Quantitative Aptitude PYQs

Explanation: 

Case 1: x < 4

In this case, both |x - 4| and |x - 7| are negative. So,

⇒ -(x - 4) - (x - 7) = 15

⇒ -2x + 11 = 15

⇒  x = -2  which belongs in the range x < 4

Case:2  4 ≤ x ≤  7

In this case, |x - 4| is nonnegative, and |x - 7| is negative.

⇒ (x - 4) - (x - 7) = 15

⇒ 7 = 15, which is not possible.

Case:3 x > 7

In this case, both |x - 4| and |x - 7| are nonnegative.

 (x - 4) + (x - 7) = 15

2x = 26, or x = 13

Therefore, there is only two real number solution.

Concept:

Codomain - A codomain is the group of possible values that the dependent variable can take.

Range - The range is all the elements from set B that have the corresponding pre-image in set A.

Calculation:

Given, \(f(x)=\frac{2 x+3}{3 x+5}, x \in A\)

For f(x) to be defined, 3x + 5 ≠ 0

⇒ x = \(-\frac{5}{3}\)

∴ A = \(R \backslash\left{-\frac{5}{3}\right}\)

Now, y = f(x) is onto

⇒ y = \(\frac{2 x+3}{3 x+5}\)

⇒ 3xy + 5y = 2x + 3

⇒ 3xy - 2x = 3 - 5y

⇒ x(3y - 2) = 3 - 5y

⇒ x = \(\frac{3-5y}{3y-2}\)

For x to be defined for \(R \backslash\left{\frac{2}{3}\right}\)

Since, for onto functions co-domain = range

∴ B = \(R \backslash\left{\frac{2}{3}\right}\)

∴ \(A=R \backslash\left{-\frac{5}{3}\right} \text { and } B=R \backslash\left{\frac{2}{3}\right}\)

Concept:

In the quadratic equation ax² + bx + c = 0 

Sum of root (α +  β) = -b/a

Product of root (αβ) = c/a

Formula used:

(x + y)² = x² + y² + 2xy

(x - y)2 = x2 + y2 - 2xy

Explanation:

We have  x2 + ax + b = 0

By using the above concept

α + β = - a            -----(1)

α.β =  b              -----(2)

To find α, we need to consider the given statements.

Statement 1: 

α + β = 0 or β = - α 

Also, α2 + β2 = 2

⇒ α2 +  (-α)2 = 2

⇒ 2α² = 2

⇒ α = ± 1

Statement 2: αβ² = -1, a = 0

From equation (1)

α + β = - a

α + β = 0               [∵ a = 0]

⇒ β = - α 

Now

⇒ αβ² = -1

⇒ α (-α)² = -1

⇒ α³ = -1

⇒ α = -1  

∴  The correct answer is option (3).

Formula used:

\({{\rm{T}}_{\left( {{\rm{r;}} + {\rm{;}}1} \right)}} = {\rm{;}}{{\rm{;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} × {{\rm{a}}^{{\rm{n}} - {\rm{r}}}} × {{\rm{b}}^{\rm{r}}}\)

ⁿC_r = \(\frac{n!}{r!(n-r)!}\) 

Calculation:

Given expression is \(\left(x^{-\frac{8}{3}}+x^2 \log _{10} x\right)^8\)

Here, r = 5, n = 8

According to the question

\(T_6= ~^ 8C_5 × {(x^{-8/3})^{8-5}} × {(x^2.log_{10}~x)}^5\)

Simplifying, we get:

T₆ = ⁸C₅ x^-8 × x¹⁰ (log₁₀x)⁵

5600 = 8C5 x²(log10x)5       [∵ T₆ = 5600]

Since, nCr = \(\frac{n!}{r!(n-r)!}\) 

5600 = 56 × x2(log10x)5 

x2(log10x)5 = 100

From the above option, let x = 10

100 × (log1010)5 = 100 

(log1010)5 = 1

1 = 1      [log₁₀10 = 1]

Therefore, the value of x that satisfies the equation is x = 10.

