DISCUSS
Two situations are given below:
1 The cost of 1kg potatoes and 2kg tomatoes was 30 on a certain day. After two days,
the cost of 2kg potatoes and 4kg tomatoes was found to be 366.
The coach of a cricket team of M.K.Nagar High School buys 3 bats and 6 balls for
3900. Later he buys one more bat and 2 balls for 1300.
Identify the unknowns in each situation. We observe that there are two unknowns in each
se.
Answers
1) The cost of 1kg potatoes and 2kg tomatoes was 30 on a certain day. After two days, the cost of 2kg potatoes and 4kg tomatoes was found to be 66 .
Answer :-
Let us assume that, cost of 1 kg potatoes is Rs.x and cost of 1 kg tomatoes is Rs.y .
so,
→ x + 2y = 30 ------ Eqn.(1) => a1x + b1y = c1
and, after two days
→ 2x + 4y = 66 Eqn.(2) => a2x + b2y = c2
comparing both Equations we get,
→ a1/a2 = 1/2
→ b1/b2 = 2/4 = 1/2
→ c1/c2 = 30/66 = 15/33 .
then,
→ 1/2 = 1/2 ≠ 15/33 .
or,
→ a1/a2 = b1/b2 ≠ c1/c2 .
therefore, we can conclude that, both the equations are parallel . so, they does not intersect each other at any point. Hence no solution is possible .
This concludes that, the cost of potato and tomatoes was different after 2 days .
2) The coach of a cricket team of M.K.Nagar High School buys 3 bats and 6 balls for 3900. Later he buys one more bat and 2 balls for 1300.
Answer :-
Let cost of a bat is Rs.x and cost of a ball is Rs.y .
so,
→ 3x+ 6y = 3900 ------ Eqn.(1) => a1x + b1y = c1
and, after two days
→ x + 2y = 1300 Eqn.(2) => a2x + b2y = c2
comparing both Equations we get,
→ a1/a2 = 3/1
→ b1/b2 = 6/2 = 3/1
→ c1/c2 = 3900/1300 = 3/1 .
then,
→ 3/1 = 3/1 = 3/1 .
or,
→ a1/a2 = b1/b2 = c1/c2 .
therefore, we can conclude that, both the lines of equations are coincident lines . So, they have infinite number of solutions possible as both the equations are equivalent .
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