DAT Practice Question Help

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PaulC

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Hello everyone,

I need some help here. I'm not sure how to solve this problem.

"If ab=2 bc=3 cd=6 then ad=?"

Not sure how to go about solving this problem. I think I'll feel stupid once someone shows me how simple it is but I have no idea how to approach it at the moment.

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I'm not sure if it's the easiest way but you can rearrange for a single variable and than substitute that into another equation. Rinse and repeat til you get your answer than you can back track and find the value of each variable. There is probably a quicker way via subtracting/adding equations.
 
Hello everyone,

I need some help here. I'm not sure how to solve this problem.

"If ab=2 bc=3 cd=6 then ad=?"

Not sure how to go about solving this problem. I think I'll feel stupid once someone shows me how simple it is but I have no idea how to approach it at the moment.

I'm going to go ahead and take a stab at this, sorry if I just completely get it wrong.

a*b = 2, so I'm going to assume integers to begin with, so I'm assuming A = 2 and B = 1
b*c = 3, so B = 1 is working out, therefore C must be 3
c*d = 6, so if C = 3, then D should equal 2...

a*d = 4, because a = 2 and d = 2 ?

It would be nice if we were able to see the actual answer and/or the multiple choice possibilities! :D

__________________________________________________
Edit: Tried it a different way, got the same answer (just follow the numbers in order...) :D

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Last edited:
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ab=2
bc=3
cd=6

My answer is similar to top one.
leave bc=3 as is
but b=2/a...(1) from ab=2
and c=6/d...(2) from cd=6

now substitute (1) and (2) into bc=3
(2/a)(6/d)=3
(12/ad)=3
12= 3ad
4=ad
 
Hello everyone,

I need some help here. I'm not sure how to solve this problem.

"If ab=2 bc=3 cd=6 then ad=?"

Not sure how to go about solving this problem. I think I'll feel stupid once someone shows me how simple it is but I have no idea how to approach it at the moment.

you want to solve for AD, so i will attempt to find what i can substitute for A and D, reducing to once common term.

I. ab=2, therefore a= 2/b
II. bc=3 therefore b = 3/c
III. cd=6, therefroe d= 6/c

now i can substitue for AD
( 2/b ) ( 6/c) =?

To come up with like terms, i will substitute for my previous vale of b, which could also be done in the opposite manner substituting for c instead in equation II.

( 2/ [3/c] ) ( 6/c) =?

( 2c/3 ) ( 6/c ) = ?

c's cancel, and you have (2/3) (6) = 12/3 = 4
 
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