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Bunuel
If a, b, c, and d are consecutive even integers and a < b < c < d, then a + b is how much less than c + d ?

A. 2
B. 4
C. 6
D. 8
E. 10

We can write the consecutive even integers in the form:
a=2n
b=2n+2
c=2n+4
d=2n+6 where n is a non-negative integer. \(n\geq{0}\)

Now a+b=4n+2 & c+d=4n+10

So, (c+d)-(a+b)=4n+10-(4n+2)=8

Ans. D
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Bunuel
If a, b, c, and d are consecutive even integers and a < b < c < d, then a + b is how much less than c + d ?

A. 2
B. 4
C. 6
D. 8
E. 10
\(a, b, c, d\)
\(2, 4, 6, 8\)
\((a+b)=(2+4)=6\)
\((c+d)=(6+8)=14\)
\((c+d)-(a+b)=(14-6)=8\)

The difference between and range of four consecutive integers will always be the same.

Test if in doubt with another number set:
\(6, 8, 10, 12\)
\((22 - 14) = 8\)

Answer D
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Bunuel
If a, b, c, and d are consecutive even integers and a < b < c < d, then a + b is how much less than c + d ?

A. 2
B. 4
C. 6
D. 8
E. 10

We can let a, b, c, and d equal 2, 4, 6, and 8 respectively.

a + b = 2 + 4 = 6

c + d = 6 + 8 = 14

Thus, a + b is 14 - 6 = 8 less than c + d.

Answer: D
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I use the substitution method in such a question:So use whatever consecutive even numbers that come to your mind.
For eg.2,4,6,8
By simple math in the mind,we get (6+8)-(2+4)=8,which is the answer.
I hope it helps
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Bunuel
If a, b, c, and d are consecutive even integers and a < b < c < d, then a + b is how much less than c + d ?

A. 2
B. 4
C. 6
D. 8
E. 10
a, b, c and d are basically x, x+2, x+4 and x+6 respectively.

Now to find: (x+4 + x+6) - (x + x+2)=8
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