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The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens
and two erasers is Rs.32.How much will three pencils, three pens and three erasers cost?
- a)25
- b)26
- c)27
- d)29
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The cost of one pencil, two pens and four erasers is Rs.22 while the c...
Solutionlet the cost of 1 pencil, 1 pen and 1 eraser are respectively x, y, z. Then according to the given condition
1x + 2y + 4z = 22
5x + 4y + 2z = 32
on adding these two equations we get 6x + 6y + 6z = 54
or 3x + 3y + 3z = 27
Most Upvoted Answer
The cost of one pencil, two pens and four erasers is Rs.22 while the c...
Given:
Cost of 1 pencil + 2 pens + 4 erasers = Rs. 22
Cost of 5 pencils + 4 pens + 2 erasers = Rs. 32
Let the cost of each pencil, pen and eraser be x, y and z respectively.
From the given information, we can form the following equations:
1x + 2y + 4z = 22 ...(1)
5x + 4y + 2z = 32 ...(2)
To find the cost of 3 pencils, 3 pens and 3 erasers, we need to find the value of 3x + 3y + 3z.
Multiplying equation (1) by 3 and subtracting equation (2) multiplied by 2, we get:
3x + 6y + 12z - (10x + 8y + 4z) = 66 - 64
-7x - 2y + 8z = 2 ...(3)
Multiplying equation (1) by 4 and subtracting equation (2) multiplied by 5, we get:
4x + 8y + 16z - (25x + 20y + 10z) = 88 - 160
-21x - 12y + 6z = -72 ...(4)
Multiplying equation (3) by 3 and adding it to equation (4) multiplied by 2, we get:
-63x - 30y + 42z = -222
3x + 3y + 3z = (222 - 42z)/3
Substituting this value in equation (1), we get:
x + 2y + 4z = 22/3
Substituting the value of x + 2y from equation (1) in equation (3), we get:
-7(x + 2y) + 8z = 2
-7(22/3 - 4z) + 8z = 2
-154/3 + 28z = 2
28z = 160/3
z = 20/7
Substituting the value of z in equation (1), we get:
x + 2y = 8/7
Substituting the values of z and x + 2y in equation (2), we get:
5x + 4y = 114/7
Solving these two equations, we get:
x = 2/7, y = 25/49
Therefore, the cost of 3 pencils, 3 pens and 3 erasers is:
3x + 3y + 3z = 27/7 = 3.86 ≈ Rs. 27 (approx)
Cost of 1 pencil + 2 pens + 4 erasers = Rs. 22
Cost of 5 pencils + 4 pens + 2 erasers = Rs. 32
Let the cost of each pencil, pen and eraser be x, y and z respectively.
From the given information, we can form the following equations:
1x + 2y + 4z = 22 ...(1)
5x + 4y + 2z = 32 ...(2)
To find the cost of 3 pencils, 3 pens and 3 erasers, we need to find the value of 3x + 3y + 3z.
Multiplying equation (1) by 3 and subtracting equation (2) multiplied by 2, we get:
3x + 6y + 12z - (10x + 8y + 4z) = 66 - 64
-7x - 2y + 8z = 2 ...(3)
Multiplying equation (1) by 4 and subtracting equation (2) multiplied by 5, we get:
4x + 8y + 16z - (25x + 20y + 10z) = 88 - 160
-21x - 12y + 6z = -72 ...(4)
Multiplying equation (3) by 3 and adding it to equation (4) multiplied by 2, we get:
-63x - 30y + 42z = -222
3x + 3y + 3z = (222 - 42z)/3
Substituting this value in equation (1), we get:
x + 2y + 4z = 22/3
Substituting the value of x + 2y from equation (1) in equation (3), we get:
-7(x + 2y) + 8z = 2
-7(22/3 - 4z) + 8z = 2
-154/3 + 28z = 2
28z = 160/3
z = 20/7
Substituting the value of z in equation (1), we get:
x + 2y = 8/7
Substituting the values of z and x + 2y in equation (2), we get:
5x + 4y = 114/7
Solving these two equations, we get:
x = 2/7, y = 25/49
Therefore, the cost of 3 pencils, 3 pens and 3 erasers is:
3x + 3y + 3z = 27/7 = 3.86 ≈ Rs. 27 (approx)
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