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VITEEE PCME Mock Test - 11 - JEE MCQ


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30 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests - VITEEE PCME Mock Test - 11

VITEEE PCME Mock Test - 11 for JEE 2024 is part of VITEEE: Subject Wise and Full Length MOCK Tests preparation. The VITEEE PCME Mock Test - 11 questions and answers have been prepared according to the JEE exam syllabus.The VITEEE PCME Mock Test - 11 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for VITEEE PCME Mock Test - 11 below.
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VITEEE PCME Mock Test - 11 - Question 1

With a given surface area, the volume of a right circular cylinder will be maximum if height is

VITEEE PCME Mock Test - 11 - Question 2

The smaller area enclosed by the circle x2 + y2 = 9 and the line x = 1 is

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VITEEE PCME Mock Test - 11 - Question 3

If n is not a multiple of 3 then ωn + ω2n =

VITEEE PCME Mock Test - 11 - Question 4

The function f(x) = tan x/x for x ≠ 0 ; f (0) = 1 at x = 0 is

VITEEE PCME Mock Test - 11 - Question 5

VITEEE PCME Mock Test - 11 - Question 6

VITEEE PCME Mock Test - 11 - Question 7

If 1 , ω , ω2  , … ωn-1 are the n, nth roots of unity and z1 and z2 are any two, complex numbers, then 

VITEEE PCME Mock Test - 11 - Question 8

VITEEE PCME Mock Test - 11 - Question 9

For solving dy/dx=4x+y+1 suitable substitution is

VITEEE PCME Mock Test - 11 - Question 10

The solution of the equation cosx cosy (dy/dx)=-sinx siny is

VITEEE PCME Mock Test - 11 - Question 11

The differential equation of family of lines which passes (1,-1) is

VITEEE PCME Mock Test - 11 - Question 12

If f : R → R is defined by f x = 2 x + | x | then f (3x) − f (-x)   − 4 x =

VITEEE PCME Mock Test - 11 - Question 13

VITEEE PCME Mock Test - 11 - Question 14

If f(x)   = e -1/x2 , x ≠ 0, and f(0) = 0 then f ′(0)    is

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 14

VITEEE PCME Mock Test - 11 - Question 15

y = A e+ B e2x + C e3x  satisfies the diferential equation

VITEEE PCME Mock Test - 11 - Question 16

∫logx dx =

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 16

 

= x log x − ∫ d x + c
= x log x − x + c

VITEEE PCME Mock Test - 11 - Question 17

∫[(1)/(4x2+9)]dx =

VITEEE PCME Mock Test - 11 - Question 18

If sin-1x − cos-1x = π/6 , then x =

VITEEE PCME Mock Test - 11 - Question 19

VITEEE PCME Mock Test - 11 - Question 20

if f : R→R be such that f(1) = 4 and f'(1) = 12 then 

VITEEE PCME Mock Test - 11 - Question 21

Directions: Study the radar chart given below and answer the following questions.
Radar chart shows the markup % and discount % on five different articles sold by a shopkeeper.

 

Note:-1. Mark up % on any article=[(Marked Price-Cost Price) of that article)/Cost Price of that article]*100
2.Discount % on any article=[(Marked Price-Selling Price) of that article)/Marked Price of that article]*100

Q. If ratio of selling price of A to that of C is 40 : 27, then find marked price of C is what percent of cost price of A?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 21

Let selling price of A and that C be Rs 40x & Rs 27x respectively.
So, marked price of C = 27x *100/90 = Rs 30x
And, cost price of A = 40x * 100/80 * 100/125 =  Rs 40x
Required% = 30x/40x *100 = 75%

VITEEE PCME Mock Test - 11 - Question 22

Direction: Study the following graph carefully to answer the questions that follow.
Number of Students (In thousands) in Two Different Universities in Six Different Years

Q. In which year was the difference between the number of students in university 1 and the number of students in university 2 highest?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 22

Difference between the students of university 1 and university 2 in 2007 = 2000 – 1000 = 1000
Difference in number of students in 2008 = 25000 – 15000 = 10000
Difference in number of students in 2009 = 35000 – 25000 = 10000
Difference in number of students in 2010 = 20000 – 15000 = 5000
Difference in number of students in 2011 = 30000 – 25000 = 5000
Difference in number of students in 2012 = 35000 – 20000 = 15000

It is clear from above that the difference between the number of students in university 1 and the number of students in university 2 is highest in the year 2012.

VITEEE PCME Mock Test - 11 - Question 23

Directions: Study the following graph carefully to answer the questions that follow.
Number of Students (In thousands) in Two Different Universities in Six Different Years

Q. What was the difference between the number of students in university 1 in the year 2010 and the number of students in university 2 in year 2012?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 23

Number of students in university 1 in 2010 = 20000
Number of students in university 2 in 2012 = 20000
Required difference = 20000 – 20000 = 0.

