Question : If $\frac{(17)^3+(7)^3}{\left(17^2+7^2-\mathrm{k}\right)}=24$, then what is the value of $\mathrm{k}?$
Option 1: 119
Option 2: 128
Option 3: 24
Option 4: 109
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: 119
Solution : We know that $a^3+b^3=(a+b)(a^2+b^2-ab)$ So, $17^3+7^3=(17+7)(17^2+7^2-17\times 7)$ Given, $\frac{17^3+7^3}{17^2+7^2-k}=24$ ⇒ $\frac{(17+7)(17^2+7^2-17\times 7)}{17^2+7^2-k}=24$ ⇒ $17^2+7^2-17\times 7=17^2+7^2-k$ $\therefore k= 119$ Hence, the correct answer is 119.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : If $\frac{4\left[(17)^3-(7)^3\right]}{\left(17^2+7^2+p\right)}=40$, then what is the value of $p$?
Question : What is the value of
Question : Which of the following statements is correct? I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+94^2$ $- 93^2+\ldots \ldots+22^2-21^2$ is 4840. II. The value of
Question : Which of the following statements is correct? I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+$ $94^2-93^2+\ldots \ldots+22^2-21^2$ is 4840. II. The value of
Question : If $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$ and $\mathrm{K}<0$, then what is the value of $\mathrm{K}^{11}+\frac{1}{\mathrm{~K}^4}$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile