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Question : If in a $\triangle$ABC as drawn in the figure, AB = AC and $\angle$ACD = 130°, then $\angle$BAC is equal to:
Option 1: 60°
Option 2: 50°
Option 3: 70°
Option 4: 80°
Correct Answer: 80°
Solution : $\angle$ACD = 130° Now, $\angle$ACD + $\angle$ACB = 180° ⇒ 130° + $\angle$ACB = 180° ⇒ $\angle$ACB = 180° – 130° = 50° Since AB = AC, ⇒ $\angle$ABC = $\angle$ACB = 50° We know that $\angle$ABC + $\angle$ACB + $\angle$BAC = 180° ⇒ 50°
Question : Directions: In the following question, some parts of the sentence have errors, and some are correct. Find out which part of the sentence has an error. The number of that part is the answer. If a sentence is error-free, your answer is "No Error".
He can be a basketball player since he is tall like a mule.
(1) He can be
(2) he is tall like a mule.
(3) No Error
(4) a basketball player since
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4
Correct Answer: 2
Solution : The phrase "like a mule" in part (2) is incorrect because it doesn't convey the intended meaning. Instead, it should be "because he is tall" to explain why he can be a basketball player. "Like a mule" is a simile that doesn't fit in this
Question : The venue of the Energy Summit, 2008 was
Option 1: Jeddah
Option 2: Tehran
Option 3: Riyadh
Option 4: Tripoli
Correct Answer: Jeddah
Solution : The correct answer is Jeddah.
The Energy Summit, focused on the rocketing oil prices, was held in 2008 in Jeddah, Saudi Arabia. The meeting was held because global oil prices hit a record high. The meeting was attended by the world's major oil producers and
Question : Direction: If Z = Y > R = M and G > H = Z = Q, then which of the following options is NOT correct?
Option 1: H = Y
Option 2: G > Q
Option 3: R > Z
Option 4: Q > R
Correct Answer: R > Z
Solution : Given: Z = Y > R = M and G > H = Z = Q, After combining the statements: G > H = Z = Q = Z = Y > R = M Let's check each option –
First option: H
Question : Comprehension:
Read the following passage and answer the questions given after it.
The excitement of bird watchers on spotting a particular species for the very first time is unparalleled. With eyes shining and pride bursting, they exclaim, “I had a life today”, or “It was a lifer for me”. For the longest time, I couldn’t fathom what all the fuss was about. I mean, at some point in their lives, even a crow or mynah would have been a lifer — seen for the very first time. So, here, I’ve done a bit of jugaad with the term and defined it as a bird that you see maybe (but not necessarily) for the first time, but which has made a lasting, life-changing impact on you. Then I recalled some of my own “lifers”.
Number 1 is the little coppersmith barbet. The first bird I saw through brand new, big and powerful binoculars — and it was solely responsible for my getting interested in birds. The fellow looked like a tubby little clown with hiccups and that just blew me away.
I will never forget the first time I saw grey hornbills aeons ago: over sullen grey skies in the Borivali National Park (now called Sanjay Gandhi National Park) — squealing as they flew high up across the sky. They looked as if they had just left Jurassic Park. Or, for that matter, their larger, more glamorous, cousins — Great pied hornbills. Tramping through a streambed in Kalagarh (near Corbett), we suddenly heard this rasping, whooshing, sound. Up there, in the clear blue, were six-seven huge black-and-white birds with colossal yellow beaks flying in tandem across the clear blue sky, their wings making a rasping sound.
Say “paradise flycatcher” and a birder’s eyes will begin to glint: “Where? When? Will it be there now?” are questions that will be shot out like machine-gun bullets. The first time I saw a full-grown milk-white-and-glossy-black male, with its glamorous 18-inch streamer tail, was at the Sultanpur National Park in Haryana. But I remember better the flycatchers, that made me run around in a tea garden in Palampur, teasingly whistling at me from one end to the other. The nesting pair in Naukuchiatal was more accommodating except that I had to stand kneedeep in the hotel’s garbage dump to get a good view of them flitting to and fro the gully nearby. To compensate, one actually flew nearly down to my feet to snatch up a blue bottle I had missed.
