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12 Aug 2021, 04:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (01:40) correct
21%
(02:00)
wrong
based on 112
sessions
History
Date
Time
Result
Not Attempted Yet
If \(a + b + c = 6\), \(a^2 + b^2 + c^2 = 14\) and \(a^3 + b^3 + c^3 = 36\), then what is the value of abc ?
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
12 Aug 2021, 05:03
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\(a+b+c = 6\)
\(a^2 + b^2 + c^2 = 14\)
\(a^3 + b^3 + c^3 = 36\)
=> \((a+b+c)^2 = 36 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
=> \(ab + bc + ca = 11\)
Now,
\(a^3 + b^3 + c^3 = (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ca) + 3abc\\
\)
=> 36 = 6*(14-11) + 3abc
=> 3abc = 18
=> abc = 6
OA should be B
\(a^2 + b^2 + c^2 = 14\)
\(a^3 + b^3 + c^3 = 36\)
=> \((a+b+c)^2 = 36 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
=> \(ab + bc + ca = 11\)
Now,
\(a^3 + b^3 + c^3 = (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ca) + 3abc\\
\)
=> 36 = 6*(14-11) + 3abc
=> 3abc = 18
=> abc = 6
OA should be B
12 Aug 2021, 05:06
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Bookmarks
Solution:
There is a methodical and a hit and trial method to do this question.
Methodical way:
For this, you must be knowing a couple of identities like \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\) and \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
We are given a+b+c=6
squaring both sides and applying the first identity we can get:
\(⇒ a^2+b^2+c^2+2(ab+bc+ca)=36\)
\(⇒ 14+2(ab+bc+ca)=36\)
\(⇒ab+bc+ca=\frac{36-14}{2}\)
\(⇒ ab+bc+ca=11\)
We know another identity: \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
Plugging the values we get: \(36-3abc=6(14-11)\)
\(⇒ 36-3abc=18\)
\(⇒ 3abc=18\)
\(⇒ abc=\frac{18}{3}\)
\(⇒ abc=6\)
Hit and trial method:
We are not given the nature of a, b and c. However we can infer that they are integers otherwise the sum of their squares wouldn't have been integers. \(a^2+b^2+c^2=14\).
Now since we have 3 integers adding to give 6 i.e., \(a+b+c=6\) let us assume \(a=b=c=2\) or \(a=1, b=2, c=3\)
When we plug in both the cases we find that \(a=1, b=2, c=3\) satisfy all the given expressions.
Thus \(abc=1\times 2\times 3=6\)
Hence the right answer is Option B.
There is a methodical and a hit and trial method to do this question.
Methodical way:
For this, you must be knowing a couple of identities like \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\) and \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
We are given a+b+c=6
squaring both sides and applying the first identity we can get:
\(⇒ a^2+b^2+c^2+2(ab+bc+ca)=36\)
\(⇒ 14+2(ab+bc+ca)=36\)
\(⇒ab+bc+ca=\frac{36-14}{2}\)
\(⇒ ab+bc+ca=11\)
We know another identity: \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
Plugging the values we get: \(36-3abc=6(14-11)\)
\(⇒ 36-3abc=18\)
\(⇒ 3abc=18\)
\(⇒ abc=\frac{18}{3}\)
\(⇒ abc=6\)
Hit and trial method:
We are not given the nature of a, b and c. However we can infer that they are integers otherwise the sum of their squares wouldn't have been integers. \(a^2+b^2+c^2=14\).
Now since we have 3 integers adding to give 6 i.e., \(a+b+c=6\) let us assume \(a=b=c=2\) or \(a=1, b=2, c=3\)
When we plug in both the cases we find that \(a=1, b=2, c=3\) satisfy all the given expressions.
Thus \(abc=1\times 2\times 3=6\)
Hence the right answer is Option B.
