Quantitative Aptitude PYQs

Given:

p(x) leaves a remainder of 2 when divided by (x - 1).

p(x) leaves a remainder of 1 when divided by (x - 2).

We need the remainder when p(x) is divided by (x - 1)(x - 2).

Formula used:

The remainder when p(x) is divided by (x - 1)(x - 2) will be a linear polynomial of the form ax + b.

From the given conditions: p(1) = 2 and p(2) = 1.

Calculations:

Assume remainder = ax + b.

From p(1) = 2:

⇒ a(1) + b = 2

⇒ a + b = 2

From p(2) = 1:

⇒ a(2) + b = 1

⇒ 2a + b = 1

Solving the two equations:

1. a + b = 2
2. 2a + b = 1

⇒ Subtract equation 1 from equation 2:

⇒ (2a + b) - (a + b) = 1 - 2

⇒ a = -1

Substitute a = -1 in equation 1:

⇒ -1 + b = 2

⇒ b = 3

Final Remainder:

Remainder = ax + b = -x + 3 = 3 - x

∴ The correct answer is option (3).

Given:

I. x + (1/x) > 1

II. (x + (1/x))² > 2

III. (x + (1/x))⁴ > 9

Formula used:

For positive real numbers, x + (1/x) ≥ 2 when x > 0.

Calculation:

Check I: x + (1/x) > 1

⇒ x > 0 ensures x + (1/x) ≥ 2 > 1

⇒ I is correct.

Check II: (x + (1/x))² > 2

⇒ x + (1/x) ≥ 2

⇒ (x + (1/x))² ≥ 2² = 4

⇒ II is correct.

Check III: (x + (1/x))⁴ > 9

⇒ (x + (1/x)) ≥ 2

⇒ (x + (1/x))⁴ ≥ 2⁴ = 16 > 9

⇒ III is correct.

∴ The correct answer is option (4).

Given:

Let q² - 5p - 4 be negative, where q > p and p, q are natural numbers.

Formula used:

For q2 - 5p - 4 < 0 , solve for p, considering q > p.

Calculation:

To find the largest p:

Let q = p + 1 (since q > p ).

⇒ (p + 1)² - 5p - 4 < 0

⇒ p² + 2p + 1 - 5p - 4 < 0

⇒ p² - 3p - 3 < 0

Factorize or test values for p:

For p = 3:

⇒ 3² - 3×3 - 3 = 9 - 9 - 3 = -3 (negative).

For p = 4:

⇒ 4² - 3×4 - 3 = 16 - 12 - 3 = 1 (positive).

Largest p is 3 where q² - 5p - 4 < 0 .

∴ The correct answer is option (1).

Given:

x, y are natural numbers < 20

x, y, (x + y), (x − y) are prime numbers

Formula used:

Prime rule: Except 2, all primes are odd

Calculation:

x + y is prime

⇒ Sum of two odd primes is even (not prime)

⇒ One number must be 2

Since (x − y) is prime

⇒ x > y

⇒ y = 2

Now x is prime < 20 and x > 2

Check values:

x = 3 ⇒ x − y = 1 (not prime) 

x = 5 ⇒ x − y = 3, x + y = 7 

x = 7 ⇒ x + y = 9 

x = 11 ⇒ x − y = 9 

x = 13 ⇒ x + y = 15 

x = 17 ⇒ x − y = 15 

x = 19 ⇒ x + y = 21 

Valid set:

(x, y, x + y, x − y) = (5, 2, 7, 3)

∴ Number of possible combinations = One (Option 1).

Given:

(x + 1)(x + p)(x² + p²) = x⁴ − 1

Formula used:

x⁴ − 1 = (x − 1)(x + 1)(x² + 1)

Calculation:

x⁴ − 1 = (x − 1)(x + 1)(x² + 1)

Given expression:

(x + 1)(x + p)(x² + p²)

Common factor: (x + 1)

⇒ Remaining factors must match

⇒ (x + p) = (x − 1)

⇒ p = −1

Check:

p² = 1

⇒ (x² + p²) = (x² + 1)

Hence LHS = (x + 1)(x − 1)(x² + 1)

= x⁴ − 1

∴ The value of p is −1.

Given:

(2 + √3)^x + (2 - √3)^x = 2

Formula used:

Let a = 2 + √3, b = 2 - √3. Then a × b = 1, so b^x = a^-x.

Calculations:

Let y = (2 + √3)^x. Then given ⇒ y + 1/y = 2.

⇒ y + 1/y = 2

⇒ y² + 1 = 2y

⇒ y² - 2y + 1 = 0

⇒ (y - 1)² = 0

⇒ y = 1 ⇒ (2 + √3)^x = 1 ⇒ x = 0

Therefore, (2 + √3)^x - (2 - √3)^x = 1 - 1 = 0

∴ The required value is 0.

Given:

\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{6}\)

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{13}{36}\)

Formula used:

\(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2 = \dfrac{1}{a^2}+\dfrac{1}{b^2}+2\cdot\dfrac{1}{ab} \)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3 = \dfrac{1}{a^3}+\dfrac{1}{b^3}+3\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\cdot\dfrac{1}{ab} \)

Calculation:

⇒ S = \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{6}\)

⇒ S² = \(\dfrac{25}{36}\)

⇒ Use S² = (1/a²+1/b²) + 2×(1/ab)

⇒ \(\dfrac{25}{36} = \dfrac{13}{36} + 2\times\dfrac{1}{ab}\)

⇒ 2×(1/ab) = \(\dfrac{12}{36}=\dfrac{1}{3}\)

⇒ \(\dfrac{1}{ab}=\dfrac{1}{6}\)

⇒ Now S³ = \(\dfrac{125}{216}\)

⇒ Use S³ = (1/a³+1/b³) + 3×S×(1/ab)

⇒ \(\dfrac{125}{216} = \left(\dfrac{1}{a^3}+\dfrac{1}{b^3}\right) + 3\times\dfrac{5}{6}\times\dfrac{1}{6}\)

⇒ 3×S×(1/ab) = 3×(5/6)×(1/6) = \(\dfrac{15}{36}=\dfrac{5}{12}=\dfrac{90}{216}\)

⇒ \(\dfrac{1}{a^3}+\dfrac{1}{b^3} = \dfrac{125}{216}-\dfrac{90}{216}=\dfrac{35}{216}\)

∴ The correct answer is \(\dfrac{35}{216}\).

Shortcut Trick

To find the remainder when a polynomial is divided by (x² + 1), substitute x² = −1 into the polynomial.

Given polynomial: x⁶

Rewrite as: (x²)³

Substitute x² = −1 ⇒ (−1)³ = −1

∴ The correct answer is −1.

Alternate Method

Given: Dividend = x⁶, Divisor = x² + 1

Formula Used: Remainder Theorem for polynomials.

⇒ Let f(x) = x⁶

⇒ We use the identity: a³ + b³ = (a + b)(a² − ab + b²)

⇒ Consider x⁶ + 1 = (x²)³ + 1³

⇒ x⁶ + 1 = (x² + 1)(x⁴ − x² + 1)

⇒ x⁶ = (x² + 1)(x⁴ − x² + 1) − 1

⇒ This is in the form: Dividend = (Divisor × Quotient) + Remainder

⇒ Comparing the terms, Remainder = −1

∴ The correct answer is −1.

Additional Information

Remainder Theorem

If a polynomial f(x) is divided by (x − a), the remainder is f(a).

Imaginary Unit Substitution

When dividing by (x² + 1), the roots are x = ±i. Substituting x² = −1 is a direct application of this property.

Degree of Remainder

The degree of the remainder is always strictly less than the degree of the divisor. For a quadratic divisor, the remainder is at most linear (ax + b).

Given:

(x + 2) is a factor of which one of the following polynomials:

Formula used:

If (x + c) is a factor of a polynomial f(x), then f(-c) = 0.

Calculation:

Here, (x + 2) is the factor, so c = -2. Substitute x = -2 into the given polynomials:

Option 1:

f(x) = x⁵ - 4x⁴ - 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ - 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 - 64 + 24 + 32 + 28 + 12

= 0

Since f(-2) = 0, (x + 2) is a factor of Option 1.

