Question

find two consecutive even numbers such that half of the larger exceeds 1/8th of the smaller by 4 find the numbers

Answer 1

Answer:

Let those two number be ‘x’ (small number) and 'x+2′ (large number), according to the question:

((Large number)/2) -([(Small number)/4)=5

Or,

((x + 2)/2) - (x/4) = 5,

((2x + 4)/4) - (x/4) = 5,

(2x - x + 4) = 20,

x + 4 = 20,

x = 16,

So first even number is 16 and second consecutive even number is 18 which was (x + 2).

Answer 2

Step-by-step explanation:

farhatr72

17.01.2018

Math

Secondary School

answered

Find 2 consecutive even numbers such that half of the larger exceeds 1/4th of the smaller by 5.

2

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RK242

RK242

Here it is..

Hope it helps

tardymanchester

tardymanchester

Answer:

Two consecutive numbers are 16 and 18.

Step-by-step explanation:

To find : 2 consecutive even numbers such that half of the larger exceeds 1/4th of the smaller by 5.

Solution :

Let the two consecutive even numbers are 2n , 2n+2

Half of the larger number \frac{2n+2}{2}

According to question,

\frac{2n+2}{2}=\frac{2n}{4}+5

n+1=\frac{n}{2}+5

n-\frac{n}{2}=5-1

\frac{n}{2}=4

n=8

So, The larger number 2n+2=2(8)+2=18

Smaller number 2n=2(8)=16

Therefore, Two consecutive numbers are 16 and 18.