Quantitative Aptitude PYQs

We can write,

⇒ 16 = 2⁴

⇒ 36 = 2² × 3²

⇒ 45 = 3² × 5

⇒ 48 = 2⁴ × 3

⇒ LCM of the numbers = Product of highest powers of prime factors = 2⁴ × 3² × 5 = 720

Now, Highest 4-digit number = 9999

On dividing 9999 by 720, remainder = 639

∴ Required number = 9999 - 639 = 9360

Given:

x = y^a, y = zb and z = x^c 

Calculation:

x = ya

Substituting value of ‘y’,

⇒ x = (z^b)^a = z^ab

Substituting value of ‘z’,

⇒ x = (x^c)^ab

⇒ x = x^abc

Equating powers of ‘x’,

⇒ abc = 1

∴ Value of abc is 1

Given,

⇒ x = 2 + 2^2/3 + 2^1/3

⇒ x - 2 = 2^2/3 + 2^1/3       ----(1)

Taking cube on both sides,

⇒ (x - 2)³ = (2^2/3 + 2^1/3)³

∵ (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ x³ - 8 - 6x² + 12x = 4 + 2 + 6(2^2/3 + 2^1/3)

Substituting from (1),

⇒ x³ - 8 - 6x² + 12x = 6 + 6(x - 2)

⇒ x³ - 6x² + 12x - 6x = 6 + 8 - 12

⇒ x³ - 6x² + 6x = 2

We can write, 33 = 3 × 11

A number is divisible by 11, if the difference between the sum of its odd digits and the sum of its even digits is either 0 or 11,

⇒ Required difference = (X + Y + X) - (X + X) = Y

∵ Y is a digit, hence, it cannot be 11,

⇒ Y = 0

Now, for a number to be divisible by 3, the sum of its digits should be divisible by 3,

⇒ Sum of digits = X + X + 0 + X + X = 4X

In order for the sum of digits to be divisible by 3, digit X should be a multiple of 3,

⇒ X is either 3, 6 or 9

⇒ Required number = 33033, 66066 or 99099

∴ 3 five-digit numbers are possible of the given form

For a number to be divisible by 3, the sum of its digits should be divisible by 3,

⇒ Sum of digits = X + Y + 2 + 3 + 5 = X + Y + 10

Also, given X + Y ≤ 5,

Thus, to satisfy the above two conditions, (X + Y) is either equal to 2 or 5,

Also, X cannot be equal to 0, otherwise the number will not be a 5-digit number

When (X + Y) = 2, possible pairs are (1, 1) and (2, 0)

When (X + Y) = 5, possible pairs are (1, 4), (2, 3), (3, 2), (4, 1) and (5, 0)

∴ Total number of p

ossible pairs of (X, Y) = 7

Note: (0, 2) and (0, 5) should not be counted because otherwise the number will not remain a five-digit number.

Given,

⇒ x² - 6x - 27 > 0

⇒ x² - 9x + 3x - 27 > 0

⇒ x(x - 9) + 3(x - 9) > 0

⇒ (x - 9)(x + 3) > 0

As we know, when ab > 0, then there are two cases:

⇒ Either a > 0 and b > 0

⇒ Or a < 0 and b < 0

Considering the first case,

⇒ (x - 9) > 0 and (x + 3) > 0

⇒ x > 9 and x > - 3

⇒ x > 9

Considering the second case,

⇒ (x - 9) < 0 and (x + 3) < 0

⇒ x < 9 and x < - 3

⇒ x < - 3

∴ x < - 3 or x > 9

Given:

The given number is 38808

Concept:

Let K be the composite number, then

K = xa × yb × zc

Number of total factors = (a + 1) × (b + 1) × (c + 1)

Calculation:

38808 = 2³ × 3² × 7² × 11

Total no. of divisors of 38808

⇒  (3 + 1)(2 + 1)(2 + 1)(1 + 1)

⇒  4 × 3 × 3 × 2 = 72

Excluding 1 and itself, required no. of divisors

⇒  72 - 2 = 70

∴ Required number of divisior is 70.

Let the other polynomial be f(x),

As we know,

Product of two numbers = Product of their LCM & HCF

⇒ (x² - 4x - 21) × f(x) = (x + 3)(x³ - 9x² - x + 105)

⇒ (x + 3)(x - 7) × f(x) = (x + 3)(x³ - 9x² - x + 105)

⇒ f(x) = (x³ - 9x² - x + 105)/(x - 7)

∴ f(x) = (x² - 2x - 15)

Given, (α + β) = - q/p and αβ = r/p

Option 1 ⇒ 1/(α + β) = 1/(- q/p) = - p/q

Option 2 ⇒ 1/α + 1/β = (α + β)/αβ = (- q/p)/(r/p) = - q/r

Option 3 ⇒ - 1/αβ = - 1/(r/p) = - p/r

Option 4 ⇒ αβ/(α + β) = (r/p)/(- q/p) = - r/q

Now,

∵ p < r

Option 1 > Option 4

∵ q < r

⇒ Option 3 > Option 1

∵ p < q

⇒ Option 3 > Option 2

∴ Option 3 ⇒ - 1/αβ is the greatest

Let A can alone complete the work in ‘x’ days.

⇒ A’s 1-day work = 1/x

∵ A and B together completed the work in 5 days.

⇒ B’s 1-day work = 1/5 - 1/x

Now, if A worked twice & B worked 1/3^rd in a day, time taken = 3 days

⇒ 2(1/x) + (1/3)(1/5 - 1/x) = 1/3

⇒ 2/x + 1/15 - 1/3x = 1/3

⇒ 5/3x = 4/15

⇒ x = 5/3 × 15/4 = 25/4 days

∴ A can alone complete the work in 25/4 days

∵ (a³ + b³) = (a + b)(a² - ab + b²)

Putting a = x², b = 1/x², we get,

⇒ (x⁶ + 1/x⁶) = (x² + 1/x²)(x⁴ - 1 + 1/x⁴)

⇒ k = (x⁴ - 1 + 1/x⁴)

Let the three consecutive natural numbers be (x - 1), x, (x + 1)

⇒ Sum of squares = (x - 1)² + x² + (x + 1)² = 110

⇒ x² - 2x + 1 + x² + x² + 2x + 1 = 110

⇒ 3x² = 110 - 2 = 108

⇒ x² = 108/3 = 36

⇒ x = √36 = 6 (neglecting negative values)

⇒ The three consecutive numbers are 5, 6 and 7

∴ Sum of their cubes = 5³ + 6³ + 7³ = 125 + 216 + 343 = 684

Given, HCF of p & q = 16

Let p = 16x and q = 16y

Product of p & q = 16x × 16y = 7168

⇒ xy = 28

⇒ xy = 2 × 2 × 7

∵ p > 60 and q > 60,

Neither x nor y will be equal to 2,

⇒ x and y can have only two possible values, either 4 or 7

Hence, either p = 64 and q = 112 or p = 112 and q = 64

∴ Sum of the numbers = 64 + 112 = 176

According to the logarithmic rules,

⇒ log_a x = (log_b x)/(log_b a)

⇒ log_a xy = log_a x + log_a y

⇒ log_a xⁿ = n log_a x

⇒ log₁₀ 10 = 1

Hence,

⇒ log₁₀₀ (0.72)

= log₁₀ (0.72)/ log₁₀ (100)

= log₁₀ (2³ × 3² × 10^{- 2})/log₁₀ (10²)

= [log₁₀ (2³) + log₁₀ (3²) + log₁₀ (10^{- 2})]/log₁₀ (10²)

= [3 log₁₀ (2) + 2 log₁₀ (3) - 2 log₁₀ (10)]/[2 log₁₀ (10)]

= [(3 × 0.3010) + (2 × 0.4771) - 2]/2

= (0.903 + 0.9542 - 2)/2

= - 0.1428/2

= - 0.0714

= - 1 + 0.9286

= \(\bar 1.9286\)            [∵ bar 1 = \(\bar 1\) represents - 1]

Suppose a^x = b^y = c^z = p

⇒ a = p^1/x

⇒ b = p^1/y

⇒ c = p^1/z

Now,

⇒ abc = 1

⇒ p^1/x × p^1/y × p^1/z = 1

⇒ p^{(1/x + 1/y + 1/z)} = p⁰

Equating powers of ‘p’,

⇒ (1/x + 1/y + 1/z) = 0

For a quadratic equation of form (ax² + bx + c) = 0,

Sum of roots = - b/a

Product of roots = c/a

Given roots of equation as ‘α’ and ‘β’,

⇒ (α + β) = - b/a

⇒ αβ = c/a

Now,

⇒ 1/(aα + b) + 1/(aβ + b)

= (aβ + b + aα + b)/(a²αβ + αab + βab + b²)

= [a(α + β) + 2b]/[a²αβ + ab(α + β) + b²)

= [a(- b/a) + 2b]/[a²(c/a) + ab(- b/a) + b²]

= (- b + 2b)/(ac - b² + b²)

= b/ac

We can write,

⇒ XYZ = 100X + 10Y + Z

⇒ YZX = 100Y + 10Z + X

⇒ ZXY = 100Z + 10X + Y

Sum of the numbers

= 100X + 10Y + Z + 100Y + 10Z + X + 100Z + 10X + Y

= 111X + 111Y + 111Z

= 111(X + Y + Z)

Hence, the sum of the numbers is divisible by both 111 and (X + Y + Z)

∴ Statement 1 and 2, both are correct.