Option 2:

f(x) = x⁵ + 4x⁴ - 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ + 4(-2)⁴ - 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 + 64 + 24 + 32 + 28 + 12

= 128 (Not 0)

Option 3:

f(x) = x⁵ - 4x⁴ + 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ + 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 - 64 - 24 + 32 + 28 + 12

= -48 (Not 0)

Option 4:

f(x) = x⁵ - 4x⁴ - 3x³ + 8x² + 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ - 3(-2)³ + 8(-2)² + 14(-2) + 12

= -32 - 64 + 24 + 32 - 28 + 12

= -56 (Not 0)

∴ The correct answer is Option 1.

Given:

log₁₀2 = 0.301

log₁₀3 = 0.477

Find number of digits in 60⁶⁰

Formula used:

The number of digits in any number ‘n’ is equal to + 1, where X is a floor function and it gives the largest integer less than or equal to x.

Calculations:

log₁₀60 = log₁₀(6 × 10) = log₁₀6 + 1

log₁₀6 = log₁₀2 + log₁₀3 = 0.301 + 0.477 = 0.778

⇒ log₁₀60 = 0.778 + 1 = 1.778

⇒ log₁₀60⁶⁰ = 60 × 1.778 = 106.68

⇒ No. of digits in (60)⁶⁰ = 106.68 + 1 = 106 + 1 = 107

∴ Number of digits = 107.

Given:

(17²⁵ + 19²⁵) ÷ 18

Formula used:

a mod m → reduce base first

Calculation:

⇒ 17 mod 18 = 17

⇒ 19 mod 18 = 1

⇒ 17²⁵ mod 18 = 17²⁵ (17 ≡ −1 mod 18)

⇒ (−1)²⁵ = −1 ⇒ remainder = 17

⇒ 19²⁵ mod 18 = 1²⁵ = 1

⇒ Total remainder = 17 + 1 = 18

⇒ 18 mod 18 = 0

∴ Remainder = 0.

Given:

HCF of x and y = H

p = (x³ + y³) ÷ (x² − xy + y²)

q = (x³ − y³) ÷ (x² + xy + y²)

Formula used:

x³ + y³ = (x + y)(x² − xy + y²)

x³ − y³ = (x − y)(x² + xy + y²)

Calculation:

p = (x + y)(x² − xy + y²) ÷ (x² − xy + y²)

⇒ p = x + y

q = (x − y)(x² + xy + y²) ÷ (x² + xy + y²)

⇒ q = x − y

HCF(p, q) = HCF(x + y, x − y)

Let x = Ha, y = Hb where HCF(a, b) = 1

⇒ p = H(a + b)

⇒ q = H(a − b)

HCF(p, q) = H × HCF(a + b, a − b)

If a and b both odd ⇒ a + b and a − b are even

⇒ HCF(a + b, a − b) = 2

⇒ HCF(p, q) = 2H

If one of a, b is even ⇒ HCF(a + b, a − b) = 1

⇒ HCF(p, q) = H

∴ Both statements I and II are correct.

Given:

x⁴ = x² + 1 , x > 0

Formula used:

Put y = x² ⇒ x⁴ = y²

Calculation:

⇒ y² = y + 1

⇒ y² − y − 1 = 0

⇒ y = (1 + √5) / 2 (positive root)

⇒ x² = (1 + √5)/2

⇒ x⁴ = y² = ((1 + √5)/2)²

⇒ x⁴ = (6 + 2√5)/4 = (3 + √5)/2

⇒ 2x⁴ = 3 + √5

∴ The correct answer is 3 + √5.

Given:

\(\dfrac{p+q}{q+r}=\dfrac{r+s}{s+p}\)

Formula used:

Cross multiplication: (p+q)(s+p) = (r+s)(q+r)

(A−B)(A+B) = A² − B²

Calculation:

(p + q)(s + p) = (r + s)(q + r)

⇒ ps + p² + qs + pq = rq + r² + sq + sr

⇒ (p² − r²) + (ps − sr) + (pq − rq) = 0

⇒ (p − r)(p + r) + s(p − r) + q(p − r) = 0

⇒ (p − r)(p + r + s + q) = 0

⇒ (p − r) = 0 or (p + r + s + q) = 0

⇒ p = r or (p + r + s + q) = 0.

∴ The correct answer is: Either p + q + r + s = 0 or p = r.

