General Aptitude - 3 - GATE Chemistry MCQ

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

The series formed is as follows,

Draw a rectangle and carved out the remaining position.

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

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

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.

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

Tracing path followed by Mr. X,

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

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.

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 

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;

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

Let us consider m = 2

In that case:

y = x² graph can be plotted as

Also y = x^0.5 is represented in graph as:

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

y = xm ⇒ increasing slope

y = x1/m ⇒ decreasing slope

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.

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.

b³ is even only when b is even.

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

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

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).

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

Concept:

Logarithmic Properties:

Calculation:

Using 

= 1

Concept:

M₁ × D₁ = M₂ × 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:

M₁ = 52,

D₁ = 10, M₂ = 52 - 12 = 40,

D₂ = ?

52 × 10 = 40 × D­₂

∴ D₂ = 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 = D_2 – D₁

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

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

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%

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.