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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
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Bunuel wrote:
If \(a + b + c + d = 12\), what is the value of \(\sqrt{a^2+b^2+c^2+d^2}\) ?

(1) \(ab = cd = ad\)

(2) \(|a| = |b| = |c| = |d|\)


 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

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(1) \(ab = cd = ad\)
\(ab = ad\) => \(b=d\)
\(cd = ad\) =>\( c=a\)
2(a+b) = 12 => a+b=6
2(c+d) = 12 => c+d=6
(a,b,c,d) = (1,5,1,5) => \(\sqrt{a^2+b^2+c^2+d^2}=\sqrt{52}\)
(a,b,c,d) = (2,4,2,4) => \(\sqrt{a^2+b^2+c^2+d^2}=\sqrt{40}\)
Insufficient

(2) \(|a| = |b| = |c| = |d|\)
(a,b,c,d) = (6,6,6,-6) => \(\sqrt{a^2+b^2+c^2+d^2}=12\)
(a,b,c,d) = (3,3,3,3) => \(\sqrt{a^2+b^2+c^2+d^2}=6\)
Insufficient

Both
(a,b,c,d) = (3,3,3,3) => \(\sqrt{a^2+b^2+c^2+d^2}=6\)
Sufficient
Ans C
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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
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Given

a + b + c + d = 12

Ques : \(\sqrt{a^2 + b^2 + c^2 + d^2}\)

Statement 1

ab = cd

ab = ad
Inference
a(b-d) = 0
a = 0 -----------(1)
b = d -----------(1.1)

cd= ad
Inference
d(c-a) = 0
d = 0 -----------(2)
c = a -----------(2.1)

Assume a = d = 0

Case 1

c = b = 6

\(\sqrt{6^2 + 6^2 } = 6\sqrt{2} \)

Case 2

c = 11 ; b = 1

\(\sqrt{11^2 + 1^2 } = \sqrt{122} \)

We are getting two different values, hence eliminate A

Statement 2

|a| = |b| = |c| = |d|

Inference

The distance of a from 0 = The distance of b from 0 = The distance of c from 0 = The distance of d from 0

Case 1

a = b = c = d = 3

\(\sqrt{a^2 + b^2 + c^2 + d^2} = \sqrt{3^2 + 3^2 + 3^2 + 3^2} = \sqrt{36}\)

Case 2

a = -6
b = c = d = 6

\(\sqrt{a^2 + b^2 + c^2 + d^2} = \sqrt{6^2 + 6^2 + 6^2 + 6^2} = \sqrt{36*4}\)

As we are getting two different values, the answer is not sufficient.

Combining

We know that the distances have to be equal, also none of the values can now be 0. Hence all of them need to be same.

a = b = c = d = 3

Hence we can get a definite answer

IMO C
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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
Asked: If \(a + b + c + d = 12\), what is the value of \(\sqrt{a^2+b^2+c^2+d^2}\) ?

\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad + bc+bd+cd)\)

(1) \(ab = cd = ad\)
\(ab = cd = ad = \sqrt{abcd}\)
\(ad = \sqrt{abcd}\)
aˆ2dˆ2 = abcd
ad = bc
ab = bc = cd = da
a = b = c = d = (a+b+c+d)/4 = 12/4 = 3
ab = ac = ad = bc = bd = cd = 3*3 = 9
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 2*6*9= 144 - 108 = 36\)
\(\sqrt{a^2+b^2+c^2+d^2} = 6\)
SUFFICIENT

(2) \(|a| = |b| = |c| = |d|\)
Case 1: a=b =c=d = 3;
ab = bc = cd = da = 3*3=9
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 2*6*9= 144 - 108 = 36\)
\(\sqrt{a^2+b^2+c^2+d^2} =6\)
Case2: a=b=c=4; d=-4
ab = bc = ca = 4*4 = 16; ad=bd=cd=-16
ab + ac + ad + bc + bd + cd = 16*3 - 16*3 = 0
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 0 = 12ˆ2\)
\(\sqrt{a^2+b^2+c^2+d^2} =12\)
NOT SUFFICIENT

IMO A
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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
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Bunuel wrote:
If \(a + b + c + d = 12\), what is the value of \(\sqrt{a^2+b^2+c^2+d^2}\) ?

