GATE Chemistry Exam  >  GATE Chemistry Tests  >  GATE Chemistry Mock Test Series  >  General Aptitude - 3 - GATE Chemistry MCQ

General Aptitude - 3 - GATE Chemistry MCQ


Test Description

20 Questions MCQ Test GATE Chemistry Mock Test Series - General Aptitude - 3

General Aptitude - 3 for GATE Chemistry 2024 is part of GATE Chemistry Mock Test Series preparation. The General Aptitude - 3 questions and answers have been prepared according to the GATE Chemistry exam syllabus.The General Aptitude - 3 MCQs are made for GATE Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for General Aptitude - 3 below.
Solutions of General Aptitude - 3 questions in English are available as part of our GATE Chemistry Mock Test Series for GATE Chemistry & General Aptitude - 3 solutions in Hindi for GATE Chemistry Mock Test Series course. Download more important topics, notes, lectures and mock test series for GATE Chemistry Exam by signing up for free. Attempt General Aptitude - 3 | 20 questions in 60 minutes | Mock test for GATE Chemistry preparation | Free important questions MCQ to study GATE Chemistry Mock Test Series for GATE Chemistry Exam | Download free PDF with solutions
General Aptitude - 3 - Question 1

The number of Hens, Ducks, and Goats in farm P are 65, 91 and 169, respectively. The total number of Hens, Ducks and Goats in a nearby farm Q is 416. The ratio of hens : ducks : goats in farm Q is 5 : 14 : 13. All the hens, ducks and goats are sent from farm Q to farm P.

The new ratio of hens : ducks : goats in farm P is ____________.

Detailed Solution for General Aptitude - 3 - Question 1

In Farm P:

Hens = 65, Ducks = 91 and Goats = 169

In Farm Q:

Total number of Hens, Ducks, and Goats = 416

The ratio of Hens : Ducks : Goats = 5 : 14 : 13

Total number of Hens = 5/32 × 416 = 65
Total number of Ducks = 14/32 × 416 = 182
Total numbers of Goats = 13/32 × 416=169

From Farm P, all the Hens, Ducks and Goats are sent to Farm Q. 

Total Hens in farm Q = 65 + 65 = 130

Total Ducks in farm Q = 91 + 182 = 273

Total Goats in farm Q = 169 + 169 = 338

New Ratio = 130 : 273 : 338 = 10 : 21 : 26

General Aptitude - 3 - Question 2

Find the missing sequence in the letter series.

B, FH, LNP, __________.

Detailed Solution for General Aptitude - 3 - Question 2

The series formed is as follows,

1 Crore+ students have signed up on EduRev. Have you? Download the App
General Aptitude - 3 - Question 3

A jigsaw puzzle has 2 pieces. One of the pieces is shown above. Which one of the given options for the missing piece when assembled will form a rectangle? The piece can be moved, rotated or flipped to assemble with the below piece.

Detailed Solution for General Aptitude - 3 - Question 3

Draw a rectangle and carved out the remaining position.

After the carved-out portion is out, make a mirror image of it.

General Aptitude - 3 - Question 4

If 
then, the value of the expression Δ 2 ⊕ 3 Δ ((4 ⊗ 2) ∇ 4) =

Detailed Solution for General Aptitude - 3 - Question 4

Concept:

Replace the symbol as per their standard sign and solve using BODMAS.

Calculation:

Given:

Δ 2 ⊕ 3 Δ ((4 ⊗ 2) ∇ 4) = + 2 - 3 + ((4 ÷ 2) × 4)

∴ + 2 - 3 + (2 × 4)

∴ + 2 - 3 + 8

∴ 10 - 3 = 7

General Aptitude - 3 - Question 5

The distribution of employees at the rank of executives, across different companies C1, C2, ..., C6 is presented in the chart given above. The ratio of executives with a management degree to those without a management degree in each of these companies is provided in the table above. The total number of executives across all companies is 10,000.

Q. The total number of management degree holders among the executives in companies C2 and C5 together is _____.

Detailed Solution for General Aptitude - 3 - Question 5

Concept:

From pie-chart, we can get the total number of executives in each company.

From the ratio table, we can get the numbers of executives with and without a management degree.

Calculation:

Given:

Total executives = 10,000

Percentage of executives in C2 = 5%, C5 = 20%

Ratio of executives with and without management degree C2 = 1 : 4, and C5 = 9 : 1.

Executives in C2 = 5/100 × 10000 = 500

Executives in C5 = 20/100 × 10000 = 2000

Executives with Management degree holders in C2 = 1/5 × 500 = 100

Executives with Management degree holders in C5 = 9/10 × 2000=1800

∴ total executives with management degree holders in C2 and C5 = 100 + 1800 = 1900.

General Aptitude - 3 - Question 6

A digital watch X beeps every 30 seconds while watch Y beeps every 32 seconds.They beeped together at 10 AM.The immediate next time that they will beep together is _____.

Detailed Solution for General Aptitude - 3 - Question 6

Concept:

The Watched will beep together at the time of their Least Common Multiple (LCM).

Calculation:

Given:

Watch X beeps at = 30 sec

Watch Y beeps at = 32 sec

They beep together at = 10.00 AM

LCM of (30 & 32) = 480 sec

And X and Y will beep together after 480 sec or 8 minutes.

So Next beep time = 10.08 AM

General Aptitude - 3 - Question 7

The front door of Mr X's house faces East. Mr X leaves the house, walking 50 m straight from the back door that is situated directly opposite to the front door. He then turns to his right, walks for another 50 m and stops. The direction of the point Mr X is now located at with respect to the starting point is ______.

Detailed Solution for General Aptitude - 3 - Question 7

Tracing path followed by Mr. X,

Hence Mr. X will be in North-West direction from the starting point.

General Aptitude - 3 - Question 8

Given below are two statements 1 and 2, and two conclusions I and II.

Statement 1: All entrepreneurs are wealthy.

Statement 2: All wealthy are risk seekers.

Conclusion I: All risk seekers are wealthy.

Conclusion II: Only some entrepreneurs are risk seekers.

Based on the above statements and conclusions, which one of the following options is CORRECT? 

Detailed Solution for General Aptitude - 3 - Question 8

The Venn diagram according to the given statements is as follows:

Conclusions:

All risk seekers are wealthy. → False (All wealthy are risk seekers but can't say vice versa is also true)

Only some entrepreneurs are risk seekers → False (All entrepreneurs are wealthy and All wealthy are risk seekers so all entrepreneurs will be risk seekers)

Hence neither conclusion I nor II is correct.

General Aptitude - 3 - Question 9

The ratio of the area of the inscribed circle to the area of the circumscribed circle of an equilateral triangle is _____

Detailed Solution for General Aptitude - 3 - Question 9

Let the radius of the circumscribed circle is R, and the radius of the inscribed circle is r.

Then area of circumscribed circle = πR2

Then the area of the inscribed circle = πr2

We know the angle between two adjacent sides in an equilateral triangle is 60°.

ΔOAB, Sin30 = r / R

R = 2r

The ratio of the area of the inscribed circle to the area of the circumscribed circle 

General Aptitude - 3 - Question 10

A box contains 15 blue balls and 45 black balls. If 2 balls are selected randomly, without replacement, the probability of an outcome in which the first selected is a blue ball and the second selected is a black ball, is ______.

Detailed Solution for General Aptitude - 3 - Question 10

Concept:

P(E) = Number of favorable outcomes / Total number of outcomes

Calculation:

Given:

Number of blue balls = 15, Number of black balls = 45

Total number of balls = 15 + 45 = 60 balls

Balls are drawn from the box at random without replacement.

The probability of drawing first blue ball and second black ball is;

General Aptitude - 3 - Question 11

Select the graph that schematically represents BOTH y = xm and y = x1/m properly in the interval 0 ≤ x ≤ 1, for integer values of m, where m > 1.

