Quantitative Aptitude PYQs

Given:

**∠**P = 4 × ∠R

∠S = 3 × ∠Q

Concept Used:

Sum of the opposite angles in a cyclic quadrilateral is 180°

∠P + ∠R = 180° 

∠Q + ∠S = 180° 

Formula Used:

Average = sum of numbers/Total numbers

Calculation:

According to question

⇒ ∠P = 4 × ∠R 

⇒ ∠P : ∠R = 4 : 1

By using the property 

⇒ ∠P + ∠R = 180° 

⇒ 4x + x = 180° 

⇒ 5x = 180° 

⇒ x = 36° 

⇒ ∠R = 36° 

Now,

⇒ ∠S = 3 × ∠Q

⇒ ∠S : ∠Q = 3 : 1

⇒ 3y + y = 180°

⇒ 4y = 180° 

⇒ y = 45° 

⇒ ∠Q = 45° 

Average of ∠R and ∠Q is

⇒ (∠R + ∠Q)/2

⇒ (36° + 45°)/2

⇒ 81°/2 = 40.5° 

∴ The average of ∠R and ∠Q is 40.5°

Given:

TIme difference = 40 min

Formula Used:

Speed(S) = Distance(D)/Time(T)

Calculation:

Let the speed of the boy be x km/h

Distance = d

Intial time = t

New time = t1

Speed = Distance/TIme

⇒ x = d/t

⇒ t = d/x

New speed = x × (9/5) ⇒ 9x/5

⇒ 9x/5 = d/t2

⇒ t2 = 5d/9x

The difference between the time is 40mins

⇒ t - t1 = 40

⇒ (d/x) - (5d/9x) = 40

⇒ (9d-5d)/9x = 40

⇒ 4d/9x = 40

⇒ d/x = 40 × (9/4)

times = distance/speed

⇒ t = 90 

∴ The usual time take by the boy to go school is 90mins

We use ratio method 

Let initial speed be s1 and new speed be s2

and initial time be t and new time be t1

⇒ s1 × (9/5) = s2

⇒ s1 : s2 = 5 : 9

As the distance is constant the reverse of the ratio speed will give as ratio of time

⇒ t : t1 = 9 : 5

⇒ t - t1 = 40

⇒ 9x - 5x = 40 

⇒ 4x = 40 ⇒ x = 10

⇒ initial time = t = 9x = 9 × 10

∴ The usual time take by the boy to go school is 90mins

Alternate Method  

Let the normal time taken by him = X

as he increases his speed by 9/5 times the time taken decreases by 5/9 times

New time taken = 5/9X

⇒ X - 5/9X = 4/9X = 40 min

⇒ X = 90 mins

∴ The usual time taken by the boy to go to school is 90mins

Formula used:

Average = Sum of observations ÷ Number of observations 

Calculation:

Average temperature on Monday, Wednesday, and Friday is 41°C

⇒ M + W + F = 41 × 3 = 123       -----(1)

Average temperature on Wednesday, Friday, and Thursday is 42°C

⇒ W + F + T = 42 × 3 = 126        -----(2)

According to the question,

Temperature on Thursday = 43°C

⇒ W + F = 126 - 43 = 83       [From Equation(2)]

⇒ M + 83 = 123                     [From Equation(1)]

⇒ M = 123 -83 = 40

∴ The temperature on Monday is 40°C.

Shortcut Trick

Equating areas of square (side a) and rectangle (L = a + 10, B = a − 5):

⇒ a² = (a + 10) × (a − 5) ⇒ a² = a² + 5a − 50

⇒ 5a = 50 ⇒ a = 10 cm

⇒ Perimeter = 2 × (L + B) = 2 × (20 + 5) = 50 cm

∴ The correct answer is 50 cm.

Alternate Method

Given:

Area of square = Area of rectangle

Length of rectangle (L) = Side of square (a) + 10

Breadth of rectangle (B) = Side of square (a) − 5

Area = a²
a
a

Area = L × B
L = a + 10
B = a − 5

=

Perimeter = 2 × (L + B)

Formula Used:

Area of square = a²

Area of rectangle = L × B

Perimeter of rectangle = 2 × (L + B)

Calculations:

⇒ a² = (a + 10) × (a − 5)

⇒ a² = a(a − 5) + 10(a − 5)

⇒ a² = a² − 5a + 10a − 50

⇒ a² = a² + 5a − 50

⇒ 5a = 50

⇒ a = 10 cm

Now, calculate the dimensions of the rectangle:

⇒ Length (L) = 10 + 10 = 20 cm

⇒ Breadth (B) = 10 − 5 = 5 cm

Calculate the perimeter of the rectangle:

⇒ Perimeter = 2 × (20 + 5)

⇒ Perimeter = 2 × 25 = 50 cm

∴ The correct answer is 50 cm.

Additional Information

Rectangle Properties

Area = Length × Breadth | Perimeter = 2 × (Length + Breadth) | Diagonal = √(L² + B²).

Square Properties

Area = a² | Perimeter = 4a | Diagonal = a√2, where a is the side length.

Relationship between Area and Perimeter

For a given constant perimeter, a square will always enclose the maximum possible area compared to any other rectangle. For a given constant area, a square has the minimum possible perimeter.

