Question

pls solve ques no 11, 13

Answer 1

Answer:

11. y = 15

13. y = 6 + 4x

Step-by-step explanation:

11) As y is directly proportional to x, the equation can be written as,

y = Kx, where 'K' is a proportionality constant.

Now, substitute given values of y = 12 and x = 4,

We get:

12 = K*4 which implies that K = 3.

This, the above assumed equation can be written as,

y = 3x.

So, when we put in x = 5 in the equation, we get

y = 3*5 = 15.

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13) Let 'y' be the variable for amount in ₹

And 'x' be the variable for kilometres.

For one kilometre i.e., x = 1, y = 10.

But for additional kilometres, y varies as 4x, since it's ₹4/km.

Thus, forming equation y = a + 4x

Substitute the values x = 1 and y = 10 in above assumed equation. We get,

10 = a + 4(1). Implying a = 6.

Therefore, the equation is y = 6 + 4x.

Now, substitute various points for x = 1, 2, 3... and obtain various 'y' values and plot them onto the graph.

Ex: For x = 1,y = 6 + 4(1) = 10

For x = 2, y = 6 + 4(2) = 14

For x = 3, y = 6 + 4(3) = 18 and so on.

Answer 2

Answer:

Step-by-step explanation:

Solution of Q ; 11

Given:

Let y varies directly as x. if y=12 when x=4 ,then write a linear equation. what is value of y when x =5

Solution:

y Varies directly as x, the eqaution will be in form of

y = kx, where k is constant of proportionality.

Putting y = 12 and x = 4, we get

=> 12 = 4k

=> k = 12/4

=> k = 3

So, y = 3x

Linear Equation : y - 3x = 0

When x = 5

y - 3*5 = 0

=> y = 15

Solution of Q ; 13:

Given:

The auto rickshaw fare in a city is charged as rs 10 for the first km and at rs 4 per km for subsequent distance covered write the linear equation to express the above statement and also draw the graph.

Solution :-

Let the total distance covered be x km

and, the total fare be Rs. y 

Since, Total fare = fare for the first kilometer + fare for the remaining kilometer of distance  ...........................(1)

We know that the fare for first kilometer is Rs. 10

Fare for the remaining kilometers i.e. (x - 1)

= 4(x - 1)

Put values in (1)

⇒ y = 10(1) + 4(x - 1)

⇒ y = 10 + 4x - 4

⇒ y = 4x + 6

When x = 0, we have, y = 4 × (0) + 6, So, y = 6

When x = - 1, we have, y = 4 × (- 1) + 6, = - 4 + 6

So, y = 2

 

When x = - 2, we have, y = 4 × (- 2) + 6 = - 8 + 6

So, y = - 2

x = 0 ; y = 6

x = - 1 ; y = 2

x = - 2 ; y = - 2

Now, you can draw the graph easily.