Question

THE COST OF 2 KG OF APPLES AND 1 KG 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...

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Answer 1

Step-by-step explanation:

Let the cost of apples per kg be Rs x

Let cost of grapes per kg be Rs y

Given that .....

2kg apples and 1 kg grapes cost rupees 160

2× cost per kg of Apple + 1 × cost per kg of grapes= 160

2x + y = 160....{1}

Also ,

4 kg apples and 2 kg grapes cost Rs 300

4× cost per kg of apples + 2 × cost per kg of grapes= 300

4x + 2y = 300

2( 2x + y ) = 2 × 150

2x + y = 150 ....{2}

Now , plotting equation

2x + y = 160 .... ( 1 )

2x + y = 150 .... ( 2 )

For equation ( 1 )

2x + y = 150

Let x = 50

2 ( 50 ) + y = 150

100 + y = 150

y = 150 - 100

y = 50

Let y = 60

2 ( 60 ) + y = 150

120 + y = 150

y = 150 - 120

y = 30

For equation ( 2 )

Let x = 50

2 ( 50 ) + y = 160

100 + y = 160

y = 160 - 100

y =60

so, x = 50 , y = 60 is a solution

Let x = 60

2 ( 60 ) + y = 160

120 + y = 160

y = 160 - 120

y = 40

So , x = 60 , y = 40 is a solution

Hope it helps you

Answer 2

$\Huge\bf \rightarrow \mid\mathcal{\underline{ \blue{Answer}}} \mid$

$x  \: 65 \:  55 \:  45 \:  35$

$y  \: 20  \: 40 \:  60 \:  80$

Let the cost of 1 kg of apples be x and that of 1 kg be y.

$\large\bf \implies\mathbb {\underline{\red{2x+y=160}}}$

$\small\bf\implies{\underline{\red{4x+2y=300⇒2x+y=150}}}$

The situation can be represented graphically by plotting these two equations.

$\small\bf\implies{\underline{\purple{2x+y=160⇒y=160−2x}}}$

$\large\bf \implies\mathbb{\underline{\pink{x \:  70  \: 60 \:  50 \:  40}}}$

$\small\bf\implies \mathbb{\underline{\green{y=160−2x  \: 20  \: 40 \:  60 \:  80}}}$

$\large\bf\implies \mathbb{\underline{\orange{2x+y=150⇒y=150−2x}}}$

$\large\bf \longrightarrow{ \underline{\mathbb {\red{ x \:  65  \: 55  \: 45  \: 35 }}}}$

‌

y=150−2x 20 40 60 80

We can see that the lines do not intersect anywhere, i.e. they are parallel.

• Hence we can not arrive at a solution.