Quantitative Aptitude PYQs

Given:

Three-digit numbers (all digits are different) which are divisible by 7 and also divisible by 7 on reversing the order of the digits

Concept Used:

For divisibility of 7 is the last term is 1, 3, 7 and 9

Calculation:

Suppose the possible number is ‘abc’ then ‘cba’ will also be divisible by 7.

We can write:

100a + 10b + c = 7x

100c + 10b + a = 7y

Subtracting both equations:

99a – 99c = 7(x – y)

99(a – c) = 7(x – y)

Here, ‘99(a – c)’ must be divisible by 7.

Since, 99 is not divisible by 7, (a – c) will be divisible by 7.

Hence, only possible values of (a – c) are 0 and 7.

If (a – c) = 0 then a = c [Which is not possible as all digits are different]

If (a – c) = 7

Possible values of a = 9, 8 & c = 2, 1 [c = 0 is not possible as it will give 2-digit number]

Hence, multiples of 7 in the form of ‘abc’ with condition a = 9, 8 & c = 2, 1 are:

168, 259 & their reverse i.e. 861, 952.

∴ Total four numbers are possible. 

Confusion Points In the question it is mentioned that the digits should not be repeated.

Given:

x and y satisfy the equation 5x + 9y = 7

Formula Used:

In A.P. 

Last term(l) = First term(a) + [Number of terms(n) - 1]⋅Common difference(d)

l = a + (n - 1)d

Then n = [(l - a)/d] + 1

Calculation:

We need to find those integer values of x which will give integer values of y when we put them in the equation.

Put x = -1 y = 12/9 (Not an integer)

Put x = -2 y = 17/9 (Not an integer)

Put x = -4 ⇒ y = 3 (Integer)

Put x = -5 y = 22/9 (Not an integer)

Put x = -13 ⇒ y = 8 (Integer)

Put x = -22 ⇒ y = 13 (Integer)

So here we can observe that values of x (starting with -1) with gap of 9 are giving the integer values for y.

So basically, we need to find the all the multiples of 9 between -500 and 500.

AP will be: -495, -486, …, 486, 495.

Number of terms = [495 + 495]/9 + 1 = 110 + 1 = 111

∴ 111 integral values of x and y will satisfy the equation 5x + 9y = 7.

Calculation:

S = XYZ + YZX + ZXY

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

S = 111X + 111Y + 111Z

S = 111 (X + Y + Z)

We can observe here:

111 is divisible by 3 and 37.

Hence, S will be divisible by (X + Y + Z), 3 and 37 always. 

∴ 1 and 3 are correct

 

On Checking statement 2

take X = 1, Y = 2 & Z = 4 

We get S = 111 (1 + 2 + 4) = 777

Which is not divisible by 9.

Hence, S will not be always divisible by 9.

∴ 2 is incorrect.

Mistake Points 

At first, we have to understand here that 

We can't take x = 0 

If we take any one of them equal to zero then we will not get a three-digit number 

like if we take x = 0, y = 2 and z = 4 

then XYZ = 024 which is a two-digit number 

Some cases also generate divisible by 9, but in all cases not satisfy the condition. the Question said always divisible by 9, but it is not possible for all cases. Like S = 111(1 + 2 + 4) = 777, or S = 111(2 + 3 + 0) = 555 both number is not divisible by 9.

Satisfy cases, S = 111(1 + 2 + 3) = 666, S = 111(2 + 3 + 4) = 999 both number also divisible. 

Confusion PointsX, Y and Z are not given equal to each other, hence we cannot consider this special case to answer the question.

Given:

Average speeds of X and Y are in the ratio 4 : 5

X takes 45 minutes more than Y to reach the destination

Formula Used:

Speed = Distance/Time

Calculation:

Hence, Ratio of time taken by X and Y to reach the destination = 5 : 4

Suppose time taken by X and Y to reach the destination are 5x and 4x respectively.

X takes 45 minutes more than Y to reach the destination.

(5x – 4x) = x = 45

So, 4x = 45 × 4 = 180

∴ Time taken by Y to reach the destination is 180 minutes.

Given: 

The sum is S and product is P.

Formula Used:

Harmonic mean of a and b = 2/(1/a + 1/b) = 2ab/(a + b)

Calculation:

The sum is S and product is P.

Then, the Harmonic mean of these two observations = 2P/S

∴ H.M. of two observations is 2P/S

Given:

Annual income of X is 20% more than that of Y;

The income of Y is less than that of X by p%.

Concept used:

If smaller value is m% less than the greater value,

then the value of m = [(greater value - smaller value)/greater value] × 100

Calculation:

Suppose annual income of Y = 100

So, annual income of X = 100 × 1.2 = 120

Income of Y is less than that of X by p%

Hence,

Required percentage (p) = [(120 – 100)/120] × 100 = \(16\frac{2}{3}\)%

∴ value of p is  \(16\frac{2}{3}\)%

Calculation:

First, we need to get the LCM of 3, 4, 5, 6, and 7.

After having lcm of it we gets 3 × 2 × 2 × 5 × 7 = 3 × 4 × 5 × 7 = 420

We see here that 2 has already it's squares but 3, 5, and 7 don't have their squares

Since we have to find the perfect square

Therefore to get the perfect square we must make 3, 5, and 7 in the form of a square.

To have the squares of these three digits we have to multiply the lcm by 3, 5 and 7

After multiplying we get 420 × 3 × 5 × 7 = 44100

Hence this will be the perfect square which will be divisible by the digits 3, 4, 5, 6, and 7.

Given:

Time in which both outlets emptied the tank together = 20 minutes

Time in which 1^st outlet emptied the tank alone = 30 minutes

Concept Used:

1/(A's time) + 1/(B's time) + ...... = 1/Total time

Calculation:

Suppose 2^ndoutlet can empty the tank alone in X minutes;

So,

1/30 + 1/X = 1/20

⇒ 1/X = 1/60

⇒ X = 60

∴ Time taken to empty the tank by second outlet alone in 60 minutes

Given:

If (x² - 1) is a factor of ax⁴ + bx³ + cx² + dx + e

Concept Used:

If (x - a) is a factor of any polynomial equation 

So. x - a = 0, x = a is satisfy that equation

Cslculation:

(x² - 1) is the factor of ax⁴ + bx³ + cx² + dx + e.

(x² - 1) = (x – 1) (x + 1)

So, x = 1, -1 will satisfy the given equation:

Putting x = -1:

a – b + c – d + e = 0

⇒ a + c + e = b + d

∴ Option (4) is corret.

Given:

\(\left( {{x^8} + \frac{1}{{{x^8}}}} \right) = 47\)

**Formula Used:**If (a + 1/a)2 = P

Then, (a + 1/a) = √(P + 2)

Calculation::

\(\left( {{x^8} + \frac{1}{{{x^8}}}} \right) = 47\)

So,

(x⁴ + 1/x⁴)² = (47 + 2)

(x4 + 1/x4)​ = 7

and

(x² + 1/x²)² = (7 + 2) 

(x2 + 1/x2) = 3

Taking cube both sides:

\(\left( {{x^6} + \frac{1}{{{x^6}}}} \right)\) + 3 × 1 × (x² + 1/x²) = 27

So,

\(\left( {{x^6} + \frac{1}{{{x^6}}}} \right)\)= 27 – 3 × 3 = 18

∴ The value of \(\left( {{x^6} + \frac{1}{{{x^6}}}} \right)\) is 18.

Given:

A wheel makes 360 revolutions in one minute

Concept Used:

In one complete revolution radian covered = 2π

Revolutions in 1 minute or 60 seconds = 360

So, revolutions in 1 second = 360/60 = 6

Calculation:

In one complete revolution radian covered = 2π

∴ In 6 revolutions radian covered = 6 × 2π = 12π

∴ Number of radians the wheel turns in one second is 12π

Given: 

25 cosec²x + sec²x

Formula Used:

Cosec²x = 1 + Cot²x

Sec²x = 1 + Tan²x

Calculation:

25 cosec2x + sec2x

⇒ 25(1 + Cot²x) + (1 + Tan²x)

⇒ 26 + (25Cot²x + Tan²x)

∴ Minimum value = 26 + 2√(25 × 1) = 26 + 10 = 36

∴ Minimum value is 36

Given:

\(\alpha = {\rm{\Pi }}_{n = 1}^{99}\cot n\theta \)

⇒ α = Cotθ × Cot2θ × Cot3θ × Cot4θ × ………….. × Cot 97θ × Cot 98θ × Cot 99θ

And 90° = 100θ

Formula Used:

Tanθ = Cot(90° – θ)

Calculation:

Tan θ = Cot (100θ – θ) = Cot 99θ

Tan 2θ = Cot (100θ – 2θ) = Cot 98θ

And so on.

Hence, we can write the above expression as:

⇒ α  = Cotθ × Cot2θ × Cot3θ × Cot4θ × …… Cot 49θ × Cot 50θ × Tan 49θ ………. × Tan 3θ × Tan 2θ × Tan θ

⇒ α  = Cot 50θ

Since, 100θ = 90°

So, 50θ = 45°

⇒ α = Cot 45° = 1

∴ The value of \(\alpha = {\rm{\Pi }}_{n = 1}^{99}\cot n\theta \) is 1.

Given:

tan6θ = cot2θ

Formula Used:

cot θ = tan (90° - θ)

Calculation:

tan6θ = tan(90° – 2θ)

⇒ 6θ = 90° – 2θ

⇒ 8θ = 90°

⇒ 4θ = 45° [Here 6θ = 67.5° which satisfies the given condition] 

∴ The valure of sec 45° is \(\sqrt 2 \).

