HO
Question : Directions: Study the given pattern carefully and select the number that can replace the question mark (?) in it.
Option 1: 47
Option 2: 28
Option 3: 24
Option 4: 19
Correct Answer: 19
Solution : Given:
The pattern can be observed running horizontally along the rows. Row 1: (140 ÷ 5) – 9 = 28 – 9 = 19 Row 2: (130 ÷ 5) – 9 = 26 – 9 =
Hi
You should aim for a percentile score of around 90-95 or higher. While the exact cutoff may vary each year based on factors like difficulty level and the number of candidates, historically, scoring in this range has generally qualified students for JEE Advanced.
Minimum Marks: While specific marks vary
Question : Directions: In the following question, a sentence is given with a blank that is to be filled in with an appropriate word. Four alternatives are suggested; choose the correct alternative out of them as your answer.
Priya _____ a taste for poetry in the company of her cousin.
Option 1: Imbibed
Option 2: Fostered
Option 3: Learnt
Option 4: Cultivated
Correct Answer: Cultivated
Solution : The correct choice is the fourth option "cultivated".
"Cultivated'' means to develop or nurture, and it fits perfectly in the sentence conveying the intended meaning that Priya developed a taste for poetry in the company of her cousin.
The meanings of the other options are
Hello,
Yes, it is possible for you to change your course from BSc to BA after failing in BSc first year. Many students go through this every year and I understand how upsetting it must be for you. However, I recommend you to connect with the college you are aiming
Question : The lateral surface area of a frustum of a right circular cone, if the area of its base is 16$\pi$ cm2, the diameter of the circular upper surface is 4 cm and the slant height is 6 cm, will be:
Option 1: $30\pi$ cm2
Option 2: $48\pi$ cm2
Option 3: $36\pi$ cm2
Option 4: $60\pi$ cm2
Correct Answer: $36\pi$ cm2
Solution : Given: Upper diameter = 4 cm Let the radius of the top surface be $r$. $\therefore r = \frac{4}{2} = 2$ cm Given that, the area of the base $=16\pi$ $⇒\pi R^2 = 16\pi$, where $R$ is the radius of base $\therefore R
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