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VITEEE PCBE Mock Test - 15 - JEE MCQ


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30 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests - VITEEE PCBE Mock Test - 15

VITEEE PCBE Mock Test - 15 for JEE 2024 is part of VITEEE: Subject Wise and Full Length MOCK Tests preparation. The VITEEE PCBE Mock Test - 15 questions and answers have been prepared according to the JEE exam syllabus.The VITEEE PCBE Mock Test - 15 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for VITEEE PCBE Mock Test - 15 below.
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VITEEE PCBE Mock Test - 15 - Question 1

Numerous vascular bundles occur scattered in the ground tissue of

VITEEE PCBE Mock Test - 15 - Question 2

Bee dances occur because of

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VITEEE PCBE Mock Test - 15 - Question 3

The drug Santonin is obtained from the flowers of

VITEEE PCBE Mock Test - 15 - Question 4
Joining of repeating units to form a macromolecules is called
VITEEE PCBE Mock Test - 15 - Question 5
Radioactive entities are used to trace certain macromolecules. If radioactive aminoacid is used which organelle it will be traced in
VITEEE PCBE Mock Test - 15 - Question 6
Who discovered the cell (actually the cell wall) in section of cork ?
VITEEE PCBE Mock Test - 15 - Question 7
Which of the following would occur as a result of the oval opening in the heart remaining open after birth?
VITEEE PCBE Mock Test - 15 - Question 8
Hardness to the cell wall is provided by
VITEEE PCBE Mock Test - 15 - Question 9
Which one is inverted pyramid?
VITEEE PCBE Mock Test - 15 - Question 10
Which one is vestigial in humans
VITEEE PCBE Mock Test - 15 - Question 11
Dinsaurs were dominant during
VITEEE PCBE Mock Test - 15 - Question 12
A vestigial organ in man is
VITEEE PCBE Mock Test - 15 - Question 13
Which of the following causes point mutation
VITEEE PCBE Mock Test - 15 - Question 14
Flow of information during translation is
VITEEE PCBE Mock Test - 15 - Question 15
Which of the following pairs is correctly matched with regard to the codon and the amino acid coded by it ?
VITEEE PCBE Mock Test - 15 - Question 16
When one gene hides the effect of another gene, the interaction factor is known as :
VITEEE PCBE Mock Test - 15 - Question 17
In a cross between a pure tall pea plant with green pos and a pure short plant with yellow pod, how many short plants out of 16 you would expect in F₂ generation?
VITEEE PCBE Mock Test - 15 - Question 18
In sex-linked traits the males are
VITEEE PCBE Mock Test - 15 - Question 19
Huntington's chorea is characterised by
VITEEE PCBE Mock Test - 15 - Question 20
Suspension of attenuated pathogen that stimulates antibody formation is
VITEEE PCBE Mock Test - 15 - Question 21

These questions are based on the following information. P, Q, R, S and T sit around a table. P sits two seats to the left of R and Q sits two seats to the right of R.

Q. If S is not sitting next to Q, who is sitting between Q and S?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 21

P sits two seats to the left of R, and Q sits two seats to the right of R. We can represent this information in the diagram below.

If S is not next to Q, then the seating arrangement is fixed as follows.

Now P is between Q and S. Choice (b)

VITEEE PCBE Mock Test - 15 - Question 22

These questions are based on the following information. P, Q, R, S and T sit around a table. P sits two seats to the left of R and Q sits two seats to the right of R.

Q. If a new person U joins the group such that the initial conditions for the seating arrangement should be observed and also a new condition that U does not sit next to R be satisfied, then which of the following statements is true?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 22

P sits two seats to the left of R, and Q sits two seats to the right of R. We can represent this information in the diagram below.

On the basis of the diagram that we drew, we find that to accommodate U we have to create a new slot between P and Q.

Hence, choice (c) is the correct answer. Choice (c)

VITEEE PCBE Mock Test - 15 - Question 23

These questions are based on the following information. P, Q, R, S and T sit around a table. P sits two seats to the left of R and Q sits two seats to the right of R.

Q. If a new person U joins the group such that the initial conditions for the seating arrangement should be observed and also a new condition that U does not sit next to P, S or T be satisfied, then who will be the neighbours of P (one on either side)?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 23

P sits two seats to the left of R, and Q sits two seats to the right of R. We can represent this information in the diagram below.

We create a new slot for the sixth person. But since U will not sit next to P, S or T, he will have to sit between R and Q. The arrangement will then look as follows:

As we can see from the diagram, the neighbours of P will be T and S.

VITEEE PCBE Mock Test - 15 - Question 24

Directions: Study the following graph and answer the question that follows.

POWER PRODUCTION OF A COMPANY (IN MILLION WATTS)


Q. What is the difference between the peak hydel power production and the least Hydel Power Production ?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 24

Difference between the peak hydel power production and the least hydel power production = 180 - 100 = 80 million watts
So, option (4) is correct.

VITEEE PCBE Mock Test - 15 - Question 25

Directions: Study the following graph and answer the question that follows.

POWER PRODUCTION OF A COMPANY (IN MILLION WATTS)

Q. In which year is the total power production (thermal + hydel) the greatest?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 25

Total power production (thermal and hydel) in 1990 = 140 + 140 = 280 million watts
Total power production (thermal and hydel) in 1991 = 140 + 180 = 320 million watts
Total power production (thermal and hydel) in 1992 = 100 + 120 = 220 million watts
Total power production (thermal and hydel) in 1993 = 150 + 130 = 280 million watts
Total power production (thermal and hydel) in 1994 = 155 + 145 = 300 million watts
Total power production (thermal and hydel) in 1995 = 153 + 141 = 294 million watts
Total power production (thermal and hydel) in 1996 = 145 + 145 = 290 million watts
The total power production is the greatest in 1991, i.e. 320 million watt.

VITEEE PCBE Mock Test - 15 - Question 26

Directions: Study the following graph and answer the question that follows.

POWER PRODUCTION OF A COMPANY (IN MILLION WATTS)

Q. What is the average thermal power production for the period 1990-1996?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 26

Required average = = 141.85 million watts

Hence, option (4) is correct.

VITEEE PCBE Mock Test - 15 - Question 27

Directions: Study the following graph carefully and answer the question given below.

Q. If the population of a city is 35,000 and the average income of a person in that city is Rs. 25,000 per annum, then what is the total amount of tax paid by all the persons of that city?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 27

Tax paid by one person of that city = 8/100 x 25,000 = Rs. 2000

Total tax paid by all the persons of that city = 2000 × 35,000 = Rs. 70,000,000

VITEEE PCBE Mock Test - 15 - Question 28

In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:

I. a + c = e,
II. b – d = d and
III. e + a = b

Which of the following options are true?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 28

We have a + c = e so possible summation 6+4=10 or 4+2 = 6.
Also b = 2d so possible values  4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .

VITEEE PCBE Mock Test - 15 - Question 29

The sum of the first 100 natural numbers, 1 to 100 is divisible by: 

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 29

The sum of the first 100 natural numbers is:

=  (n * (n + 1)) / 2
=  (100 * 101) / 2
=  50 * 101

101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.

VITEEE PCBE Mock Test - 15 - Question 30

In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?

Detailed Solution for VITEEE PCBE Mock Test - 15 - Question 30

Let the 4 digit no. be xyzw.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.

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