Advertisements
Advertisements
Question
An eraser costs Rs. 1.50 less than a sharpener. Also, the cost of 4 erasers and 3 sharpeners is Rs.29. Taking x and y as the costs (in Rs.) of an eraser and a sharpener respectively, write two equations for the above statements and find the value of x and y.
Advertisements
Solution
Cost of an eraser = Rs. x
Cost of a sharpener = Rs. y
According to given information, we have
x = y - 1.50
⇒ x - y = -1.50 ....(i)
And, 4x + 3y = 29 ....(ii)
Multiplying eqn. (i) by 3, we get
3x - 3y = -450 ....(iii)
Adding eqns. (ii) and (iii), we get
7x = 24.50
⇒ x = 3.50
⇒ 3.50 - y = -1.50
⇒ y
= 3.50 + 1.50
= 5
Thus, the cost of an eraser is Rs.3.50 and that of a sharpener is Rs.5.
APPEARS IN
RELATED QUESTIONS
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
2x − 3y − 3 = 0
`[2x]/3 + 4y + 1/2` = 0
Solve for x and y:
4x = 17 - `[ x - y ]/8`
2y + x = 2 + `[ 5y + 2 ]/3`
Solve :
11(x - 5) + 10(y - 2) + 54 = 0
7(2x - 1) + 9(3y - 1) = 25
Solve the following simultaneous equations:
41x + 53y = 135
53x + 41y = 147
Solve the following pairs of equations:
`(2)/(x + 1) - (1)/(y - 1) = (1)/(2)`
`(1)/(x + 1) + (2)/(y - 1) = (5)/(2)`
`(3)/x - (2)/y` = 0 and `(2)/x + (5)/y` = 19, Hence, find a if y = ax + 3.
Can the following equations hold simultaneously?
7y - 3x = 7
5y - 11x = 87
5x + 4y = 43
If yes, find the value of x and y.
The length of a rectangle is twice its width. If its perimeter is 30 units, find its dimensions.
If 1 is added to the denominator of a fraction, the fraction becomes `(1)/(2)`. If 1 is added to the numerator of the fraction, the fraction becomes 1. Find the fraction.
The present ages of Kapil and Karuna are in the ratio 2 : 3. Six years later, the ratio will be 5 : 7. Find their present ages.
