Quantitative Aptitude PYQs

Concept:

The highest power of the prime number p in n! = \(\displaystyle \frac{n}{p^1}+\frac{n}{p^2}+\frac{n}{p^3}.....\)

Calculation:

(30! + 35!)

⇒ (30! + 35 × 34 × 33 × 32 × 31 × 30!) 

⇒ 30! (1 + 35 × 34 × 33 × 32 × 31)

⇒ 30! (1 + 7 × 5 × 17 × 2 × 3 × 11 × 2⁵ × 31)

Now, the number of fives(5) in 30! is given by \(\displaystyle \frac{30}{5^1}+\frac{30}{5^2}+\frac{30}{5^3}= 6+1+0=7\)

So, 30! has 7 fives, and (1 + 7 × 5 × 17 × 2 × 3 × 11 × 25 × 31) has zero 5. 

∴ The max number of 5's is 7 i.e. n = 7

Concept:

1.1! + 2.2! + 3.3! + 4.4! + ..................... + n.n! = (n+1)! - 1

Calculation:

2(2 × 1) + 3(3 × 2 × 1) + 4(4 × 3 × 2 × 1) + 5(5 × 4 × 3 × 2 × 1) + ................. + 9(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) + 2

⇒ 2.2! + 3.3! + 4.4! + 5.5! + 6.6! + 7.7! + 8.8! + 9.9! + 2

⇒ (3 - 1).2! + (4 - 1).3! + (5 - 1).4! + (6 - 1).5! + (7 - 1).6! + (8 - 1).7! + (9 - 1).8! + (10 - 1).9! + 2

⇒ 3.2! - 2! + 4.3! - 3! + 5.4! - 4! + 6.5! - 5! + 7.6! - 6! + 8.7! - 7! + 9.8! - 8! + 10.9! - 9! + 2

⇒ 3! - 2! + 4! - 3! + 5! - 4! + 6! - 5! + 7! - 6! + 8! - 7! + 9! - 8! + 10! - 9! + 2

⇒ (3! + 4! + 5! + 6! + 7! + 8! + 9! + 10!) - (2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!) + 2

⇒ 10! - 2! + 2

⇒ 10! - 2 + 2

⇒ 10!

∴ 2.2! + 3.3! + 4.4! + 5.5! + 6.6! + 7.7! + 8.8! + 9.9! + 2 = 10!

Concept:

Power set B, of any set A, is a set of all the subsets of A and the number of elements of B will be 2ⁿ where n is the number of elements of A.

Calculation:

Given:

A = {(1, 2, 3)} here n = 1

∴ Number of elements in the power set of A = 2¹ = 2

Concept:

1). The general form of terms of a GP is a, ar, ar², ar³, and so on. Here, a is the first term and r is the common ratio.

       The nth term of a GP is T_n = ar^n-1

       Common ratio = r = T_n/ T_n-1

2). If a, b and c are three quantities in GP, then and b is the geometric mean of a and c. This can be written as b² = ac or b =√ac

Calculation:

Given:

a, b, c are in GP where a > 0, b > 0, c > 0

Statement I: a2, b2, c2 are in GP

As b is G.M. of a and c, b² = ac

Squaring on both sides 

⇒ b⁴ = a²c²

⇒ (b²)² = (a²)(c²) 

⇒ b² is G.M. of a² and c²

a², b², and c² are in GP

Statement II:  \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in GP

Let the common ratio be r.

⇒ b = ar

⇒ c = br = ar²

Now, \(\displaystyle \frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}=\frac{a}{ar}=\frac{1}{r}\)

⇒ \(\displaystyle \frac{\frac{1}{c}}{\frac{1}{b}}=\frac{b}{c}=\frac{ar}{ar^2}=\frac{1}{r}\)

1/a,1/b,1/c are in G.P.

Statement III:  \(√ {a}, √{b}, √{c} \) are in GP

As b is G.M. of a and c, ​\({b} = √{ac}\)​

Taking square roots on both sides 

⇒ \( √{b} = √{√{a} √{c}} \)

⇒ √b is GM of √a and √c

 \(√ {a}, √{b}, \ and\ √{c} \) are in GP

∴ Statement I, II, and III are correct.