Given:

n ∈ N, n < 7

Find number of n for which gcd(12n + 2, 8n + 1) = 1

Formula used:

Use Euclid: gcd(a, b) = gcd(b, a − q×b)

Calculations:

⇒ gcd(12n + 2, 8n + 1) = gcd(8n + 1, (12n + 2) − (8n + 1))

⇒ gcd(8n + 1, 4n + 1)

⇒ gcd(4n + 1, (8n + 1) − 2×(4n + 1))

⇒ gcd(4n + 1, −1) = 1

⇒ gcd(12n + 2, 8n + 1) = 1 for every n

n values = 1, 2, 3, 4, 5, 6 ⇒ count = 6

∴ Number of values = 6.

Given:

f = x³ + y³ + 3xy − 1

g = (x − y)⁴ − 1

Formula used:

HCF(f, g) = greatest common divisor of polynomials. Factor both and compare common factors.

Calculation:

⇒ Factor g: (x + y)⁴ − 1 = (x + y − 1)(x + y + 1)(x² + 2xy + y² + 1)

⇒ Factor f: x³ + y³ + 3xy − 1 = (x + y − 1)(x² − x y + x + y² + y + 1)

⇒ Common factor: (x + y - 1)

∴ HCF = (x + y - 1).

Given:

x = n(n + 1)(n + 2)

n is an even natural number

Formula used:

Three consecutive numbers always include:

• one multiple of 3

• at least two even numbers if n is even

Calculation:

Let n = 2k

⇒ x = 2k(2k + 1)(2k + 2)

⇒ x = 2k(2k + 1)2(k + 1)

⇒ x = 4k(k + 1)(2k + 1)

From k and (k + 1), one is even 

⇒ x has factor 8

Among k, (k + 1), (2k + 1) one is divisible by 3 

⇒ x is always divisible by 24

Check Statement I:

48 = 16 × 3

x has only 8 × 3

Example: n = 2 ⇒ x = 2 × 3 × 4 = 24

⇒ Not divisible by 48 

Statement I is false

Check Statement II:

x is divisible by 24

⇒ x² is divisible by 24² = 576

144 = 16 × 9 ⊂ 576

⇒ x² is always divisible by 144 

∴ Only Statement II is correct.

Correct option: II only.

Given:

Find LCM of A = x⁴ + x²y² + y⁴, B = x³y + y⁴, C = x⁴y² − x³y³

Formula used:

LCM = product of distinct irreducible factors with highest powers present in A, B, C

Calculations:

⇒ A = x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² − xy + y²)

⇒ B = x³y + y⁴ = y(x³ + y³) = y(x + y)(x² − xy + y²)

⇒ C = x⁴y² − x³y³ = x³y²(x − y)

Highest powers needed: x³ (from C), y² (from C), factors (x − y), (x + y), (x² + xy + y²), (x² − xy + y²)

⇒ Combine: LCM = x³ y² (x − y)(x + y)(x² + xy + y²)(x² − xy + y²)

⇒ (x − y)(x + y)(x² + xy + y²)(x² − xy + y²) = x⁶ − y⁶

⇒ LCM = x³ y² (x⁶ − y⁶)

The correct answer is Option (2).

Given:

Let XYZ be a 3-digit number. Let D be the difference between XYZ and ZYX.

We need to find the remainder when D is divided by 99.

Calculation:

Let XYZ = 100X + 10Y + Z and ZYX = 100Z + 10Y + X

D = (100X + 10Y + Z) - (100Z + 10Y + X)

⇒ D = 100X - X + 10Y - 10Y + Z - 100Z

⇒ D = 99X - 99Z

⇒ D = 99(X - Z)

When D is divided by 99:

⇒ Remainder = 0

∴ The correct answer is option (1).

Given:

p, q are natural numbers

(p + q)^p+q is divisible by 512

Formula used:

512 = 2⁹

If a = 2^k × odd, then aⁿ contains 2^k×n

Calculation:

Let s = p + q

Then (p + q)^p+q = s^s

Check options:

s = 4

4 = 2²

⇒ s^s = 2^2×4 = 2⁸ < 2⁹ 

s = 6

6 = 2 × 3

⇒ s^s contains 2⁶ < 2⁹ 

s = 8

8 = 2³

⇒ s^s = 2^3×8 = 2²⁴ ≥ 2⁹ 

s = 12

12 > 8 but not least

∴ The least value of (p + q) is 8.