find 2 consecutive even numbers such that 3/4of the first exceeds 2/5 of the second by 9......please show the working
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let a be first number a+2 be second number
givena(3/4)-(a+2)((2/5)=9
(3a/4)-(2a+4/5)=9
on solving this equation we get a as 28
hence these two numbers are 28 and 30
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Let the two consecutive even numbers be x and x + 2
Given that 2 consecutive even numbers such that (3/4) of the first number exceeds (2/5) of the second by 9.
= > (3/4) of x - (2/5) of (x + 2) = 9
= > (3x/4) - (2x + 4)/5 = 9
= > 5(3x) - 4(2x + 4)/20 = 9
= > 15x - 8x - 16 = 20 * 9
= > 7x - 16 = 180
= > 7x = 196
= > x = 28.
So,
The 1st number = x = 28
The number = x + 2 = 30.
Therefore, the numbers are 28 and 30.
Hope this helps!
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