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JEE Exam  >  JEE Tests  >  VITEEE PCBE Mock Test - 14 - JEE MCQ

VITEEE PCBE Mock Test - 14 - JEE MCQ


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30 Questions MCQ Test - VITEEE PCBE Mock Test - 14

VITEEE PCBE Mock Test - 14 for JEE 2024 is part of JEE preparation. The VITEEE PCBE Mock Test - 14 questions and answers have been prepared according to the JEE exam syllabus.The VITEEE PCBE Mock Test - 14 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for VITEEE PCBE Mock Test - 14 below.
Solutions of VITEEE PCBE Mock Test - 14 questions in English are available as part of our course for JEE & VITEEE PCBE Mock Test - 14 solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt VITEEE PCBE Mock Test - 14 | 125 questions in 150 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
VITEEE PCBE Mock Test - 14 - Question 1
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Hollow trunks in trees lead to

VITEEE PCBE Mock Test - 14 - Question 2
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Honey mainly consists of:

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VITEEE PCBE Mock Test - 14 - Question 3
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A outer covering membrane is absent over

VITEEE PCBE Mock Test - 14 - Question 4
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Which of the following mineral is found in haemoglobin ?
VITEEE PCBE Mock Test - 14 - Question 5
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Endoplasmic reticulum remains attached to
VITEEE PCBE Mock Test - 14 - Question 6
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What is true ?
VITEEE PCBE Mock Test - 14 - Question 7
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Saffron is obtained from
VITEEE PCBE Mock Test - 14 - Question 8
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Outer and inner membranes of mitochondria are
VITEEE PCBE Mock Test - 14 - Question 9
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Which of the following pairs is a sedimentary type of biogeochemical cycle?
VITEEE PCBE Mock Test - 14 - Question 10
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Interacting populations are
VITEEE PCBE Mock Test - 14 - Question 11
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A set of homologous organs is
VITEEE PCBE Mock Test - 14 - Question 12
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Animals of cold countries have shorter ears. This is
VITEEE PCBE Mock Test - 14 - Question 13
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Synthesis of DNA over RNA template/teminism occurs in
VITEEE PCBE Mock Test - 14 - Question 14
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Triplet codon refers to sequence of three bases on
VITEEE PCBE Mock Test - 14 - Question 15
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Biologically functional tryosine suppressor tRNA of Escherichia coli synthesised by Khorana in 1979 had
VITEEE PCBE Mock Test - 14 - Question 16
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Mirabilis jalapa is an example of
VITEEE PCBE Mock Test - 14 - Question 17
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How many types of gametes are found in F₁ progeny of cross between AABBCC and aabbcc
VITEEE PCBE Mock Test - 14 - Question 18
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ABO incompatibility differs from Rh incompatibility in that it
VITEEE PCBE Mock Test - 14 - Question 19
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Recessive trait disorders are characterised by
VITEEE PCBE Mock Test - 14 - Question 20
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A kind of allergy is
VITEEE PCBE Mock Test - 14 - Question 21
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Directions: Study the following graph carefully and answer the question given below.

Q. What is the respective ratio of total number of units manufactured by companies Q and R together in the year 2001 to those manufactured by companies S and T together in the year 2003?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 21

The total number of units manufactured by companies Q and R together in the year 2001 = 6.5 + 6.5 = 13 thousands
The total number of units manufactured by companies S and T together in the year 2003 = 16.4 + 12.5 = 28.9 thousands
Required ratio = 13 : 28.9
= 130 : 289

VITEEE PCBE Mock Test - 14 - Question 22
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Directions: Study the following graph carefully and answer the question given below.

Q. What is the average number of hard disks produced by Company 'X' over all the years together?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 22

Total number of hard disks produced by Company 'X' over all the years together = 6 + 10 + 18 + 15 + 30 + 28 + 38 + 35 = 180 (in thousand) = 1,80,000

Required average = 1, 80, 000/8 = 22,500

VITEEE PCBE Mock Test - 14 - Question 23
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Directions: Study the following graph carefully and answer the question given below.

Q. What is the difference between the total number of hard disks and mouses produced by Company 'X' in the year 2010 and the total number of keyboards and pen drives produced by Company 'X' in the year 2015?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 23

Total number of hard disks and mouses produced by Company 'X' in the year 2010 = 6 + 30 = 36 thousand = 36,000
Number of keyboards and pen drives produced by Company 'X' in the year 2015 = 55 + 36 = 91 thousand = 91,000
Required difference = 91,000 - 36,000 = 55,000

VITEEE PCBE Mock Test - 14 - Question 24
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Directions: Study the following graph carefully and answer the question given below.

Q. What was the respective ratio between the number of pen drives produced by Company 'X' in the years 2010, 2011, 2014 and 2015 and the number of hard disks produced by Company 'X' in the years 2011, 2012, 2015 and 2017?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 24

The number of pen drives produced by Company 'X' in years 2010, 2011, 2014 and 2015 = 10 + 15 + 22 + 36 = 83 (in thousands)
The number of hard disks produced by Company 'X' in years 2011, 2012, 2015 and 2017 = 10 + 18 + 28 + 35 = 91 (in thousands)
Required ratio = 83 : 91

VITEEE PCBE Mock Test - 14 - Question 25
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Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 25

Let us assume that the number with which Anita has to perform the multiplication is 'x'.

Instead of finding 35x, she calculated 53x.

The difference = 53x - 35x = 18x = 540

Therefore, x = 540/18 = 30

So, the new product = 30 x 53 = 1590.

VITEEE PCBE Mock Test - 14 - Question 26
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A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 26

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.

VITEEE PCBE Mock Test - 14 - Question 27
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Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 27

For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.

VITEEE PCBE Mock Test - 14 - Question 28
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In a stream, Q lies in between P and R such that it is equidistant from both P and R. A boat can go from P to Q and back in 6 hours 30 minutes while it goes from P to R in 9 hours. How long would it take to go from R to P?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 28

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4

So Option A is correct

VITEEE PCBE Mock Test - 14 - Question 29
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If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 29

A and B complete a work in = 15 days
⇒ One day's work of (A + B) = 1/ 15

B complete the work in = 20 days
⇒ One day's work of B = 1/20

⇒ A's one day's work = 1/15 − 1/20 = (4−3)/6 = 1/60

Thus, A can complete the work in = 60 days.

So Option A is correct

VITEEE PCBE Mock Test - 14 - Question 30
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A and B together can do a piece of work in 50 days. If A is 40% less efficient than B, in how many days can A working alone complete 60% of the work?
Detailed Solution for VITEEE PCBE Mock Test - 14 - Question 30

Given:

A and B together can do a piece of work in 50 days.

A is 40% less efficient than B

Concept used:

Total work = Efficiency of the workers × time taken by them

Calculation:

Let the efficiency of B be 5a

So, efficiency of A = 5a × 60%

⇒ 3a

So, total efficiency of them = 8a

Total work = 8a × 50

⇒ 400a

Now,

60% of the work = 400a × 60%

⇒ 240a

Now,

Required time = 240a/3a

⇒ 80 days

∴ A can complete 60% of the work working alone in 80 days.

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