Formula used:

Number of term in the (x + y)ⁿ = ^n+2-1C_2-1

Here, 2 is used as there are two terms x & y.

Calculation:

We have (3x - y)4(x + 3y)4

This can be written as 

[(3x - y)(x + 3y)]⁴

[3x² + 9xy - xy - 3y²]⁴

[3x² + 8xy - 3y²]⁴

Here, n = 4 & r = 3

Number of terms ^4+3-1C_3-1 = ⁶C2

⇒ (6 × 5)/(2 × 1) = 15

∴ Number of term = 15

Concept used:

Four terms of AP are: a -3d, a -d, a + d, a + 3d

Calculation:

Let, p = a - 3d, q = a - d, r = a + d, s = a + 3d

Difference between largest and smallest numbers

|(a - 3d) - (a + 3d)| = 6d

Given that  p + s = 8, 

⇒ 2a = 8

⇒ a = 4

Also, qr = 15

⇒ (a - d)(a + d) = 15

⇒ a² - d² = 15

⇒ 16 - d² = 15

⇒ d² = 16 - 15

⇒ d = 1   (absolute value)

⇒ 6d = 6 × 1 = 6

∴ The difference between the largest and smallest numbers is 6.

Concept:

(x + y)n = nC0 xn yo + nC1 xn-1 y1 +  nC2 xn-2 y2 +...... + nCn x0 yn

Where, n ≥ 0 and 

nCr = \(\frac{n!}{r!(n-r)!}\) = Binomial coefficient  

Case: I n is even, greatest binomial coefficient is given by ⁿC_n/2 

It means, n = 2r

Case: II n be odd, the greatest binomial coefficient is given by

ⁿC_(n+1)/2 or ⁿC_(n-1)/2

It means,  n = 2r - 1 & n = 2r + 1 

Explanation: 

For n = 2r:

n will be the even number of every positive integer r.

Hence statement I is correct.

For n =  n = 2r - 1 & 2r + 1:

n will be odd for every positive value of r. 

Hence statement II is correct. 

∴ Correct answer is option (3)

Formula used:

If m parallel lines in a plane are intersected by n parallel lines, what is the number of parallelograms that are formed

Number of parallelograms = mn(m-1)(n-1)/4

Calculation:

By using the above concept

mn(m-1)(n-1)/4 = 60

mn(m-1)(n-1) = 240

m.(m - 1).n.(n - 1) = 240

Here, we have one equation & two variables.

So, to find the value of m & n by by hit & trial method.

m.(m - 1).n.(n - 1) = 5 × 3 × 2 × 2 × 2 × 2

This can be re-written as  

m.(m - 1).n.(n - 1) = 5 × 4 × 4 × 3 

On comparing both sides,

m = 5 & n = 4

∴ m + n = 5 + 4 = 9  

Thus, the value of (m + n) is 9.

Given that

x is the number of permutations of the word "PERMUTATIONS "

y is the number of permutations of the word "COMBINATIONS"

The number of permutations in any word is given by

= factorial of total number of letters in word/ number of repetitive letters!

Calculation:

 

In the word PERMUTATIONS, the letter "T" is repetitive i.e., it repeats 2 times.

So, the number of permutations for the word PERMUTATIONS  would be 

= factorial of the total number of letters in PERMUTATIONS/ 2!

= 12!/(2!)

x = 12!/2 ....(1)

Similarly, in the word COMBINATIONS,

I repeat 2 times,

O repeats 2 times

and N repeats 2 times.

the total number of permutations will be =  12!/(2! 2! 2!)

y = 12!/8 ....(2)

From 1 and 2

x = 4y

The correct answer is an option (3)

Calculation:  

Firstly, we need to determine the total number of 5-digit numbers that can be created with the digits 0, 1, 2, 4, 5 without repetition.

As the first position cannot be 0 (as that would result in a 4-digit number), there are 4 choices (1, 2, 4, or 5) for the first digit.

For each of the remaining positions, we have one fewer choice because we can't repeat digits.

