Staff Selection Commission Sub Inspector Exam
Question : If [n] denotes the greatest integer < n and (n) denotes the smallest integer > n, where n is any real number, then the value of $(1\frac{1}{5})×[1\frac{1}{5}]-(1\frac{1}{5})+[\frac{1}{5}]+(1.5)$ is:
Option 1: 1.5
Option 2: 2
Option 3: 2.5
Option 4: 3.5
Correct Answer: 2
Solution : $(1\frac{1}{5}) × [1\frac{1}{5}] – (1\frac{1}{5}) + [\frac{1}{5}] + (1.5)$ = ($\frac{6}{5}) × [\frac{6}{5}] – (\frac{6}{5}) + [\frac{1}{5}] + (1.5)$ = (1.2) × [1.2] – (1.2) + [0.2] +(1.5) Applying floor and ceiling functions = 2 × 1 – 2 + 0 + 2 = 2
Question : The value of $x$ in the following equation is: 0.3 + 0.6 + 0.7 + 0.8 = $x$
Option 1: $5.3$
Option 2: $2\frac{3}{10}$
Option 3: $2\frac{2}{5}$
Option 4: $2.35$
Correct Answer: $2\frac{2}{5}$
Solution : Given: $0.3 + 0.6 + 0.7 + 0.8 = x$ $⇒x = 2.4 = \frac{24}{10} = \frac{12}{5}= 2\frac{2}{5}$ Hence, the correct answer is $2\frac{2}{5}$.
Question : The masculine aspect of the Manipuri classical dance is known as:
Option 1: Ras
Option 2: Choloms
Option 3: Pareng
Option 4: Maiba
Correct Answer: Choloms
Solution : The correct answer is Choloms.
In Manipur, the dance form known as Sankirtana is accompanied by congregational singing called Kirtan. Male dancers actively participate in this tradition, playing the Pung (drum) and Kartal (cymbals) while dancing. The masculine facet of this dance, referred to
Question : The ratio of the third proportional to 16 & 40 and the mean proportional between 10 & 40 is:
Option 1: 5 : 1
Option 2: 1 : 5
Option 3: 4 : 1
Option 4: 1 : 4
Correct Answer: 5 : 1
Solution : The third proportional to 16 and 40 is, = $\frac{(40)^2}{16}$ = 100 The mean proportional between 10 and 40 is, = $\sqrt{10 \times 40}$ = 20 The required ratio = 100 : 20 = 5 : 1 Hence, the correct answer is 5
Question : P and Q together can do a work in 12 days. P alone can do the same work in 18 days. In how many days can Q alone complete two-thirds part of the same work?
Option 1: 30
Option 2: 21
Option 3: 24
Option 4: 36
Correct Answer: 24
Solution : Time taken by P alone to complete a work = 18 days Part of work done by P alone in a day = $\frac{1}{18}$ Let $x$ be the time Q alone takes to complete a work. Part of work done by Q alone in a
Question : A boat can travel 104 km downstream in 8 hours. If the speed of the stream is 2 km/hr, then at what time will it be able to cover 13 km upstream?
Option 1: $2 \frac{2}{3}$ hours
Option 2: $1 \frac{4}{9}$ hours
Option 3: $2 \frac{1}{2}$ hours
Option 4: $1 \frac{2}{3}$ hours
Correct Answer: $1 \frac{4}{9}$ hours
Solution : Given, The boat can travel 104 km downstream in 8 hours. Speed of stream = 2 km/hr Let the speed of the boat in still water be $x$. So, $\frac{104}{(x + 2)} = 8$ ⇒ $x + 2 = 13$ $\therefore x= 11$
Question : Directions: The following table gives details about the percentage change in the population in a particular town for given years. Go through the chart given and answer the questions that follow:
Which year out of these 8 years has the highest population?
Option 1: 2008
Option 2: 2005
Option 3: 2010
Option 4: 2011
Correct Answer: 2011
Solution : The population will be highest in the year 2011. Because it is the last year of the bar graph and all the increased population will be counted in that year. Hence, the correct answer is 2011.
Question : Study the pie chart and answer the question.
Distribution (degree-wise) of the number of employees of a company working in 5 departments A, B, C, D, and E. Total number of employees = 3200 If the total number of employees working in departments A and B exceeds the number of employees in department C by $x$, then $x$ lies between:
Option 1: 120 and 140
Option 2: 80 and 100
Option 3: 100 and 120
Option 4: 140 and 160
Correct Answer: 120 and 140
Solution : The central angle formed by the number of employees in departments A and B = 79.2$^\circ$ + 57.6$^\circ$ = 136.8$^\circ$ The central angle formed by the number of employees in department C = 122.4$^\circ$ The number of employees in A and B is
Question : The simple interest on a sum of money at 10% per annum for 2 years is 8,100. Compounded annually, what would be the compound interest (in ) on the same sum for the same period at the same rate of interest?
Option 1: 8,100
Option 2: 8,505
Option 3: 8,715
Option 4: 9,000
Correct Answer: 8,505
Solution : Simple interest = $\frac{\text{Principal × Rate × Time}}{100}$ According to the question $⇒ 8100 = \frac{\text{Principal}×10×2}{100}$ $\therefore$ Principal $= 40500$ Now, $\text{Total Amount}$ $=\text{Principal}×(1+\frac{\text{Rate}}{100})^{\text{Time}}$ $ = 40500({1+ \frac{10}{100})^2}$ $ = 40500({ \frac{11}{10})^2}$ $ = 40500×{ \frac{121}{100}}$ $=49005$ $\therefore$ Compound Interest = Amount – Principal =
Question : Directions: The sequence of folding a piece of paper and how the folded paper has been cut is shown below. Choose the figure that would most closely resemble the unfolded form of the paper.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution : When the given folded paper is unfolded, it will look like –
Hence, the third option is correct.
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