Staff Selection Commission Combined Graduate Level Exam
Question : ABC is an isosceles right-angle triangle. $\angle ABC = 90 ^{\circ}$ and AB = 12 cm. What is the ratio of the radius of the circle inscribed in it to the radius of the circle circumscribing $\triangle ABC$?
Option 1: $6–\sqrt{2}: 3 \sqrt{2}$
Option 2: $2–\sqrt{2}: \sqrt{2}$
Option 3: $6–3 \sqrt{2}: 1 \sqrt{2}$
Option 4: $6–3 \sqrt{2}: 6 \sqrt{2}$
Correct Answer: $2–\sqrt{2}: \sqrt{2}$
Solution : Given: ABC is an isosceles right-angle triangle. $\angle ABC = 90 ^{\circ}$ and AB = 12 cm. In $\triangle ABC$, $H^2=B^2+P^2$ (Using Pythagoras theorem). ⇒ $H^2=12^2+12^2$ ⇒ $H^2=144+144=288$ ⇒ $H=\sqrt{288}=12\sqrt2$ cm Inradius of inscribed circle of right angled triangle = $\frac{(P + B –
Question : Directions: Study the given pattern carefully and select the number that can replace the question mark (?) in it. (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.) (2, 2, 4); (1, 6, 6); (7, 2, ?)
Option 1: 14
Option 2: 2
Option 3: 7
Option 4: 12
Correct Answer: 14
Solution : Given: (2, 2, 4); (1, 6, 6); (7, 2, ?)
Like, (2, 2, 4)→2 × 2 = 4 (1, 6, 6)→1 × 6 = 6 Similarly, (7, 2, ?)→7 × 2 = 14
So, 14 is the required answer. Hence, the first option is correct.
Question : What is the value of $\frac{7}{2}+\frac{11}{3}+\frac{7}{6}+\frac{11}{15}+\frac{7}{12}+\frac{11}{35}+\ldots \ldots+\frac{7}{156}+\frac{11}{575}$?
Option 1: $\frac{3917}{355}$
Option 2: $\frac{3816}{325}$
Option 3: $\frac{3714}{345}$
Option 4: $\frac{3216}{315}$
Correct Answer: $\frac{3816}{325}$
Solution : Given: The series is $\frac{7}{2}+\frac{11}{3}+\frac{7}{6}+\frac{11}{15}+\frac{7}{12}+\frac{11}{35}+\ldots \ldots+\frac{7}{156}+\frac{11}{575}$. Determine the value of the series by solving two distinct series independently. The sum of the first series = $\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+...+\frac{7}{156}$ ⇒ Sum = $7[1–(\frac{1}{2})+(\frac{1}{2}–\frac{1}{3})+(\frac{1}{3}–\frac{1}{4})+...+(\frac{1}{12}–\frac{1}{13})]$ ⇒ Sum = $7[1–\frac{1}{13}]$ ⇒ Sum = $7\times\frac{12}{13}=\frac{84}{13}$ The sum of the second series =
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Option 1: $3\sqrt{5}$
Option 2: $3\sqrt{5}-7$
Option 3: $3\sqrt{5}+7$
Option 4: $8$
Correct Answer: $3\sqrt{5}-7$
Solution : Given: $x+[\frac{1}{(x+7)}]=0$ Adding both sides 7, we get, ⇒ $x+7+[\frac{1}{(x+7)}]=7$ Squaring both sides, we get, ⇒ $(x+7)^{2}+(\frac{1}{x+7})^{2}+2=49$ ⇒ $(x+7)^{2}+(\frac{1}{x+7})^{2}=47$ Subtracting 2 from both sides, $(x+7)^{2}+(\frac{1}{x+7})^{2}–2=47-2$ ⇒ $[(x+7)-(\frac{1}{x+7})]^{2}=45$ ⇒ $[(x+7)-(\frac{1}{x+7})]=\sqrt{45}$ ⇒ $[(x-\frac{1}{x+7})]=\sqrt{45}-7=\sqrt{9 \times5}-7=3\sqrt{5}–7$ Hence, the correct answer is $3\sqrt{5}-7$.
Question : What will be the remainder when $27^{27}+27$ is divided by 28?
Option 1: 28
Option 2: 27
Option 3: 25
Option 4: 26
Correct Answer: 26
Solution : $27^{27}+27$ can written as $27^{27}+1^{27}-1+27=(27^{27}+1^{27})+26$ We know that An + Bn is divisible by A + B when n is odd. Here n = 27 is an odd number. When $(27^{27}+1^{27})$ is divided by 28, the remainder will be 0. And, when $(27^{27}+1^{27})+26$
Question : Directions: From the given option figures, select the one in which the question figure is embedded.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution : As there is no restriction on the rotation of the figure, we will check all possible angles where the question figure can fit itself in all the given option figures. The question figure is embedded only in the first option figure.
Hence, the first option is
Question : Directions: Select the option that is related to the third word in the same way as the second word is related to the first word. (The words must be considered meaningful English words and must not be related to each other based on the number of letters/number of consonants/vowels in the word.) Weight : Pound :: Height : ?
Option 1: Dozen
Option 2: Metres
Option 3: High
Option 4: Kilograms
Correct Answer: Metres
Solution : Given: Weight : Pound :: Height : ?
Like, a Pound is a unit used to measure Weight. Similarly, the unit used to measure Height is metre.
Hence, the second option is correct.
Question : Direction: In the following question a series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series.
156, 506, ?, 1806
Option 1: 1056
Option 2: 856
Option 3: 1456
Option 4: 1506
Correct Answer: 1056
Solution : Given:
Here,
122 + 12 = 156
222 + 22 = 506
322 + 32 = 1056
422 + 42 = 1806
So, the missing number is 1056.
Hence, the first option is correct.
Question : Select the most appropriate synonym of the given word.
Keen
Option 1: Happy
Option 2: Enthusiastic
Option 3: Wild
Option 4: Imitate
Correct Answer: Enthusiastic
Solution : The correct choice is the second option.
The word keen has a few potential synonyms, but the most appropriate one among the options provided would be enthusiastic.
The meanings of other words are as follows:
Question : The number of students in section A and section B of a class are 40 and 52, respectively. The average score in mathematics of all the students is 75. If the average score of the students in A is 20 % more than that of students in B then what is the average score of students in B?
Option 1: 71
Option 2: 65
Option 3: 69
Option 4: 63
Correct Answer: 69
Solution : Total number of students = 40 + 52 = 92 ⇒ $75 = \frac{\text{(Sum of scores of all students)}}{92}$ ⇒ Total score in mathematics = 75 × 92 = 6900 Now, Let the average score of section B students be $x$ ⇒ Average score of
The Question containing Inaapropriate or Abusive Words
Question lacks the basic details making it difficult to answer
Topic Tagged to the Question are not relevant to Question
Question drives traffic to external sites for promotional or commercial purposes
The Question is not relevant to User
And never miss an important update