GATE Numerical Ability Questions (Quant & Analytical) -2 - Mechanical Engineering MCQ

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

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

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.

Concept:

Price paid by customer at end = initial cost × (increased percentage)ⁿ

Where,

n = no. of levels

Calculation:

Given:

Cost of product at first level P = 120 Rupees.

Also,

It is increased by 25% in each level.

Hence,

Initial cost = 120 Rupees.,

Increased percentage = 25%,

n = 5

Now,

Price paid by customer at end = 120 × 1.25⁵ Rupees

∴ Price paid by customer at end = 366.21 Rupees.

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.

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%

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