Concept:

Arithmetic Progression:

If a, b and c are in AP then

b - a = c - b or 2b = a + c

Geometric Progression:

If a, b and c are in GP then

b/a = c/b  or b² = ac

Harmonic Progression:

If a, b and c are in HP then 

1/a, 1/b, and 1/c will be on AP.

Calculation:

Given:

\(\displaystyle \frac{a+b}{2}, b, \frac{b+c}{2}\) are in HP

⇒ \(\displaystyle \frac{2}{a+b}, \frac{1}{b}, \frac{2}{b+c}\) are in AP

⇒ \(\displaystyle \frac{1}{b}- \frac{2}{a+b}=\frac{2}{b+c}-​​​​\frac{1}{b}\)

⇒ \(\displaystyle \frac{1}{b}+\frac{1}{b}=\frac{2}{b+c}\ +​​​\frac{2}{a+b}\)

⇒ \(\displaystyle \frac{2}{b}=2\left[\frac{1}{b+c}\ +​​​\frac{1}{a+b}\right] \)

⇒ \(\displaystyle \frac{2}{b}=2\left[\frac{a+b+b+c}{(a+b)(b+c)}\right] \)

⇒ \(\displaystyle \frac{1}{b}=\left[\frac{a+2b+c}{(a+b)(b+c)}\right] \)

⇒ (a + b)(b + c) = ab + 2b² + bc

⇒ ab + ac + bc + b² = ab + 2b2 + bc

⇒ b2 = ac

Since b2 = ac, b is the geometric mean of a and c.

∴ a, b, c are in GP

Concept:

tan 15° = 2 - √3

cot 15° = 2 + √3

Formula used:

(a + b)² = a² + 2ab + b²

(a - b)2 = a2 - 2ab + b2

Calculation:

Let the required value is y.

⇒ y = cot2 15° + tan2 15°

⇒ y = (2 + √3)² + (2 - √3)²

Using the formula given above

⇒ y = (4 + 4√3 + 3) + (4 - 4√3 + 3) 

⇒ y = 8 + 6 = 14

∴ cot2 15° + tan2 15° = 14

Formula used :

sin 2θ = 2 sin θ cos θ 

cos C + cos D = 2cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))

Calculaton :

sin A - cos B - cos C = 0

⇒ sin A = cos B + cos C

Using the formula given above

⇒ 2 sin(A/2) cos(A/2) = 2cos(\(\frac{B+C}{2}\)) cos(\(\frac{B-C}{2}\))

We know that, for a ΔABC, ∠ A + ∠ B + ∠ C = π 

⇒ sin(A/2) cos(A/2) = cos(\(\frac{π - A}{2}\)) cos(\(\frac{B-C}{2}\))

Since, cos(π/2 - θ) = sin θ 

⇒ sin(A/2) cos(A/2) =  sin(\(\frac{ A}{2}\)) cos(\(\frac{B-C}{2}\))

⇒ cos(A/2) =  cos(\(\frac{B-C}{2}\))

⇒ A/2 = (B - C)/2

⇒ A = B - C

⇒ B = A + C

But ∠ A + ∠ B + ∠ C = π 

⇒ B = π - B

⇒ 2B = π 

⇒  B = π/2

∴ Angle  B is equal to \(\frac{\pi}{2}\).

Formula used:

\(tan(A-B) = \frac{tan\ A - tan \ B}{1 + tanA\ tanB}\)

\( tan\ 2β = \frac{2tanβ}{1-tan^2β}\)

Calculation:

Given that, 2 tan α = 1

⇒ tan α = 1/2           ------(1)

Also, we have \(α + β = \frac{π}{4}\)

⇒ \(β = \frac{π}{4}-α\)

\(⇒ tan\ β = tan(\frac{π}{4}-α)\)

\(⇒ tan\ β = \frac{tan(\frac{\pi}{4})-tan\alpha}{1+tan(\frac{\pi}{4})tan\alpha}\)       

By using the formula given above

\(⇒ tan\ β = \frac{1 -tanα}{1+tanα}\)                                    [∵ tan(π/4) = 1]

Put the value of tanα from equation (1)

\(⇒ tan\ β = \frac{1 -\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}\)

⇒ tan β = 1/3        -----(2)

We know that, 

 \( tan\ 2β = \frac{2tanβ}{1-tan^2β}​​\)

