Class 10 Maths
Question 1: Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.
Solution: Let us assume Aftab’s current age = x and his daughter’s current age = y
Seven years ago: Aftab’s age `= x - 7` and daughter’s age `= y - 7`
As per question;
`x – 7 = 7(y – 7)`
Or, `x – 7 = 7y – 49`
Or, `x = 7y – 49 + 7`
Or, `x = 7y – 42`
Or, `7y – x – 42 = 0` ……(1)
This equation gives following values for x and y
x | - 35 | - 28 | - 21 | - 14 | - 7 | 0 | 7 | 14 | 21 | 28 | 35 | 42 | 49 |
y | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
Three years from now:
Aftab’s age `= x + 3` and daughter’s age `= y + 3`
As per question;
`x + 3 = 3(y + 3)`
Or, `x + 3 = 3y + 9`
Or, `x = 3y + 9 – 3`
Or, `x = 3y + 6`
Or, `3y – x + 6 = 0` ……..(2)
This equation gives following values for x and y:
x | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 | 39 | 42 | 45 |
y | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
The following graph is plotted for the given pair of linear equations:
Daughter’s age = 12 years and Aftab’s age = 42 years
Question 2: The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Solution: Let us assume that price of one bat = x and price of one ball = y. Following equations can be written as per the question:
`3x + 6y = 3900`
Or, `x + 2y = 1300` ……..(1)
This equation will give following values for x and y
x | 1500 | 1400 | 1300 | 1200 | 1100 |
y | - 100 | - 50 | 0 | 50 | 100 |
`x + 3y = 1300` ……….(2)
This equation will give following values for x and y
x | 1600 | 1450 | 1300 | 1150 | 1000 |
y | - 100 | - 50 | 0 | 50 | 100 |
The following graph is plotted for the given pair of linear equations.
Price of one bat = Rs. 1300 and Price of one ball = zero
Question 3: The cost of 2 kg apples and 1 kg grapes was found to be Rs. 160. After a month, the cost of 4 kg apples and 2 kg grapes is Rs. 300. Represent this situation algebraically and graphically.
Solution: Let us assume that cost of 1 kg apple = x and cost of 1 kg grapes = y. Following equations can be written as per the question.
`2x + y = 160` ……..(1)
This equation gives following values for x and y
x | 70 | 60 | 50 | 40 |
y | 20 | 40 | 60 | 80 |
`4x + 2y = 300`
Or, `2x + y = 150` ……..(2)
This equation will give following values for x and y:
x | 65 | 55 | 45 | 35 |
y | 20 | 40 | 60 | 80 |
The following graph is plotted for the given pair of linear equations.
Since we get parallel lines so there will be no solution for this pair of linear equations.
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