Quantitative Aptitude PYQs

Given:

The number is 26381

Concept used:

Factor:- A number or quantity that when multiplied with another produces a given number or expression.

In simple words, The prime factor is finding which prime numbers multiply together to make the original number.

Example: The prime factors of 15 are 3 and 5 (because 3 ×  5 = 15, and 3 and 5 are prime numbers)

Calculation:

Prime factorisation of 26381

⇒ 23 × 31 × 37

If we check all the options we can see only option 1 and option 4 is giving unit digit 1 when numbers are multiplied

But, 29 × 17 × 37 = 18241

∴ The answer is option 4 which is 23, 31, 37

Given:

Polynomial is (x – 1) (x – 2) (x – 4) – 90

Concept used:

If (x – a) is a factor of f(x) then f(a) = 0

Factoring a polynomial is the process of decomposing a polynomial into a product of two or more polynomials of smaller degree.

For example, f(x) = x² + 5x + 6 

f(x) = (x + 3) (x + 2)

Calculation:

Let f(x) = (x – 1) (x – 2) (x – 4) – 90

With help of options we can get,

f(7) = (7 - 1)(7 - 2)(7 - 4) – 90

⇒ 6 × 5 × 3 – 90

⇒  90 – 90

⇒ 0

⇒ f(7) = 0

∴ x – 7 is a factor of the polynomial (x – 1) (x – 2) (x – 4) – 90.

Given:

The number is (23 – 4√15)

Formula used:

(a - b)² = a² + b² - 2ab

Calculation:

23 - 4√15

⇒ 3 + 20 – 4 √15

⇒ (√3)² + 4 × 5 – 2 × √3 × (2 √5)

⇒ (√3)² + 4 × (√5)² – 2 × √3 × (2 √5)

⇒ (√3)² + ( 2√5 )² – 2 × √3 × (2 √5)

⇒ (√3 – 2√5)²

Square root = √3 – 2√5

∴ The square root of (23 - 4√15) is (√3 – 2√5).

Given:

Number is 371000 

Concept used:

Divisibility Rule of 9:

If the sum of digits of the number is divisible by 9, then the number itself is divisible by 9

Calculation:

(37)1000

⇒ (36 + 1)¹⁰⁰⁰

Since 36 is divisible by 9

Remainder from 36 = 0

Now,

1¹⁰⁰⁰ divided by 9 will always give you the remainder 1

∴ The required remainder is 1

Given:

LCM + HCF = 1484

LCM – HCF = 1428

Concept used:

a × b = LCM of (a, b) × HCF of (a, b)

Where a and b are two numbers

Calculation:

LCM + HCF = 1484

LCM – HCF = 1428

⇒ 2LCM = 2912

⇒ LCM = 1456

⇒ HCF = 28

Let the other numbers be x

112 × x = LCM × HCF

⇒ 112x = 1456 × 28

⇒ x = (1456 × 28)/112

⇒ x = 364

∴ The other number is 364.

Given:

Initially number of men = 20

Time allotted = 10 days

Formula used:

M₁ D₁/ W₁ = M₂ D₂/ W₂

Here,

M = Number of Men, D = Number of Days, W = Work

Calculation:

According to the question

(20 × 5)/(1/4) = (M₂ × 5)/(3/4)

⇒ 20 × 3 = M₂

⇒ M₂ = 60

More men needed = 60 – 20 = 40

∴ 40 more men they need to employ to finish the work on time.

Given:

Condition is x is a negative real number

Calculation:

Case 1.

If we multiply a natural number with a negative real number we will always get a negative number 

eg. 1 × -2 = -2 

Hence, Case 1 is wrong

Case 2.

If we take x = – 1

(– 1)² + (– 1)

⇒ 1 – 1 = 0 which is not ≥ 0

Hence, Case 2 is also wrong

Case 3.

If we take x = – 1

2 × (– 1) < (– 1) < –(– 1)

–2 < – 1 < 1

Hence, Case 3 is correct

Case 4.

If we take real number x = -3^1/4 then we get 

x² = 3^1/2 which is not a rational number

Hence, Case 4 is also wrong

∴ 1, 2, and 4 are not correct

Given:

2x2 + xy - 3y2

Concept used:

The standard form for linear equations in two variables

ax + by + c = 0

Calculation:

2x² + xy – 3y² 

⇒ 2x² + 3xy – 2xy – 3y² 

⇒ x(2x + 3y) – y(2x + 3y)

⇒ (x – y)(2x + 3y)

Here we have two linear factors are x - y and 2x + 3y 

Now, we have to find the sum of factors 

(x – y) + (2x + 3y)

⇒ 3x + 2y

∴ The required sum of the linear factors is 3x + 2y.

Conecpt used:

When D = b² – 4ac < 0, Roots are not real

When D = b2 – 4ac = 0, Roots are real and equal 

When D = b2 – 4ac > 0, Roots are real and unequal

Calculation:

From Option 1^st

Given equation is 2x2 + 16x + 3 = 0

a = 2, b = 16, c = 3

D = (16)² – 4 × 2 × 3

⇒ 256 – 24 = 232 > 0 

Hence, the above equation has real roots.

From Option 2^nd

Given equation is 2x2 + 10x - 1 = 0

a = 2, b = 10, c = 1

D = (10)2 – 4 × 2 × 1

⇒ 100 – 8 = 92 > 0 

Hence, the above equation has real roots.

From Option 3^rd

Given equation is x2 - 8x + 1 = 0

a = 1, b = 8, c = 1

D = (8)2 – 4 × 1 × 1

⇒ 64 – 4 = 60 > 0

Hence, the above equation has real roots.

From Option 4^th

Given equation is 4x2 + 9x + 6 = 0

a = 4,b = 9, c = 6

D = (9)2 – 4 × 4 × 6

⇒ 81 – 96

⇒ – 15 < 0

Hence, the above equation has not real roots.

∴ Required answer is option 4

Given:

Sum of the roots = 7

Product of the roots = 12

Concept used:

x² – (sum of roots)x + product of roots = 0

Calculation:

Let the roots of the given quadratic equation are α and β 

According to the question

α + β = 7   ----(i)

and α.β = 12    ----(ii)

On solving the above equations, we get

α = 4, β = 3

According to the question

New roots are α' = 4/2 = 2, β' = 3 × 2 = 6

⇒ α' + β' = 2 + 6 = 8

⇒ α'.β' = 2 × 6 = 12

So, Resulting quadratic equation is

x2 – (α' + β')x + α'.β' = 0

⇒ x² – 8x + 12 = 0

∴ The resulting quadratic equation is x2 – 8x + 12 = 0

Concept used:

For real and distinct solutions, D = b² – 4ac > 0

Calculation:

Given equation is x2 – kx + 2 = 0

According to the equation,

a = 1, b = – k, c = 2

For real and distnict roots, 

D = b² – 4ac > 0

⇒ (– k)² – 4 × 1 × 2 > 0

⇒ k² – 8 > 0

⇒ k2 – (2√2)² > 0

We know that, a² - b² = (a + b)(a - b)

⇒ (k - 2√2) (k + 2√2) > 0

This can be possible in two ways,

Case:1 (k - 2√2) > 0 and (k + 2√2) > 0

⇒ k > 2√2 and k > -2√2

⇒ k > 2√2

Case:2 (k - 2√2) < 0 and (k + 2√2) < 0

⇒ k < 2√2 and k < -2√2

⇒ k < - 2√2

∴ k > 2√2 or k < – 2√2

Given:

α + β + γ = αβ + βγ + γα     ----(i)

Calculation:

(1 – α)(1 – β)(1 – γ)

⇒ (1 – α – β – αβ)(1 – γ)

⇒ (1 – α – β – γ + αβ + αγ + βγ – αβγ)

⇒ {1 – (α + β + γ ) + αβ + αγ + βγ – αβγ}

From (i), we get

⇒ {1 – (αβ + αγ + βγ) + αβ + αγ + βγ – αβγ}

⇒ (1 – αβ – αγ – βγ  + αβ + αγ + βγ – αβγ)

⇒ 1 – αβγ

∴ Required answer is (1 – αβγ)

Given:

log10x + log10x2 = 2 log10x + 1

Formula used:

log a + log b = log ab

bloga = log a^b

Calculation:

log10x + log10x2 = 2 log10x + 1

⇒ log₁₀x³ = log₁₀x² + log₁₀10

⇒ log10x3 = log1010x2

On comparing both sides

x³ = 10x²

⇒ x = 10

∴ The value of x is 10.

