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VITEEE PCME Mock Test - 2 - JEE MCQ


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30 Questions MCQ Test - VITEEE PCME Mock Test - 2

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

The maximum value of sinx(1 + cos x) will be at

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 1

VITEEE PCME Mock Test - 2 - Question 2

The area, in square unit, bounded by the curves y = x3, y = x2 and the ordinates x = 1, x = 2 is

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VITEEE PCME Mock Test - 2 - Question 3

If z₁ and z₂ are two non-zero complex numbers such that |z₁ + z₂| = |z₁| + |z₂|, the value of arg(z₁) - arg(z₂) is

VITEEE PCME Mock Test - 2 - Question 4

If n is any positive integer ,then the value of
[i(4n+1) - i(4n-1)] /2 equals

VITEEE PCME Mock Test - 2 - Question 5

 for x ≠ 5 and f is continuous at x = 5 then f(5) =

VITEEE PCME Mock Test - 2 - Question 6

 is a singular matrix, then x =

VITEEE PCME Mock Test - 2 - Question 7

VITEEE PCME Mock Test - 2 - Question 8

If x ≠ 0  , then x =

VITEEE PCME Mock Test - 2 - Question 9

If the function f is defined by f(x)   = x / 1 + | x | then at what points is f differentiable

VITEEE PCME Mock Test - 2 - Question 10

The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant is

VITEEE PCME Mock Test - 2 - Question 11

The slope of the tangent at (x,y) to a curve passing thro' a point (2,1) is x2+y2/2xy, then the eqution of the curve is

VITEEE PCME Mock Test - 2 - Question 12

If x(dy/dx) = y(log y - log x + 1), then the solution of the equation is

VITEEE PCME Mock Test - 2 - Question 13

The solution of differential equation (x+2y3) dy/dx = y is

VITEEE PCME Mock Test - 2 - Question 14

The interval in which y=x2ex is increasing with respect to x, is

VITEEE PCME Mock Test - 2 - Question 15

Let f : R → R be defined by  


Then the value of f(-1.75) + f(0.5) + f(1.5) is

VITEEE PCME Mock Test - 2 - Question 16

∫secx dx=

VITEEE PCME Mock Test - 2 - Question 17

VITEEE PCME Mock Test - 2 - Question 18

If cot-1[(cos α)1/2] - tan-1[(cot α)1/2] = x, then sin x =

VITEEE PCME Mock Test - 2 - Question 19

VITEEE PCME Mock Test - 2 - Question 20

VITEEE PCME Mock Test - 2 - Question 21

Directions: Read the table given below and answer the questions accordingly.

Shares traded in Chandigarh, Punjab, Haryana (In Rupees)

Q. The average for the high rates of Chandigarh for all the companies was

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 21

The average for all the high involved with the Chandigarh region was

(438 + 148 + 28 + 190.5 + 160)/5 which gives 192.9.

VITEEE PCME Mock Test - 2 - Question 22

Directions: Read the table given below and answer the questions accordingly.

Shares traded in Chandigarh, Punjab, Haryana (In Rupees)

Q. Total of low value of the Punjab region exceeds the total of the low value of the region of Haryana by how much?

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 22

Total value for the low of the Punjab region for the given companies is 395 + 180 + 23 + 177.5 + 140.5 = 916

whereas the total for the low of the Haryana region is 413.75 + 180 + 24.5 + 177.5 + 102.5 = 898.25.

So it exceeds by 916 – 898.25 = 17.75.

VITEEE PCME Mock Test - 2 - Question 23

Directions: Read the table given below and answer the questions accordingly.

Shares traded in Chandigarh, Punjab, Haryana (In Rupees)

Q. Low for RCOM for all the given states/UT form what percent of the low for ZARA for the same?

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 23

The total of lows for the RCOM for all the given states/ UT are given to be 532 and the ZARA has the total for their lows to be 500.

Thus ATQ (532/500)*100 = 106.4%

VITEEE PCME Mock Test - 2 - Question 24

Directions: Answer the given question based on the following line graph:
Calculate the increase in stock of cereal 1 for 1992

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 24

Within a year, increase/decrease in stock for any cereal cannot be determined from the given data.

VITEEE PCME Mock Test - 2 - Question 25

Directions: Study the given information carefully and answer the following question.

The graph below represents sales and net profit in Rs. crore of IVP Ltd. for five years from 1994-95 to 1998-99. During this period, the sales increased from Rs. 100 crore to Rs. 680 crore. Correspondingly, the net profit increased from Rs. 2.5 crore to Rs. 12 crore. Net profit is defined as the excess of sales over total costs.

Q. Defining profitability as the ratio of net profit to net sales, it can be concluded that

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 25

It can be seen that profitability did not follow a fixed pattern as the first three statements try to generalise it. Therefore, they are not applicable.

VITEEE PCME Mock Test - 2 - Question 26

Directions: Study the given information carefully and answer the following question.

The graph below represents sales and net profit in Rs. crore of IVP Ltd. for five years from 1994-95 to 1998-99. During this period, the sales increased from Rs. 100 crore to Rs. 680 crore. Correspondingly, the net profit increased from Rs. 2.5 crore to Rs. 12 crore. Net profit is defined as the excess of sales over total costs.

Q. The highest percentage growth in net profit as compared to the previous year was seen in

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 26

From the graph, we can calculate the growth in profits.

The highest percentage growth in net profit as compared to the previous year was seen in 1995-96.

VITEEE PCME Mock Test - 2 - Question 27

Directions: Study the given information carefully and answer the following question.

The graph below represents sales and net profit in Rs. crore of IVP Ltd. for five years from 1994-95 to 1998-99. During this period, the sales increased from Rs. 100 crore to Rs. 680 crore. Correspondingly, the net profit increased from Rs. 2.5 crore to Rs. 12 crore. Net profit is defined as the excess of sales over total costs.

Q. The highest percentage of growth in sales over the previous year occurred in

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 27

From the graph, we know that the percentage growth in sales were:

It is but obvious from the above table that the maximum percentage increase in sales as compared to the previous year occurred in 1995-96.

VITEEE PCME Mock Test - 2 - Question 28

Sheldon had to cover a distance of 60 km. However, he started 6 minutes later than his scheduled time and raced at a speed 1 km/h higher than his originally planned speed and reached the finish at the time he would reach it if he began to race strictly at the appointed time and raced with the assumed speed. Find the speed at which he travelled during the journey described.

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 28

Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins

Thus, 25 km/h is the correct answer. 

So Option A is correct

VITEEE PCME Mock Test - 2 - Question 29

Two sprinters run the same race of100 m One runs at a speed of 10 m/s and the other runs at 8 m/s. By what time will the first sprinter beat the other sprinter?

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 29

Correct option is C
Time taken by first sprinter 
= 100/10 = 10sec
Time taken by second sprinter 
= 100/80 = 12.5sec
Difference = 12.5 - 10 = 2.5 sec

VITEEE PCME Mock Test - 2 - Question 30

X can do a piece of work in 20 days. He worked at it for 5 days and then Y finished it in 15 days. In how many days can X and Y together finish the work?

Detailed Solution for VITEEE PCME Mock Test - 2 - Question 30
  • X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
  • This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.  
  • X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.

So Option C is correct

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