Therefore,

Total possible 5-digit numbers = 4 × 4 × 3 × 2 × 1 = 96

Next, we need to determine how many of these 5-digit numbers are greater than 50,000.

The first digit of such a number must be either 5 (1 choice) because 0, 1, 2, and 4 would result in a number less than 50,000.

After the first digit has been decided, we can choose any of the remaining 4 digits for the second position, then 3 for the third, 2 for the fourth, and 1 for the fifth

Total possible 5-digit numbers are greater than 50,000

 1 × 4 × 3 × 2 × 1 = 24.

∴ Required % = \(\frac{24}{96} \times 100=25 %\).

Concept:

GM of a & b is √(ab)

Formula used:

sin²x + cos² = 1

cos 2x = 1 - 2sin²x

Calculation:

Given that, sin β is the GM of sin α & cos α then

sin β = √( sin α × cos α)

sin²β = (sin α × cos α)

2sin²β = 2sin α × cos α

1 - 2sin2β = 1 - 2sin α × cos α

By using the above formula

cos 2β = sin²α + cos²α - 2sin α × cos α

Since, (a - b)² = a² + b² - 2ab

∴ cos 2β = (cosα - sin α)²

Concept:

AM of two number a & b is (a + b)/2

Formula used:

1. 1 + tan² θ =  sec² θ
2. sin ²θ + cos²θ = 1
3. sec θ = 1/cos θ
4. 2 cos² θ - 1 = cos 2 θ

Calculation:

Since tan γ is the arithmetic mean (AM) of sin α and cos α,

\(tan γ =\frac{sin α + cos α }{2}\)

⇒ tan²γ = (sin α + cos α)²/4

⇒ tan2γ = (sin α + cos α)2/4

tan2γ = (sin²α + cos²α + 2sin α.cos α)2/4

By using the formula (1)

⇒ tan2γ = (1 + sin 2α)/4

Adding 1 on both sides

⇒ 1 + tan2γ = 1 + [(1 + sin 2α)/4]

⇒ 1 + tan2γ = (4 + 1 + sin 2α)/4

⇒ 1 + tan2γ = (5 + sin 2α)/4

But 1 + tan2 θ = sec2 θ

⇒ sec2γ = (5 + sin 2α)/4

But, sec θ = 1/cos θ

​⇒ cos2γ = 4/(5 + sin 2α)

​⇒ 2cos2γ = 2 × 4/(5 + sin 2α)

​⇒ 2cos2γ = 8/(5 + sin 2α)

​⇒ 2cos2γ - 1 = 8/(5 + sin 2α) - 1

​⇒ cos²γ = [8/(5 + sin 2α)] - 1

​⇒ cos²γ = [8 - (5 + sin 2α)] / (5 + sin 2α)

​⇒ cos²γ = [3 - sin 2α)] / (5 + sin 2α)

⇒ 1/cos²γ = (5 + sin 2α)/(3 - sin 2α)

But, sec θ = 1/cos θ

**∴ sec²γ = (5 + sin 2α)/(3 - sin 2α)**​​.

Formula used:

tan θ = Perpendicualr / Base

tan (x + y) = (tan x + tan y)/(1 - tan x.tan y)

Calculation:

Given that 

α = tan-10.5

⇒ tan α = 1/2      ------(1)

In an ΔPAB

tan θ = 10/x       ------(2)

In an ΔPAC

tan (α + θ) = (20 + 10)/x

⇒ tan (α + θ) = 30/x

By using the above formula 

(tan α + tan θ)/(1 - tan α.tan θ) = 30/x

Using equations (1) & (2)

⇒ [1/2 + 30/x]/[1 - (1/x2)(30/x)]

⇒ \(\frac{1/2 + 10/x}{1 - (1/2)(10/x)}=30/x\)

⇒ (x -20)(2x - 10) = 30/x

⇒ x² - 20x = 60x - 300

⇒ x² - 40x + 300 = 0

⇒ x² - 30x - 10x + 300 = 0

⇒ x(x - 30) - 10(x - 30) = 0

⇒ (x - 10)(x - 30) = 0

⇒ x = 10 & 30

Statement:1 x can take two values which are in the ratio 1 : 3

Clearly, the ratio of the value of x = 1/3

Statement:2 x can be equal to the height of the flagstaff.

x = 10 & 30. So x ≠ 20

∴ Only statement 1 is correct.

