Question

find two consecutive even no.s such that 3/4 of the first exceeds two fifth of the other no. by 9 .
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Answer 1

Answer:

Required consecutive numbers are a i.e. 28 and a + 2 i.e. 28 + 2 = 30.

Step-by-step explanation:

It is given that the required numbers are the two consecutive even numbers, they can be negative too.

So, let the required numbers are a and ( a + 2 ) .

According to the question : -

= > 3 / 4 of the first exceeds two fifth of the other number by 9.

= > 3 / 4 of first number = 2 / 5 of 2nd number + 9

Case 1 : Where the required numbers are positive.

= > 3 / 4 x a = [ 2 / 5 x ( a + 2 ) ] + 9

= > 3a / 4 = 2( a + 2 ) / 5 + 9

= > 3a / 4 = { 2a + 4 + 45 } / 5

= > 3a / 4 = ( 2a + 49 ) / 5

= > 15a = 4( 2a + 49 )

= > 15a = 8a + 196

= > 15a - 8a = 196

= > 7a = 196

= > a = 28

Hence the required consecutive numbers are a i.e. 28 and a + 2 i.e. 28 + 2 = 30.

Answer 2

$\textbf{\underline{\underline{Answer :-}}}$

The two numbers are 28 and 30.

$\textbf{\underline{\underline{Explanation :-}}}$

$\textsf{\underline{\underline{Given :}}}$

The numbers are consecutive and even.

$\dfrac{3}{4}$ of first is = $\dfrac{2}{5}$ of other number by 9

$\textsf{\underline{\underline{To find :}}}$

The numbers

$\textsf{\underline{\underline{Solution :}}}$

Consider the 1st number as x

Consider the 2nd number as x + 2

★ Equation :

$\boxed{\tt{ \frac{3}{4} \: of \: x =\frac{2}{5} \: of \: (x + 2) + 9}}$

$\tt{\implies} \: \dfrac{3}{4} \times x =\dfrac{2}{5} \times( x + 2) + 9$

$\tt{\implies} \: \dfrac{3x}{4} =\dfrac{2(x + 2)}{5} + 9$

$\tt{\implies} \: \dfrac{3x}{4} =\dfrac{2x + 4}{5} + 9$

Now, we need to make $\dfrac{2x+4}{5}$ and $\dfrac{9}{1}$ like fractions.

LCM of 1 and 5 = 5

$\tt{\implies} \: \dfrac{3x}{4} =\dfrac{2x + 4}{5} +\dfrac{9 \times 5}{1 \times 5}$

$\tt{\implies} \: \dfrac{3x}{4} = \dfrac{2x + 4 + 45}{5}$

$\tt{\implies} \: 5 \times 3x = 4(2x + 49)$

$\tt{\implies} \: 15x = 8x + 196$

$\tt{\implies} \: 15x - 8x = 196$

$\tt{\implies} \: 7x = 196$

$\tt{\implies} \: x =\dfrac{196}{7}$

$\tt{\implies}\red{ \: x = 28}$

$\therefore$ One number = 28

Value of x + 2

$\tt{\implies} \: 28 + 2$

$\tt{\implies}\red{ \: 30}$

$\therefore$ The two numbers are 28 and 30.