Question

3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be 160. After a
month, the cost of 4 kg of apples and 2 kg of grapes is 300. Represent the situation
algebraically and geometrically.​

Answer 1

Answer:

Step-by-step explanation:

Let the cost of apples per kg be Rs X

Let the cost of grapes per kg be Rs Y

2kg apples and 1kg grapes cost Rs 160

2 * Cost per kg of apples + 1 * cost per kg of grapes = 160

->2x+y=160

->2x+y-160 = 0 ....(1)

4kg apples and 2kg grapes cost Rs 300

4 * Cost per kg of apples + 2 * cost per kg of grapes = 300

->4x+2y=300

->4x+2y-300=0

->2(2x+y-150)=0

->(2x+y-150)=0/2=0 ....(2)

Now, plotting equations,

2x+y-160 = 0

2x+y-150 = 0

Solving (1),

2x+y-160 = 0

y = 150-2x

Let x = 50 ,

y=150-2(50)

y=150-100

y=100

So, x = 50, y = 50 is a solution  

i.e., (50,50) is a solution.

Let x = 60 ,

y=150-2(60)

y=150-120

y=30

So, x = 60, y = 30 is a solution  

i.e., (60,30) is a solution.

Solving (2),

2x+y-160

y = 160-2x

Let x = 50 ,

y=160-2(50)

y=160-100

y=60

So, x = 50, y = 60 is a solution  

i.e., (50,60) is a solution.

Let x = 60 ,

y=160-2(60)

y=160-120

y=40

So, x = 60, y = 40 is a solution  

i.e., (60,40) is a solution.

Find the attachment for geometric situation.

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