Formula used:

tan θ = Perpendicualr / Base

tan (x + y) = (tan x + tan y)/(1 - tan x.tan y)

Calculation:

Given that 

α = tan-10.5

⇒ tan α = 1/2      ------(1)

In an ΔPAB

tan θ = 10/x       ------(2)

In an  ΔPAC

tan (α + θ) = (20 + 10)/x

⇒ tan (α + θ) = 30/x

By using the above formula 

(tan α + tan θ)/(1 - tan α.tan θ) = 30/x

Using equations (1) & (2)

⇒ [1/2 + 30/x]/[1 - (1/x2)(30/x)]

⇒ \(\frac{1/2 + 10/x}{1 - (1/2)(10/x)}=30/x\)

⇒ (x -20)(2x - 10) = 30/x

⇒ x2 - 20x = 60x - 300

⇒ x2 - 40x + 300 = 0

⇒ x2 - 30x - 10x + 300 = 0

⇒ x(x - 30) - 10(x - 30) = 0

⇒ (x - 10)(x - 30) = 0

⇒ x = 10 & 30

From equation (2)

tan θ = 10/10 & 10/30

⇒ tan θ =1 & 1/3

∴ Option 3 is correct.

Concept used:

Sine Rule: In the triangle ABC, where sides BC, CA & AB are a, b & c

respectively, we can write 

sin A/a = sin B/b = sin C/c

Where A, B & C are the angles opposite to the sides a, b & c

Formula used:

Arithmetic Mean of three number p, q & r is (P + q + r)/3

The perimeter of the triangle is the sum of three sides.

Calculation:

Let the side of the triangle be BC, CA & AB are a, b & c which are opposite to

its angle A, B & C respectively.

According to the question

a + b + c = 6[(sin A + sin B + sin C)/3]

a + b + c = 2(sin A + sin B + sin C)     ----(1)

But from sine rule

sin A/a = sin B/b = sin C/c = k (let)

sin A = ak, sin B = bk, sin C = ck

From equation (1)

a + b + c = 2(ak + bk + ck) 

a + b + c = 2k(a + b + c)

k = 1/2 

But side BC (i.e. a) = √3 & CA (i.e. b) = 1

Since, sin A = ak

sin A = (√3) × (1/2) = √3/2 

A = 60° 

Similarly,  sin B = bk

sin B = 1/2 

B = 30° 

As, A + B + C = 180°

60° + 30° + C = 180° 

C = 90° 

From equation (1)

a + b + c = 2(sin 60° + sin 30° + sin 90°) 

a + b + c = 2(√3/2 + 1/2 + 1) 

∴ a + b + c = 3 + √3

Concept used:

Sine Rule: In the triangle ABC, where sides BC, CA & AB are a, b & c

respectively, we can write 

sin A/a = sin B/b = sin C/c

Where A, B & C are the angles opposite to the sides a, b & c

Formula used:

Arithmetic Mean of three number p, q & r is (P + q + r)/3

The perimeter of the triangle is the sum of three sides.

Calculation:

Let the side of the triangle be BC, CA & AB are a, b & c which are opposite to

its angle A, B & C respectively.

According to the question

a + b + c = 6[(sin A + sin B + sin C)/3]

a + b + c = 2(sin A + sin B + sin C)     ----(1)

But from sine rule

sin A/a = sin B/b = sin C/c = k (let)

sin A = ak, sin B = bk, sin C = ck

From equation (1)

a + b + c = 2(ak + bk + ck) 

a + b + c = 2k(a + b + c)

k = 1/2 

But side BC (i.e. a) = √3 & CA (i.e. b) = 1

Since, sin A = ak

sin A = (√3) × (1/2) = √3/2 

A = 60° 

Similarly,  sin B = bk

sin B = 1/2 

B = 30° 

As, A + B + C = 180°

60° + 30° + C = 180° 

C = 90° 

Now, 

Statement: 1 ABC is a right-angled triangle.

