Mathematics Test 271

We are given the compound inequality:

(5x - 1) < (x + 1)² < (7x - 3)

Let's break it into two separate inequalities and solve each one.

First inequality:

(5x - 1) < (x + 1)²

Expand the square on the right-hand side: (5x - 1) < (x² + 2x + 1)

Simplify: 5x - 1 < x² + 2x + 1

Move all terms to one side: 0 < x² - 3x + 2

Factor the quadratic expression: 0 < (x - 1)(x - 2)

This inequality holds when: x < 1 or x > 2

Second inequality:

(x + 1)² < (7x - 3)

Expand the square on the left-hand side: (x² + 2x + 1) < 7x - 3

Move all terms to one side: x² - 5x + 4 < 0

Factor the quadratic expression: (x - 1)(x - 4) < 0

This inequality holds when: 1 < x < 4

Combining both inequalities:

From the first inequality, we have: x < 1 or x > 2

From the second inequality, we have: 1 < x < 4

Now, combining the two inequalities:

x < 1 or x > 2 (from the first inequality),

1 < x < 4 (from the second inequality).

The overlap of these two conditions is: 2 < x < 4

Thus, the possible integer values for x in this range are: x = 3

Conclusion:

There is 1 integer value of x that satisfies both inequalities.

Correct Answer: A: 1.

Natural number (N): Natural numbers include all the whole numbers excluding the number 0.

Given:

Initial ratio = 5 ∶ 6

Final ratio should be less than 17 ∶ 22

Calculation:

Let the least whole number that is needed to be subtracted be a.

According to the question,

(5 - a)/(6 - a) < 17/22

⇒ 5 × 22 - 22a < 17 × 6 - 17a 

⇒ 110 - 22a < 102 - 17a 

⇒ 110 - 102 < - 17a + 22a 

⇒ 8 < 5a 

⇒ 8/5 = 1.6 < a 

∴ The least whole number must be 2.

Given:

Equation₁ = ax + 9y = 1

Equation₂ = 9y - x - 1 = 0 

Concept used:

If linear equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Here, the equations have an infinite number of solutions, if

a₁/a₂ = b₁/b₂ = c₁/c₂

Calculation:

We have equations,

ax + 9y = 1 

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a₁ = a, b₁ = 9, c₁ = -1

and, a₂ = 1, b₂ = -9, c₂ = 1

As we know that 

a1/a2 = b1/b2 = c1/c₂

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.

⇒ 6x + 2(6 - x) > 2x - 2

⇒ 6x + 12 - 2x > 2x - 2

⇒ 2x > - 14

⇒ x > - 7     ----(1)

⇒ 2x - 2 < 5x/2 - 3x/4

⇒ 2x - 2 < 7x/4

⇒ 8x - 7x < 8

⇒ x < 8     ----(2)

From (1) and (2),

- 7 < x < 8

∴ x = 5 satisfies the given conditions from the above options.

Given:

If 2nC3 : nC2 = 12 : 1

Formula used :

ⁿC_r = n!/r!(n - r)!

Calculation:

If 2nC3 : nC2 = 12 : 1      ----(1)

Using the formula

²ⁿC₂ = {2n!/3! (2n - 3)!}

⇒ 2n(2n - 1)(2n - 2)(2n - 3)!/(2n - 3)! × 3 × 2

⇒ n(2n - 1)2(n - 1)/3      ----(2)

ⁿC₂ = {n!/2! (n - 2)!

⇒ n(n - 1)(n - 2)!/(n - 2)! × 2

⇒ n(n - 1)/2      ----(3)

Putting equation 2 and 3 in equation 1

⇒ {n(2n - 1)2(n - 1)/3}/{n(n - 1)/2} = 12/1

⇒ 2n - 1 = 9

⇒ 2n = 10 

⇒ n = 5

∴ The value of n is 5.

⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are:

33, 35, 37, 53, 55, 57, 73, 75, 77 

Given: 

(7 men + 6 women) 5 persons are to be chosen for a committee.

Formula used: ⁿC_r = n!/(n - r)! r!

Calculation:

Ways in which at least 3 men are selected;

⇒ 3 men + 2 women

⇒ 4 men + 1 woman 

⇒ 5 men + 0 woman 

Number of ways = ⁷C₃ × ⁶C₂ + ⁷C₄ × ⁶C₁ + ⁷C₅ × ⁶C₀

⇒ 7!/(3! × 4!) × 6!/(2! × 4!) + 7!/(4! × 3!) × 6!/(1! × 5!) + 7!/(5! × 2!) × 6!/(6!× 0!)

⇒ 35 × 15 + 35 × 6 + 21 

⇒ 735 + 21 = 756

∴ The required no of ways = 756.

Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels(EOA), all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now, 

The required number of ways = 2520 × 6 

⇒ 15120

∴ The required number of ways is 15120.

Given:

The number of ways of arrangements of 10 persons in four chairs  

Formula used:

ⁿP_r = n!/(n – r)!

Where, n = Number of persons

r = Number of chairs

Calculation:

According to the question

nPr = n!/(n – r)!

⇒ 10!/(10 – 4)!

⇒ 10!/6!

⇒ (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(6 × 5 × 4 × 3 × 2 × 1)

⇒ (10 × 9 × 8 × 7)

⇒ 5040

∴ The required value is 5040