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19 Aug 2022, 07:39
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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) \(|a| = |b| = |c| = |d|\)
(1) \(ab = cd = ad\)
(2) \(|a| = |b| = |c| = |d|\)
19 Aug 2022, 07:39
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Official Solution:
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 \(a + b + c + d = 0 \neq {12}\). 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
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 \(a + b + c + d = 0 \neq {12}\). 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
