Question

Frame any two linear equations based on daily life situations in one variable and find solution

Answer 1

Answer:

1) Problem :

Bilal have some money.He give half of it to his brother then spend half of remaining on shopping.Now he has 10 rupees.How many money did he has initially.

Solution:

Let

Bilal have money = x

Then

according to condition given in the question

Bilal given half of it money to his brother = x/2

Bilal has remaining money = x - x/2 = x/2

Now

He spend half of remaining on shopping = (x/2)/2 = x/4

Now

Bilal remaining money = x/2 - x/4 = x/4

Now according to final condition

Bilal has money = 10 = x/4

⇒   x = 10×4 =40

So

Bilal has initially 40 rupees.

2) Problem :

There are two identical jugs , A and B . Jug A is 3/7 full of water and Jug B is 96/11 full . What fraction of the capacity of a jug of water should be poured from B to A so that they both have the same volume of water ?​

Solution:

We are given A and B are two jugs and have equal volume

Let say both have volume = y m³

The water in Jug A = 3/7 of total volume of A = 3y/7

The water in Jug B = 96/11 of total volume of B = 96y/11

Suppose x is subtracted form B and added in A such that volume to A and B become equal.

New volume of B = 96y/11 - x

New volume of A = 3y/7 + x

According to condition

         New volume of B = New volume of A

Putting values we get

           96y/11 - x = 3y/7 + x

⇒  x + x = 96y/11 - 3y/7 = ( 672y - 33y ) / 77 = 639y / 77

⇒      2x = 639y / 77

⇒        x = 639y / 154

So

639y/154 is added in A to balance the volumes of A and B

Answer 2

Answer:

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Step-by-step explanation:

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