Question : If $a^2+b^2+c^2=14$ and $a+b+c=6$, then the value of $(ab+bc+ca)$ is:
Option 1: 11
Option 2: 12
Option 3: 13
Option 4: 14
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 11
Solution : Given: $a^2+b^2+c^2=14$ and $a+b+c=6$. On squaring the given equation $a+b+c=6$, we get, $(a+b+c)^2=6^2$ ⇒ $a^2+b^2+c^2+2(ab+bc+ca)=36$ Substitute the value of $a^2+b^2+c^2=14$ in the above equation and we get, ⇒ $14+2(ab+bc+ca)=36$ ⇒ $2(ab+bc+ca)=22$ ⇒ $(ab+bc+ca)=11$ Hence, the correct answer is 11.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $a+b+c$ = 6 and $ab+bc+ca$ = 11, then the value of $bc(b+c)+ca(c+a)+ab(a+b)+3abc$ is:
Question : If $\frac{a^{2} - bc}{a^{2}+bc}+\frac{b^{2}-ca}{b^{2}+ca}+\frac{c^{2}-ab}{c^{2}+ab}=1$, then the value of $\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ac}+\frac{c^{2}}{c^{2}+ab}$ is:
Question : If $\frac{2+a}{a}+\frac{2+b}{b}+\frac{2+c}{c}=4$, then the value of $\frac{ab+bc+ca}{abc}$ is:
Question : If $(a+b+c)=14$ and $\left(a^3+b^3+c^3-3 a b c\right)=98$, find the value of $(ab+bc+ca)$.
Question : What is the value of ${a}^3+{b}^3+{c}^3$ if $(a+b+c)=0$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile