Quantitative Aptitude PYQs

Concept:

If a cube of a number is negative then the number is also negative,

For example, x = - 3

Therefore,  x3 = - 3 × - 3 × - 3 = -27

Calculation :

x3 = - 29.8        

⇒ x = - ve

Now, using equation (2), we get

⇒ y³ = 3x

We know that value of x is negative, so

⇒ y³ = 3 (- ve) = - ve

So, y is also negative

Now, if both the terms x and y are negative

Sum of negative terms is always negative

⇒ 3 (-ve)  + 2 (-ve)

⇒ - ve

⇒ 3x + 2y is negative.

∴ Both Statement - I and Statement - Il are together sufficient to answer the question.

Concept :

ax2 + bx + c = 0        

Discriminant (D) = b² - 4ac      

Product of roots = c/a 

If both roots are of opposite sign then c/a < 0

Calculations :

ax2 + bx + c = 0

Question : Does the equation ax2 + bx + c = 0 have real roots of opposite sign?

Mathematically, the question is saying

b² - 4ac > 0 & c/a < 0   (is correct or not)

Statement:I D > 0

⇒  b2 - 4ac > 0

⇒ b² > 4ac 

With this information, we can not determine the sign of a & c.

Statement: II  c/a > 0 

Therefore, roots will be of the same sign as well.

ax² + bx + c = 0 has real root of same sign. 

So we are able to answer the question.

∴ Both Statement – I and Statement – II are together sufficient to answer the question

Given :

a2 + b2 + c2 – ab – bc – ca

Where a, b and c are distinct real numbers (a ≠ b ≠ c)

Formula used :

a2 + b2 + c2 – ab – bc – ca = \(\frac{1}{2}\)[(a - b)² + (b - c)² + (c - a)²]    

Calculations :

It is given a, b and c are distinct real numbers 

Let us take the value of a = 1, b = 2 and c = 3 (a < b < c) 

We know that,

a2 + b2 + c2 – ab – bc – ca = \(\frac{1}{2}\)[(a - b)2 + (b - c)2 + (c - a)2]

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

⇒ (1/2) (1 + 1 + 4)

⇒ (6/2) = 3

⇒ \(\frac{1}{2}\)[(a - b)2 + (b - c)2 + (c - a)2] is always positive

Therefore, (a2 + b2 + c2 – ab – bc – ca) not depend on the statement - I and statement - II

∴ Neither Statement – I nor Statement – II is required to answer the question

Given :

 \(\frac{{{x^6}, + ,{y^6}}}{{{x^4}, + ,{y^4}}}\) always greater than \(\frac{{{x^4}, + ,{y^4}}}{{{x^2}, + ,{y^2}}}\) (x ≠ y ≠ 0)

Calculations :

⇒ \(\frac{{{x^6}, + ,{y^6}}}{{{x^4}, + ,{y^4}}} > \frac{{{x^4}, + ,{y^4}}}{{{x^2}, + ,{y^2}}}\)

⇒ (x⁶ + y⁶)(x² + y²) > (x⁴ + y⁴)(x4 + y4)

⇒ x⁸ + x⁶y² + x²y⁶ + y⁸ > x⁸ + y⁸ + 2x⁴y⁴

⇒ x²y²(x⁴ + y⁴) > 2x⁴y⁴

⇒ (x4 + y4) > 2x2y²

⇒ (x4 + y4) - 2x2y² > 0

⇒ (x² - y²)² > 0      ---- (1)

The above equation (1) can also be written as

⇒ (y² - x²)2 > 0

Thus, x is not always greater than y.

Statement I is not sufficient.

Using equation (1), we get

⇒ {(x + y)(x - y)}² > 0 

⇒ x + y > 0 and x - y > 0

Thus, Statement II is also not sufficient

∴ Neither Statement – I nor Statement – II is required to answer the question

Given:

In a quadratic equation

Sum of roots = Product of roots 

Concept Used:

The general equation of quadratic equation is 

ax² + bx + c = 0 ; a ≠ 0

Sum of roots = -b/a

Product of roots = c/a

Calculation:

Let the quadratic equation be ax2 + bx + c = 0 

Sum of roots = -b/a

Product of roots = c/a

According to the question 

Sum of roots = Product of roots 

⇒ -b/a = c/a

⇒ -b = c or b = -c 

i.e b and c are the additive inverse of each other 

Here, we can say that there can be infinite many quadratic equation possible when a ≠ 0 and b = -c

eg: x² - 4x + 4 = 0,  2x2 - 9x + 9 = 0, 5x2 - 16x + 16 = 0................ etc.

∴ The correct answer is option 4.

Shortcut Trick 

 

Given x = \(\frac{{{\rm{1 - cos}},{\rm{θ }},{\rm{ + }}\sin {\rm{θ }}}}{{1 + \sin {\rm{θ }},}}\)

Take θ = 45° 

We get x = \(\frac{1 - \frac{1}{√2} + \frac{1}{√2} }{1 + \frac{1}{√2}}\) = \(\frac{√2}{ √2 + 1}\) = 2 - √2   ----(i)

To find \(\frac{{{\rm{sin}},{\rm{θ }},{\rm{ + }},\cos {\rm{θ }} - 1}}{{\cos {\rm{θ }},}}\) = y (let)

Put the same θ = 45° here, we get 

y = \(\frac{ \frac{1}{√2} + \frac{1}{√2} - 1}{ \frac{1}{√2}}\) = 2 - √2   ----(ii)

Form (i) and (ii), we get 

x = y 

∴ The value of \(\frac{{{\rm{sin}},{\rm{θ }},{\rm{ + }},\cos {\rm{θ }} - 1}}{{\cos {\rm{θ }},}}\) is x.

 

Alternate Method

Given x = \(\frac{{{\rm{1 - cos}},{\rm{θ }},{\rm{ + }}\sin {\rm{θ }}}}{{1 + \sin {\rm{θ }},}}\)

On rationalizing we get 

x = \(\frac{{{\rm{1 - cos}},{\rm{θ }},{\rm{ + }}\sin {\rm{θ }}}}{{1 + \sin {\rm{θ }},}} \times \frac{1 - sinθ}{1 - sinθ}\)

⇒ x = \(\frac{1 - sinθ - cosθ + sinθ. cosθ + sinθ - sin^2θ}{1 - sin^2θ }\)

⇒ x = \(\frac{cos^2θ - cosθ + sinθ. cosθ }{cos^2θ }\)

Taking cosθ common from N and D, we get

⇒ x = \(\frac{cos\theta - 1 + sin\theta}{cos\theta }\) 

The R.H.S we got here is the required value 

∴ The value of \(\frac{{{\rm{sin}},{\rm{θ }},{\rm{ + }},\cos {\rm{θ }} - 1}}{{\cos {\rm{θ }},}}\) is x.

Given:

cos (x + y) = 0

sin (x - y) = \(\frac{1}{2}\)

Concept Used:

sin A = sin B then A = B

cos A = cos B then A = B

Where 0 ≤ A, B ≤ 90° 

Formula Used:

cos 90° = 0, sin 30° = 1/2, cot 90° = 0

Calculation:

We have 

cos (x + y) = 0

⇒ cos (x + y) = cos 90°   

⇒ x + y = 90°       -----(i)

And sin (x - y) = \(\frac{1}{2}\)

⇒ sin (x - y) = sin 30°

⇒ x - y = 30°        -----(ii)

On adding equation (i) and (ii), we get 

2x = 120° 

⇒ x = 60° 

Now, From (i)

60° + y = 90°  

⇒ y = 30° 

Now, We have to find cot (2x - y)

So, cot (2 × 60° - 30°) = cot 90° = 0

∴ The value of cot (2x - y) is 0.

Given:

Trigonometric Expression is 

sin4 θ + cos4 θ - 2sin2 θ cos2θ

Concept Used:

-1 ≤ cosθ ≤ 1, 0 ≤ cos²θ ≤ 1

-1 ≤ sinθ ≤ 1, 0 ≤ sin2θ ≤ 1

cos2 θ - sin2 θ = cos 2θ

Calculation:

We have sin4 θ + cos4 θ - 2sin2 θ cos2θ

⇒ (cos2 θ)² + (sin2 θ)2 - 2. sin2 θ. cos2θ

Using formula a² + b² - 2ab = (a - b)²

⇒ (cos2 θ - sin2 θ)2

⇒ (cos 2θ)²

⇒ cos² 2θ

According to the concept used 

The minimum value of cos θ is 0

∴ The minimum value of cos2 2θ is 0.

Shortcut TrickGiven cos θ + sec θ - 2 = 0

⇒ cos θ + 1/cos θ - 2 = 0

We know that the maximum value of cosθ = 1 at θ = 0° 

So, the above equation is valid only when θ = 0°

Now we have to find cos4 θ + sec4 θ - 2

At θ = 0° 

cos4 0° + sec4 0° - 2

⇒ 1 + 1 - 2 = 0

∴ The required value of cos4 θ + sec4 θ - 2 is 0.

 

Alternate Method 

Given:

cos θ + sec θ - 2 = 0

Formula used:

sec θ = 1/cos θ

Calculation:

Let cos θ be x    -----(i)

We have cos θ + sec θ - 2 = 0

⇒ cos θ + 1/cos θ - 2 = 0

⇒ x + 1/x - 2 = 0     [ From (i) ]

⇒ x + 1/x = 2 

Squaring both side we get 

⇒ x2 + 1/x2 + 2 = 4

⇒ x2 + 1/x2  = 2

Again squaring both side, we get 

⇒ x4 + 1/x4 + 2 = 4

⇒ x4 + 1/x4 - 2 = 0      -----(ii)

Now, from (i) and (ii), we get 

cos4 θ + sec4 θ - 2 = 0 

∴ The required value of cos4 θ + sec4 θ - 2 is 0.

Given:

y = cos2 x + sec2 x; 0 ≤ x < \(\frac{π}{2}\)

Concept Used:

• A.M ≥ G.M
• If a and b are two numbers then (a + b)/2 ≥ √ab
• cos2 x × sec2 x = 1

Calculation:

We have y = cos2 x + sec2 x     ----(i)

Let the two numbers be cos2 x and sec2 x 

According to the concept used

A.M ≥ G.M

⇒ [cos2 x + sec2 x]/2 ≥ √[cos2 x × sec2 x]

⇒ [cos2 x + sec2 x]/2 ≥ 1

⇒ [cos2 x + sec2 x]  ≥ 2

Now, from (i)

⇒ y ≥ 2

∴ The correct relation is y ≥ 2.

Given:

A, B, C are acute angles

sin(B + C - A) = cos(C + A - B) = tan(A + B - C) = 1

Concept used:

Calculation:

sin(B + C - A) = 1

So, sin(B + C - A) = sin90°

⇒ (B + C - A) = 90°      ----(i)

cos(C + A - B) = 1

So, cos(C + A - B) = cos0°

⇒ (C + A - B) = 0°      ----(ii)

tan(A + B - C) = 1

So, tan(A + B - C) = tan45°

⇒ (A + B - C) = 45°      ----(iii)

By adding (i), (ii), and (iii) we get

B + C - A + C + A - B + A + B - C = 90° + 0° + 45°

⇒ A + B + C = 135°

∴ A + B + C is equal to 135°

Given:

A, B, C, D are the angles of a cyclic quadrilateral

\(\sin \left( {\frac{{{\rm{A}},{\rm{ + }},{\rm{C}}}}{2}} \right), + ,\sin \left( {\frac{{{\rm{B}},{\rm{ + }},{\rm{D}}}}{2}} \right)\)

Concept used:

The sum of the opposite angles of a cyclic quadrilateral is supplementary.

Calculation:

According to the concept,

∠A + ∠C = 180°

∠B + ∠D = 180°

So,

\(\sin \left( {\frac{{{\rm{A}},{\rm{ + }},{\rm{C}}}}{2}} \right), + ,\sin \left( {\frac{{{\rm{B}},{\rm{ + }},{\rm{D}}}}{2}} \right) = \sin \left( {\frac{{\rm180^o}}{2}} \right), + ,\sin \left( {\frac{\rm180^o}{2}} \right)\)

⇒ \(\sin \left( {\frac{{{\rm{A}},{\rm{ + }},{\rm{C}}}}{2}} \right), + ,\sin \left( {\frac{{{\rm{B}},{\rm{ + }},{\rm{D}}}}{2}} \right) = \sin {{\rm90^o}}, + ,\sin {{\rm90^o}}\)

⇒ \(\sin \left( {\frac{{{\rm{A}},{\rm{ + }},{\rm{C}}}}{2}} \right), + ,\sin \left( {\frac{{{\rm{B}},{\rm{ + }},{\rm{D}}}}{2}} \right) = 1 +1\)

⇒ \(\sin \left( {\frac{{{\rm{A}},{\rm{ + }},{\rm{C}}}}{2}} \right), + ,\sin \left( {\frac{{{\rm{B}},{\rm{ + }},{\rm{D}}}}{2}} \right) =2\)

∴ Required answer is 2

Given:

Distance from the point of observation = 10 km

The aeroplane is at a height of 8 km above the ground

Concept used:

sinθ = P/H

Here P = Perpendicular, H = Hypotenuse

Calculation:

According to the question,

AC = 10 km

So, sinθ = AB/AC

⇒ sinθ = 8/10 = 4/5 = 0.8

Since,

sin45° = 1/√2 ≈ 0.71

sin60° = √3/2 ≈ 0.87

As, 0.71 ≤ sin θ  ≤ 0.87

⇒ 45° ≤ θ ≤ 60° 

∴ Required answer is 45°< θ < 60°

Given:

The angle of elevation of the top of a tower of height x meter from a point on the ground is found to be 60°

By going y metre away from that point, it becomes 30°

Concept used:

tanθ = P/B

Here P = perpendicular, B = base

Calculation:

According to the question,

tan60° = AB/BC

⇒ √3 = x/BC

⇒ BC = x/√3           -------(1)

Again, tan30° = AB/BD

⇒ 1/√3 = AB/BD

⇒ 1/√3 = x/BD

⇒ BD = √3x

BD = BC + CD

⇒ √3x = x/√3 + y        [From equation (1)]

⇒ √3x = (x + √3y)/√3

⇒ 3x = x + √3y

⇒ 2x = √3y

∴ Required answer is Option 3

Shortcut Trick

Since the condition for α and β are not given 

So, we can take any value of α and β 

Let α = β = 45° we can take 0° also but not 90° because tan90° = ∞ 

Now, On putting α = β = 45° in the given equation, we get

(sec245° + tan 45° · tan 45° - tan2 45°)2 + (tan 45° - tan45°)2 - sec2 45° · sec245°

⇒ [(√2)² + 1 × 1 - 1]² + (1 - 1)2 - (√2)2 × (√2)2 

⇒ 4 + 0 - 4 = 0 

∴ The required answer is 0.

 

Alternate Method

Given:

(sec2α + tan α · tan β - tan2 α)2 + (tan α - tan β)2 - sec2 α · sec2 β

Formula Used:

sec²θ - tan²θ = 1

sec2θ = tan2θ + 1

Calculation:

We have (sec2α + tan α · tan β - tan2 α)2 + (tan α - tan β)2 - sec2 α · sec2 β

⇒ (sec2α - tan2 α + tan α · tan β)2 + (tan α - tan β)2 -  (tan2α  + 1)·(tan2β + 1)

⇒ (1 + tan α . tan β)2 + (tan α - tan β)2 -  (tan2α  + 1)·(tan2β + 1)

⇒ 1 + tan2α . tan2β + 2tan α·tan β + tan2α + tan2β - 2tan α · tan β -  tan2α . tan2β - tan2α - tan2β - 1

⇒ 0

∴ The required result is 0.

Given:

tan θ + sec θ = 3

Formula Used:

sec²θ - tan²θ = 1

(secθ + tan θ) (secθ - tan θ) = 1

Calculation:

We have 

tan θ + sec θ = 3      ----(i)

According to the formula used

(secθ + tan θ) (secθ - tan θ) = 1

⇒ 3 (secθ - tan θ) = 1

⇒ (secθ - tan θ) = 1/3      ----(ii)

On adding (i) and (ii), we get 

2secθ = 3 + 1/3 = 10/3

⇒ secθ = 5/3     ----(iii)

On putting the value of secθ in (i), we get

tanθ + 5/3 = 3 

⇒ tanθ = 3 - 5/3 = 4/3     ----(iv)

Now, we have to find the vlaue of 3 tan θ + 9 sec θ

So, from (iii) and (iv), we get 

3 × 4/3 + 9 × 5/3

⇒ 4 + 15 = 19

∴ The required value of 3 tan θ + 9 sec θ is 19.

Given:

1. \(\sqrt {{{\sec }^2}{\rm{θ }},{\rm{ + }},{\rm{cose}}{{\rm{c}}^2}{\rm{θ }}} , = ,\tan {\rm{θ }} + \cot {\rm{θ }}\) 2. \(\sqrt {{{\tan }^2}{\rm{θ }},{\rm{ + }},{\rm{co}}{{\rm{t}}^2}{\rm{θ }},{\rm{ + }},{\rm{4}}} , = ,\sec {\rm{θ }} + {\rm{cosec}},{\rm{θ }}\)

Concept used:

sec2θ - tan2θ = 1

cosec2θ - cot2θ = 1

cosecθ = 1/sinθ

(a + b)² = a² + b² + 2ab

Calculation:

Statement 1:

\(\sqrt {{{\sec }^2}{\rm{θ }},{\rm{ + }},{\rm{cose}}{{\rm{c}}^2}{\rm{θ }}} , = ,\tan {\rm{θ }} + \cot {\rm{θ }}\)

⇒ sec²θ + cosec²θ = (tanθ + cotθ)²

We know that, (a  + b)² = a² + 2ab + b²

⇒ sec2θ + cosec2θ = tan²θ + cot2θ + 2tanθ.cotθ

⇒ sec2θ - tan2θ + cosec2θ - cot2θ = 2tanθ × 1/cotθ

⇒ 1 + 1 = 2

⇒ 2 = 2 [Satisying]

Statement 2:

\(\sqrt {{{\tan }^2}{\rm{θ }},{\rm{ + }},{\rm{co}}{{\rm{t}}^2}{\rm{θ }},{\rm{ + }},{\rm{4}}} , = ,\sec {\rm{θ }} + {\rm{cosec}},{\rm{θ }}\)

⇒ tan2θ + cot2θ + 4 = (secθ + cosecθ)²

⇒ tan2θ + cot2θ + 4 = sec2θ + cosec2θ + 2secθ × cosecθ

⇒ tan2θ - sec2θ + cot2θ - cosec2θ + 4 = 2secθ × cosecθ

⇒ - 1 - 1 + 4 = 2secθ × cosecθ

⇒ 2 = 2secθ × cosecθ

⇒ 1 = secθ × cosecθ

⇒ sin θ cos θ = 1 which is not possible because 0 < θ < 90°

∴ Only 1 is the correct statement.

Given:

The radius of the circle is 1 unit 

Angle subtends by the chord at the center = 2θ 

Concept Used:

Sinθ = Perpendicular/Hypotenuse

Calculation:

Construction: Draw a line OM ⊥ AB

Now, In ΔAOM and ΔBOM

OM = OM    [common side]

∠AMO = ∠BMO = 90° 

And OA = OB  [Hypotanuse of triangles]

So, ΔAOM ≅ ΔBOM   by R-H-S Rule of congruency

So, ∠AOM = ∠BOM = 2θ/2 = θ 

And AM = BM = AB/2    ----(i)

Now, In ΔBOM

sinθ = BM/OB = BM/1

⇒ BM = sinθ 

Now, from (i), we get 

⇒ AB/2 = sinθ 

⇒ AB = 2sinθ 

∴ The length of chord is 2sinθ.

Shortcut TrickGiven equation is tan2 θ + 3 sec θ - 9 = 0

How to think about trick:

Since R.H.S is a rational number 

So, L.H.S should be rational for this 3 secθ must be rational because tan2 θ and 9 are rational number

Now, we have to think that on which θ, secθ is rational so we get θ = 60° 

We can't take 0° or 90° according to the question 

Process:

Put θ = 60**°** in the given equation 

We get, LHS = tan2 60°  + 3 sec 60° - 9 

⇒ 3 + 3 × 2 - 9 = 0 = RHS

Now, We have to find the value of 12 cot2 θ + 3cosec θ, at θ = 60° 

⇒ 12 cot2 60° + cosec 60°

⇒ 12 (1/3) + 3 × 2/√3

⇒ 4 + 2√3 = (√3)² + (1)² + 2.(√3) .(1)

Using formula a² + b² + 2ab = (a + b)²

⇒ (√3 + 1)²

∴ The required value is (√3 + 1)2.

Given:

\(\frac{{\cos ,{\rm{θ }}}}{{\text{cosec},{\rm{θ }}, + ,1}}, + ,\frac{{\cos ,{\rm{θ }}}}{{\text{cosec},{\rm{θ }}, - ,1}}, = ,2\)

Formula Used:

Cosec²θ - 1 = cot²θ 

Calculation:

We have \(\frac{{\cos ,{\rm{θ }}}}{{\text{cosec},{\rm{θ }}, + ,1}}, + ,\frac{{\cos ,{\rm{θ }}}}{{\text{cosec},{\rm{θ }}, - ,1}}, = ,2\)

⇒ \(\frac{cosθ \ (cosecθ \ - \ 1) \ + \ cosθ \ (cosecθ \ + \ 1) }{cosec^2θ - 1} = 2\)

⇒ \(\frac{cosθ \ cosecθ \ - \ cosθ \ + \ cosθ \ cosecθ \ + \ cosθ }{cot^2θ } = 2\)

⇒ \(\frac{2cosθ \ cosecθ }{cot^2θ } = 2\)

⇒ cosθ × (1/sinθ) = cot²θ

⇒ cotθ = cot2θ

⇒ cotθ = 1 

⇒ θ = 45° 

Now, We have to find the value of sin4 θ + cos4 θ

Putting θ = 45°, we get 

⇒ sin4 45° + cos4 45°

⇒ (1/4) + (1/4) = 1/2

∴ The value of sin4 θ + cos4 θ is 1/2.