Question : Find the sum of $\left (1-\frac{1}{n+1} \right) + \left (1-\frac{2}{n+1} \right) + \left (1- \frac{3}{n+1} \right)+.....\left ( 1- \frac{n}{n+1} \right)$.
Option 1: $n$
Option 2: $\frac{n}{2}$
Option 3: $(n+1)$
Option 4: $\frac{(n+1)}{2}$
Latest: SSC CGL 2024 final Result Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: $\frac{n}{2}$
Solution : $\left (1-\frac{1}{n+1} \right) + \left (1-\frac{2}{n+1} \right) + \left (1- \frac{3}{n+1} \right)+.....\left (1- \frac{n}{n+1} \right)$ = $\frac{n}{n+1}+\frac{n-1}{n+1}+\frac{n-2}{n+1}+...........+\frac{1}{n+1}$ $= \frac{1}{(n+1)}\frac{[n(n+1)]}{2}$ $=\frac{n}{2}$ Hence, the correct answer is $\frac{n}{2}$.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : If $M =\left ( \frac{3}{7} \right ) ÷ \left ( \frac{6}{5} \right ) ×\left ( \frac{2}{3} \right ) + \left ( \frac{1}{5} \right ) ×\left ( \frac{3}{2} \right )$ and $N = \left ( \frac{2}{5} \right ) × \left ( \frac{5}{6} \right ) ÷ \left ( \frac{1}{3} \right ) + \left ( \frac{3}{5} \right ) × \left ( \frac{2}{3} \right ) ÷ \left ( \frac{3}{5} \right )$, then what is the value of $\frac{M}{N}$?
Option 1: $\frac{207}{560}$
Option 2: $\frac{339}{1120}$
Option 3: $\frac{113}{350}$
Option 4: $\frac{69}{175}$
Question : The value of $\left(2 \frac{6}{7}\right.$ of $\left.4 \frac{1}{5} \div \frac{2}{3}\right) \times 5 \frac{1}{9} \div\left(\frac{3}{4} \times 2 \frac{2}{3}\right.$ of $\left.\frac{1}{2} \div \frac{1}{4}\right)$ is:
Option 1: 25
Option 2: 19
Option 3: 23
Option 4: 21
Question : The value of $\left(1 \frac{1}{3} \div 2 \frac{6}{7}\right.$ of $\left.5 \frac{3}{5}\right) \times\left(6 \frac{2}{5} \div 4 \frac{1}{2}\right.$ of $\left.5 \frac{1}{3}\right) \div\left(\frac{3}{4} \times 2 \frac{2}{3} \div \frac{5}{9}\right.$ of $\left.1 \frac{1}{5}\right)=k$, where $\mathrm{k}$ lies between:
Option 1: 0.07 and 0.08
Option 2: 0.007 and 0.008
Option 3: 0.0007 and 0.0008
Option 4: 0.7 and 0.8
Question : The value of $9 \div\left\{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6} \div\left(\frac{3}{4}-\frac{1}{3}\right)\right.$ of $\left.\frac{2}{9}\right\}$ is:
Option 1: $\frac{540}{173}$
Option 2: $\frac{340}{173}$
Option 3: $\frac{480}{173}$
Option 4: $\frac{2540}{173}$
Question : Directions: Which of the following numbers will replace the question mark (?) in the given series? $\left [ \frac{1}{176} \right ],\left [ \frac{1}{88} \right ],?,\left [ \frac{1}{22} \right ],\left [ \frac{1}{11} \right ]$
Option 1: $\left[\frac{1}{32}\right]$
Option 2: $\left\lceil\frac{1}{44}\right]$
Option 3: $\left\lceil\frac{1}{48}\right]$
Option 4: $\left\lceil\frac{1}{42}\right]$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile