Quantitative Aptitude PYQs

A month of 30 days beginning with Saturday will have : 5 Sundays

Total number of visitors in that month = 5 × 510 + 25 × 240 = 2550 + 6000 = 8550

∴ Average = 8550/30 = 285

\(\frac{{36}}{{11}} = 3; + ;\frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}}\)

\(\Rightarrow 3; + ;\frac{3}{{11}} = 3; + ;\frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}}\)

\( \Rightarrow \frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}} = \frac{3}{{11}}\)

\(\Rightarrow \frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}} = \frac{1}{{\frac{{11}}{3}}}\)

\(\Rightarrow \frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}} = \frac{1}{{3; + ;\frac{2}{3}}}\)

\(\Rightarrow \frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}} = \frac{1}{{3; + ;\frac{1}{{\frac{3}{2}}}}}\)

\( \Rightarrow \frac{1}{{x; + ;\frac{1}{{y; + ;\frac{1}{z}}}}} = \frac{1}{{3; + ;\frac{1}{{1; + ;\frac{1}{2}}}}}\)

Hence, x = 3, y = 1 and z = 2

∴ (x + y + z) = 3 + 1 + 2 = 6

Given :

Selling price of two items = Rs 990

Profit = 10%

Loss = 10%

Formula used :

Selling price = \(\frac{(100 \pm Profit/loss%)}{100} × Cost \ price\)  ----- (1)

Loss = \(\frac{CP - SP}{CP} \times 100\) ----- (2)

Calculations :

Let the CP of both article be 100x

When profit is 10%, using equation (1)

⇒ SP of first item = \(\frac{110}{100} × 100x = 110x\)       ---- (2)

When loss is 10%, using equation (1)

⇒ SP of second item = \(\frac{90}{100} × 100x = 90x\)         ----- (3)

We know that SP of both article is Rs 990

So, according to the question

SP of first item = 990

⇒ 110x = 990

⇒ x = 9

CP of first item = 100x

⇒ 100 × 9 = 900

SP of second item = 90x

⇒ 90x = 990

⇒ x = 11

CP of second item = 100x

⇒ 100 × 11 = Rs 1100

Total Cost price of both item = 900 + 1100

⇒ Rs 2000

Total Selling price of both item = Rs 1980

Using equation (2), we get

⇒ \(\frac{2000 - 1980}{2000} \times 100\)

⇒ \(\frac{20}{2000} \times 100\)

⇒ 1% loss

∴ The the combined percentage of loss for the two items is 1%.

Suppose the speeds of car and train are ‘x’ and ‘y’, respectively.

According to the question

480/x + 120/y = 11      ---- (I)

And

400/x + 200/y = 35/3      ---- (II)

Multiply equation I by 5 and II by 3 and subtract them :

⇒ 2400/x – 1200/x = 55 – 35

⇒ 1200/x = 20

⇒ x = 60

And y = 40

Hence, x : y = 3 : 2

Given:

(x – x²) is maximum when 0 < x < 1.

Hence, only possible option is x = 0.5 or 1/2

6²³ × 75⁹ × 105²

⇒ 2²³ × 3²³ × 5¹⁸ × 3⁹ × 5² × 3² × 7²

⇒ 2²³ × 3³⁴ × 5²⁰ × 7²

⇒ 2³ × 3³⁴ × 10²⁰ × 7²

Since 10ⁿ divides the expression.

Largest possible value of n = 20

3⁹⁸ – 3⁸⁹

⇒ 3⁸⁹ (3⁹ – 1)

Here, unit digit of 3⁸⁹ = Unit digit of 3¹ = 3

And

Unit digit of (3⁹ – 1) = Unit digit of (3¹ – 1) = 2

Hence, the digit in the unit’s place of the number represented by 3⁹⁸ – 3⁸⁹ = 3 × 2 = 6

Given: Sum of the squares of four consecutive natural numbers is 294.

Let the first number be x.

x² + (x + 1)² + (x + 2)² + (x + 3)² = 294

⇒ x² + x² + 1 + 2x + x² + 4 + 4x + x² + 9 + 6x = 294

⇒ 4x² + 12x + 14 = 294

⇒ 4x² + 12x – 280 = 0

⇒ x² + 3x – 70 = 0

⇒ (x + 10) (x – 7) = 0

Hence, x = 7 and x can’t be negative.

Hence, sum of numbers = 7 + 8 + 9 + 10 = 34

Concept:

Quadratic Equation:

The Standard Form of a Quadratic Equation is ax2 + bx + c = 0

• Sum of roots = −b/a
• Product of roots = c/a

The quadratic equation of root p and q can be written as x² - (p + q)x + pq = 0

 

Calculation:

The quadratic equation of roots equal to p and q is

x2 + px + q = 0               -----(1)

From the using the above concept,

Equation can be written as

x² - (p + q)x + pq = 0     -----(2)

By compairing both equations, we get

p = -(p + q)

⇒ 2p = -q             -----(3)

Also,

q = pq

⇒ p = 1

Put this value in equation (3)

⇒ 2 × 1 = -q

⇒ q = -2

∴  The values of p and q respectively are 1 and -2.

Formula used:

(a +b)(a - b) = a² - b² 

Calculation:

x² – y² = 35

⇒ (x + y) (x – y) = 5 × 7

Hence, x = 6 and y = 1

And,

(x + y) (x – y) = 1 × 35

Hence, x = 18 and y = 17

Hence, 2 pair is possible.

Given that:

(b – 6) is one root of the quadratic equation x² – 6x + b = 0

x² – 6x + b = 0

Putting x = (b – 6)

⇒ b² + 36 – 12b – 6b + 36 + b = 0

⇒ b² – 17b + 72 = 0

⇒ (b – 9) (b – 8) = 0

⇒ b = 9, 8

Hence, maximum value of b² = 9² = 81

Given:

\(a; = \sqrt {(7; + ;4√ 3) }\)

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

7 + 4√3 = 4 + 3 + 4√3

⇒ 2² + (√3)² + 2 × 2 × √3

⇒ (2 + √3)²

Hence, a = \(\sqrt {7; + ;4√ 3 }\) = (2 + √3)

Now, 1/a = \(\frac{1}{2 \ + \ √3}\)

⇒ \(\frac{1}{2 \ + \ √3} \ \times \ \frac{2 \ - \ √3}{2 \ - \ √3}\)

⇒ \(\frac{2 \ - \ √3}{4 \ - \ 3 }\)

⇒ 1/a = 2 - √3

a + 1/a = 2 + √3 + 2 – √3 = 4

∴ The correct answer is 4.

Concept:

Minimum value of ax² + bx + c = 0

When a > 0  then

y_min = (4ac - b²)/4a

Calculation:

Minimum value of expression x² + 5x + 10 

= (4ac – b²)/4a     (∵ a > 0)

⇒  (40 – 25)/4 = 15/4

Hence, maximum value of the expression \(\frac{1}{{{x^2}; + ;5x; + ;10}} = \frac{4}{{15}}\)

Calculation:

According to the question:

(x + 2)(x – 3) : (x + 4)(x – 2) = 3 : 4

⇒ (x² – x – 6) : (x² + 2x – 8) = 3 : 4

⇒ 4x² – 4x – 24 = 3x² + 6x – 24

⇒ x² = 10x

⇒ x = 10

Hence, x = 10.

Product of an even number with any number even or odd gives an even number only.

Hence, the statement in option 3 is not correct.

All other statements are correct.

∴ Required answer is Option 3

Suppose n² + 19n + 92 = k²

Multiplying both sides by 4 we get,

4n² + 76n + 368 = 4k²

⇒ (2n)² + 2 × 19 × 2n + 361 + 7 = 4k²

or (2n + 19)² = (4k² – 7)

Suppose 2n + 19 = x

Hence, 4k² – 7 = x²

⇒ 4k² – x² = 7

⇒ (2k – x) (2k + x) = 7

Factors of 7 are 7 and 1 as it is a prime number.

2k – x = 7 and 2k + x = 1      ---- (1)

2k – x = 1 and 2k + x = 7     ---- (2)

From equation (1), k = 2

From equation (2), k = 2

Hence the value of k = 2

Now,

(2n + 19)² = (4k² – 7) = 16 – 7 = 9

⇒ (2n + 19)² = 9

⇒ (2n + 19) = ± 3

Now,

If (2n + 19) = 3

⇒ n = -8

If (2n + 19) = -3

⇒ n = -11

Hence, sum of all integer values of n = -8 – 11 = -19

(x³ + 3x² + 3x + 1) = (x + 1)³

(x³ + 5x² + 5x + 4) = (x + 4) (x² + x + 1)

(x² + 5x + 4) = (x + 4) (x + 1)

Hence, LCM = (x + 1)³(x + 4)(x² + x + 1)

Formula used:

(a³ + b³ + c³ - 3abc) = (a + b + c)(a² + b² + c² - ab - bc - ac)

Calculation:

We know that, if a + b + c = 0,

Then a³ + b³ + c³ = 3abc

Since, [(x – y) + (y – z) + (z – x)] = 0

[(x – y)³ + (y – z)³ + (z – x)³] = 3 (x – y)(y – z)(z – x)

Hence,

\(\frac{{{{\left( {x - y} \right)}^3}; + ;{{\left( {y - z} \right)}^3}; + ;{{\left( {z - x} \right)}^3}}}{{9\left( {x - y} \right)\left( {y - z} \right)\left( {z - x;} \right)}} = \frac{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x;} \right)}}{{9\left( {x - y} \right)\left( {y - z} \right)\left( {z - x;} \right)}} = \frac{1}{3}\)

X = {a, {b}, c}, Y = {{a}, b, c} and Z = {a, b, {c}}

(X ∩ Y) = {c}

∴ (X ∩ Y) ∩ Z = φ

Formula used:

If α and β are the roots of quadratic equation ax2 + bx + c = 0

Sum of root α + β = -b/a

Product of root αβ = c/a

Calculation:

px² + 3x + 2q = 0

Sum of roots = -6

⇒ -3/p = -6

⇒ p = 1/2      ------(1)

Product of roots = -6

⇒ 2q/p = -6

⇒ 4q = -6      [From equation (1)]

⇒ q = -3/2

Hence, (p – q) = 1/2 + 3/2 = 2