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JEE Exam  >  JEE Tests  >  VITEEE PCBE Mock Test - 12 - JEE MCQ

VITEEE PCBE Mock Test - 12 - JEE MCQ


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30 Questions MCQ Test - VITEEE PCBE Mock Test - 12

VITEEE PCBE Mock Test - 12 for JEE 2024 is part of JEE preparation. The VITEEE PCBE Mock Test - 12 questions and answers have been prepared according to the JEE exam syllabus.The VITEEE PCBE Mock Test - 12 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for VITEEE PCBE Mock Test - 12 below.
Solutions of VITEEE PCBE Mock Test - 12 questions in English are available as part of our course for JEE & VITEEE PCBE Mock Test - 12 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 - 12 | 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 - 12 - Question 1
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Largest number of chloroplasts in the leaf is in

VITEEE PCBE Mock Test - 12 - Question 2
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Tassar silk is extracted from

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VITEEE PCBE Mock Test - 12 - Question 3
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Glycolysis can not occur in the absence of

VITEEE PCBE Mock Test - 12 - Question 4
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Based on its function, the heart is often referred to as a “double pump”. Which of the following would explain this?
VITEEE PCBE Mock Test - 12 - Question 5
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One of the following is native to West Asia
VITEEE PCBE Mock Test - 12 - Question 6
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Autolysis is connected with which one of the following organelle?
VITEEE PCBE Mock Test - 12 - Question 7
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Chloroplasts are self replicating units as they possess
VITEEE PCBE Mock Test - 12 - Question 8
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Carboniferous period of coal deposition was
VITEEE PCBE Mock Test - 12 - Question 9
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Experiments on Acetabularia by Hammerling proved the role of
VITEEE PCBE Mock Test - 12 - Question 10
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In comparative study of grassland and pond ecosystems
VITEEE PCBE Mock Test - 12 - Question 11
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Evolution of a species or group can be studied through
VITEEE PCBE Mock Test - 12 - Question 12
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The earliest fossil form in the phylogeny of horse is
VITEEE PCBE Mock Test - 12 - Question 13
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Khorana et al synthesised RNA with copolymer of UGUGUGUGUG... It produced a peptide with alternate cysteine and valine. The codons for the two are
VITEEE PCBE Mock Test - 12 - Question 14
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Coded information in nucleic acid depends upon
VITEEE PCBE Mock Test - 12 - Question 15
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Meselson and Stahl proved by their experiment about DNA, that it has
VITEEE PCBE Mock Test - 12 - Question 16
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Main reason for success of Mendel was
VITEEE PCBE Mock Test - 12 - Question 17
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In a certain plant, red colour flower (R) is dominant over white colour flower (r). when a heterozygous Rr plant is selfed, 64 offsprings are obtained. The number of white offsprings are
VITEEE PCBE Mock Test - 12 - Question 18
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A human male has one one barr body in his somatic cell nuclei. The chromosome constitution must be
VITEEE PCBE Mock Test - 12 - Question 19
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Which would result in haemolysis in foetus ?
VITEEE PCBE Mock Test - 12 - Question 20
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Cornea transplantation has become common due to
VITEEE PCBE Mock Test - 12 - Question 21
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Directions: Study the table carefully and answer the given question.

The following table shows the number of students in different engineering departments at seven IITs. Some of the data is missing from the table.

The above table is given for year 2014. In 2015, the intake of Mechanical Engineering at all the IITs together was increased by 10%, intake of Mineral Engineering was increased by 10% and intake of Chemical Engineering was increased by 8%. The intake of other departments remained the same as that of previous year.

Q. In 2014, all the students (in case total number of students was 55 or less) or only top 55 students from every department at each IIT received a scholarship of Rs. 2000. Out of the remaining, all the students (in case number of remaining students was 45 or less) or only the next 45 students from every department at each IIT received a scholarship of Rs. 1200. What is the total amount of scholarship awarded to students studying in Mining Engineering department at all the IITs together in 2014, if the total number of students in Mining Engineering at all the IITs together in 2015 was 600?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 21

Total number of Mining Engineering students in 2014 at all the IITs = 600
Total number of Mining Engineering students at IIT Madras and IIT Kanpur = 600 – (95 + 155 + 65 + 90 + 140) = 55
So, all the Mining Engineering students at IIT Madras and IIT Kanpur received a scholarship of Rs. 2000 each. (As the number of students at each of the two IITs is less than 55)
Then, total scholarship amount awarded to Mining Engineering students = 55  6  2000 + (40 + 45 + 10 + 35 + 45)1200 = 6,60,000 + 2,10,000 = Rs. 8,70,000

VITEEE PCBE Mock Test - 12 - Question 22
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Directions: Study the table carefully and answer the given question.

The following table shows the number of students in different engineering departments at seven IITs. Some of the data is missing from the table.

The above table is given for year 2014. In 2015, the intake of Mechanical Engineering at all the IITs together was increased by 10%, intake of Mineral Engineering was increased by 10% and intake of Chemical Engineering was increased by 8%. The intake of other departments remained the same as that of previous year.

Q. What was the ratio of the total number of students in Mechanical Engineering at IIT Delhi and IIT Roorkee in 2014, if the total number of students in Mechanical Engineering at all the IITs in 2014 was 50% more than the total number of students at IIT Delhi in 2015 and IIT Delhi had 29 students more in Mechanical Engineering as compared to Mechanical students at IIT Roorkee in 2014?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 22

Suppose the total numbers of Mechanical Engineering students at IIT Delhi and IIT Roorkee in 2014 are y and z, respectively.
So, y - z = 29 ...(1)
Then, total number of students at IIT Delhi in 2014 = y + 174 + 225 + 95 + 80 = y + 574
And total number of students at IIT Delhi in 2015 = 1.1y + 174 + 225  1.08 + 95 + 80  1.1
= 1.1y + 174 + 243 + 95 + 88 = 1.1y + 600 ...(2)
Total number of Mechanical Engineering students at all the seven IITs in 2014 = y + z + 100 + 150
+ 220 + 190 + 220 = y + z  + 880 ...(3)
Then, from equations (2) and (3),
1.5(1.1y + 600) = y + z + 880
⇒ 1.65y + 900 = y + z + 880 
⇒ 0.65y - z = -20
⇒ 0.65y - y + 29 = -20 (from equation 1)
⇒ - 0.35y = -49
⇒ y = 140 and z = 111
Hence, ratio of the total number of students in Mechanical Engineering at IIT Delhi and IIT Roorkee in 2014 = 140 : 111

VITEEE PCBE Mock Test - 12 - Question 23
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Directions: The following table shows some incomplete information about the runs scored by four batsmen in four innings.

Additional information:
1. Average runs scored in the 1st innings for all the players were 67.5.
2. Paramjeet's average runs for all 4 innings were 1.5 times the runs scored by Jeetu in 1st innings.
3. Sachin's average runs were 72.5 in all four innings.
4. Total runs scored by Dhooni in all the four innings were 300.
5. The runs scored in 2nd innings were 10 less than the runs scored in 1st innings for all the players.
6. Paramjeet and Dhooni scored the same runs in 4th innings.
7. Sum of runs scored by all the players in the 4th innings is 270.

Q. Who scored the least in the 2nd innings?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 23

 (i) Average runs scored in the 1st innings = 67.5
Total score = 67.5 × 4 = 270.0
(ii) Average runs of Sachin = 72.5
Total runs of Sachin = 72.5 × 4 = 290
So, the runs made by Sachin in 3rd innings = 290 - 80 - 70 - 80 = 60
(iii) Total score in the 4th innings = 270 (from 7)
Now, let Paramjeet`s, Dhooni`s and Jeetu`s scores in 4th innings be x, x and y.
=> 2x + y + 80 = 270
=> 2x + y = 190 ..…(i)
Let Jeetu`s and Dhooni`s scores in 1st innings be p and q, respectively.
=> p + q + 80 + 60 = 270
=> p + q = 130 ..…(ii)
From 2:
i.e. 6p - x = 170 ..….(iii)
From 5: Jeetu`s and Dhooni`s scores in 2nd innings are p - 10 and q - 10, respectively.
=> Jeetu`s total runs = 220
=> p + p - 10 + 60 + y = 220
2p + y = 170 ..….(iv)
Solving (i) and (iv), we get 2x - 2p = 20 ……(v)
Solving (iii) and (v), we get x = 46 and p = 36
From (i) and (ii),
y = 98 and q = 94
Thus,

In 2nd innings, Jeetu Malik scored the lowest.

