Quantitative Aptitude PYQs

Given:

\(2 + \sqrt {2 + \sqrt {2 + \sqrt { \ldots \ldots \ldots } } } \)

Calculation:

\( ⇒ 2 + \sqrt {2 + \sqrt {2 + } } = x\)

⇒  2 + √x = x

⇒  √x = x – 2

Squaring both sides

⇒  x = x² – 4x + 4

⇒ x² – 5x + 4 = 0

⇒ x² – 4x – x + 4 = 0

⇒ x(n - 4) – 1(n - 4) = 0

⇒ (x - 4) (n - 1) = 0

⇒ x = 1, 4

but 1 cannot be the answer because sum is greater than 2

∴  x = 4

Given:

52% of candidates failed in English and 42% failed in only Mathematics.

17% failed in both the subjects

Concept used:

We will be using the concept of the Venn diagram.

Solution:

From Venn diagram

So, percentage failed in exam = 35 + 17 + 42 = 94%

Hence, percentage of passed = 100 - 94 = 6%

Mistake Points

Here in question, failed percentage in only mathematics given not the total failed in mathematics. For both cases, the answer will be different.

Wife’s share = x

each son’s share = y

Each daughter’s share = z

Each son receives 3 times as much as each daughter

⇒ y = 3z       ----(1)

Each daughter receives 2 twice as much as their mother

⇒ z = 2x (2)

total sum is Rs. 3,90,000

⇒ x + 5y + 4z = 3,90,000       ----(3)

3 equation, 3 unknowns

⇒ x = 10,000

so, their mother receives Rs. 10,000

Let’s say the sum of money be Rs. X

Then amount \( = P{\left( {1 + \frac{R}{n}} \right)^{nt}}\)

\(P = x,;R = \frac{{40}}{{100}}& ;n = 1\)

\( \Rightarrow Amount = x\left( {1 + \frac{{40}}{{100}}} \right)\)

Amount = x × (1.4) ^t

Now put, t = 2

⇒ Amount = x × (1.4)²

= 1.96 x

Now put, t = 3

Amount = x (1.4)³

= 2.74 x

so, it’s more than double

• Now put t = 4

Amount = x (1.4)⁴

= 3 .84 x

So, amount is more the tripled

So complete four years for amount to be more than  trebled

Let’s say person divides it in x, y & z parts

Then x + y + z = 17, 2000      ----(1)

after two year they get same interest from all three

⇒ simple interest \(= \frac{{P; \times ;R; \times ;T}}{{100}}\)

⇒ x × 0.05 × 2 = y × 0.06 × 2

⇒ 5x = 6y      ----(2) 

⇒ y × 0.06 × 2 = z × 0.09 × 2

⇒ 6y = 9z      ----(3)

3 equation 3 unknown

= > x = 7200, y = 6000 & z = 4000

so, money invested at 9% is 4000.

As we know that,

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

then let a = x-y

b = y – z

c = z – x

& we know a + b + c = x – y + y – z + z – x = 0

Then RHS term is zero

⇒ (x - y)³ + (y - z)³ + (z - x)³ – 3(x - y) (y - z) (z - x) = 0

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

a^x = b^y = c^z = r

then a = r ^1/x, b = r^1/y, c = r^1/z

b² = ac

⇒ r^2/y = r^1/x. r^1/z

\({r^{\frac{2}{y}}} = {r^{\left( {\frac{1}{x}; + ;\frac{1}{z}} \right)}}\)

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

Concept:

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:

P & q are the roots of the equation x² – 15x + r = 0

the sum of roots = p + q = 15      ----(1)

p – q = 1 (given)

⇒ p = 8 & q = 7

Then product of roots = r = p × q

r = 8 × 7 = 56

∴ The value of r is 56.

Calculation:

x² – 7x + 12 > 0

x² – 4x – 3x + 12 > 0

x (x - 4) – 3 (x - 4) > 0

We have,

(x - 4) (x - 3) > 0

Which represents the product of two terms is greater than zero. 

This will be possible in two cases:

Case:1  (x - 4) > 0 and  (x - 3) > 0

⇒ x > 4 & x > 3 

⇒ x > 4 or x ∈ (4, ∞)

Case:2  (x - 4) < 0 and  (x - 3) < 0

(x - 4) < 0 and  (x - 3) < 0

⇒ x < 4 & x < 3 

⇒ x < 3 or x ∈ (-∞, 3)

Hence, corrct naswer is 

x ∈ (-∞, 3) ∪ (4, ∞)

Given:

The given expression is 5²ⁿ – 2³ⁿ

Shortcut Trick

52n – 23n

put n = 1

⇒ 5² – 2³ 

⇒ (25 – 8) = 17

∴ The required factor is 17.

Alternate MethodConcept:

aⁿ - bⁿ always have a common factor (a - b)

Calculation:

We have,

(5²)ⁿ  - (2³)ⁿ

Using the above concept, 

(5² - 2³) will always be a factor 

⇒ 25 - 8 = 17

∴ The required factor is 17.

Given:

tan x = 1 for 0 < x < 90°

Formula used:

sin 45° = \(\frac{1}{{\sqrt2}}\)

cos 45° = \(\frac{1}{{\sqrt2}}\)

Calculation:

tan x = 1 for 0 < x < 90°

then, x = 45°

⇒ 2 sin x cos x = 2 × sin 45° cos 45°

\(= 2 \times \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = 1\)

∴ The required value is 1.

As we know that

sin (A + B) = sin A cos B + cos A sin B

& given expression is sin 46° cos 44° + cos 46° sin 44°

by equating we get

A = 46°, B = 44°

so, sin (46 + 44) = sin 90°

= 1

Given trigonometric expression is 4 sin²θ + 1

We know that,     0 ≤  sinθ ≤  1, [∵ 0 ≤  θ ≤  90°]

⇒ 0 ≤ sin²θ ≤ 1

⇒ 0 ≤ 4sin2θ ≤ 4

⇒ 1≤ 4sin2θ + 1 ≤ 5

Now, we can write the value of (4 sin2θ + 1) belongs to 1 to 5.

Now, check by option 

Option 1^st is not possible because the vale of (4 sin2θ + 1) can lie between 1 and 2 also.

Option 3^rd is also not possible because at θ = 0° 

The value of 4cosθ = 4 and the value of (4 sin2θ + 1) is 1

⇒ 4cosθ > (4 sin2θ + 1)

Option 4^th is also not possible because at θ  90° 

4 tan θ = ∞ and the value of (4 sin2θ + 1) = 5

⇒ 4 tan θ > (4 sin2θ + 1)

∴ 2^nd option is correct.

Regular hexagon has side “x” let’s say tower at B has height H_B & height at C has height H_C

\(\Rightarrow \tan 30^\circ = \frac{{{H_B}}}{x}\)        ----(1)

angle of elevation for A to the top of tower C is 45°

\(\tan 45^\circ = \frac{{{H_c}}}{y}\)       ----(2)

two sides & included angle are given

then y² = x² + x² - 2 * x * x * cos 120°

y²= 2x² - 2x² cos 120°

y² = 2x² (1 - cos 120°)

y² = 2x² (1 + 0.5)

y² = 3x²

\(y = \sqrt 3 x\)       ----(3)

So, form equation 1, 2 & 3

\(\tan 30^\circ = \frac{{{H_B}}}{x} \Rightarrow x = \frac{{{H_B}}}{{\tan 30^\circ }}\)

\(\tan 45^\circ = \frac{{{H_c}}}{y} \Rightarrow \sqrt 3 x = \frac{{{H_c}}}{{\tan 45^\circ }}\)

equate above 2 equation

\(\frac{1}{{\sqrt 3 }} = \frac{{{H_B}}}{{{H_C}}} \times \frac{{\tan 45^\circ }}{{\tan 30^\circ }}\)

\(\Rightarrow \frac{{{H_B}}}{{{H_C}}} = \frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} \times 1 = \frac{1}{3}\)

so, H_B : H_C = 1 : 3

As we know \(\tan \theta = \frac{1}{{\cot \theta }}\)

& tan (90 - θ) = cot θ

Then

⇒ tan 1° tan 2° tan 3°-------- tan 45° -------- tan 84° tan 88° tan 99°

⇒ tan 1° tan 2° tan 3°-------- tan 45° -------- tan (90 - 3)° tan (90 - 2)° tan (90 - 1)°

⇒ tan 1° tan 2° tan 3° -------- tan 45° ------- cot 3° cot 2° cot 1°

Replace cot θ by \(\frac{1}{{\tan \theta }}\)

Then by cancelling all terms we get tan 45° & tan 45° = 1

So persons is standing on first street at point P facing towards north. Now he takes 150° turn & travels for 15 minutes at the steel of 20 km/hr.

So distance travelled = speed × time

\(= 20\frac{{km}}{{hr}} \times \frac{1}{4}hr\)

= 5 km

So, he travels 5 km & reaches point ‘q’.

Now, he takes 60° to the left & travels for 20 minutes at the speed of 30 km/hr & reaches point ‘r’.

So, distance travelled

\(= 30\frac{{km}}{{hr}} \times \frac{1}{3}hr\)

So, distance between p & q is 5 km & q & r is 10 km. Now point p is at street first & point r is at street 2^nd for distance \(sq = \sin 30^\circ = \frac{{sq}}{5}\)

⇒ sq = 2.5 km

So, distance between two streets are sq + qr

= 2.5 + 10

= 12.5 km

3 tan θ = cot θ

\(\Rightarrow 3\tan \theta = \frac{1}{{\tan \theta }}\)

⇒ 3 tan² θ = 1

\(\Rightarrow \tan \theta = \frac{1}{{\sqrt 3 }}\)

\(\Rightarrow \theta = \frac{\pi }{6}\)

Formula used:

sin²θ + cos²θ = 1

Calculation:

sin2 25° + sin2 65°

⇒ sin² 25° + sin² (90° - 25°)

⇒ sin² 25° + cos² 25°

⇒ 1

∴ The required value is 1.

As we know that

(A + B)³ = A³ + B³ + 3AB (A + B)

put A = sin²θ. & B = cos²θ

\(\Rightarrow {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^3} = {1^3}\left( {L.H.S} \right)\)

RHS term. = sin⁶θ + cos⁶θ + 3 sin²θ cos²θ (sin²θ + cos²θ)

= sin⁶θ + cos⁶θ + 3 sin²θ cos²θ

⇒ sin⁶θ + cos⁶θ + 3 sin²θ cos²θ = 1

so, value of expression is zero

1. sec α  Range is (- ∞, 1] ⋃ [1, ∞)

sec ∝ = ¼ (Not possible)

2. tanβ range is [-∞, ∞]

tanβ = 20 (possible)

3. cosec γ Range is R - (-1, 1)        [Where R = Real number]

cosec γ = ½ (Not possible)

4. cos δ Range is [-1, 1]

cos δ = 2 (Not possible)

So, 3 statements are not possible