\( ⇒ tan\ 2β = \frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}=\frac{\frac{2}{3}}{1-\frac{1}{9}}\)

∴  tan 2β = \(\frac{3}{4}\)

Formula used:

tan(A + B) = \(\displaystyle \frac{tan\ A+\ tan\ B}{1\ -\ tan\ A.\ tan\ B}\)

sin 2θ = 2 sin θ cos θ   

Calculation:

tan (45° + θ) = 1 + sin 2θ

⇒ \(\displaystyle \frac{tan\ 45^o+\ tan\ θ}{1\ -\ tan\ 45^o.\ tan\ θ}\) = 1 + sin 2θ

⇒ \(\displaystyle \frac{1+\ tan\ θ}{1\ -\ tan\ θ}\) = 1 + sin 2θ

Since, tan θ = sin θ/cos θ 

⇒ \(\displaystyle \frac{cos\ θ+\ sin\ θ}{cos\ θ-\ sin\ θ}\) = 1 + sin 2θ

⇒ cos θ + sin θ = cos θ - sin θ + sin 2θ(cos θ - sin θ)

⇒ 2 sin θ = sin 2θ(cos θ - sin θ)

We know that, sin 2θ = 2 sin θ cos θ   

⇒ 2 sin θ = 2 sin θ. cos θ (cos θ - sin θ)

⇒ 1 = cos θ (cos θ - sin θ)

⇒ 1 = cos²θ - cos θ.sin θ

⇒ sin2θ = - cos θ.sin θ

⇒ sin2θ + cos θ.sin θ = 0

⇒ sin θ(sin θ + cos θ) = 0

⇒ sin θ = 0 or sin θ = - cos θ

⇒ sin θ = 0 or tan θ = -1

⇒ θ = 0 or θ = \(\displaystyle -\frac{\pi}{4}\)

As θ lies in \(\displaystyle -\frac{\pi}{4}< θ <\frac{\pi}{4},\) 

⇒ θ = 0^o

cos 2θ = cos (2 × 0) = 1

∴ cos 2θ = 1

Concept:

If A + B = 90° then

sin A = cos B

Calculation:

Given that, sin 2θ = cos 3θ

We know that sin A = cos B then A + B = 90° 

⇒ 2θ + 3θ = 90°  

⇒ 5θ = 90° 

⇒ θ = 18° 

Hence, required value 

1 + 4 sin θ  

⇒ 1 + 4 sin 18°   

⇒ 1 + 4((√5 - 1)/4)       [∵ \(\sin 18^\circ =\frac{√ 5 - 1}{4}\)]

⇒ 1  + √5 - 1 = √5

∴ 1 + 4 sin θ = √5

Concept:

Calculation:

Given that, 

\(\tan θ = - \frac{5}{12}\) = perpendicular/Base

In triangle ABC, using the Pythagoras theorem

AC² = AB² + BC²

⇒ AC² = 5² + 12² = 13²

⇒ AC = 13 

Hence, 

sin θ = perpendicular/hypotenuse

⇒ sin θ = 5/13

Since tan θ is negative in 2^nd and 4^th quadrants.

sin θ in 2^nd quadrants = 5/13

sin θ in 4^th quadrants = - 5/13

∴ Both 5/13 and -5/13 are possible.

Concept:

Formula used:

1. cos 2θ = 2cos²θ - 1
2. cos(2π - θ) = cos θ
3. cos(π + θ) = - cos θ
4. (a - b)² = a² + b² - 2ab
5. (a + b)²  = a² +b² + 2ab

 

Calculation:

Let the required value is y.

⇒ y = \(\cos^4\frac{7π}{8}+\cos^4\frac{5π}{8}\)

⇒ y =  \((\cos^2\frac{7π}{8})^2+(\cos^2\frac{5π}{8})^2\)

Using the formula discussed above

⇒ y = \(\Big(\frac{\cos( \frac{2\times7π }{8})+1}{2}\Big)^2+\Big(\frac{\cos (\frac{2\times5π }{8})+1}{2}\Big)^2\)

⇒ y = \((\frac{\cos( \frac{7π }{4})+1}{2})^2+(\frac{\cos (\frac{5π }{4})+1}{2})^2\)

