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KEAM Paper 2 Mock Test - 2 - JEE MCQ


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13 Questions MCQ Test - KEAM Paper 2 Mock Test - 2

KEAM Paper 2 Mock Test - 2 for JEE 2024 is part of JEE preparation. The KEAM Paper 2 Mock Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The KEAM Paper 2 Mock Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for KEAM Paper 2 Mock Test - 2 below.
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KEAM Paper 2 Mock Test - 2 - Question 1

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KEAM Paper 2 Mock Test - 2 - Question 2

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KEAM Paper 2 Mock Test - 2 - Question 3

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KEAM Paper 2 Mock Test - 2 - Question 4

The approximate value of square root of 25.2 is

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KEAM Paper 2 Mock Test - 2 - Question 5

If the tangent to the curve  at any point on it cuts the axes OX and OY at P and Q respectively, then OP + OQ is 

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KEAM Paper 2 Mock Test - 2 - Question 6

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KEAM Paper 2 Mock Test - 2 - Question 7

If A and B  are two given sets, then A ∩  (A∩B)is equal to

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KEAM Paper 2 Mock Test - 2 - Question 8

 is equal to

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KEAM Paper 2 Mock Test - 2 - Question 9

 is equal to

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KEAM Paper 2 Mock Test - 2 - Question 10

The coefficient of x in the equation x2 + px + q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

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KEAM Paper 2 Mock Test - 2 - Question 11

If the sets A and B are defined as
A = {(x, y) : y =  1/x, 0 ≠ x ∈ R}
B = {(x, y) : y = -x, x ∈ R}, then

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KEAM Paper 2 Mock Test - 2 - Question 12

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KEAM Paper 2 Mock Test - 2 - Question 13

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

Detailed Solution for KEAM Paper 2 Mock Test - 2 - Question 13
  • n(A) = 40% of 10,000 = 4,000
  • n(B) = 20% of 10,000 = 2,000
  • n(C) = 10% of 10,000 = 1,000
  • n(A ∩ B) = 5% of 10,000 = 500
  • n(B ∩ C) = 3% of 10,000 = 300
  • n(C ∩ A) = 4% of 10,000 = 400
  • n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]
= 4000 - [500 + 400 - 200] = 4000 - 700 = 3300

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