Students can track their progress and improvement through regular use of ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(d).
S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(d)
Question 1.
A man buys 8 dozen mangoes, 10 dozen apples and 4 dozen bananas. Mangoes cost ₹ 18 per dozen, apples ₹ 9 per dozen and bananas ₹ 6 per dozen. Represent the quantities bought by a row matrix and the prices by a column matrix and hence obtain the total cost.
Solution:
Let Q be the qunatity matrix and is of order 1 x 3 & P be the price matrix and is of order 3 x 1
Thus Q = [8 10 4] & P = \(\left[\begin{array}{c}
18 \\
9 \\
6
\end{array}\right]\)
∴ required total cost = QP
= \(\left[\begin{array}{lll}
8 & 10 & 4
\end{array}\right]\left[\begin{array}{c}
18 \\
9 \\
6
\end{array}\right]\)
[8 x 18 + 10 x 9 + 4 x 6] = [250]
∴ required total cost be ₹ 258.
Question 2.
A store has in stock 20 dozen shirts, 15 dozen trousers and 25 dozen pairs of socks. If the selling prices are ₹ 50 per shirt, ₹ 90 per trousers and ₹ 12 per pair of socks, then find the total amount the store owner will get after selling all the items in the stock.
Solution:
Let S be the stock matrix and is of order 1 x 3 and P be the price matrix and is of order 3 x 1 and both matrices are represented as under :
S = [20 x 12 15 x 12 25 x 12]
= [240 180 300] & P = \(\left[\begin{array}{l}
50 \\
90 \\
12
\end{array}\right]\)
∴ Total amount get by owner
= SP = [240 180 300]\(\left[\begin{array}{l}
50 \\
90 \\
12
\end{array}\right]\)
= [240 x 50 + 180 x 90 + 300 x 12]
= [1200 + 16200 + 3600]
= [31800]
Hence required total amount get by store owner by selling all the items be ₹ 31800.
Question 3.
A shopkeeper has 10 dozen Physics books, 8 dozen Chemistry books and 5 dozen Mathematics books. If their selling prices are 65.70, ₹ 43.20 and ₹ 76.50 each respectively, find by matrix method the total amount of the sale if all the books are sold.
Solution:
Let Q be the quantity matrix and is of order 1 x 3 and S be the selling matrix and is of order 3 x 1. Both matrices represented as under :
Q = [10 x 12 8 x 12 5 x 12]
= [120 96 60] & S = \(\left[\begin{array}{l}
65 \cdot 70 \\
43 \cdot 20 \\
76 \cdot 50
\end{array}\right]\)
Thus, amount raised by sale = QS
= [120 96 60] \(\left[\begin{array}{l}
65 \cdot 70 \\
43 \cdot 20 \\
76 \cdot 50
\end{array}\right]\)
= [120 x 65.70 + 96 x 43.20 + 60 x 76.50]
= [7884 + 4147.20 + 4590]
= [16621.20]
Hence, total get by shopkeeper if all books are sold be ₹ 16621.20.
Question 4.
A manufacturer produces three products A, B, C which he sells in the market. Annual sale volumes are indicated as follows:
| Markets | Products | ||
| A | B | C | |
| I | 8,000 | 10,000 | 15,000 |
| II | 10,000 | 2,000 | 20,000 |
If unit sale prices of A, B and C are ₹ 2.25, ₹ 1.50 and ₹ 1.25 respectively, find the total revenue in each market with the help of matrices.
Solution:
The given data in the form of table be as under :
| Markets | Products | ||
| A | B | C | |
| I
II |
8,000 10,000 |
10,000
2,000 |
15,000 20,000 |
| Price per unit | 2.25 | 1.50 | 1.25 |
Thus, sale matrix
Thus, total revenue from market I be ₹ 51750 and from market II be ₹ 50500 respectively.
Question 5.
Man invests X 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 X 50,000 among the two types of bonds so as to obtain an annual total interest of ₹ 2780.
Solution:
Let ₹ x be invested in first type of bonds and ₹ (50000 – x) in second type bonds. The value of both bonds can be written in the form of row matrix A = [ x 50000 – x] and the amounts received as interest from there two types of bonds can be written in column matrix
B is given as under : B = \(\left[\begin{array}{c}
\frac{5}{100} \\
\frac{6}{100}
\end{array}\right]\)
Since interest per year received by a single number i.e. a matrix of order of 1 x 1 which can be obtained by product matrix.
AB = \(\left[\begin{array}{ll}
x & 50000-x
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{6}{100}
\end{array}\right]\)
= \(\left[\frac{5 x}{100}+(50000-x) \frac{6}{100}\right]\)
Also total interest is given to be ₹ 2780.
∴ \(\left[\frac{5 x}{100}+\frac{(50000-x) 6}{100}\right]\) = [2780]
⇒ 5x + (50000 – x)6 = 2780 x 100
⇒ 300000 – x = 278000
⇒ x = 22000
Thus, ₹ 22000 be invested in bond I and ₹ (50000 – 22000) i.e. ₹ 28000 be invested in bond II.
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 o, ads of stones and 10, 5 and 8 truck loads of sand respectively. If the cost of one truck load of one and sand is ₹ 1200 and ₹ 500 respectively, then find the total amount paid by the contractor each of these firms A, B, C respectively.
Solution:
The given data can be put in table form as under :
| Firms | No. of trucks loads of stones | No. of trucks loads of sand |
| A B C |
40 35 25 |
10 5 8 |
| Cost of price of loading per truck | 1200 | 500 |
Thus, quantity matrix = \(\left[\begin{array}{cc}
40 & 10 \\
35 & 5 \\
25 & 8
\end{array}\right]_{3 \times 2}\)
& cost matrix = \(\left[\begin{array}{c}
1200 \\
500
\end{array}\right]_{2 \times 1}\)
∴ amount paid by contractor
= \(\left[\begin{array}{cc}
40 & 10 \\
35 & 5 \\
25 & 8
\end{array}\right]\left[\begin{array}{c}
1200 \\
500
\end{array}\right]\)
= \(\left[\begin{array}{c}
40 \times 1200+10 \times 500 \\
35 \times 1200+5 \times 500 \\
25 \times 1200+8 \times 500
\end{array}\right]\)
= \(\left[\begin{array}{l}
53000 \\
44500 \\
34000
\end{array}\right]\)
Thus, requiared amount paid by contractor to firm A = ₹ 53,000, amount paid by contractor to firm B = ₹ 44,500 & amount paid by contractor to firm C = ₹ 34000.