Detailed Solution for General Aptitude - 3 - Question 11

Given y = xm where m > 1 & y = x1/m and 0 ≤ x ≤ 1

Let us consider m = 2

In that case:

y = x2 graph can be plotted as

Also y = x0.5 is represented in graph as:

∴ y = xm & y = x1/m is correctly represented by option 1) i.e.

y = xm ⇒ increasing slope

y = x1/m ⇒ decreasing slope

General Aptitude - 3 - Question 12

M and N had four children P, Q, R and S. Of them, only P and R were married. They had children X and Y respectively. If Y is a legitimate child of W, which one of the following statements is necessarily FALSE?

Detailed Solution for General Aptitude - 3 - Question 12

Using the above symbols we get the following family tree:

From the tree diagram, we can see that R and W are the parents of Y as Y is a legitimate child of W.

Therefore it is possible that R is the father of Y and W is the wife of R. Hence the statements in options 2 and 3 are not necessarily false.

Also, from the tree diagram, we can conclude that M and N are the grandparents of Y and hence the statement in option 1 is not necessarily false.

As Y is a legitimate child of W, W is the spouse of R and not P.

Hence the statement 'W is the wife of P' is necessarily false.

General Aptitude - 3 - Question 13

Given that a and b are integers and a + a2 b3 is odd, which one of the following statements is correct?

Detailed Solution for General Aptitude - 3 - Question 13

Given (a + a2b3) is odd

⇒ a (1 + ab3) is odd

As the multiplication of two numbers is odd only when those two numbers are odd.

That means ‘a’ is odd and (1 + ab3) is also odd

⇒ ab3 is even

As ‘a’ is an odd number, b3 should be even number to get ab3 as even.

bis even only when b is even.

Therefore, ‘a’ is odd and ‘b’ is even.

General Aptitude - 3 - Question 14

A person divided an amount of Rs. 100,000 into two parts and invested in two different schemes. In one he got 10% profit and in the other he got 12%. If the profit percentages are interchanged with these investments he would have got Rs.120 less. Find the ratio between his investments in the two schemes.

Detailed Solution for General Aptitude - 3 - Question 14

Let Rs. 100,000 is divided into two parts x and y respectively.

Part I: Scheme – 1 (10% profit)

After profit in Scheme – 1, he will get 1.10x rupees

Scheme – 2 (12% profit)

After profit in Scheme – 2, he will get 1.12y rupees

Total amount that he will get from Scheme 1 and Scheme 2 after applying profits of 10% and 12% respectively = 1.10x + 1.12y     …1)

Part II: When profit percentages are interchanged

Scheme – 1: (12% profit), Scheme – 2: (10% profit)

Total amount after applying these profit percentages = 1.12x + 1.10y     …2)

Given that,

1.10x + 1.12y – 1.12x – 1.10y = 120

0.02y – 0.02x = 120

2y – 2x = 12000

y – x = 6000     …3)

y + x = 100000     …4)

Solving 3) and 4)

2y = 106000

Y = 53000, x = 47000

Ratio between the investments = x/y = 47/53

General Aptitude - 3 - Question 15

Consider the following three statements:

(i) Some roses are red.

(ii) All red flowers fade quickly.

(iii) Some roses fade quickly.

Q. Which of the following statements can be logically inferred from the above statements?

Detailed Solution for General Aptitude - 3 - Question 15

Consider the following Venn diagram:

If all red flowers fade quickly and some roses are red, then these roses being red flowers will also fade quickly.

Hence, it can be inferred that some rose fade quickly if the statements (i) and (ii) are true.

Whereas if any of the statements (i) and (ii) are false, nothing can be inferred about statement (iii).

General Aptitude - 3 - Question 16

A rectangle becomes a square when its length and breadth are reduced by 10 m and 5 m, respectively. During this process, the rectangle loses 650 m2 of area. What is the area of the original rectangle in square meters?

Detailed Solution for General Aptitude - 3 - Question 16

Let the length and breadth of the rectangle is L and B respectively.

The area of the rectangle = LB

It becomes square when its length and breadth are reduced by 10 m and 5 m, respectively.

Now, the length and breadth of the square become (L – 10) and (B – 5) respectively.

As in a square the length and breadth are the same, L – 10 = B – 5

⇒ L = B + 5

The area of the square = (L – 10) (B – 5)

Given that, the rectangle loses 650 m2 of area

⇒ LB – 650 = (L – 10) (B – 5)

⇒ LB – 650 = LB + 50 – 10B – 5L

⇒ 10B + 5L = 700

⇒ 10B + 5(B + 5) = 700

⇒ B = 45 m

⇒ L = B + 5 = 50 m

Area of the original rectangle = LB = 50 × 45 = 2250 m2

General Aptitude - 3 - Question 17

The value of the expression

Detailed Solution for General Aptitude - 3 - Question 17

Concept:

Logarithmic Properties:

Calculation:

Using 

= 1

General Aptitude - 3 - Question 18

It was estimated that 52 men can complete a strip in a newly constructed highway connecting cities P and Q in 10 days. Due to an emergency, 12 men were sent to another project. How many numbers of days, more than the original estimate, will be required to complete the strip?

Detailed Solution for General Aptitude - 3 - Question 18

Concept:

M× D= M× D2

Where,

M1 = Number of men in initial condition, M2 = Number of men in final condition

D1 = Number of days required in the initial condition,

D2 = Number of days required in the final condition

Calculation:

Given:

M= 52,

D= 10, M= 52 - 12 = 40,

D= ?

52 × 10 = 40 × D­2

 D= 13 days

Now,

Numbers of days more than the original estimate is equal to:

Number of days required in the final condition - Number of days required in the initial condition

∴ Numbers of days more than the original estimate = D2  D1

∴ Numbers of days more than the original estimate = 13 – 10

∴ Numbers of days more than the original estimate = 3 days

General Aptitude - 3 - Question 19

The bar graph shows the data of the students who appeared and passed in an examination for four school P, Q, R and S. The average of success rates (in percentage) of these four schools is _____.

Detailed Solution for General Aptitude - 3 - Question 19

Concept:

Average success rate of one school = Total no of passed students/Total no. of students who appeared

The average success rates of these four schools = Sum of Average success rate of all four schools/Total no. of schools

Therefore, The average success rates of these four schools will be 

Calculation:

The average success rates of these four schools will be 

= 59%

∴ The average success rates of these four schools = 59.0%

General Aptitude - 3 - Question 20

What is the least multiple of 7, which when divided by 2, 3, 4, 5 and 6, leaves the remainders 1, 2, 3, 4, 5 respectively?

Detailed Solution for General Aptitude - 3 - Question 20

L.C.M. Of 2, 3, 4, 5,6 = 60
Other numbers divisible by 2, 3, 4, 5, 6 are 60k, where k is a positive integer.
Since, 2 - 1 = 1, 3 - 2 = 1, 4-3=1, 5 - 4 = 1 and 6-5=1, the remainder in each case is less than the divisor by 1, the required number = 60k - 1 = (7 x 8k) + (4k - 1 ) .
Now this number is to be divisible by 7. Whatever may be the value of k the portion 7 x 8k is always divisible by 7. Hence, we must choose the least value of k which will make (4k - 1 ) divisible by 7. Putting k equal to 2 value is divisible by 7.
Therefore, the required number = 60k - 1 = 6 0 x 2 - 1 = 119.

18 docs|37 tests
Information about General Aptitude - 3 Page
In this test you can find the Exam questions for General Aptitude - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for General Aptitude - 3, EduRev gives you an ample number of Online tests for practice

Top Courses for GATE Chemistry

Download as PDF

Top Courses for GATE Chemistry