If A + B = 5, B + C = 4, and C + D = 6, then AB + CD + AC + BD is:

Lets bring down what the question is finally asking about
AB + CD + AC + BD
A(B+C) + D(B+C)
(A+D)(B+C)
We know B+C = 4
So 4(A+D)

(A+B)-(B+C) = 5-4
(A-C)=1
(A-C) + (C+D) = (A+D) = 1+6 =7

Hence 4(A+D) = 4×7 = 28
Ans: Option D

Posted from my mobile device

A + B = 5
B + C = 4
C + D = 6
There are 3 equations with 4 unknowns. This means there isn't any unique solution. We can assign values to A,B,C,D as long as the equations are valid:

For example: A=1,B=4,C=0,D=6

AB + CD + AC + BD = 4+0+0+24 = 28

Answer D

Posted from my mobile device

AB+CD+AC+BD
A(B+C)+D(B+C)
(A+D)(B+C)
A+B=5 -
B+C=4
difference between those two gives A-C=1
C + D=6
Adding those two we will get A +D =7
then (A+D)(B+C)
7*4 = 28

Step 1: rearrange the question stem, Factor by Grouping

AB + AC + CD + BD = ?

A (B + C) + D (C + B) = ?

(A + D) (B + C) = ?

Given (B + C) = 4

(4) (A + D) = ?

Step 2: COMBINE the 3 equations - ADD

Notice that B and C appear twice in the 3 equations and A and D only ONCE

A + B + B + C + C + D = 5 + 4 + 6

A + D + 2B + 2C = 15

A + D = 15 - 2B - 2C

A + D = 15 - 2 (B + C)


Substitute the value of B + C = 4

A + D = 15 - 2 (4)

A + D = 7

*Q* what is the value of: (4) (A + D) = ?

(4) (7) = 28


Method 2: all the answer choices are positive integer values. The product of each the variables is most likely an integer value ——-> which means each individual variable is most likely an integer value

Try A = 1 —> B = 4 —-> C = 0 ——-> D = 6

AB + CD + AC + BD = ?

4 + 0 + 0 + 24 = 28


Try A = 2 ——> B = 3 —-> C = 1 —-> D = 5

6 + 5 + 2 + 15 = 28


28 is the answer

Posted from my mobile device

So to make things easier we toggle with the main question itself.
AB+BD+AC+CD can be factorized into (A+D)(B+C).
We know B+C=4
Therefore the question is what is the value of 4(A+D)
Now (A+B)-(B+C)=5-4 ------> A-C=1
(A-C) + (C+D)=6+1 ---------> A+D=7
Therefore answer= 7*4= 28 (D)

A + B = 5, B + C = 4, and C + D = 6, then AB + CD + AC + BD

Adding all three equations

=>A+2B+2C+D=15
=>A+D+2(B+C)=15
=>A+D=15-8=7

AB + CD + AC + BD
=>AB+AC+CD+BD
=>A(B+C)+D(B+C)
=>(A+D)(B+C)
=>7*4=28

Hence D

If A + B = 5, B + C = 4, and C + D = 6, then AB + CD + AC + BD is:

(A) 11
(B) 12
(C) 15
(D) 28
(E) 48

Solution
A+B=5,,,,,,,,,,,,1
B+C=4,,,,,,,,,,,,2
C+D=6,,,,,,,,,,,,3
add all there
A+B+C+D=11
A+D=7
AB + CD + AC + BD
AB+AC+CD+BD
A(B+C)+D(C+B)
(A+D)(B+C)
(7)(4)
28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.