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


 


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


 


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Bunuel please explain why the OA should not be A. Where I was wrong?


Kinshook
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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Kinshook
Bunuel please explain why the OA should not be A. Where I was wrong?


Kinshook
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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