VITEEE PCBE Mock Test - 12 - Question 24
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Directions: The following table shows some incomplete information about the runs scored by four batsmen in four innings.

Additional information:
1. Average runs scored in the 1st innings for all the players were 67.5.
2. Paramjeet's average runs for all 4 innings were 1.5 times the runs scored by Jeetu in 1st innings.
3. Sachin's average runs were 72.5 in all four innings.
4. Total runs scored by Dhooni in all the four innings were 300.
5. The runs scored in 2nd innings were 10 less than the runs scored in 1st innings for all the players.
6. Paramjeet and Dhooni scored the same runs in 4th innings.
7. Sum of runs scored by all the players in the 4th innings is 270.

Q. Who scored the lowest runs in the 1st innings?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 24

(i) Average runs scored in the 1st innings = 67.5
Total score = 67.5 × 4 = 270.0
(ii) Average runs of Sachin = 72.5
Total runs of Sachin = 72.5 × 4 = 290
So, the runs made by Sachin in 3rd innings = 290 - 80 - 70 - 80 = 60
(iii) Total score in the 4th innings = 270 (from 7)
Now, let Paramjeet`s, Dhooni`s and Jeetu`s scores in 4th innings be x, x and y.
=> 2x + y + 80 = 270
=> 2x + y = 190 ..…(i)
Let Jeetu`s and Dhooni`s scores in 1st innings be p and q, respectively.
=> p + q + 80 + 60 = 270
=> p + q = 130 ..…(ii)
From 2:
i.e. 6p - x = 170 ..….(iii)
From 5: Jeetu`s and Dhooni`s scores in 2nd innings are p - 10 and q - 10, respectively.
=> Jeetu`s total runs = 220
=> p + p - 10 + 60 + y = 220
2p + y = 170 ..….(iv)
Solving (i) and (iv), we get 2x - 2p = 20 ……(v)
Solving (iii) and (v), we get x = 46 and p = 36
From (i) and (ii),
y = 98 and q = 94
Thus,

Lowest runs were scored in 1st innings by Jeetu Malik.

VITEEE PCBE Mock Test - 12 - Question 25
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1 ’s are given 100 times, 2 ’s are given 100 times and 3’s are given 100 times. Now numbers are made by arranging these 300 digits in all possible ways. How many of these numbers will be perfect squares?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 25

However you arrange the digits, you will observe that the sum of digits is always equal to (100)(1+2+3) = 600.

Notice that 600 is divisible by 3 but not by 9.

Hence the original number is divisible by 3 but not by 3^2 which is 9.

ORIGINAL NUMBER = 3(p) where p is an integer does not have the prime factor 3. But 3 is not a perfect square!

Hence there will be no number which has 100 1’s, 100 2’s and 100 3’s which will be a perfect square.

VITEEE PCBE Mock Test - 12 - Question 26
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Find the remainder when 73 * 75 * 78 * 57 * 197 * 37 is divided by 34.
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 26

Given:

73 × 75 × 78 × 57 × 197 × 37 is divided by 34

Calculation:

73 × 75 × 78 × 57 × 197 × 3734

We have taken individual remainder like

When 73 is divided by 34 gives remainder is 5

Similarly

VITEEE PCBE Mock Test - 12 - Question 27
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Four bells ring together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 27

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Calculation:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours. So the correct option is B

Note: In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

VITEEE PCBE Mock Test - 12 - Question 28
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Look at this series: 53, 53, 40, 40, 27, 27,...so on. What number should come next?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 28

A number is repeated. Then it is reduced by 13 and the result is repeated.

► The next number is 27-13=14.

VITEEE PCBE Mock Test - 12 - Question 29
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Look at this series: 4, 11, 19, 41, ?, 161...so on. What number should come?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 29

Pattern:

► 4 × 2 + 3 = 11,
► 11 × 2 – 3 = 19,
► 19 × 2 + 3 = 41,
► 41 × 2 – 3 = 79,
► 79 × 2 + 3 = 161.

VITEEE PCBE Mock Test - 12 - Question 30
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Look at this series: 6, 12, 48, 264,1560...so on. What number should come next?
Detailed Solution for VITEEE PCBE Mock Test - 12 - Question 30

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