CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 32

Short Answer Type Questions

Question 1. Fourteen times a two-digit number is equal to eight times the number obtained by reversing the order of its digits. If the difference of the digits is 4, determine the number.
Answer:
Let the tens digit be x and the units digit be y, so the number is \( 10x + y \) and the reversed number is \( 10y + x \).
\( 14(10x + y) = 8(10y + x) \Rightarrow 140x + 14y = 80y + 8x \Rightarrow 132x = 66y \Rightarrow y = 2x \).
The digits differ by 4. Since \( y = 2x \) and x is not zero, y is the larger digit, so \( y - x = 4 \Rightarrow 2x - x = 4 \Rightarrow x = 4 \) and \( y = 8 \).
Check: \( 14 \times 48 = 672 = 8 \times 84 \) and \( 8 - 4 = 4 \).
Hence the number is 48.
In simple words: Turn the two statements into two linear equations in the digits and solve them.

Teacher's Note:
1. Write a two-digit number as 10x + y, never as xy.
2. Use the relation y = 2x to decide which digit is larger before using the difference.

Question 2. A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of Rs. 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs. 1028. Find the cost price of the saree and the list price (price before discount) of the sweater.
Answer:
Let the cost price of the saree be Rs. x and the list price of the sweater be Rs. y.
First case: \( 1.08x + 0.9y = 1008 \Rightarrow 108x + 90y = 100800 \Rightarrow 6x + 5y = 5600 \) ... (i)
Second case: \( 1.10x + 0.92y = 1028 \Rightarrow 110x + 92y = 102800 \Rightarrow 55x + 46y = 51400 \) ... (ii)
From (i), \( y = \frac{5600 - 6x}{5} \). Putting this in (ii): \( 275x + 46(5600 - 6x) = 257000 \Rightarrow 275x + 257600 - 276x = 257000 \Rightarrow x = 600 \).
Then \( y = \frac{5600 - 3600}{5} = 400 \).
Check: \( 1.08 \times 600 + 0.9 \times 400 = 648 + 360 = 1008 \) and \( 1.10 \times 600 + 0.92 \times 400 = 660 + 368 = 1028 \).
Hence the cost price of the saree is Rs. 600 and the list price of the sweater is Rs. 400.
In simple words: A profit adds a percentage to the cost price; a discount takes a percentage off the list price.

Teacher's Note:
1. Profit is calculated on the cost price and discount on the marked (list) price.
2. Reduce each equation by its common factor before eliminating a variable.

Long Answer Type Questions

Teacher's Note:
1. Choose a scale (here 1 unit = Rs. 100) so that all the points fit on the graph paper.
2. Read the coordinates of the intersection carefully and convert back to rupees.

Question 4. For what values of m and n does the following system of linear equations have infinitely many solutions?
\( 3x + 4y = 12 \) and \( (m + n)x + 2(m - n)y = 5m - 1 \)
Answer:
For infinitely many solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), so \( \frac{3}{m + n} = \frac{4}{2(m - n)} = \frac{12}{5m - 1} \).
From the first two ratios: \( 3 \times 2(m - n) = 4(m + n) \Rightarrow 6m - 6n = 4m + 4n \Rightarrow m = 5n \) ... (i)
From the first and third ratios: \( 3(5m - 1) = 12(m + n) \Rightarrow 15m - 3 = 12m + 12n \Rightarrow m = 4n + 1 \) ... (ii)
From (i) and (ii): \( 5n = 4n + 1 \Rightarrow n = 1 \) and \( m = 5 \).
Check: the second equation becomes \( 6x + 8y = 24 \), which is twice \( 3x + 4y = 12 \), so the two equations represent the same line.
Hence \( m = 5 \) and \( n = 1 \).
In simple words: Infinitely many solutions means the two equations are multiples of each other, so all three ratios must be equal.

Teacher's Note:
1. Equate the three ratios in pairs to get two equations in m and n.
2. Always check that the second equation is a constant multiple of the first.

Teacher's Note:
1. Find the point where the sloping line meets each vertical line to get the vertices.
2. The line 2x - y - 4 = 0 meets the x-axis at x = 2, which lies outside the strip between x = 3 and x = 5.