Question

The denominator of the fraction is 1 more twice its numerator if 1 is added to numerator and the denominator respectively the ratio of numerator to denominator is 1:2 find the fraction

Answer 1

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Q) The denominator of the fraction is 1 more than twice its neumerator and if 1 is added to neumerator and the denominator respectively, the ratio of neumerator to denominator is 1:2 . Find the fraction .

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★ Given Conditions :

• denominator is 1 more than twice of neumerator.
• after adding 1 to both neumerator and denominator , the ratio of neumerator and denominator becomes 1:2

★ To Find :

• The fraction = ?

★ Solution :

• Let the neumerator be n
• and the denominator be d

$\boxed{ \sf \: fraction = \frac{n}{d} }$

According to the first condition -

denominator = 2×neumerator + 1

$\mapsto\sf{d = 2n + 1}...$

$\leadsto \sf \: \underline{d - 2n = 1}....(i)$

According to the second condition -

( neumerator + 1 ) : ( denominator + 1 ) = 1 : 2

( n + 1 ) : ( d + 1 ) = 1 : 2

$\mapsto \sf \: \frac{n + 1}{d + 1} = \frac{1}{2}$

$\mapsto \sf \: 2(n + 1) = d + 1$

$\mapsto \sf \: 2n + 2 = d + 1$

$\mapsto \sf \: 2 - 1 = d - 2n$

$\leadsto \sf \underline{d - 2n = 1}....(ii)$

As you can see , eq(i) is equal to eq(ii)

so ,

they may have Infinitely many solutions...

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# Some solution :

Let n = 1

put in eq(i)

$\mapsto \rm\: d - 2 \times 1 = 1$

$\mapsto \rm \: d - 2 = 1$

$\leadsto \rm \: d = 3$

so ,

$\boxed{ \bf \: fraction = \frac{1}{3} }$

or

Let n = 2

$\mapsto \rm \: d - 2 \times 2 = 1$

$\mapsto \rm \: d - 4 = 1$

$\leadsto \rm \: d = 5$

$\boxed{ \bf \: fraction = \frac{2}{5} }$

Similarly u can find more solutions ...

Answer 2

Answer:

ANSWER

We have,

Let the fraction

y

x

So, Numerator =y

Denominator=x

Then, According to given question,

y=2x+1

y−2x=1......(1)

So,

According to second condition,

y+1

x+1

=

2

1

⇒2x+2=y+1

⇒2x−y=−1......(2)

Solve from (1) and (2) to, and we get,

x=4andy=9

The fraction is

y

x

=

9

4

.

Step-by-step explanation:

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