Quantitative Aptitude PYQs

Given:

2 × 3 × 5 × 7 × 11 + 1

Concept Used:

Prime number: A number that is divisible by either 1 or itself is called a prime number.

Composite number: A number that is divisible by number\s other than 1 or itself is called a composite number.

Calculation:

Given number = 2 × 3 × 5 × 7 × 11 + 1 = 2310 + 1 = 2311

The number 2311 is only divisible by 1 and itself,

∴ The given number is a prime number

Given: 

HCF of the numbers in a pair = 12

Concept Used:

Mean = (First term + last term)/2

Calculation:

Let, the numbers be ‘12p’ and ‘12q’

Product of numbers = 2160

⇒ 12p × 12q = 2160

⇒ pq = 15

This is only possible when the values of p and q are either 1 and 15 or 3 and 5

Hence, the two pairs of numbers are (12, 180) and (36, 60)

⇒ Mean of first pair = x = (12 + 180)/2 = 96

⇒ Mean of second pair = y = (36 + 60)/2 = 48

Mean of x and y = (96 + 48)/2 = 72

∴ Mean of x and y is 72.

Given:

(54)¹⁰

Concept Used:

loga xn = n loga x

loga xy = loga x + loga y

Calculation:

The number of digits in any number ‘n’ is equal to \({\rm{lo}}{{\rm{g}}_{10}}{\rm{n}}\)+ 1, where X is a floor function and it gives the largest integer less than or equal to x

Now,

log₁₀ (54)¹⁰

= 10 log₁₀ 54                                                    [∵ log_a xⁿ = n log_a x]

= 10 log₁₀ (2 × 3³)

= 10 log₁₀ 2 + 3 × 10 log₁₀ 3                             [∵ log_a xy = log_a x + log_a y]

= (10 × 0.301) + (30 × 0.477)

= 3.01 + 14.31

= 17.32

No. of digits in (54)¹⁰ = 17.32+ 1 = 17 + 1 = 18

∴ No. of digits in (54)¹⁰ is 18.

Given: 

\({{\rm{x}}^{\sqrt {\rm{x}} }} = \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}}\)

Concept Used:

log a^b = b⋅log a

Calculation:

Taking logarithm on both sides, we get,

⇒ √x log x = (1/n) log x^x

⇒ √x log x = (x/n) log x

⇒ n √x log x = x log x

⇒ n √x log x – x log x = 0

⇒ (√x log x)(n – √x) = 0

Equating first term to 0,

⇒ √x log x = 0

⇒ log x = 0 = log 1

⇒ x = 1

Equating second term to 0,

⇒ n – √x = 0

⇒ √x = n

⇒ x = n²

∴ {1, n²} is a set of solution of the given equation

Given: 

n(Physics) = 50, n(Mathematics) = 75, n(Physics ∩ Mathematics) = 35

Formula Used:

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Calculation:

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ n(Physics ∪ Mathematics) = 50 + 75 – 35 = 90

⇒ No. of students who registered either for Physics, or Mathematics or both = 90

Total no. of students = 250

No. of students who registered neither for Physics nor for Mathematics = 250 – 90 = 160

∴ No. of students who registered neither for Physics nor for Mathematic is 160.

Given: 

Number = 10ⁿ – 1

Calculation:

Putting n = 1, number = 10 – 1 = 9

Putting n = 2, number = 100 – 1 = 99

Putting n = 3, number = 1000 – 1 = 999

This means that the given number has ‘n’ digits each of which is equal to 9,

⇒ Sum of digits of number = 9n

But, the sum of digits = 3798

So, n = 3798/9 = 422

∴ Value of n is 422

Calculation:

We can write,

⇒ 2222 < 4096

⇒ 2222 < 2¹²

⇒ (2222)² < (2¹²)²

⇒ (2222)² < 2²⁴

Hence, 2222² < 2²²²²

Similarly,

⇒ 222 < 256

⇒ 222 < 2⁸

⇒ (222)²² < (2⁸)²²

⇒ (222)²² < 2¹⁷⁶

Hence, (222)²² < 2²²²²

Also,

⇒ 22 < 32

⇒ 22 < 2⁵

⇒ (22)²²² < (2⁵)²²²

⇒ (22)²²² < 2¹¹¹⁰

Hence, (22)²²² < 2²²²²

∴ Among the given numbers, 2²²²² is the largest number

Given:

m is the number of prime numbers between 0 and 50 

n is the number of prime numbers between 50 and 100

Concept Used:

Prime number: A number that is divisible by either 1 or itself is called a prime number.

Calculation:

Prime numbers between 0 and 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

⇒ m = 15

Prime numbers between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

⇒ n = 10

So, (m – n) = 15 – 10 = 5

∴ The value of m - n is 5

Given:

\(\frac{{\rm{a}}}{{\rm{b}}} = \frac{1}{3},\frac{{\rm{b}}}{{\rm{c}}} = 2,\frac{{\rm{c}}}{{\rm{d}}} = \frac{1}{2},\frac{{\rm{d}}}{{\rm{e}}} = 3\) and \(\frac{{\rm{e}}}{{\rm{f}}} = \frac{1}{4}\)

Calculation:

We can write,

\(\frac{{abc}}{{def}}\)

\( = \frac{a}{b} \times \frac{{{b^2}}}{{{c^2}}} \times \frac{{{c^3}}}{{{d^3}}} \times \frac{{{d^2}}}{{{e^2}}} \times \frac{e}{f}\)

\( = \frac{1}{3} \times {\left( 2 \right)^2} \times {\left( {\frac{1}{2}} \right)^3} \times {\left( 3 \right)^2} \times \frac{1}{4}\)

\( = \frac{3}{8}\)

∴ The value of \(\frac{{abc}}{{def}}\) is \( \frac{3}{8}\).

Given:

Expression: 3x + 3x + 1 + 3x + 2 

Calculation:

3^x + 3^{x + 1} + 3^{x + 2} = 3^x + (3^x)(3) + (3^x)(3²) = 3^x (1 + 3 + 9) = 3^x (13)

Since, x is any natural number, we can have x = 1, 2, 3 , ... so on

For x = 1, we  get ,  3x (13) = 3 × 13 = 39.

This could be the largest. But, in the options, we can see that there are numbers greater than 39. So we check for the next natural number.

So for x = 2

3x (13) = 32 × 13 = 117

We know,

3x + 3x + 1 + 3x + 2 = 3x + (3x)(3) + (3x)(32) = 3x (1 + 3 + 9) = 3x (13)

Now, we can see that the resultant expression is a multiple of 3 and 13. 

So, let us check from the options which is the largest multiple of both 3 and 13. 

As we can see, 117 is the largest multiple of 3 and 13.

∴ The largest divisor of the given expression from the given options is  117.

Given:

A two-digit number is 9 more than four times of the number obtained by interchanging its digits.

The product of digits in the two-digit number is 8

Calculation:

Let the ten’s digit and the unit digit of the number be ‘x’ and ‘y’ respectively

Given, product of digits = xy = 8      ---- (1)

⇒ Number = 10x + y

On interchanging the digits, number obtained = 10y + x

⇒ 10x + y = 9 + 4(10y + x)

⇒ 10x + y = 9 + 40y + 4x

⇒ 6x – 39y – 9 = 0

⇒ 2x – 13y – 3 = 0

Multiplying by ‘x’,

⇒ 2x² – 13xy – 3x = 0

Substituting from (1),

⇒ 2x² – 13(8) – 3x = 0

⇒ 2x² – 3x – 104 = 0

⇒ 2x² – 16x + 13x – 104 = 0

⇒ (x – 8)(2x + 13) = 0

⇒ x = 8, – 13/2 (neglect negative & fractions)

⇒ x = 8

⇒ y = 8/8 = 1

Required number = 10x + y = 80 + 1 = 81

∴ The required number is 81.

Given:

α and β are the roots of the quadratic equation x2 + kx – 15 = 0,

Formula Used:

For any quadratic equation of form (ax² + bx + c),

Sum of roots = – b/a

Product of roots = c/a

Calculation:

α and β are the roots of the quadratic equation x² + kx – 15 = 0,

⇒ α + β = – k

⇒ k = – (α + β)      ---- (1)

Also, αβ = – 15      ---- (2)

But, (α – β) = 8

Substituting from (2),

⇒ α + 15/α = 8

⇒ α² – 8α + 15 = 0

⇒ (α – 3)(α – 5) = 0

⇒ α = 3, 5

Substituting in (2),

⇒ β = – 5, – 3

Substituting in (1),

When α = 3, β = – 5, k = – (3 – 5) = 2

When α = 5, β = – 3, k = – (5 – 3) = – 2

 ∴ Positive value of k is 2

Given:

(x – 1) and (x + 2) divide the polynomial (x3 + 4x2 + px + q),

Concept Used:

According to the factor theorem, if (x – a) divides the polynomial f(x), then remainder = f(a) = 0

Calculation:

(x – 1) and (x + 2) divide the polynomial (x³ + 4x² + px + q),

⇒ f(1) = (1)³ + 4(1)² + p(1) + q = 0

⇒ 1 + 4 + p + q = 0

⇒ p + q = – 5      ---- (1)

Similarly,

⇒ f(– 2) = (– 2)³ + 4(– 2)² + p(– 2) + q = 0

⇒ – 8 + 16 – 2p + q = 0

⇒ 2p – q = 8      ---- (2)

Adding (1) and (2), we get,

⇒ p + q + 2p – q = – 5 + 8

⇒ 3p = 3

⇒ p = 1

⇒ q = – 5 – 1 = – 6

∴ The values of p and q are 1, – 6

Concept used:

HCF: Product of the smallest power of each common prime factor in the numbers.

