Staff Selection Commission Combined Graduate Level Exam
Question : The length of the altitude of an equilateral triangle is $6 \sqrt{3}\;m$. The perimeter of the equilateral triangle (in m) is:
Option 1: $12 \sqrt{2}$
Option 2: $36 \sqrt{2}$
Option 3: $36$
Option 4: $24 \sqrt{3}$
Correct Answer: $36$
Solution : Let $a$ be the side of an equilateral triangle. Altitude of an equilateral triangle is $6 \sqrt{3}\;m$. Since altitude $=\frac{\sqrt 3 a}{2}$ So, $6 \sqrt{3} = \frac{\sqrt 3 a}{2}$ ⇒ $a = 6×2 = 12$ m Therefore, the perimeter = $3a = 3×12 = 36$
Question : The Appointments Committee of the Cabinet (ACC) extended the tenure of the current Research and Analysis Wing (RAW) chief for another year, until June 30, 2023. Who is the current chief?
Option 1: Mukul Rohatgi
Option 2: Anil Chauhan
Option 3: Samant Kumar Goel
Option 4: Sunil Barthwal
Correct Answer: Samant Kumar Goel
Solution : The correct answer is Samant Kumar Goel.
It is Samant Kumar Goel's second extension of service. He is the third Research and Analysis Wing (RAW) chief with more than three years of term. Before him, RAW founder R. N. Kao served for
Question : If $\sin (a+b)=1$ and $\cos (a-b)=\frac{1}{2}$, find $a$.
Option 1: $75^{\circ}$
Option 2: $30^{\circ}$
Option 3: $15^{\circ}$
Option 4: $45^{\circ}$
Correct Answer: $75^{\circ}$
Solution : $\sin (a+b)=1$ $⇒\sin (a+b)=\sin 90^{\circ}$ $⇒(a+b)=90^{\circ}$ ...(i) $\cos (a-b)=\frac{1}{2}$ $⇒\cos (a-b)=\cos 60^{\circ}$ $⇒(a-b)=60^{\circ}$ ... (ii) Solve the above equations $2a=150^{\circ}$ $⇒a=75^{\circ}$ Hence, the correct answer is $75^{\circ}$.
Question : Directions: Kritika walks 40 m towards the south. Then, turning to her right, she rides 30 m. Then, turning to her left, she rides 50 m. Again, she turns to her left and rides 30 m. How far (in m) is she from her initial position?
Option 1: 65 m
Option 2: 70 m
Option 3: 80 m
Option 4: 90 m
Correct Answer: 90 m
Solution : Firstly, we will draw the diagram as per the given instructions –
Now, we have to find the distance between starting and end point –
Therefore, the distance between Kritika's final point and starting point = 40 + 50 = 90 m. Hence, the
Question : The sum of 10 terms of the arithmetic series is 390. If the third term of the series is 19, find the first term:
Option 1: 3
Option 2: 5
Option 3: 7
Option 4: 8
Correct Answer: 3
Solution : Given: The sum of 10 terms of the arithmetic series is 390. So, $n=10$ The third term of the series is 19. Let the first term of the series be $a$. We know, sum of arithmetic progression (A.P.) = $\frac{n}{2}[2a+(n-1)d]$ $n^{th}$ term = $a+(n-1)d$ So,
Question : Direction: In this question, some equations are solved on the basis of a certain system. Find out the correct alternative for the unsolved equation on that basis.
4 x 5 x 8 =584
7 x 3 x 9 = 397
9 x 7 x 3 = ?
Option 1: 397
Option 2: 793
Option 3: 973
Option 4: 739
Correct Answer: 739
Solution : Here, the given equations follows a pattern in which the first number, the middle number and the last number in LHS becomes the units place digit, the hundreds place digit and the tens place digit in the answer part in RHS respectively.
⇒ 4 x
Question : Directions: In the following question below are given some statements followed by some conclusions based on those statements. Taking the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows the given statements. Statements: All R is P. All R is Q. Conclusions: I. No P is Q. II. No R is P.
Option 1: Neither conclusion (I) nor (II) follows
Option 2: Only conclusion (II) follows
Option 3: Both conclusions (I) and (II) follow
Option 4: Only conclusion (I) follows
Correct Answer: Neither conclusion (I) nor (II) follows
Solution : The possible Venn diagram, according to the given statements is as follows –
Let's analyse the conclusions – Conclusion (I): No P is Q – From the Venn diagram, it is evident that the two circles representing P and Q
Question : Directions: One or two statements are given, each followed by two conclusions or assumptions I and II. You have to consider the statements to be true, even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions or assumptions, if any, follows from the given statements. Statements: I. Irregularity is a cause for failure in exams. II. Some regular students fail in the examinations. Conclusions: I. All failed students are regular. II. All successful students are not regular.
Option 1: Only conclusion I follows
Option 2: Only conclusion II follows
Option 3: Both conclusion I and conclusion II follow
Option 4: Neither conclusion I nor conclusion II follows
Correct Answer: Neither conclusion I nor conclusion II follows
Solution : Conclusion I: All failed students are regular. According to the given statements, irregularity is a cause of failure and some regular students fail in the examination. Therefore, all failed students are regular can not be the correct conclusion. Conclusion
Question : Fill in the blank with the correct collocation. Holm, a Danish traveller, had made a/an ________ replica of the tablet, which in 1908 was deposited in the Metropolitan Museum of Art, New York.
Option 1: wild
Option 2: obscure
Option 3: corrupt
Option 4: exact
Correct Answer: exact
Solution : The correct option is the fourth option.
Explanation: An exact replica means a precise copy, identical in every detail to the original. In this context, it indicates that Holm created an accurate duplicate of the tablet.
The meanings of the other options are as follows:
Question : The average mark of 50 students in an examination was 65. It was later found that the marks of one student had been wrongly entered as 83 instead of 38. The correct average is:
Option 1: 63.9
Option 2: 64.5
Option 3: 64.7
Option 4: 64.1
Correct Answer: 64.1
Solution : Average marks of 50 students = 65. Total marks of the students = 50 × 65 = 3250 Difference by the wrong entry = 83 – 38 = 45 So, the actual total marks of the students = 3250 – 45 = 3205 Average =
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