⇒ y = \((\frac{\cos(2\pi- \frac{π}{4})+1}{2})^2+(\frac{\cos(\pi+\frac{π}{4})+1}{2})^2\)

Using the formula (2) & (3)

⇒ y = \((\frac{\cos (\frac{π}{4})+1}{2})^2+(\frac{1-\cos(\frac{π}{4})}{2})^2\)

Since, cos(π/4) = 1/√2

⇒ y = \((\frac{\ \frac{1}{\sqrt 2}+1}{2})^2+(\frac{1-\frac{1}{\sqrt 2}}{2})^2\)

⇒ y = \((\frac{\sqrt 2 +1}{2\sqrt2})^2+(\frac{\sqrt 2-1 }{2\sqrt2})^2\)

Using the formula (4) & (5)

⇒ y = \(\frac{ (2 + 1+2\sqrt 2)+(2 + 1 - 2\sqrt 2)}{8}\)

⇒ y = 6/8 = 3/4

∴ \(\cos^4\frac{7π}{8}+\cos^4\frac{5π}{8}\) = 3/4

Formula used:

cos(\(\frac{\pi}{2}-θ\)) = sin θ 

cos²θ - sin²θ = cos 2θ 

Calculation:

Let the required value be y.

⇒ y = \(\sin^2\left(\frac{\pi}{4}+θ\right)-\sin^2\left(\frac{\pi}{4}-θ\right)\)

Since cos(\(\frac{\pi}{2}-θ\)) = sin θ 

⇒ y = \(\cos^2\left(\frac{\pi}{2}-(\frac{\pi}{4}+θ)\right)-\sin^2\left(\frac{\pi}{4}-θ\right)\)

⇒ y = \(\cos^2\left(\frac{\pi}{4}-θ\right)-\sin^2\left(\frac{\pi}{4}-θ\right)\)

We know that, cos2θ - sin2θ = cos 2θ 

⇒ y = cos 2(\(\frac{\pi}{4}-θ\))

⇒ y = cos (\(\frac{\pi}{2}-2θ\)) = sin 2θ 

∴ \(\sin^2\left(\frac{\pi}{4}+θ\right)-\sin^2\left(\frac{\pi}{4}-θ\right)\) = sin 2θ

Given:

A flagstaff of height h is surmounted on a verticle tower.

Angles of elevation of the bottom and top of the flagstaff are θ and 2θ respectively.

Formula used:

1. tan θ = Perpendicular/Base

2. tan 2θ = \(\frac{(2tan θ)}{(1 - tan^2θ)} \)

3. cos 2θ = \( \frac{1-tan^2θ}{1+tan^2θ}\)

 

Calculation:

Let the height of the tower be x.

In right triangle ABC

tan θ = AC/AB = x/AB

⇒ AB = x/tan θ          ------(1)

In right triangle ABD

tan 2θ = AD/AB = (x + h)/AB

⇒ \(\frac{(2tan θ)}{(1 - tan^2θ)} \) = (x + h)/AB   [Using formula (2)]

⇒ \(\frac{(2tan θ)}{(1 - tan^2θ)} \) = \(\frac{ (x + h)tan\ θ}{ x} \)     [From equation (1)]

⇒ \(\frac{(2)}{(1 - tan^2θ)} \) = \(1+\frac{ h}{ x} \) 

⇒ \(\frac{h}{x} = \frac{2}{1-tan^2θ}-1\)

⇒  \(\frac{h}{x} = \frac{2-(1-tan^2θ)}{1-tan^2θ}\)

⇒  \(\frac{h}{x} = \frac{1+tan^2θ}{1-tan^2θ}\) 

⇒ \(\frac{x}{h} = \frac{1-tan^2θ}{1+tan^2θ}\) = cos2θ       [Using formula (3)]

⇒ x = hcos 2θ 

∴ The height of the tower is h cos 2θ

Formula used:

1. tan θ = Perpendicular/Base

2. tan θ increases with the increase of θ when θ ∈ (0, 90°)

• tan 30° = 1/√3 ≈ 0.57
• tan 45° = 1
• tan 60° = √3 ≈ 1.732

 

Calculation:

In Δ ABC, 

tan 60° = AB/BC = √3x/BC

⇒ √3 = √3x/BC

⇒ BC = x        -----(1)