As, ∠C = 90°. therefore, ΔABC is a right-angled triangle.

Statement:

1 The angles of the triangle are in the AP

The angles of the triangle are 30°, 60° & 90° which are in AP.

∴ Both statements are correct.

Formula used:

Minimum value of x + 1/x is 2 & it comes at x = 1

Calculation:

Given that 

\(x=\frac{\sin ^2 A+\sin A+1}{\sin A}\) where 0 < A ≤ \(\frac{π}{2}\)

This can be written as 

x = sin A + 1/sin A + 1

For a minimum value of x,

⇒ sin A = 1

⇒ x_min= 1 + 1/1 + 1 = 3

∴ The required answer is 3.

Formula used:

Minimum value of x + 1/x is 2 & it comes at x = 1

Calculation:

Given that 

\(x=\frac{\sin ^2 A+\sin A+1}{\sin A}\) where 0 < A ≤ \(\frac{π}{2}\)

This can be written as 

x = sin A + 1/sin A + 1

For a minimum value of x,

⇒ sin A = 1

⇒ A = π/2

∴ The required answer is π/2.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = \(\frac{b^2+c^2-a^2}{2bc}\)

cos B = \(\frac{a^2+c^2-b^2}{2ac}\)

cos C = \(\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Using cosine rule, b² + c² - a² = 2bc cos A ...(i)

a² + c2 - b2 = 2ac cos B ...(ii)

a2 + b2 - c2 = 2ab cos C ...(iii)

Adding (i), (ii) and (iii), we get:

a² + b² + c² = 2ab cos C + 2bc cos A + 2ca cos B

According to the question, a2 + b2 + c2 = ac + √3bc

Comparing the coefficients of both equations:

2cos C = 0 ⇒ cos C = 0 ⇒ C = 90^∘

2cos B = 1 ⇒ cos B = 1/2 ⇒ B = 60^∘ 

2cos A = √3 ⇒ cos A = √3 /2 ⇒ A = 30^∘

⇒ The angles of triangles are A = 30^∘, B = 60^∘, C = 90^∘

∴ The triangle is a Right angled triangle.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = \(\frac{b^2+c^2-a^2}{2bc}\)

cos B = \(\frac{a^2+c^2-b^2}{2ac}\)

cos C = \(\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Using cosine rule,

b2 + c2 - a2 = 2bc cos A ...(i)

a2 + c2 - b2 = 2ac cos B ...(ii)

a2 + b2 - c2 = 2ab cos C ...(iii)

Adding (i), (ii) and (iii), we get:

a2 + b2 + c2 = 2ab cos C + 2bc cos A + 2ca cos B

According to the question, a2 + b2 + c2 = ac + √3bc

Comparing the coefficients of both equations:

2cos C = 0 ⇒ cos C = 0 ⇒ C = 90∘

2cos B = 1 ⇒ cos B = 1/2 ⇒ B = 60∘ 

2cos A = √3 ⇒ cos A = √3 /2 ⇒ A = 30∘

⇒ The angles of triangles are A = 30∘, B = 60∘, C = 90∘

Given, c = 8

sin A = sin 30^∘ = \(\frac{a}{c}\)

⇒ \(\frac{1}{2}\) = \(\frac{a}{c}\)

⇒ a = 4

sin B = sin 60^∘ = \(\frac{b}{c}\)

⇒ \(\frac{√{3}}{2}\) = \(\frac{b}{c}\)

⇒ b = 4√3

∴ Area of triangle = \(\frac{1}{2}\) × a × b

= \(\frac{1}{2}\) × 4 × 4√3

= 8√3

∴ Area of the triangle is 8√3.