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Question : Krishna travels from point A to point B in 6 hours and returns to point A from point B in 5 hours. Point A and Point B are 220 miles apart along a straight highway. What is the average speed of Krishna for the whole journey?
Option 1: 44 miles/hr
Option 2: 40 miles/hr
Option 3: 50 miles/hr
Option 4: 22 miles/hr
Correct Answer: 40 miles/hr
Solution : $\text{Average speed}=\frac{\text{Total distance}}{{\text{Total time}}}=\frac{220+220}{6+5}=\frac{440}{11}=40$ miles/hr Hence, the correct answer is 40 miles/hr.
Question : Directions: From the given answer figures, select the one in which the question figure is hidden/embedded.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution : Because there is no restriction on the rotation of the figure, we will check all the possible orientations of the question figure in which of the given option figures it can fit itself. By comparison of all option figures, the question figure is embedded only in
Question : Directions: In the following question, select the related word from the given alternatives. Frame : Picture :: ? : ?
Option 1: Box : Lid
Option 2: Chair : Cushion
Option 3: Cover : Book
Option 4: Window : Screen
Correct Answer: Cover : Book
Solution : Given: Frame : Picture :: ? : ? (Here, the first term is used to protect the second term.)
Let's check each option – First option: Box : Lid; Box is not used to protect lid. Second option: Chair : Cushion; Cushion is
Question : If $\cos^{2}\alpha-\sin^{2}\alpha=\tan^{2}\beta$, then the value of $\cos^{2}\beta-\sin^{2}\beta$ is:
Option 1: $\cot^{2}\alpha$
Option 2: $\cot^{2}\beta$
Option 3: $\tan^{2}\alpha$
Option 4: $\tan^{2}\beta$
Correct Answer: $\tan^{2}\alpha$
Solution : Given: $\cos^{2}\alpha–\sin^{2}\alpha=\tan^{2}\beta$ We know that: $1+\tan^{2}\alpha=\sec^{2}\alpha$ So, $\cos^{2}\alpha-\sin^{2}\alpha=\sec^{2}\beta-1$ ⇒ $\cos^{2}\alpha-\sin^{2}\alpha+1=\sec^{2}\beta$ ⇒ $\cos^{2}\beta=\frac{1}{\cos^{2}\alpha–\sin^{2}\alpha+(\sin^{2}\alpha+\cos^{2}\alpha)}$ ⇒ $\cos^{2}\beta=\frac{1}{2\cos^{2}\alpha}$ Also, $\sin^{2}\beta=1-\cos^{2}\beta=1-\frac{1}{2\cos^{2}\alpha}$ So, $\cos^{2}\beta-\sin^{2}\beta$ $=\frac{1}{2\cos^{2}\alpha}-(1-\frac{1}{2\cos^{2}\alpha})$ $= \frac{1}{\cos^{2}\alpha}-1$ $= \sec^{2}\alpha-1$ $= \tan^{2}\alpha$ Hence, the correct answer is $\tan^{2}\alpha$.
Question : Which country hosted the first-ever Commonwealth Games in 1930?
Option 1: Australia
Option 2: New Zealand
Option 3: England
Option 4: Canada
Correct Answer: Canada
Solution : The correct option is Canada.
The inaugural Commonwealth Games were held in Hamilton, Ontario, Canada, from August 16 to 23, 1930. The event was originally known as the British Empire Games. 11 countries participated in this game.
Question : Dadabhai Naoroji was a_____.
Option 1: Businessman
Option 2: Doctor
Option 3: Soldier
Option 4: British Officer
Correct Answer: Businessman
Solution : The correct answer is Businessman.
Dadabhai Naoroji, born into a Parsi family in Bombay, British India, initially pursued a business career. He was a partner in the firm Cama & Co., marking the establishment of the first Indian company in Britain. Additionally, Naoroji was actively
Question : The process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats) is called ______.
Option 1: adjustive radiation
Option 2: adjustive evolution
Option 3: adaptive evolution
Option 4: adaptive radiation
Correct Answer: adaptive radiation
Solution : The correct answer is adaptive radiation.
The process of evolution of different species in a given geographical area, starting from a point and radiating to other areas of geography (habitats), is called "adaptive radiation." Adaptive radiation occurs when a single ancestor species gives rise
Question : Stomatal opening is based on which of the following?
Option 1: Exosmosis
Option 2: Endosmosis
Option 3: Plasmolysis in guard cells
Option 4: Decrease in concentration of cell sap
Correct Answer: Endosmosis
Solution : The correct option is Endosmosis.
Endosmosis is the mechanism by which plant guard cells open their stomata. The passage of water over a semi-permeable membrane from a low solute concentration area to a high solute concentration area is osmosis. Water travels into guard cells
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. As a rule, you see, I'm not lugged into family rows. On the occasions when Aunt is calling Aunt like mastodons bellowing (1)_______. Primaeval swamps and Uncle James's letter about Cousin Mabel's peculiar behaviour is being shot (2)_____ the family circle ('Please read this carefully and send it on Jane') the clan has a tendency (3)______. ignore me. It's one of the advantages I get from (4)_______. a bachelor and, according to my nearest and dearest, practically a half-witted bachelor (5)________that.
Question: Select the most appropriate option to fill in the blank number 4.
Option 1: not being
Option 2: removing
Option 3: being
Option 4: becoming
Correct Answer: being
Solution : The third option is correct.
Explanation: In this sentence, the gerund being would be used after the preposition from to indicate the source or origin of the advantage. The words from being connect the preposition with its object, a bachelor. This construction implies that
Question : On day one, with speed $v$, R covers a distance $x$, in t time. On the next day, he covers a distance of 2.5 $x$ in 0.75 $t$ time. What is his speed in the next day?
Option 1: $3.5v$
Option 2: $\frac{10}{3}v$
Option 3: $4.5v$
Option 4: $\frac{5}{3}v$
Correct Answer: $\frac{10}{3}v$
Solution : Given: $\frac{x}t=v$ Now, distance travelled by R on the second day = $2.5x$ Time taken by him to cover that distance = $0.75t$ So, speed on second day = $\frac{2.5x}{0.75t}=\frac{10}{3}v$ Hence, the correct answer is $\frac{10}{3}v$.
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