50) + (180 \times 90) + (300 \times 12) \end{bmatrix} \)
\( = \begin{bmatrix} 12000 + 16200 + 3600 \end{bmatrix} \)
\( = \begin{bmatrix} 31800 \end{bmatrix} \)
Therefore, the store owner will get a total of Rs 31800 after selling all the items. This method efficiently combines quantities and prices.
In simple words: First, we find the exact number of each item. Then, we make a list of all items and a list of all their prices. Multiplying these two lists in a special way (matrix multiplication) tells us the total money earned.

🎯 Exam Tip: Always convert dozens to individual units before creating the matrices if the unit price is for single items. This avoids errors in calculation. Ensure matrices are compatible for multiplication.

 

Question 3. A shopkeeper has 10 dozen Physics books, 8 dozen Chemistry books and 5 dozen Mathematics books. If their selling prices are Rs 65.70, Rs 43.20 and Rs 76.50 each respectively, find by matrix method the total amount of the sale if all the books are sold.
Answer: First, convert the dozens into individual book counts.
Number of Physics books \( = 10 \times 12 = 120 \)
Number of Chemistry books \( = 8 \times 12 = 96 \)
Number of Mathematics books \( = 5 \times 12 = 60 \)
Let Q be the quantity matrix and S be the selling price matrix.
The quantity matrix Q is a \( 1 \times 3 \) matrix, and the selling price matrix S is a \( 3 \times 1 \) matrix.
Quantity matrix:
\( Q = \begin{bmatrix} 120 & 96 & 60 \end{bmatrix} \)
Selling price matrix:
\( S = \begin{bmatrix} 65.70 \\ 43.20 \\ 76.50 \end{bmatrix} \)
To find the total amount raised by the sale, we multiply the quantity matrix by the selling price matrix.
Amount raised by sale \( = QS \)
\( = \begin{bmatrix} 120 & 96 & 60 \end{bmatrix} \begin{bmatrix} 65.70 \\ 43.20 \\ 76.50 \end{bmatrix} \)
\( = \begin{bmatrix} (120 \times 65.70) + (96 \times 43.20) + (60 \times 76.50) \end{bmatrix} \)
\( = \begin{bmatrix} 7884 + 4147.20 + 4590 \end{bmatrix} \)
\( = \begin{bmatrix} 16621.20 \end{bmatrix} \)
So, the total amount received by the shopkeeper if all books are sold is Rs 16621.20. This matrix method helps organize and calculate total revenue efficiently.
In simple words: We list the total number of each type of book and their individual prices. Then, we use matrix multiplication to add up the money from selling all books.

🎯 Exam Tip: Ensure all values are in the correct units (e.g., single items instead of dozens) before forming the matrices to avoid calculation errors. Decimal points must be handled carefully in multiplication.

 

Question 4. A manufacturer produces three products A, B, C which he sells in the market. Annual sale volumes are indicated as follows:

If unit sale prices of A, B and C are Rs 2.25, Rs 1.50 and Rs 1.25 respectively, find the total revenue in each market with the help of matrices.
Answer: We can represent the sales data as a matrix and the unit prices as another matrix.
The sales quantity matrix (S) from the table is:
\[ S = \begin{bmatrix} 8000 & 10000 & 15000 \\ 10000 & 2000 & 20000 \end{bmatrix}_{2 \times 3} \]
The unit price matrix (P) is a column matrix:
\[ P = \begin{bmatrix} 2.25 \\ 1.50 \\ 1.25 \end{bmatrix}_{3 \times 1} \]
To find the total revenue in each market, we multiply the sales quantity matrix S by the price matrix P.
Total Revenue \( = SP \)
\[ = \begin{bmatrix} 8000 & 10000 & 15000 \\ 10000 & 2000 & 20000 \end{bmatrix} \begin{bmatrix} 2.25 \\ 1.50 \\ 1.25 \end{bmatrix} \]
Now, we perform the matrix multiplication:
\[ = \begin{bmatrix} (8000 \times 2.25) + (10000 \times 1.50) + (15000 \times 1.25) \\ (10000 \times 2.25) + (2000 \times 1.50) + (20000 \times 1.25) \end{bmatrix} \]
\[ = \begin{bmatrix} 18000 + 15000 + 18750 \\ 22500 + 3000 + 25000 \end{bmatrix} \]
\[ = \begin{bmatrix} 51750 \\ 50500 \end{bmatrix} \]
So, the total revenue from Market I is Rs 51750, and the total revenue from Market II is Rs 50500. This calculation helps a manufacturer understand the earnings from different markets.
In simple words: We put how many of each product were sold in each market into one big list (matrix) and the price of each product into another list. When we multiply these lists, we get the total money earned from each market.

🎯 Exam Tip: When setting up matrices for revenue, ensure the sales matrix has markets as rows and products as columns, and the price matrix is a column vector of prices per product. This ensures correct multiplication, leading to a revenue matrix where each row represents a market's total earnings.

