Question : The value of $\frac{p^{2}- (q - r)^{2}}{(p + r)^{2} - (q)^{2}}$ + $\frac{q^{2} - (p - r)^{2}}{(p + q)^{2} - (r)^{2}}$ + $\frac{r^{2}- (p - q)^{2}}{(q + r)^{2} - (p)^{2}}$ is:
Option 1: 1
Option 2: 2
Option 3: 0
Option 4: 3
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: 1
Solution : $\frac{p^{2} - (q - r)^{2}}{(p + r)^{2} - (q)^{2}}$ + $\frac{q^{2} - (p - r)^{2}}{(p + q)^{2} - (r)^{2}}$ + $\frac{r^{2} - (p – q)^{2}}{(q + r)^{2} - (p)^{2}}$ = $\frac{(p + q - r)(p - q + r)}{(p + q + r)(p - q + r)} + \frac{(p + q - r)(-p + q + r)}{(p + q + r)(p + q - r)} + \frac{(p - q + r)(–p+ q + r)}{(p + q + r)(–p + q + r)}$ = $\frac{p + q - r - p + q + r + p -q + r}{p + q+ r}$ = $\frac{p + q + r}{p + q + r}$ = $1$ Hence, the correct answer is 1.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : The value of $ \frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}$, where $p \neq q \neq r$, is equal to:
Question : If $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$, find the value of $(pa+qb+rc)$.
Question : If $\frac{1}{p}+\frac{1}{q}=\frac{1}{p+q}$, the value of $\left (p^{3}-q^{3}\right )$ is:
Question : The value of $\frac{1}{(p-n)(n-q)}+\frac{1}{(n-q)(q-p)}+\frac{1}{(q-p)(p-n)}$ is:
Question : If $p=9, q=\sqrt{17}$, then the value of $(p^2-q^2)^{-\frac{1}{3}}$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile