Staff Selection Commission Combined Higher Secondary Level Exam
Question : Directions: In the following question, select the missing number from the given responses.
Option 1: 8
Option 2: 13
Option 3: 4
Option 4: 12
Correct Answer: 13
Solution : Like in, the first tree, 4 × 3 = 12; 12 × 3 = 36 And in, the third tree, 8 × 2 = 16; 16 × 2 = 32 Similarly, in the second tree 13 × 1 = 13; 13 × 1 = 13
Question : Which of the following is mentioned in Article 39A of the Indian Constitution?
Option 1: To secure adequate means of livelihood.
Option 2: To promote equal justice and to provide free legal aid.
Option 3: Equal pay for equal work.
Option 4: Prevention of concentration of wealth.
Correct Answer: To promote equal justice and to provide free legal aid.
Solution : The correct option is To promote equal justice and provide free legal aid.
Article 39A of the Indian Constitution pertains to the promotion of equal justice and the provision of free legal aid. It states:
Question : Directions: IKLN is related to FHIK in a certain way based on the English alphabetical order. In the same way, GIJL is related to DFGl. To which of the following is MOPR related, following the same logic?
Option 1: JMOP
Option 2: JKMO
Option 3: JLMO
Option 4: JLNO
Correct Answer: JLMO
Solution : Given: IKLN is related to FHIK –
Subtract 3 from the place value of each letter of IKLN to obtain the required code – I – 3 = F; K – 3 = H; L – 3 = I; N – 3 = K Thus,
Question : The three sides of a triangle are 7 cm, 9 cm, and 8 cm. What is the area of the triangle?
Option 1: $12 \sqrt{3} \;\mathrm{Sq} . \mathrm{cm}$
Option 2: $10\sqrt{3} \;\mathrm{Sq} . \mathrm{cm}$
Option 3: $12 \sqrt{5} \;\mathrm{Sq} . \mathrm{cm}$
Option 4: $2 \sqrt{5} \;\mathrm{Sq} . \mathrm{cm}$
Correct Answer: $12 \sqrt{5} \;\mathrm{Sq} . \mathrm{cm}$
Solution : Using the formula: $s = \frac{(a + b + c)}{2}$ $s = \frac{(7 + 9 + 8)}{2}$ $s = \frac{24}{2}$ $s = 12$ cm Now, plug the values of a, b, c, and s into Heron's formula to find the area
Question : ________diseases last for only shorter periods.
Option 1: Acute
Option 2: Chronic
Option 3: Hereditary
Option 4: Genetic
Correct Answer: Acute
Solution : The correct answer is Acute.
Acute diseases last for only shorter periods. The conditions within acute diseases typically have a rapid onset and short duration, often resolving without prolonged medical intervention. Examples of acute diseases include the common cold, flu, gastroenteritis, and some respiratory
Questions : '$a$' divides 228 leaving a remainder 18. The biggest two-digit value of '$a$' is:
Option 1: 70
Option 2: 21
Option 3: 35
Option 4: 30
Correct Answer: 70
Solution : Given: Number = 228, divisor = $a$, and remainder = 18 Subtract 18 from 228 = 210 Factors of 210 = 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, and 210 $\therefore$ The biggest two-digit value of $a$
Question : Comprehension: In the following passage, some words have been deleted. Read the passage carefully and select the most appropriate option to fill in each blank. Our school organised a graduation day to award certificates to the students. The graduation day highlights the overall achievements of students. All the students who are being graduated are invited formally with their parents. Students wore their attire according to the dress code given to them by the school administration. High school graduation____________(1) is one of the most awaited and last event of any school. The function began with ____________(2) the lamp, followed by the distribution of certificates in a very systematic way. Our principal made a long and__________(3) speech and blessed us for the future___________(4). On behalf of our graduating class, our class representative extended a vote of thanks to show ___________(5) to all staff members. One of the parents expressed their experiences and views. At last, all the students, parents, and staff members had lunch together. Select the most appropriate option to fill in the blank number 2.
Option 1: lighting
Option 2: lightening
Option 3: sparking
Option 4: glittering
Correct Answer: lighting
Solution : The correct option will be the first option.
Lighting the lamp is, in a way, an invocation of the waves of knowledge that are activated on stage. It also means keeping the waves of knowledge generated in the knowledge-imparting stage constantly active through the
Question : Directions: The diagram shows the age distribution of patients admitted to a hospital on a particular day. Study the diagram and answer the question.
About 11% of the patients who got admitted to the hospital on that particular day were of age:
Option 1: either between 35 years and 40 years or between 55 years and 60 years
Option 2: between 60 years and 65 years
Option 3: between 35 years and 40 years
Option 4: between 35 years and 40 years and between 55 years and 60 years
Correct Answer: between 35 years and 40 years and between 55 years and 60 years
Solution : Total patients = 35 From the given graph, About 11% of the patients were admitted. ⇒ 11% of 35 $=\frac{11}{100} \times 35=3.85 \approx 4$ Here, The number of patients between 35 and 40
Question : Select the option that expresses the given sentence in passive voice.
A sudden noise frightened the horse.
Option 1: The horse has been frightened by a sudden noise.
Option 2: The horse was frightened by a sudden noise.
Option 3: The horse had frightened by a sudden noise.
Option 4: The horse got frightened by a sudden noise.
Correct Answer: The horse was frightened by a sudden noise.
Solution : The second option is the correct choice.
Passive voice is the voice in which the object experiences an action rather than the person who performs the action. Hence, the object in the active sentence becomes the subject in
Question : If $p^2+\frac{1}{p^2}=14$, then find the value of $\left(p^3+\frac{1}{p^3}\right)$.
Option 1: 56
Option 2: 60
Option 3: 48
Option 4: 52
Correct Answer: 52
Solution : Given: $p^2+\frac{1}{p^2}=14$ we know that, $p^2+\frac{1}{p^2}=(p+\frac{1}{p})^2-2\times p\times \frac{1}{p}$ $⇒(p+\frac{1}{p})^2-2\times p\times \frac{1}{p}=14$ $⇒(p+\frac{1}{p})^2=14+2$ $⇒(p+\frac{1}{p})^2=16$ $⇒(p+\frac{1}{p})=4$ Now cubing both sides, we get: $⇒(p+\frac{1}{p})^3=4^3$ $⇒p^3+\frac{1}{p^3}+3\times p\times \frac{1}{p}(p+ \frac{1}{p})=4^3$ $⇒p^3+\frac{1}{p^3}+3\times4=64$ $⇒p^3+\frac{1}{p^3}=64-12$ $⇒p^3+\frac{1}{p^3}=52$ Hence, the correct answer is 52.
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