GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the

Question

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|

M38-09

paragamrutkar · Answer

a + b + c + d = 12  - Given Statement

Statement 1 only   - 
 ab = cd = ad

b = d 
 a = c 
 
 2 * (a + b) = 12 
 a + b = 6

\sqrt{a^2+b^2+c^2+d^2} = \sqrt{2 * (a^2+b^2)} 
If a = 1, b = 5
 \sqrt{2 * (a^2+b^2)} = \sqrt{2 * (1 + 25)} = \sqrt{52}

If a = 2, b = 3 
 \sqrt{2 * (a^2+b^2)} = \sqrt{2 * (4 + 9)} = \sqrt{26}

Different answers. Hence,  INSUFFICIENT

Statement 2 only   - 
 |a| = |b| = |c| = |d|

From the given statement and Statement 2, we can infer that there can be 2 types of sets.

Set 1  -
{|a|, |a|, |a|, |a|} -> all +ve
 4|a| = 12 . Therefore,  a = 3

\sqrt{a^2+b^2+c^2+d^2}= \sqrt{4*9} = 6

Set 2  -
{|a|, |a|, |a|, -|a|} -> 3 +ve and 1 -ve
 2|a| = 12 . Therefore,  a = 6

\sqrt{a^2+b^2+c^2+d^2}= \sqrt{4*36} = 12

Different answers. Hence,  INSUFFICIENT

1 & 2 Together   -

Since a = c & b = d, Set 2 is not possible. 
Only Set 1 is possible. Hence  a = 3  &  \sqrt{a^2+b^2+c^2+d^2}= \sqrt{4*9} = 6

SUFFICIENT

Hence, the answer is C.

desertEagle · Answer

(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

gmatophobia · Answer

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

Kinshook · Answer

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

VelvetThunder · Answer

Kinshook · Answer

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

Bunuel · Answer

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} .

Bunuel · Answer

GMAT CLUB Official Explanation:

If  a + b + c + d = 12 , what is the value of  \sqrt{a^2+b^2+c^2+d^2}  ? 
 
(1)  ab = cd = ad 
 
It's so tempting to think that  a=b=c=d=3 . In this case  \sqrt{a^2+b^2+c^2+d^2}=\sqrt{36}=6  It's certainly a possibility but is it the only one? Let's check:
 
Well, notice that if any three of the unknowns is 0, then  ab = cd = ad  holds and from  a + b + c + d = 12  it would follow that the fourth unknown must be 12. For example,  a=b=c=0  and  d=12 . In this case  \sqrt{a^2+b^2+c^2+d^2}=\sqrt{12^2}=12  
 
So, this statement is not sufficient.
 
(2)  |a| = |b| = |c| = |d| 
 
We can have the following two ases: 
 
 I.  If all four of the unknowns are positive, then we'd have  a = b = c = d . In this case  a + b + c + d = 4a=12 , which gives  a = b = c = d=3  and thus  \sqrt{a^2+b^2+c^2+d^2}=\sqrt{36}=6 .
 
 II.  If three of the unknowns are positive and the fourth one is negative, then we'd have  a = b = c = -d  (it does not matter which three are positive, so we can assign any). In this case  a + b + c + d = 2a=12 , which gives  a = b = c = -d=6  and thus  \sqrt{a^2+b^2+c^2+d^2}=\sqrt{4*36}=12 .
 
Two different answers. Not sufficient. 
  
 FYI:   If two of the unknowns are positive and the remaining two are negative, then we'd have  a = b = -c = -d . In this case  a + b + c + d = 0 
eq {12} . So, this case is NOT possible. Also, all four of the unknowns cannot be negative because the sum of four negative numbers cannot be 12.
   
(1)+(2) Case II from (2) cannot be true because if one of the unknowns is negative and the other three are positive, then  ab = cd = ad  won't hold true: in this case  ab ,  cd , or  ad  will be negative and the remaining two expressions will be positive. Thus, we have case I from (2), which means that  a = b = c = d=3  and thus  \sqrt{a^2+b^2+c^2+d^2}=\sqrt{36}=6 . Sufficient.

Answer: C

GMAT Club World Cup 2022 (DAY 3): If a + b + c + d = 12, what is the

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