(1) \(ab = cd = ad\)

(2) \(|a| = |b| = |c| = |d|\)


 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

Compete, Get Better, Win prizes and more

 


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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
Bunuel please explain why the OA should not be A. Where I was wrong?


Kinshook wrote:
Asked: If \(a + b + c + d = 12\), what is the value of \(\sqrt{a^2+b^2+c^2+d^2}\) ?

\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad + bc+bd+cd)\)

(1) \(ab = cd = ad\)
\(ab = cd = ad = \sqrt{abcd}\)
\(ad = \sqrt{abcd}\)
aˆ2dˆ2 = abcd
ad = bc
ab = bc = cd = da
a = b = c = d = (a+b+c+d)/4 = 12/4 = 3
ab = ac = ad = bc = bd = cd = 3*3 = 9
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 2*6*9= 144 - 108 = 36\)
\(\sqrt{a^2+b^2+c^2+d^2} = 6\)
SUFFICIENT

(2) \(|a| = |b| = |c| = |d|\)
Case 1: a=b =c=d = 3;
ab = bc = cd = da = 3*3=9
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 2*6*9= 144 - 108 = 36\)
\(\sqrt{a^2+b^2+c^2+d^2} =6\)
Case2: a=b=c=4; d=-4
ab = bc = ca = 4*4 = 16; ad=bd=cd=-16
ab + ac + ad + bc + bd + cd = 16*3 - 16*3 = 0
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 0 = 12ˆ2\)
\(\sqrt{a^2+b^2+c^2+d^2} =12\)
NOT SUFFICIENT

IMO A
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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
Expert Reply
Kinshook wrote:
Bunuel please explain why the OA should not be A. Where I was wrong?


Kinshook wrote:
Asked: If \(a + b + c + d = 12\), what is the value of \(\sqrt{a^2+b^2+c^2+d^2}\) ?

\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad + bc+bd+cd)\)

(1) \(ab = cd = ad\)
\(ab = cd = ad = \sqrt{abcd}\)
\(ad = \sqrt{abcd}\)
aˆ2dˆ2 = abcd
ad = bc
ab = bc = cd = da
a = b = c = d = (a+b+c+d)/4 = 12/4 = 3
ab = ac = ad = bc = bd = cd = 3*3 = 9
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 2*6*9= 144 - 108 = 36\)
\(\sqrt{a^2+b^2+c^2+d^2} = 6\)
SUFFICIENT

(2) \(|a| = |b| = |c| = |d|\)
Case 1: a=b =c=d = 3;
ab = bc = cd = da = 3*3=9
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 2*6*9= 144 - 108 = 36\)
\(\sqrt{a^2+b^2+c^2+d^2} =6\)
Case2: a=b=c=4; d=-4
ab = bc = ca = 4*4 = 16; ad=bd=cd=-16
ab + ac + ad + bc + bd + cd = 16*3 - 16*3 = 0
\(a^2+b^2+c^2+d^2 = (a + b + c + d)ˆ2 - 2(ab+ac+ad+bc+bd+cd) = 12ˆ2 - 0 = 12ˆ2\)
\(\sqrt{a^2+b^2+c^2+d^2} =12\)
NOT SUFFICIENT

IMO A


For (1) we can have many other cases. For example, a = c = d = 0 and b = 12, gives \(\sqrt{a^2+b^2+c^2+d^2}=12\). Or, a = d = 0 and b = c = 6, gives \(\sqrt{a^2+b^2+c^2+d^2}=6\sqrt{2}\).
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Re: GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the [#permalink]
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