Staff Selection Commission Combined Graduate Level Exam
Question : Jawahar Tunnel one of the largest in India is located in the State of
Option 1: Jammu and Kashmir
Option 2: Maharashtra
Option 3: Karnataka
Option 4: Himachal Pradesh
Correct Answer: Jammu and Kashmir
Solution : The correct option is Jammu and Kashmir.
The Jawahar Tunnel is a road tunnel located in Jammu and Kashmir, India. It is a vital transit route in the area. The tunnel connects the villages of Banihal and Qazigund by passing through the Pir
Question : Direction: In this question, a word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in the two matrices, given below. The columns and rows of Matrix (I) are numbered from 0 to 4 and that of Matrix (II) are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, e.g. A can be represented by 01,13, etc. and B can be represented by 58, 69, etc. Similarly, you have to identify the set for the word FINE.
Option 1: 00, 04, 02, 56
Option 2: 12, 10, 13, 67
Option 3: 24, 59, 31, 78
Option 4: 31, 32, 33, 87
Correct Answer: 00, 04, 02, 56
Solution : Given – FINE
Option first: 00, 04, 02, 56
FINE
The first set represents the word FINE.
Option second: 12, 10, 13, 67
FIAE
Option third: 24, 59, 31, 78
ATAT
Option fourth: 31, 32, 33, 87
FINT
Hence, third option is
Question : If $x=\frac{6pq}{p+q}$, then the value of $\frac{x+3p}{x–3p}+\frac{x+3q}{x–3q}$ is:
Option 1: 6
Option 2: 8
Option 3: 2
Option 4: 3
Correct Answer: 2
Solution : Given: $x=\frac{6pq}{p+q}$ ⇒ $x=\frac{6pq}{p+q}$ and $x=\frac{6pq}{p+q}$ ⇒ $\frac{x}{3p}=\frac{2q}{p+q}$ and $\frac{x}{3q}=\frac{2p}{p+q}$ Applying componendo and dividendo in the above expressions, we get, ⇒ $\frac{x+3p}{x–3p}=\frac{2q+p+q}{2q–(p+q)}$ and $\frac{x+3q}{x–3q}=\frac{2p+p+q}{2p–(p+q)}$ ⇒ $\frac{x+3p}{x–3p}=\frac{p+3q}{q–p}$ and $\frac{x+3q}{x–3q}=\frac{3p+q}{p–q}$ The value of the given expression $\frac{x+3p}{x–3p}+\frac{x+3q}{x–3q}$ is as follows, $\frac{x+3p}{x–3p}+\frac{x+3q}{x–3q}=\frac{p+3q}{q–p}-\frac{3p+q}{q–p}$ $=\frac{p+3q–3p–q}{q–p}$ $=\frac{2(q-p)}{q-p}$ $=2$ Hence, the correct
Question : Which of the following festivals is not associated with Sikkim?
Option 1: Losar
Option 2: Lohri
Option 3: Sakewa
Option 4: Yenya
Correct Answer: Lohri
Solution : The correct option is Lohri.
Lohri is not associated with the state of Sikkim. Lohri is a winter festival celebrated in North India, particularly in the Punjab region. It is usually observed on January 13th every year.
Question : A shopkeeper sells an item at a profit of 15% and uses a weight which is 20% less. Find his actual profit percentage.
Option 1: 42.5%
Option 2: 50%
Option 3: 40%
Option 4: 43.75%
Correct Answer: 43.75%
Solution : Given: A shopkeeper sells an item at a profit of 15%. False weight used = 20% less than the actual weight Let the cost price of 100 gm = Rs. 100 if the cost price of 1 gm is Re. 1 Selling price of 100
Question : Direction: The following bar graph shows the production of table fans in a factory for one week. Study the bar graph and answer the given question.
The maximum production exceeds the minimum production by:
Option 1: 400
Option 2: 420
Option 3: 500
Option 4: 540
Correct Answer: 420
Solution : The provided bar graph shows that the maximum production is 540 and the minimum production is 120, respectively. Difference = Maximum production – Minimum production = 540 – 120 = 420 Hence, the correct answer is 420.
Question : A, B, and C are employed to do a piece of work for INR 5,290. A and B together are supposed to do $\frac{19}{23}$ of the work and B and C together $\frac{8}{23}$ of the work. Then A should be paid:
Option 1: INR 4,250
Option 2: INR 3,450
Option 3: INR 1,950
Option 4: INR 2,290
Correct Answer: INR 3,450
Solution : Part of work done by A and B together = $\frac{19}{23}$ of total work Part of work done by B and C together = $\frac{8}{23}$ of total work Part of the work done by A alone = Total work – Part of work done
Question : If $m-n=2$ and $mn=15,(m,n>0)$ , then the value of $(m^2-n^2)(m^3-n^3)$ is:
Option 1: 1856
Option 2: 1658
Option 3: 1586
Option 4: 1568
Correct Answer: 1568
Solution : Given: $m-n=2$ and $mn=15,(m,n>0)$ $(m-n)^2=2^2$ ⇒ $m^2+n^2-2mn=4$ ⇒ $m^2+n^2=4+(2×15)=34$ ⇒ $m^2+n^2+30=34+30$ [adding 30 to both sides] ⇒ $m^2+n^2+2mn=64$ [as, $mn=15$] ⇒ $(m+n)^2=8^2$ ⇒ $m+n=8$ Now, $(m^2-n^2)(m^3-n^3)$ = $(m+n)(m-n) (m-n)(m^2+mn+n^2)$ = $8×2×2×(34+15)$ = $1568$ Hence, the correct answer is 1568.
Question : Select the option that can be used as a one-word substitute for the given group of words.
One who runs away from the law.
Option 1: Fatalist
Option 2: Convict
Option 3: Fugitive
Option 4: Lunatic
Correct Answer: Fugitive
Solution : The third option is the correct answer.
The term fugitive is the most appropriate, as it specifically refers to a person who is fleeing or evading legal authorities to escape punishment or arrest.
The meanings of the other options are as follows:
Question : Which of the following is equal to $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$?
Option 1: $\frac{1+\sin \theta}{\cos \theta}$
Option 2: $\frac{1+\tan \theta}{\cot \theta}$
Option 3: $\frac{1+\cot \theta}{\tan \theta}$
Option 4: $\frac{1+\cos \theta}{\sin \theta}$
Correct Answer: $\frac{1+\sin \theta}{\cos \theta}$
Solution : Given: The given trigonometric expression is $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$. Use the trigonometric identity, $\sec^2\theta–\tan^2\theta=1$ ⇒ $[\frac{\tan \theta+\sec \theta–(\sec^2\theta–\tan^2\theta)}{\tan \theta–\sec \theta+1}]=[\frac{(\tan \theta+\sec \theta)(1–\sec\theta+\tan \theta)}{\tan \theta–\sec \theta+1}$ $=\tan \theta+\sec \theta=\frac{1+\sin \theta}{\cos \theta}$ Hence, the correct answer is $\frac{1+\sin \theta}{\cos \theta}$.
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