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Question : Directions: In each of the following questions, which one set of letters/numbers when sequentially placed at the gaps in the given letter series shall complete it? bb_aab_caab_ca_
Option 1: abac
Option 2: bcab
Option 3: cbba
Option 4: acab
Correct Answer: cbba
Solution : Given: bb_aab_caab_ca_
To fill the series we have to divide the series – bb_aa / b_caa / b_ca_ Let's check each option – First option: abac; bbaaa / bbcaa / bacac (No repeated pattern has been found.) Second
Question : The base of a right prism is a triangle whose sides are 8 cm, 15 cm and 17 cm, and its lateral surface area is 480 cm$^2$. What is the volume (in cm$^3$) of the prism?
Option 1: 540
Option 2: 600
Option 3: 720
Option 4: 640
Correct Answer: 720
Solution : Given: Lateral surface area of prism = 480 cm2 Lateral surface area of prism = perimeter of base × height Perimeter of triangle = 8 + 15 + 17 = 40 cm Let the height of the prime is $h$ cm ⇒ $40 ×
Question : If $ \frac{k-k \cot ^2 30^{\circ}}{1+\cot ^2 30^{\circ}}=\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ}$, then the value of k (correct to two decimal places) is:
Option 1: 5.55
Option 2: – 6.83
Option 3: – 5.58
Option 4: 6.83
Correct Answer: – 6.83
Solution : Computing RHS, $\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ} = (\frac{\sqrt{3}}{2})^2+4\times1^2-(\frac{2}{\sqrt{3}})^2$ ⇒ $\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ} = \frac{3}{4}+4-\frac{4}{3}$ ⇒ $\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ} = \frac{41}{12}$ Computing LHS, $ \frac{k-k \cot ^2 30^{\circ}}{1+\cot ^2 30^{\circ}} = k(\frac{1- (\sqrt{3})^2}{1+(\sqrt{3})^2})$ ⇒
Question : Directions: How many quadrilaterals are there in the given figure?
Option 1: 2
Option 2: 3
Option 3: 4
Option 4: 5
Correct Answer: 2
Solution : The given figure can be labeled as shown below –
There are 2 quadrilaterals in the above figure. They are ABFE and ABCE.
Hence, the first option is correct.
Question : Select the word that is OPPOSITE in meaning to the given word.
Adversity
Option 1: Fortune
Option 2: Affliction
Option 3: Brilliance
Option 4: Penury
Correct Answer: Fortune
Solution : The correct choice is the first option.
Adversity refers to difficulties, misfortune, or hardship that someone faces in life.
On the other hand, fortune stands in opposition to adversity. It refers to good luck, success, wealth, or favourable circumstances.
The meanings of the other
Question : Which of the following nations was a part of a coalition called "Allied Powers" in World War II?
Option 1: USSR
Option 2: Germany
Option 3: Italy
Option 4: Japan
Correct Answer: USSR
Solution : The correct answer is the USSR
The Soviet Union was part of the coalition called the "Allied Powers" in World War II. The Allied Powers, also known as the Allies, included several nations, with major contributors being the United States, the Soviet Union, the United
Question : The marked price of a watch is Rs. 1600. The shopkeeper gives successive discounts of 10% and $x$% to the customer. If the customer pays Rs. 1224 for the watch, the value of $x$ is:
Option 1: 5%
Option 2: 10%
Option 3: 15%
Option 4: 20%
Correct Answer: 15%
Solution : The first discount of 10% on Rs. 1600. Discount = 10% of Rs. 1600 = $\frac{10}{100}$ × 1600 = Rs. 160 Price of watch after 1st discount = Rs. 1600 – Rs. 160 = Rs. 1440 As the customer pays Rs. 1224 after the second
Question : The number of students in a class has increased by 20% and the number has now become 66. Initially, the number was:
Option 1: 45
Option 2: 50
Option 3: 55
Option 4: 60
Correct Answer: 55
Solution : Given: The number of students in a class has increased by 20% and the number has now become 66. Let the initial number of students be $x$. According to the question, ⇒ $\frac{120x}{100} = 66$ ⇒ $x = 55$ Hence, the correct answer is 55.
Question : The table given below shows the number of students from five classes playing different games.
Which of the following statements is correct? I. The average number of students playing A and B in the 9th class is 32. II. The average number of students playing A and C in the 8th class is 75. III. The total number of students playing all three games in the 10th class is 77.
Option 1: II and III
Option 2: I, II and III
Option 3: I and III
Option 4: I and II
Correct Answer: I and III
Solution : I. The average number of students playing A and B in 9th class = $\frac{38+26}{2}$ = $\frac{64}{2}$ = $32$ II. The average number of students playing A and C in the 8th class = $\frac{40+90}{2}$ = $\frac{130}{2}$ = $65$ III. The total number
Question : Directions: Arrange the given words in the sequence in which they occur in the dictionary: i. Examination ii. Explicit iii. Expenditure iv. Experience
Option 1: i, iv, iii, ii
Option 2: i, ii, iii, iv
Option 3: i, iv, ii, iii
Option 4: i, iii, iv, ii
Correct Answer: i, iii, iv, ii
Solution : Given: i. Examination ii. Explicit iii. Expenditure iv. Experience
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