In Δ  ABD

tan θ = AB/BD = √3x/(BC + x)

⇒ tan θ = √3x/2x

⇒ tan θ = √3/2  ≈ 0.86  

Since, 0.57 < 0.86 < 1

∴ 30° < θ < 45°

Concept:

tan-1x + tan-1y equals to

Case:1 \(tan^{-1}(\frac{x+y}{1-xy})\) if xy < 1

Case:2  \(π\ +\ tan^{-1}(\frac{x + y}{1-xy} )\) if x, y > 0 and xy > 1

Case:3  \(-π\ +\ tan^{-1}(\frac{x + y}{1-xy} )\)  if x, y < 0 and xy > 1

Calculation:

Here, \(\frac{1}{2}\times \frac{x}{3}\) < 1 as 0 < x < 6

Hence, we can follow the case 1

⇒ \(\tan^{-1} \left(\frac{1}{2}\right)+\tan^{-1} \left(\frac{x}{3}\right)=\frac{\pi}{4}\)

⇒ \(tan^{-1}(\frac{\frac{1}{2}+\frac{x}{3}}{1-\frac{1}{2}\times \frac{x}{3}}) = \frac{\pi}{4}\)

⇒ \((\frac{\frac{3+2x}{6}}{\frac{6 - x}{6}}) = tan(\frac{\pi}{4})\)     (∵ tan^-1x = θ ⇒ x = tan θ) 

⇒ \(\frac{3+2x}{6-x} = 1\)     (∵ tan\(\frac{\pi}{4}\) = 1)

⇒ 3 + 2x = 6 - x

⇒ 3x = 3 

∴  x = 1

Concept :

sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]

Calculation:

Given:  3 sin-1 x + cos-1 x = π 

 ⇒ 3 sin-1 x + cos-1 x = 2 sin-1 x + [sin-1 x + cos-1 x] = π 

As we know that, sin-1 x + cos-1 x =\(\frac{\pi}{2}\), x ∈ [-1, 1]

⇒ 2 sin-1 x + [\(\frac{\pi}{2}\)] = π

⇒ 2 sin-1 x = π - \(\frac{\pi}{2}\)

⇒ 2 sin-1 x = \(\frac{\pi}{2}\)

⇒ sin-1 x = \(\frac{\pi}{4}\)

∴ x = sin \(\frac{\pi}{4}\) = \(\frac{1}{\sqrt2}\)

Formula used:

tan(A + B) = \(\frac{tan \ A + tan\ B }{1-tan\ A\ tan\ B}\)

Calculation:

Given that,

tan α + tan β = 1 - tan α.tan β     -----(1)

By using the above formula, we can write

tan(α + β) = \(\frac{tan \ α + tan\ β }{1-tan\ α\ tan\ β}\)

⇒ tan(α + β)(1 - tanα tan β) = tan α + tanβ 

Using equation (1)

⇒ tan(α + β) (tan α + tan β) = (tan α + tanβ) 

⇒ tan(α + β) = 1

⇒ (α + β) = \(\frac{\pi}{4}\)

∴ The values of (α + β) is \(\frac{\pi}{4}\).

Concept:

If A + B = 45° then

(1 + tan A)(1 + tan B) = 2

Calculation:

Given that,

(1 + tan θ) (1 + tan 9θ) = 2

We know that, if A + B = 45° then

(1 + tan A)(1 + tan B) = 2

⇒ θ + 9θ = π/4

⇒ 10θ = π/4

Hence, the required value

tan 10θ = tan(π/4)

∴  tan 10θ = 1

Additional Information 

∵ tan (A + B) = tan 45° = (tan A + tan B) / (1 – tan A tan B)

⇒ 1 – tan A tanB = tan A + tan B

⇒ 1 + 1 = 1 + tan A  + tan B + tan A tan B

⇒ (1 + tan A) (1 + tan B) = 2

Formula used:

sin(360° - θ) = - sin θ  

Calculation:

sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360°

⇒ sin 0° + sin 10° + ....+ sin 180° + sin (360 - 170°) +  sin (360 - 160°) + .....+ sin (360 - 10°) + sin (360 - 0°)  

By using the  above formula

⇒ sin 0° + sin 10° + ......- sin 10° - sin 0° = 0

∴ sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360° = 0