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Bunuel
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Could you please elaborate on Statement 1 and your math??


Ted21
ST 1 A+C=B
Draw a line parallel to side-measure 1 from C TO A
Now for the right angle triangle 1^2 + (a-c)^2 = b^2
on solving this we'll get the value of ac ST 1 SUFF
ST 2 A-C=1
Answer = 1
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kelvind13
Could you please elaborate on Statement 1 and your math??


Ted21
ST 1 A+C=B
Draw a line parallel to side-measure 1 from C TO A
Now for the right angle triangle 1^2 + (a-c)^2 = b^2
on solving this we'll get the value of ac ST 1 SUFF
ST 2 A-C=1
Answer = 1

Statement 1:
a+c=b
square both sides of the above equation. This then becomes a^2 + 2ac + c^2 = b^2

a^2 + 2ac + c^2 = b^2
-(a^2 - 2ac + c^2 + 1 = b^2)

When we subtract the two equations it becomes

a^2 + 2ac + c^2 = b^2
-a^2 + 2ac - c^2 - 1 = -b^2

which equals
4ac-1=0
4ac=1
ac=(1/4)

Hopefully this helps
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Bunuel,

Can you please post the complete solution?

Thanks
Akash
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AkashKashyap
Bunuel,

Can you please post the complete solution?

Thanks
Akash

Akash

Please refer the fig.attached
(1) a+c=b
sq. both sides
a^2+c^2+2ac=b^2-----------(A)
also from fig the cutout upper right angled triangle having hypotenuse b we can write
(a-c)^2+1^2=b^2
a^2+c^2-2ac+1=b^2-----------(B)
now equate (A) & (B)

a^2+c^2+2ac=a^2+c^2-2ac+1
ac=1/4
suff...

(2) a-c=1
we can only calculate b=sq rt 2 nothing else
insuff...

Ans A
Attachments

lnzFUe1NsWpRHYetrW_BAMHT3M0ml9ssJU6sexynIY0SxO2MrXLRX71d4EjIqD-ITXHoVyRCXvhKECDrtvrdwtt0Q9b1h5-W_LSClrh9liw7pUognufJTS-OvSXIC20UfQ.png
lnzFUe1NsWpRHYetrW_BAMHT3M0ml9ssJU6sexynIY0SxO2MrXLRX71d4EjIqD-ITXHoVyRCXvhKECDrtvrdwtt0Q9b1h5-W_LSClrh9liw7pUognufJTS-OvSXIC20UfQ.png [ 4.96 KiB | Viewed 6962 times ]

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Bunuel

In the trapezoid above, with sides a, b, c, and 1, what is the value of ac ?

(1) a + c = b

(2) a – c = 1


Kudos for a correct solution.

Attachment:
lnzFUe1NsWpRHYetrW_BAMHT3M0ml9ssJU6sexynIY0SxO2MrXLRX71d4EjIqD-ITXHoVyRCXvhKECDrtvrdwtt0Q9b1h5-W_LSClrh9liw7pUognufJTS-OvSXIC20UfQ.png

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There are 3 variables and 1 hidden equation from Pythagoras theorem, which is \((a-c)^2 + 1 = b^2\). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. However, this question is not about a single variable, but about ab. We need to look into any relation about \(ab\).

Condition 1)

\((a+c)^2 = b^2 = (a-c)^2 + 1\)
\((a+c)^2 = (a-c)^2 + 1\)
\(a^2 + 2ac + c^2 = a^2 -2ac + c^2 + 1\)
\(2ac = -2ac + 1\)
\(4ac = 1\)
\(ac = 1/4\)

This is sufficient.

Condition 2)

We have \((a-c)^2 + 1 = b^2\).
\(b^2 = 2\)
We can not declare the value of \(ac\).

Thus, this is not sufficient.

Therefore, the correct answer is A.
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Bunuel

In the trapezoid above, with sides a, b, c, and 1, what is the value of ac ?

(1) a + c = b

(2) a – c = 1


Kudos for a correct solution.

Attachment:
lnzFUe1NsWpRHYetrW_BAMHT3M0ml9ssJU6sexynIY0SxO2MrXLRX71d4EjIqD-ITXHoVyRCXvhKECDrtvrdwtt0Q9b1h5-W_LSClrh9liw7pUognufJTS-OvSXIC20UfQ.png

Refer to the figure given by rohit8865 above: https://gmatclub.com/forum/in-the-trape ... l#p1718294

That is how we can link b with a and c.
So we know that \(b^2 = 1 + (a-c)^2\) (By Pythagorean theorem)

(1) a + c = b

\((a+c)^2 = 1 + (a-c)^2\)
ac = 1/4
Sufficient alone

(2) a – c = 1
Then, b = sqrt(2) but we do not know anything about a and c. We know that the diff between them is 1 so if a = 4, b = 3 and if a = 6, b = 5 etc but we don't know what a and c are. Imagine that we hold the trapezoid from the top and can stretch it up as much as we wish. The values of a and c will keep changing but all else will stay the same. Hence we cannot get the value of ac. Not sufficient.

Answer (A)
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