Question

7 is added to a certain number and the sum is multiplied by 5 the product obtained is divided by 9 and 3 is subtracted from the quotient now the result is 12 find two times of the number



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Answer 1

Answer:

There are 2 answers to this question: 20 and 21.

How?

Let's see.

Let the initial number be n.

When 7 is added to it, we have n+7

When multiplied by 5, we get 5(n+7)

When divided by 5(n+7)/9

Let the quotient of the above division be q

Now, when 3 is subtracted from the quotient of the above division (q) we are left with 12. So,

q−3=12

So we know that q=15

Now, since q was the quotient, there may or may not have been a remainder. So we now know that 5(n+7)/9 lies in the range [15,16)

That means the fraction has a value greater than or equal to 15, but less than 16.

This means that 5(n+7)

lies in the range [135,144) and n+7 lies in the range [27,28.8)

Now, since n

is an integer, n+7 can either be 27 or 28.

Thus we get 2 values of n, 20 and 21.

P.S.: You can recalculate from top to bottom using both these values and verify the result.

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Answer 2

Answer:

There are 2 answers to this question: 20 and 21.

How?

Let's see.

Let the initial number be n.

When 7 is added to it, we have n+7

When multiplied by 5, we get 5(n+7)

When divided by 5(n+7)/9

Let the quotient of the above division be q

Now, when 3 is subtracted from the quotient of the above division (q) we are left with 12. So,

q−3=12

So we know that q=15

Now, since q was the quotient, there may or may not have been a remainder. So we now know that 5(n+7)/9 lies in the range [15,16)

That means the fraction has a value greater than or equal to 15, but less than 16.

This means that 5(n+7)

lies in the range [135,144) and n+7 lies in the range [27,28.8)

Now, since n

is an integer, n+7 can either be 27 or 28.

Thus we get 2 values of n, 20 and 21.

P.S.: You can recalculate from top to bottom using both these values and verify the result.

Please mark me as brainlist thx