Of course, there have been rarities: the highlight of the regular Bharatpur (the Keoladeo National Park) visits was the darshan of VIP Siberian cranes. Then they stopped coming, which was the first indication of their slow extinction — even if it was just “local” to our area. The gloriously uppity Great Indian bustards in the Karera Sanctuary (Madhya Pradesh) were another unforgettable sighting. The sheer disdain with which they flounced away from our howling, jolting jeep and took to their wings was a lesson in being put in your place. Now, not only does the sanctuary not exist anymore but those magnificent muscular birds are crashing to total extinction.
Question: Which pair of birds did the author see in Naukuchiatal?
Option 1: Coppersmith barbet
Option 2: Paradise flycatcher
Option 3: Hornbill
Option 4: Great Indian bustard
Correct Answer: Paradise flycatcher
Solution : The correct choice is the second option.
According to the fourth paragraph of the passage, the author saw paradise flycatcher. The author mentions in the paragraph that he saw a pair of resting paradise flycatchers in Naukuchiatal.
Question : $\mathrm{O}$ is the centre of a circle and $\mathrm{A}$ is a point on a major arc $\mathrm{BC}$ of the circle. $\angle \mathrm{BOC}$ and $\angle \mathrm{BAC}$ are the angles made by the minor arc $\mathrm{BC}$ on the centre and circumference, respectively. If $\angle \mathrm{ABO}=40^{\circ}$ and $\angle \mathrm{ACO}=30^{\circ}$, then find $\angle \mathrm{BOC}$.
Option 1: $130^{\circ}$
Option 2: $140^{\circ}$
Option 3: $120^{\circ}$
Option 4: $150^{\circ}$
Correct Answer: $140^{\circ}$
Solution :
Given: That $O$ is the centre of a circle and $A$ is a point on a major arc $BC$ of the circle. $\angle BOC$ and $\angle BAC$ are the angles made by the minor arc $BC$ on the centre and circumference, respectively. Also $\angle ABO=40^{\circ}$
Question : Which Indian shuttler won the men's singles final at the Scottish Open in Glasgow, Scotland on 24 November 2019?
Option 1: Sameer Verma
Option 2: Lakshya Sen
Option 3: Arvind Bhatt
Option 4: Anand Pawar
Correct Answer: Lakshya Sen
Solution : The correct answer is Lakshya Sen.
Indian badminton player Lakshya Sen secured victory at the Scottish Open men's singles championship by defeating Brazilian Ygor Coelho. As the top seed, world number 41 Lakshya Sen clinched his fourth title in three months with an
Question : Directions: In the following question, some parts of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If the sentence is free from error, select "No Error."
You will come (1) / to my sister's wedding tomorrow, (2) / isn't it? (3) / No Error (4)
Option 1: (1)
Option 2: (2)
Option 3: (3)
Option 4: (4)
Correct Answer: (3)
Solution : The error lies in the third part of the sentence.
The question tag "isn't it" is not appropriate in this context. The sentence is structured as a statement followed by a question to seek confirmation. The contracted and negative version of the helping verb, followed
Question : Directions: If A ÷ B means that A is the brother of B, A × B means that A is the sister of B, and A – B means that A is the father of B then which of the following expressions shows that P is the father of R?
Option 1: P – Q × R
Option 2: P × Q – R
Option 3: P ÷ Q × R
Option 4: P – Q – R
Correct Answer: P – Q × R
Solution : Let's check the options – First option: P – Q × R According to the given expression, the family tree will be as follows – Here, the quadrilateral represents the male, and the circular figure represents the female in the figure.
Question : In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. What is the measure of $\angle$ADC (in degrees)?
Option 1: 15
Option 2: 30
Option 3: 45
Option 4: 60
Correct Answer: 30
Solution : Given: In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. Given that AD = BD. Then, $\angle$ABD = $\angle$BAD = 15° Now, in $\triangle$ABD ⇒ $\angle$ABD + $\angle$BAD + $\angle$ADB = 180° ⇒ $\angle$ADB = 180° –
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