Given, 1/m + 1/n - 1/mn = 2/5

⇒ (n + m - 1)/mn = 2/5

⇒ 5n + 5m - 5 = 2mn

Putting m = 1,

⇒ 5n + 5 - 5 = 2n

⇒ n = 0 (Not valid as m & n are positive integers)

Putting m = 2,

⇒ 5n + 10 - 5 = 4n

⇒ n = - 5 (Not valid as m & n are positive integers)

Putting m = 3,

⇒ 5n + 15 - 5 = 6n

⇒ n = 10

Similarly if we take n = 3 we get m = 10

Putting m = 4,

⇒ 5n + 20 - 5 = 8n

⇒ 3n = 15

⇒ n = 5

Similarly if we take n = 4 we get m = 5

Hence, possible values of (m, n) ⇒ (3, 10), (10, 3), (4, 5), (5, 4)

∴ Number of possible values = 4

Considering option 1:

Suppose (a^{q - r} b^{r - p} c^{p - q}) = abc

Equating powers, we get,

⇒ q - r = 1

⇒ r - p = 1

⇒ p - q = 1

Adding the three equations, we get,

⇒ 0 = 3, which is not valid

Considering option 2:

Suppose (a^{q - r} b^{r - p} c^{p - q}) = xyz

Substituting for a, b & c, we get,

⇒ (xy^{p - 1})^{q - r} × (yz^{q - 1})^{r - p} × (zx^{r - 1})^{p - q} = xyz

⇒ (x)^{[q - r + (r - 1)(p - q)]} × (y)^{[(p - 1)(q - r) + r - p]} × (z)^{[(q - 1)(r - p) + p - q)]} = xyz

⇒ (x)^{[q - r + pr - qr - p + q]} × (y)^{[pq - pr - q + r + r - p]} × (z)^{[qr - pq - r + p + p - q)]} = xyz

⇒ (x)^{2q - p - r + pr - qr)} × (y)^{(2r - p - q + pq - pr)} × (z)^{(2p - q - r + qr - pq)} = xyz

Equating powers, we get,

⇒ 2q - p - r + pr - qr = 1

⇒ 2r - p - q + pq - pr = 1

⇒ 2p - q - r + qr - pq = 1

Adding the three equations, we get,

⇒ 0 = 3, which is also not valid

Considering option 3:

Suppose (a^{q - r} b^{r - p} c^{p - q}) = 0

This is not possible as raising a non-zero number to any power will never be equal to 0

∴ The correct answer is none of the above.

Alternate solution∶

a = xy^{p - 1}, b = yz^{q - 1}, c = zx^{r - 1},

⇒ a^{q - r} b^{r - p} c^{p - q} = (xy^{p - 1})^{(q - r)}.(yz^{q - 1})^{(r - p)}.(zx^{r - 1})^{(p - q)}

⇒ x^{(q - r + r - p + p - q)} y^{(pq - pr - q + r + qr - pq - r + p + pr - qr - p + q)}

⇒ x⁰.y⁰

⇒ 1

∴ The correct answer is none of the above.

Let the number of sides of the two polygons be ‘5x’ and ‘4x’

∵ Measure of interior angle of regular polygon with ‘n’ sides = (n - 2) × 180°/n

Given, difference between interior angles = 9°

⇒ [(5x - 2) × 180°/5x] - [(4x - 2) × 180°/4x] = 9°

⇒ [(5x - 2) × 36°/x] - [(4x - 2) × 45°/x] = 9°

⇒ x180° - 72° - x180° + 90° = x9°

⇒ 18° = x9°

⇒ x = 2

⇒ One of them is a decagon (10 sides) and other is an octagon (8 sides)

Now,

Sum of exterior angles of a regular polygon = 360°

⇒ Sum of exterior angles of the two polygons = 360° + 360° = 720°

∴ Statements 2 and 3 are correct.