LCM: Product of the greatest power of each prime factor, involved in the numbers

Calculation:

⇒ (x² + 5x + 6) = (x + 2)(x + 3)

⇒ (x² + 8x + 15) = (x + 3)(x + 5)

Using the above concept,

⇒ HCF = (x + 3)

According to the question, HCF = (x + k)

⇒ (x + 3) = (x + k)

∴ k = 3

Given: 

5^{x + 1} – 5^{x – 1} = 600

Concept Used:

a^{b + c} = a^b × a^c

Calculation:

5x + 1 – 5x – 1 = 600

⇒ (5^x × 5) – (5^x × 1/5) = 600

⇒ 5^x (5 – 1/5) = 600

⇒ 5^x = (5/24) × 600

⇒ 5^x = 125 = 5³

⇒ x = 3

10^2x = 10²⁽³⁾ = 10⁶ = 1000000

∴ The value of 10^2x is 1000000.

Given:

A number divides 12288, 28200 and 44333 so as to leave the same remainder

Calculation:

Let the number be ‘x’ and the remainder be ‘r’

⇒ (12288 – r), (28200 – r) and (44333 – r) are exactly divisible by ‘x’

As we know, if two numbers are divisible by a number, then their difference is also divisible by the same number

⇒ Difference of first two numbers = 28200 – r – 12288 + r = 15912

⇒ Difference of last two numbers = 44333 – r – 28200 + r = 16133

Hence, 15912 and are also divisible by ‘x’

⇒ 15912 = 2³ × 3² × 13 × 17

⇒ 16133 = 13 × 17 × 73

⇒ HCF of 15912 and 16133 = 13 × 17 = 221

∴ The largest number that divides 12288, 28200 and 44333 leaving the same remainder is 221

Given:

x² + 9y² = 6xy

Formula Used:

(a2 – 2ab + b2) = (a – b)2

Calculation:

x2 + 9y2 = 6xy

⇒ x² – 6xy + 9y² = 0                         (∵ (a² – 2ab + b²) = (a – b)²)

⇒ (x – 3y)² = 0

⇒ (x – 3y) = 0

⇒ x = 3y

∴ y ∶ x is 1 ∶ 3

Given: 

mⁿ = 1331

Calculation:

As we know, 11³ = 1331

Comparing, we get,

⇒ m = 11, n = 3

⇒ (m – 1)^{n – 1} = (11 – 1)^{3 – 1} = (10)² = 100

∴ The value of (m - 1)^{n - 1} is 100

Given:

\(\frac{1}{{{{\rm{a}}^{{\rm{m}} - {\rm{n}}}} - 1}} + \frac{1}{{{{\rm{a}}^{{\rm{n}} - {\rm{m}}}} - 1}}\)

Concept Used:

x^-1 = 1/x

Calculation:

Let, x = a^{m – n}

⇒ a^{n – m} = a^{– (m – n)} = x^{– 1} = 1/x

Hence, we can write,

\(\frac{1}{{{a^{m - n}} - 1}} + \frac{1}{{{a^{n - m}} - 1}}\)

\(\frac{1}{{x - 1}} + \frac{1}{{\frac{1}{x} - 1}}\)

\( = - \frac{1}{{1 - x}} + \frac{x}{{1 - x;}}\)

\( = - \left( {\frac{{1 - x}}{{1 - x}}} \right)\)

= -1

∴ The value of \(\frac{1}{{{{\rm{a}}^{{\rm{m}} - {\rm{n}}}} - 1}} + \frac{1}{{{{\rm{a}}^{{\rm{n}} - {\rm{m}}}} - 1}}\) is - 1.

Given: 

\({\rm{x}} = \sqrt 2 ,{\rm{;y}} = \sqrt[3]{3}\) and \({\rm{z}} = \sqrt[6]{6}\)

Calculation:

x = √2 = (2)^1/2

⇒ x² = 2

Taking cube of both sides,

⇒ (x²)³ = (2)³

⇒ x⁶ = 8

⇒ x = ⁶√8

Similarly,

⇒ y = ³√3 = (3)^1/3

⇒ y³ = 3

Taking square on both sides,

⇒ (y³)² = (3)²

⇒ y⁶ = 9

⇒ y = ⁶√9

Now, we can write,

⇒ 3 < 8 < 9

⇒ ⁶√3 < ⁶√8 < ⁶√9

∴ z < x < y

x = √2 ,  y = 3√3 , z = ⁶√6

Taking power 6 to all three, we get

⇒ x⁶ = (2^1/2)⁶ , y⁶ = (3^1/3)⁶, z⁶ = (6^1/6)⁶

⇒ x6 = 8, y6 = 9, z6 = 6

∴ z < x < y