Question

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Answer 1

Answer:

Let the number be ×

Thrice a number = 3×

Increased by 8 = 50

3× + 8 = 50

3× = 50-8= 42

× = 42÷3 = 14...

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Answer 2

Topic

Linear Equations

Given

1) Thrice a number when increased by 8 gives 50.

2) The difference of 12 and a number is 15.

3) A number whose 6th part exceeds its 9th part by 1.

To Find

The required number in each question.

Solution

1) Let number be 'x'.

It is given that thrice the number when increased by 8 gives 50.

3x + 8 = 50

3x = 50 - 8

3x = 42

x = 42/3

x = 14

So, required number is 14.

2) Let number be 'x'.

It is given that the difference of 12 and a number is 15.

12 - x = 15

x = 12 - 15

x = -3.

So, the required number is -3.

3) Let number be 'x'.

It is given that the 6th part of a number exceeds its 9th part by 1.

$\frac{1}{6}x - \frac{1}{9}x = 1$

$\frac{ (3 - 2) }{18}x = 1$

$\frac{1}{18}x = 1$

Cross multiply,

x = 18

So, the required number is 18.

Answer

So, the answer of respective questions are

• 14

• -3 and

• 18