 

Question 5. Man invests Rs 50,000 into two types of bonds. The first bond pays 5% interest per year and the second bond pays 6% interest per year. Using matrix multiplication, determine how to divide Rs 50,000 among the two types of bonds so as to obtain an annual total interest of Rs 2780.
Answer: Let \( x \) be the amount invested in the first type of bond (paying 5% interest).
Then, the amount invested in the second type of bond (paying 6% interest) will be \( Rs (50000 - x) \).
We can represent the amounts invested as a row matrix A:
\( A = \begin{bmatrix} x & 50000 - x \end{bmatrix} \)
The annual interest rates (expressed as decimals) can be represented as a column matrix B:
\( B = \begin{bmatrix} \frac{5}{100} \\ \frac{6}{100} \end{bmatrix} \)
To find the total annual interest, we multiply matrix A by matrix B.
Total interest \( = AB \)
\[ = \begin{bmatrix} x & 50000-x \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{6}{100} \end{bmatrix} \]
\[ = \begin{bmatrix} x \left( \frac{5}{100} \right) + (50000-x) \left( \frac{6}{100} \right) \end{bmatrix} \]
We are given that the total annual interest is Rs 2780.
So, we set the result of the matrix multiplication equal to this total interest:
\[ \begin{bmatrix} \frac{5x}{100} + \frac{6(50000-x)}{100} \end{bmatrix} = \begin{bmatrix} 2780 \end{bmatrix} \]
This means:
\( \frac{5x}{100} + \frac{6(50000-x)}{100} = 2780 \)
Multiply the entire equation by 100 to remove the denominators:
\( 5x + 6(50000 - x) = 2780 \times 100 \)
\( 5x + 300000 - 6x = 278000 \)
Combine the \( x \) terms:
\( 300000 - x = 278000 \)
Now, solve for \( x \):
\( -x = 278000 - 300000 \)
\( -x = -22000 \)
\( x = 22000 \)
So, Rs 22000 should be invested in the first bond. The remaining amount will go to the second bond.
Amount in second bond \( = 50000 - 22000 = 28000 \)
Thus, Rs 22000 should be invested in bond I and Rs 28000 in bond II. This method ensures the total interest target is met.
In simple words: We imagine how much money goes into each bond type. Then, we use matrix math to set up an equation that shows how much total interest we will get. Solving this equation tells us exactly how to split the money between the two bonds to reach the goal interest.

🎯 Exam Tip: When using matrix multiplication for investment problems, ensure the investment amounts are in a row matrix and the interest rates are in a column matrix. This setup allows the dot product to correctly calculate the total interest. Remember to convert percentages to decimals.

 

Question 6. In a development plan of a city, a contractor has taken a contract to construct certain houses for which he needs building materials like stones, sand etc. There are three firms A, B, C that can ripply him these materials. At one time these firms A, B, C supplied him 40, 35 and 25 truck loads of stones and 10, 5 and 8 truck loads of sand respectively. If the cost of one truck load of one and sand is Rs 1200 and Rs 500 respectively, then find the total amount paid by the contractor each of these firms A, B, C respectively.
Answer: We need to organize the quantities supplied by each firm and the cost per truck load into matrices.
The quantity matrix (Q) shows the truck loads of stones and sand supplied by each firm:

So, the quantity matrix Q is:
\[ Q = \begin{bmatrix} 40 & 10 \\ 35 & 5 \\ 25 & 8 \end{bmatrix}_{3 \times 2} \]
The cost matrix (C) represents the price per truck load for stones and sand:
\[ C = \begin{bmatrix} 1200 \\ 500 \end{bmatrix}_{2 \times 1} \]
To find the total amount paid to each firm, we multiply the quantity matrix Q by the cost matrix C.
Amount paid by contractor \( = QC \)
\[ = \begin{bmatrix} 40 & 10 \\ 35 & 5 \\ 25 & 8 \end{bmatrix} \begin{bmatrix} 1200 \\ 500 \end{bmatrix} \]
Now, we perform the matrix multiplication:
\[ = \begin{bmatrix} (40 \times 1200) + (10 \times 500) \\ (35 \times 1200) + (5 \times 500) \\ (25 \times 1200) + (8 \times 500) \end{bmatrix} \]
\[ = \begin{bmatrix} 48000 + 5000 \\ 42000 + 2500 \\ 30000 + 4000 \end{bmatrix} \]
\[ = \begin{bmatrix} 53000 \\ 44500 \\ 34000 \end{bmatrix} \]
Therefore, the contractor paid Rs 53,000 to firm A, Rs 44,500 to firm B, and Rs 34,000 to firm C. This matrix approach streamlines the calculation of costs from multiple suppliers.
In simple words: We make a list of how many trucks of stones and sand each company gave. Then, we make another list of how much one truck of stones and one truck of sand costs. Multiplying these lists helps us find out the total money paid to each company.

🎯 Exam Tip: Clearly define your quantity and cost matrices, paying attention to their dimensions. Ensure that the number of columns in the quantity matrix matches the number of rows in the cost matrix for valid multiplication.