Given:

Height = 15 m

Makes angle = 30° 

Formula Used:

sin θ = perpendicular/hypotenius

sin 30° = 1/2

Calculation:

From the figure:

In triangle ABC:

Sin30° = AB/BC

⇒ 1/2 = AB/BC

⇒ BC = 2AB

Total height of the tree = AB + BC = 15 m

Hence, 3AB = 15

⇒ AB = 5 m

∴ The height from the ground to the point where tree is broken is 5 m

Given:

Two vertical towers distance =100 feet

The distance of the pole from the shorter tower = 75

The shorter tower length = 40 feet 

 A pole length = 6 feet

Formula Used:

tan θ = Perpendicular/Base

Calculation:

In triangle FGH:

GH = 40 – 6 = 34 feet

HF = DE = 75 feet

So, tanx = GH/HF = 34/75

In triangle ABF:

BF = CE = 100 + 75 = 175 feet

So, tanx = AB/BF = AB/175

Hence,

34/75 = AB/175

⇒ AB = 7/3 × 34 = 79.33 feet

∴ Height of taller tower = AC = 79.33 + 6 = 85.33 or 85 feet (Approximately)

Shortcut Trick

If the angles of elevation from two points at distances p and q from the base of a tower are complementary (θ₁ + θ₂ = 90°), the height (h) is always the geometric mean of the distances.

Given angles: 27° and 63° ⇒ Sum = 27° + 63° = 90°.

Shortcut Formula: h = √(p × q)

∴ The correct answer is √pq.

Alternate Method

Given: Distances from base are p and q. Angles are 27° and 63°.

Formula Used: tan θ = Perpendicular / Base

⇒ In first triangle: tan 27° = h / q

⇒ In second triangle: tan 63° = h / p

⇒ Since 63° = 90° − 27°, tan 63° = cot 27°

⇒ cot 27° = h / p ⇒ tan 27° = p / h

⇒ Equating tan 27°: h / q = p / h

⇒ h² = p × q

⇒ h = √pq

∴ The correct answer is √pq.

Additional Information

Complementary Angles Rule

If α + β = 90°, then tan α × tan β = 1. This property is extensively used in height and distance problems involving two points.

Geometric Mean in Trigonometry

When the product of two trigonometric ratios is constant, the unknown side (height) often results in a geometric mean √ab of the given lengths.

Trigonometric Identities

tan(90° − θ) = cot θ and cot θ = 1 / tan θ are the core identities for solving complementary angle problems.

Given:

sin2 6° + sin2 12° + sin2 18° + … + sin2 84° + sin2 90°

Formula Used:

sin (90° - θ) = cos θ

Calculation:

sin² 6° + sin² 12° + sin² 18° + … + sin² 78° + sin² 84° + sin² 90°

⇒ sin² 6° + sin² 12° + sin² 18° + … + sin² (90 – 12) + sin² (90 – 6) + sin² 90°

⇒ sin² 6° + sin² 12° + sin² 18° + … + Cos² 12° + Cos² 6° + sin² 90°

⇒ (sin² 6° + cos² 6°) + (sin² 12° + cos² 12°) + ………… (sin² 42° + cos² 42°) + sin² 90°

⇒ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

⇒ 8

∴ The value of sin2 6° + sin2 12° + sin2 18° + … + sin2 84° + sin2 90° is 8.

Given:

\(\frac{{\cos θ }}{{1 + \sin θ }} + \frac{1}{{\cot θ }}\)

Formula USed:

cot θ = cos θ/sin θ 

sin² θ + cos² θ = 1

Calculation:

\(\frac{{\cos θ }}{{1 + \sin θ }} + \frac{1}{{\cot θ }}\)

⇒ \(\frac{{\cos θ }}{{1 + \sin θ }} + \frac{{sin;θ }}{{\cos θ }}\)

⇒ \(\frac{{co{s^2}θ + sinθ + ;si{n^2}θ }}{{(1 + \sin θ );cosθ }}\)

⇒ \(\frac{{1 + sinθ }}{{(1 + \sin θ );cosθ }}\)

⇒ 1/cosθ

⇒ secθ

∴ The value of \(\frac{{\cos θ }}{{1 + \sin θ }} + \frac{1}{{\cot θ }}\) is sec θ.

Given:

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

Calculation:

Let x = \(\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} \) and y = \( \frac{{\sin \theta + 1}}{{\cos \theta }}\)

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

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

x = \(\frac{{(\sin \theta+ 1)^2 - \cos^2\theta }}{{(\sin \theta + \cos \theta)^2- 1^2}}\)

x = \(\frac{{(\sin^2 \theta+ 1+ 2\sin \theta) - \cos^2\theta }}{{(\sin^2 \theta + \cos^2 \theta+ 2\sin \theta \cos\theta)- 1}}\)

x = \(\frac{{(\sin^2 \theta+ 1+ 2\sin \theta) -(1- \sin^2\theta )}}{{(1+ 2\sin \theta \cos\theta)- 1}}\)

x = \(\frac{{2\sin^2 \theta+2\sin \theta}}{{ 2\sin \theta \cos\theta}}\)

x = \(\frac{{2\sin \theta(1+\sin \theta)}}{{ 2\sin \theta \cos\theta}}\)

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

and y = \( \frac{{\sin \theta + 1}}{{\cos \theta }}\)

∴ x - y = \(\frac{{1+\sin \theta}}{{ \cos\theta}}\) - \( \frac{{\sin \theta + 1}}{{\cos \theta }}\) = 0

∴ The value of \(\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} - \frac{{\sin \theta + 1}}{{\cos \theta }}\) is 0.