VITEEE PCME Mock Test - 11 - Question 24

Directions: Study the following graph carefully to answer the questions that follow.
Number of Students (In thousands) in Two Different Universities in Six Different Years

Q. If 25% of the students in university 2 in the year 2010 were females, what was the number of male students in the university 2 in the same year?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 24

Total students in university 2 in 2010 = 15000
∴  Number of girls = 15000 × 25/100 = 3750
and number of boys = 15000 – 3750 = 11250.

VITEEE PCME Mock Test - 11 - Question 25

Directions: In this type of questions, some particular words are assigned certain substituted names. Then a question is asked that is to be answered in the substituted code language.

If 'light' is called 'morning', 'morning' is called 'dark', 'dark' is called 'night', 'night' is called 'sunshine', and 'sunshine' is called 'dusk', when do we sleep?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 25

We sleep in the 'night'. but 'night' is called 'sunshine'. So, we sleep in the 'sunshine'.
Hence, option B is correct.

VITEEE PCME Mock Test - 11 - Question 26

Directions: In this type of questions, some particular words are assigned certain substituted names. Then a question is asked that is to be answered in the substituted code language.

If 'orange' is called 'butter', 'butter' is called 'soap', 'soap' is called 'ink', 'ink' is called 'honey' and 'honey' is called 'orange', which of the following is used for washing clothes ?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 26

Clearly, 'soap' is used for washing the clothes. But a 'soap' is called 'ink'. So, 'ink' is used for washing the clothes.

Hence, option E is correct

VITEEE PCME Mock Test - 11 - Question 27

Directions: In this type of questions, some particular words are assigned certain substituted names. Then a question is asked that is to be answered in the substituted code language.

If 'sky' is 'star', 'star' is 'cloud', 'cloud' is 'earth', 'earth' is 'tree', and 'tree' is 'book', then where do the birds fly?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 27

Birds fly in the 'sky' and as given 'sky' is 'star'. So, birds fly in the 'star'.

Hence, option C is correct.

VITEEE PCME Mock Test - 11 - Question 28

In a class of 60, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. In an English test, the average mark scored by students majoring in Mathematics is 45 and that of students majoring in Biology is 60. However, the combined average mark in English, of students of these two majors, is 50. What is the maximum possible number of students who major ONLY in Physics?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 28

Let Tm and Tb be total scores of the students majoring in Mathematics and Biology respectively.

According to the given conditions,

Tm = (M + 6) × 45

Tb = (B + 15) × 60

Also, (Tm + Tb) = (B + P + 21) × 50 

45(M + 6) + 60(B + 15) = 50M + 50B + 1050

10B – 5M + 270 + 900 = 1050

10B – 5M = –120

M = 2B + 24

Here, we need to determine the maximum value of P. 

∴ We need to minimize the value of B. Minimum value of B can be 0.

∴ M = 24

Again, we know that, 

M + B + P + 21 = 60 

⇒ 24 + 0 + P + 21 = 60

⇒ P = 15

Hence, option (A).

VITEEE PCME Mock Test - 11 - Question 29

Out of 60 families living in a building, all those families which own a car own a scooter as well. No family has just a scooter and a bike. 16 families have both a car and a bike. Every family owns at least one type of vehicle and the number of families that own exactly one type of vehicle is more than the number of families that own more than one type of vehicle. What is the sum of the maximum and minimum number of families that own only a bike?

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 29

From the information given in the question, the following Venn Diagram can be constructed:

So, in order to maximize the number of families that own only a bike, we can put the remaining 44 families in ‘only bike’ region.

Similarly, in order to minimize the number of families that own only a bike, we can put the remaining 44 families in ‘only scooter’ region.

So, the maximum number of families that own only a bike is 44 and the minimum number of families that own only a bike is 0.

So, sum = 44 + 0 = 44

VITEEE PCME Mock Test - 11 - Question 30

400 students were admitted to the 2018-19 MBA batch. 200 of them did not choose “Business Statistics”. 100 of them did not choose “International Management’. There were 80 students who did not choose any of the two subjects. Find the number of students who chose both Business Statistics and International Management.

Detailed Solution for VITEEE PCME Mock Test - 11 - Question 30

Number of students who chose Business statistics = 400 − 200 = 200

Number of students who chose International Management = 400 − 100 = 300

Number of students who chose at least one of the two subjects = 400 − 80 = 320

∴ Number of students who chose both the subjects = 200 + 300 − 320 = 500 − 320 = 180

Hence, option (b).

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