Given:

LCM of two primes p and q = 2231

p > q

Concept used:

Product of two numbers = HCF × LCM

Calculation:

p and q are prime numbers

So, HCF of p and q = 1

p × q = 2231

By using the unit digit method,

1 can be got by 7 × 3, 1 × 1, 9 × 9

2231 is close to 2300

Now,

97 × 23 = 2231

So, p = 97 and q = 23

Now, p – q = 97 – 23 = 74

∴ The value of p - q is 74.

Given:

Three runners take 20, 30, and 35 minutes to complete one round.

Concept used:

LCM method to find the meeting time

Calculation:

LCM of 20 minutes, 30 minutes, and 35 minutes = 420 minutes

420 minutes = 7 hours

∴ They next meet at the starting point after 7 hours

Given:

Cost of 5 pencils, 6 notebooks and 7 erasers = Rs. 250

Cost of 6 pencils, 4 notebooks and 2 erasers = Rs. 180

Calculation:

Let cost of 1 pencil, 1 notebook and 1 eraser be p, n and e respectively

According to the question:

5p + 6n + 7e = 250     

By multiplying with 6:

30p + 36n + 42e = 1500     ----eq.1

Also, 6p + 4n + 2e = 180

By multiplying by 5:

30p + 20n + 10e = 900     ----eq.2

By subtracting eq.2 from eq.1

16n + 32e = 600

⇒ 2n + 4e = 75

∴ Cost of 2 notebooks and 4 erasers is Rs. 75

Concept:

The number of zeroes is counted by the power of 10 and (5 × 2) 

Calculation:

150 × 249 × 348 × .... × 50¹

It contains 10⁴¹, 20³¹, 30²¹, 40¹¹, 50¹

⇒ 10¹⁰⁵

It also contains 5⁴⁶, 15³⁶, 25²⁶, 35¹⁶, 45⁶, 5 (of 50) and 2⁴⁹, 12³⁹, 22²⁹, 32¹⁹, 42⁹

⇒ 546, 1536, 552, 3516, 456, 5 (of 50) and 249, 4⁴⁷, 1239, 2229, 3219, 429  there will be many number in which 2 exist.

⇒ (5 × 2)¹⁵⁷; here only power of 5s will be counted as power of 2s are in greater number

∴ Number of zeroes in the product of 150 × 249 × 348 × .... × 50¹ = 105 + 157 = 262.

Given:

p and q are roots of x2 - 60x + 899 = 0

p > q

Calculation:

x2 - 60x + 899 = 0

⇒ (x - 31) (x - 29) = 0

⇒ roots of the equation(p, q) are 31 and 29

⇒ p = 31, q = 29

Checking option 1):

31 - 29 - 1 = 1, hence it is false

Checking option 2:

31 - 58 + 27 = 0, hence it is correct.

∴ Option 2 is correct.

Given:

Roots of x2 - 4x - log10N = 0 are real

Formula used:

Roots of quadratic equation ax² + bx + c = 0 are real 

if D ≥ 0  i.e b² - 4ac ≥ 0 

And lf logaN = x

Then N = a^x

Calculation:

We have the quadratic equation x2 - 4x - log10N = 0

On comparing this equation with ax2 + bx + c = 0, we get 

a = 1, b = -4 and c = - log10N

Now, According to the question 

D = b2 - 4ac  ≥ 0

⇒ (- 4)² - 4 × 1 × (- log10N) ≥ 0

⇒ 16 + 4log10N ≥ 0

⇒ 4log10N ≥ -16

⇒ log10N ≥ - 4

⇒ N ≥ 10^-4

⇒ N ≥ 0.0001

∴ The minimum value of N is 0.0001.

Given:

5x-1 = (2.5)log105

Formula Used:

If a^x = n then x = log_aⁿ

log_a^b = log_e^b/log_e^a

Calculation:

We have 5x-1 = (2.5)log105

⇒ (2.5)log105 = 5x-1 

⇒ log₁₀⁵  = log_2.5^5x-1 

⇒ log105  = (x - 1) log2.55

⇒ (x - 1) = (log105)/(log2.55)

⇒ (x - 1) = log₁₀^2.5

⇒ x = log102.5 _{+ 1}

⇒ x = log102.5 ₊ log₁₀¹⁰

⇒ x = log1010 × 2.5

⇒ x = log1025

⇒ x = log105²

⇒ x = 2log10​5

∴ The value of x is 2log10​5.