A shopkeeper sold 3 pencils and 4 pens to a customer for Rs.44.A little later another customer purchased 6 pencils and 8 pens for Rs.90.Form a pair of linear equations to represent this situation.
Answers
Given : A shopkeeper sold 3 pencils and 4 pens to a customer for Rs.44.
A little later another customer purchased 6 pencils and 8 pens for Rs.90.
To Find : Form a pair of linear equations to represent this situation.
Solution:
Let say pencil cost = Rs x
and Pen cost = Rs y
shopkeeper sold 3 pencils and 4 pens to a customer for Rs.44.
=> 3x + 4y = 44
6 pencils and 8 pens for Rs.90.
=> 6x + 8y = 90
a pair of linear equations to represent this situation.
3x + 4y = 44
6x + 8y = 90
additional info : no solution exist for this linear pair of equation so can be mistake in data in Question
as 6x + 8y = 90 => 3x + 4y = 45
while another equation is 3x + 4y = 44
Hence no solution
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Given :-
- A shopkeeper sold 3 pencils and 4 pens to a customer for Rs.44.
- A little later another customer purchased 6 pencils and 8 pens for Rs.90.
Solution :-
Let us assume that , SP of 1 pencil is Rs.x and SP of 1 pen is Rs.y .
so,
→ SP of 3 pencil + 4 pens = Rs.44
→ 3 * x + 4 * y = 44
→ 3x + 4y = 44 ---------- Equation (1)
similarly,
→ SP of 6 pencil + 8 pens = Rs.90
→ 6 * x + 8 * y = 44
→ 6x + 8y = 90 ---------- Equation (2)
now, Equation (2) can be written as,
→ 2(3x + 4y) = 90
→ 3x + 4y = 45
but first Equation also says that,
→ 3x + 4y = 40
therefore,
→ 3x + 4y = 40
→ 3x + 4y = 45
→ 40 ≠ 45 .
which is not possible .
But we can represent both equations a pair of linear equations as ,
→ 3x + 4y = 44 .
→ 6x + 8y = 90 .