Question : x, y, and z are distinct prime numbers where x < y < z. If x + y + z = 70, then what is the value of z?
Option 1: 29
Option 2: 43
Option 3: 31
Option 4: 37
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: 37
Solution : Given: x, y, and z are distinct prime numbers where x < y < z. x + y + z = 70 The sum indicates that at least one integer is even because odd + odd + odd = odd. There is just one even prime number that exists and that is 2. The smallest prime number is 2. This means x = 2. Option first: z = 29 ⇒ y = 68 – 29 = 39 (y is a non-prime number) Option second: z = 43 ⇒ y = 68 – 43 = 25 (y is a non-prime number) Option third: z = 31 ⇒ y = 68 – 31 = 37 (z < y) Option fourth: z = 37 ⇒ y = 68 – 37 = 31 (y < z. Also, it satisfies the conditions.) Hence, the correct answer is 37.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79}$, where x, y, and z are natural numbers, then the value of $(2 x+3 y-z)$ is:
Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Question : $x,y,$ and $z$ are real numbers. If $x^3+y^3+z^3 = 13, x+y+z = 1$ and $xyz=1$, then what is the value of $xy+yz+zx$?
Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where $x \neq y \neq z$, is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile