Staff Selection Commission Combined Graduate Level Exam
Question : Directions: Select the set in which the numbers are related in the same way as the numbers of the given set. (NOTE: Operations should be performed on the whole numbers, without breaking down the numbers into their constituent digits. E.g. 13 – operations on 13 such as adding/subtracting/multiplying etc to 13 can be performed. Breaking down 13 into 1 and 3 and then performing mathematical operations on 1 and 3 is not allowed.) (15, 45, 90) (8, 24, 48)
Option 1: (9, 27, 45)
Option 2: (7, 21, 24)
Option 3: (6, 18, 36)
Option 4: (12, 36, 68)
Correct Answer: (6, 18, 36)
Solution : Given: (15, 45, 90); (8, 24, 48)
Like, (15, 45, 90)→15 × 3 = 45 and 15 × 6 = 90 (8, 24, 48)→8 × 3 = 24 and 8 × 6 = 48
Let's check the options – First option: (9, 27,
Question : Akham Lakshmi Devi, a recipient of the Sangeet Natak Akademi award, is renowned for which of the following dances?
Option 1: Kathak
Option 2: Odissi
Option 3: Sattriya
Option 4: Manipuri
Correct Answer: Manipuri
Solution : The correct option is Manipuri.
Akham Lakshmi Devi is renowned for the Manipuri dance form. She was a renowned Manipuri dancer and advocate, showcasing this traditional dance style from the northeastern state of Manipur. Akham Lakshmi Devi was conferred with the Sangeet Natak Akademi Award
Question : What will be the value of $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$?
Option 1: $\frac{1}{\sqrt{2}}$
Option 2: $\sqrt{3}$
Option 3: $1$
Option 4: $\frac{1}{\sqrt{3}}$
Correct Answer: $1$
Solution : Given: $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$ = $\frac{\sin (90^{\circ}-60^{\circ}) \sin (90^{\circ}-50^{\circ}) \sin (90^{\circ}-40^{\circ}) \sin (90^{\circ}-30^{\circ})}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$ = $\frac{\cos 60^{\circ} \cos 50^{\circ} \cos 40^{\circ} \cos 30^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos
Question : A is twice as fast as B and B is thrice as fast as C. The journey covered by C in $1\frac{1}{2}$ hour will be covered by A in:
Option 1: 15 minutes
Option 2: 30 minutes
Option 3: 1 hour
Option 4: 10 minutes
Correct Answer: 15 minutes
Solution : Given: A is twice as fast as B and B is thrice as fast as C. The journey is covered by C in $1\frac{1}{2}$ hour. Let the speed of C be $x$ metre/min. So, the speed of B will be $3x$ metre/min. And the
Question : Which of the following chemicals is used as a preservative to slow browning and discolouration in foods and beverages during preparation, storage, and distribution?
Option 1: Nitrous oxide
Option 2: Phosgene
Option 3: Sulphites
Option 4: Chlorine
Correct Answer: Sulphites
Solution : The correct answer is Sulphites.
Sulphites is a sulphur-containing compound that has been used for centuries to prevent discolouration and slow browning during the storage and distribution of many foods. When some fruits are cut, the exposure of the flesh to oxygen results in
Question : A, B and C together can complete a work in 12 days. A and B alone can do the same work in 36 days. In how many days can C alone complete the same work?
Option 1: 48
Option 2: 25
Option 3: 12
Option 4: 36
Correct Answer: 36
Solution : Given: A and B alone can do the same work in 36 days. Let the total work = LCM of (12, 36, 36) = 36 units Efficiency of A $=\frac{36}{36} = 1$ unit per day Efficiency of B $=\frac{36}{36} = 1$ unit per day Efficiency
Question : Direction: Select the missing number from the given responses.
Option 1: 14
Option 2: 16
Option 3: 21
Option 4: 22
Correct Answer: 16
Solution : Given:
The middle number in a row is the product of the first number and the square root of the third number.
In row 1, the square root of 4 = 2 and 14 = 7
Question : G is the centroid of the equilateral triangle ABC. If AB = 10 cm, then the length of AG (in cm) is:
Option 1: $\frac{5 \sqrt3}{3}$
Option 2: $\frac{10 \sqrt3}{3}$
Option 3: $5 \sqrt3$
Option 4: $10\sqrt 3$
Correct Answer: $\frac{10 \sqrt3}{3}$
Solution : AB = 10cm Since AD is the perpendicular bisector of BC, ⇒ BD = 5cm and $\angle$ADB = 90° ⇒ AD = $\sqrt{\text{AB}^{2}-\text{BD}^{2}}$ = $\sqrt{10^{2}-5^{2}}$ = $\sqrt{75}$ = $5\sqrt{3}$ cm Since G is the centroid, AG : GD = 2 : 1 ⇒ AG
Question : If $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=k+\tan ^2 \theta+\cot ^2 \theta$, then the value of $k$ is equal to:
Option 1: 7
Option 2: 2
Option 3: 5
Option 4: 9
Correct Answer: 7
Solution : Given: $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=k+\tan ^2 \theta+\cot ^2 \theta$ ⇒ $\sin^{2}θ + 2\sinθ\operatorname{cosec}θ + \operatorname{cosec}^{2}θ +\cos^{2}θ + 2\cos θ \secθ + \sec^{2}θ=k+\tan ^2 \theta+\cot ^2 \theta$ ⇒ $k = \sin^{2}θ + \operatorname{cosec}^{2}θ + \cos^{2}θ+\sec^{2}θ + 2(1+1) - \tan ^2 \theta-\cot ^2 \theta$ ⇒ $k
Question : The following table shows the production of different types of two-wheelers from 1993 to 1998. Number of two-wheelers (in 1000s)
What is the percentage increase in the total production of all types of two-wheelers in 1998 in comparison to 1996 (rounded off to the nearest integer)?
Option 1: 27%
Option 2: 25%
Option 3: 24%
Option 4: 26%
Correct Answer: 25%
Solution : Percentage increase in the total production of all types of two-wheelers in 1998 in comparison to 1996 = $\frac{334-268}{268}×100$ = 24.63% $\approx$ 25% Hence, the correct answer is 25%.
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