Staff Selection Commission Combined Higher Secondary Level Exam
Question : Directions: Select the option that is related to the third term on the same basis as the second term is related to the first term. Electric current : Ampere :: Thermodynamic Temperature : ?
Option 1: Mole
Option 2: Kelvin
Option 3: Pascal
Option 4: Candela
Correct Answer: Kelvin
Solution : Given: Electric current : Ampere :: Thermodynamic Temperature : ?
Like, the unit used to measure electric current is an ampere. Similarly, the unit used to measure thermodynamic temperature is kelvin.
Hence, the second option is correct.
Question : In 1929, under the presidency of _______, the Lahore Congress formalised the demand of Purna Swaraj.
Option 1: Sardar Vallabhbhai Patel
Option 2: Mahatma Gandhi
Option 3: Subhas Chandra Bose
Option 4: Jawaharlal Nehru
Correct Answer: Jawaharlal Nehru
Solution : The correct answer is Jawaharlal Nehru.
The Lahore Session of the Indian National Congress was held in 1929 and passed the historic Purna Swaraj, which means the total independence resolution under the presidency of Jawaharlal Nehru. The Indian National Congress selected January 26, 1930, as the day to celebrate as ‘’Independence Day’, and on the same day, the Constitution of India was enacted in 1950.
Question : Ram Babu donated 3% of his income to a charity and deposited 12% of the rest in the bank. If he has Rs. 12,804, then his income was:
Option 1: Rs. 17,640
Option 2: Rs. 15,000
Option 3: Rs. 7,500
Option 4: Rs. 14,550
Correct Answer: Rs. 15,000
Solution : Let us assume Ram Babu's income is $x$. According to the question, $x \times\frac{97}{100}\times \frac{88}{100} = 12804$ ⇒ $x$ = 15,000 Hence, the correct answer is Rs. 15,000.
Question : Directions: In a certain code language, READER is written as 123421 and DIRTY is written as 49178. How is DEARER written in that code language?
Option 1: 432121
Option 2: 423212
Option 3: 423121
Option 4: 412312
Correct Answer: 423121
Solution : Given: READER is written as 123421 and DIRTY is written as 49178.
On combining the letters of both the given words, the codes for each letter are – Like, READER is written as 123421; R→1; E→2; A→3; D→4; E→2; R→1 And, DIRTY is written as 49178; D→4; I→9; R→1; T→7; Y→8 Similarly, DEARER; D→4; E→2; A→3; R→1; E→2; R→1
So, DEARER is coded as 423121 in the code language. Hence, the third option is correct.
Question : Directions: Arrange the given words in the sequence in which they occur in the dictionary. i. Irrelevant ii. Invincible iii. Irresistible iv. Invariable v. Investigate
Option 1: iv, ii, v, iii, i
Option 2: iv, v, i, iii, ii
Option 3: ii, v, iv, i, iii
Option 4: iv, v, ii, i, iii
Correct Answer: iv, v, ii, i, iii
Solution : Given: i. Irrelevant ii. Invincible iii. Irresistible iv. Invariable v. Investigate
Step 1: Compare the first letter of each word. Since all the words start with the same letter I, we move on to the next letter. Step 2: Let's move to the second letter of each word. The second letter of each word is r, n, r, n, n. Based on the alphabetical order of these letters, we can arrange them: Invincible, Invariable, Investigate Irrelevant, Irresistible. Step 3: Since the third letter of (Invincible, Invariable, Investigate) is the same, move on to the next letter. Compare the fourth letter of (Invincible, Invariable, Investigate). Invariable will come before Investigate and then Invincible in the sequence as a comes before e, and e comes before i in the alphabetical system. Step 4: Compare the fifth letter of (Irrelevant, Irresistible). Irrelevant will come before Irresistible in the sequence as i comes before s in the alphabetical system.
So, the sequence is Invariable, Investigate, Invincible, Irrelevant, Irresistible, or (iv), (v), (ii), (i), (iii). Hence, the fourth option is correct.
Question : Directions: Choose the related code from the given alternatives. AF : BE :: ? : DH
Option 1: CH
Option 2: CG
Option 3: DG
Option 4: CI
Correct Answer: CI
Solution : Given: AF : BE :: ? : DH
Subtract and add 1 alternatively to the letters of the second letter pair to get the first term. Like, B – 1 = A; E + 1 = F Similarly, D – 1 = C; H + 1 = I
So, DH is related to CI. Hence, the fourth option is correct.
Question : What is the value of (a + b)3 – a3 – b3?
Option 1: – 3ab(a – b)
Option 2: 3ab(a + b)
Option 3: – 3ab(a + b)
Option 4: 3ab(a – b)
Correct Answer: 3ab(a + b)
Solution : (a + b)3 – a3 – b3 = a3 + b3 + 3ab(a + b) – a3 – b3 = 3ab(a + b) Hence, the correct answer is 3ab(a + b).
Question : If in $\triangle PQR$ and $\triangle DEF, \angle P=52^{\circ}, \angle Q=74^{\circ}, \angle R=54^{\circ}, \angle D=54^{\circ}, \angle E=74^{\circ}$ and $\angle F=52^{\circ}$, then which of the following is correct?
Option 1: $\triangle \mathrm{PQR} \sim \triangle \mathrm{FED}$
Option 2: $\triangle \mathrm{RQP} \sim \triangle \mathrm{FED}$
Option 3: $\triangle \mathrm{PRQ} \sim \Delta \mathrm{FED}$
Option 4: $\triangle \mathrm{PQR} \sim \triangle \mathrm{DEF}$
Correct Answer: $\triangle \mathrm{PQR} \sim \triangle \mathrm{FED}$
Solution : $\because$ $\angle P=\angle F=52^0$ $\angle Q=\angle E=74^0$ And $\angle R=\angle D=54^0$ By AAA criteria, triangles are similar. So, $\triangle \mathrm{PQR} \sim \triangle \mathrm{FED}$. Hence, the correct answer is $\triangle \mathrm{PQR} \sim \triangle \mathrm{FED}$.
Question : Who among the following has been accused of committing the judicial murder of Raja Nand Kumar?
Option 1: Lord Wellesley
Option 2: Lord Dalhousie
Option 3: Lord Cornwallis
Option 4: Warren Hastings
Correct Answer: Warren Hastings
Solution : The correct option is Warren Hastings.
Warren Hastings, the first Governor-General of British India, faced accusations of orchestrating the judicial murder of Raja Nand Kumar during the late 18th century. The trial of Raja Nand Kumar was marked by allegations of political motives and irregularities, contributing to the historical controversy surrounding Hastings' governance and the East India Company's influence in India.
Question : If $\tan \theta=\frac{8}{19}$, find the value of $\sec ^2 \theta$.
Option 1: $\frac{11}{19} \\$
Option 2: $1 \frac{8}{19} \\$
Option 3: $\frac{297}{361} \\$
Option 4: $1 \frac{64}{361}$
Correct Answer: $1 \frac{64}{361}$
Solution : Given: $\tan \theta=\frac{8}{19}$ We know, $\sec ^2 \theta=1+\tan^2 \theta$ $⇒\sec ^2 \theta=1+(\frac{8}{19})^2$ $⇒\sec ^2 \theta=1+\frac{64}{361}$ $⇒\sec ^2 \theta=\frac{425}{361}$ $⇒\sec ^2 \theta=1\frac{64}{361}$ Hence, the correct answer is $1\frac{64}{361}$.
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