Given:

In ΔABC,

∠A = 68°

Formula Used:

As we know incentre is the meet point of all the angle bisectors

 Incentre of a triangle = 90° + ∠A/2

Calculation:

Substituting the values in the formula 

 ∠BIC= 90°  + 68/2

⇒ ∠BIC = 124°

∴ ∠BIC = 124°

Concept used:

Calculation:

• 7 ÷ [5 + 1 ÷ 2 - {4 + (4 of 2 ÷ 4) + (4 ÷ 4 of 2)}]

⇒ - 7 ÷ [5 + 1 ÷ 2 - {4 + (8 ÷ 4) + (4 ÷ 8)}]

⇒ - 7 ÷ [5 + 1 ÷ 2 - {4 + 2 + 1/2)}]

⇒ - 7 ÷ [5 + 1/2 - 6.5]

⇒ - 7 ÷ [5.5 - 6.5]

⇒ -7/-1

∴ Answer is 7

Formula used:

a³ + b³ = (a + b) × (a² + b² – a × b)

Calculation:

[(4.35)³ + (3.25)³]/[(43.5)² + (32.5)² – 43.5 × 32.5]

⇒ [(4.35 + 3.25) × {(4.35)² +(3.25)² - (4.35 × 3.25)}]/[100 × {(4.35)² + (3.25)² – (4.35 × 3.25)}]

⇒ [4.35 + 3.25]/100

⇒ 7.6/100

⇒ 0.076

∴ The answer is 0.076

Total number of cars in 2018 

80 + 75 + 95 + 85 + 75 + 70 = 480

Total number of cars in 2019

110 + 105 + 65 + 95 + 95 + 80 = 550

Different between 2018 and 2019 

550 - 480 = 70 

Percentage less than the total sales cars from all the branches in the 2019 

⇒ \(\frac{{70}}{{550}} \times 100\)

⇒ 12.7

Shortcut Trick

Total Capacity (LCM of 15, 30, 40) = 120 units.

Efficiency: A = 8, B = 4, C = 3.

Work done by A (6 am to 10 am = 4 hrs) = 8 × 4 = 32 units.

Work done by B (8 am to 10 am = 2 hrs) = 4 × 2 = 8 units.

Remaining work at 10 am = 120 − (32 + 8) = 80 units.

Time from 10 am = 80 ÷ (8 + 4 + 3) = 80 ÷ 15 = 5 hours 20 minutes → 3:20 p.m.

∴ The correct answer is 3:20 p.m.

Alternate Method

Given: Filling times: A = 15 hrs, B = 30 hrs, C = 40 hrs. Opening times: A = 6 a.m., B = 8 a.m., C = 10 a.m.

Formula Used: Work = Time × Efficiency | Efficiency = Total Work ÷ Individual Time

⇒ Between 6 a.m. and 8 a.m. (2 hrs), only A is open.

⇒ Work done by A = 8 × 2 = 16 units.

⇒ Between 8 a.m. and 10 a.m. (2 hrs), A and B both are open.

⇒ Work done by A + B = (8 + 4) × 2 = 12 × 2 = 24 units.

⇒ Total work completed by 10 a.m. = 16 + 24 = 40 units.

⇒ Remaining work to be done after 10 a.m. = 120 − 40 = 80 units.

⇒ After 10 a.m., all three pipes A, B and C are open. Combined efficiency = 8 + 4 + 3 = 15 units/hr.

⇒ Additional time taken = 80 ÷ 15 = 5.33 hours = 5 hours (0.33 × 60) mins = 5 hours 20 minutes.

⇒ Final time = 10:00 a.m. + 5 hours 20 minutes = 3:20 p.m.

∴ The correct answer is 3:20 p.m.

Additional Information

Pipes and Cisterns - Negative Work

If an outlet pipe (leak) is present, its efficiency is taken as negative. Total efficiency = ∑(Filling efficiencies) − ∑(Emptying efficiencies).

Alternate Filling

If pipes are opened for 1 hour each in turn, calculate the work done in one full cycle (sum of efficiencies) and then find the number of cycles to reach total work.

Total Capacity Rule

Taking the LCM of individual times as the total capacity of the tank simplifies calculations by converting fractional work into whole number units of efficiency.

Fractional Time Conversion

To convert decimal or fractional hours to minutes, multiply the fractional part by 60. Example: 0.33 hours = 1/3 × 60 = 20 minutes.

GIven

x + y + z = 13,

x2 + y2 + z2 = 133

x3 + y3 + z3 = 847

Formula used :

 x3 + y3 + z3 - 3xyz = (x + y + z) ( x² + y² + z² - (xy + yz + zx) )      ----(1)

 (x + y + z)^{2 =} x2 + y2 + z²+ 2(xy + yz + zx)      ----(2)

Calculation:

From Equation (2) 

⇒ 13² = 133 + 2(xy + yz + zx)  

⇒ 169 - 133 = 2(xy + yz+ zx)

⇒ (xy + yz + zx) = 36/2

⇒ (xy + yz + zx) = 18

Put all the values in Equation (1); 

 x3 + y3 + z3 - 3xyz = (x + y + z) ( x2 + y2 + z2 - (xy + yz + zx) ) 

⇒ 847 - 3xyz = 13(133 - 18)

⇒ 847 - 115×13 = 3(xyz)

⇒ - 648 = 3(xyz)

⇒ - 216 = xyz 

⇒ - (216)1/3 = (xyz)^1/3

⇒ - (216)^1/3 = ∛xyz

 So, the value of   ∛xyz = - 6

Identity used:

(a³ + b³) = (a + b) × (a² + b² – ab)

Calculation:

\(\frac{{5.35 × 5.35 × 5.35 + 3.65 × 3.65 × 3.65}}{{53.5 × 53.5 + 36.5 × 36.5 - 53.5 × 36.5}}\)

\(\frac{({5.35 × 5.35 × 5.35 + 3.65 × 3.65 × 3.65)}}{{(53.5 × 53.5 + 36.5 × 36.5 - 53.5 × 36.5)}}\)

\(\frac{({5.35 × 5.35 × 5.35 + 3.65 × 3.65 × 3.65)}}{{(53.5 × 53.5 + 36.5 × 36.5 - 53.5 × 36.5)}}\)

\(⇒ {\frac{{(5.35 )^3 + (3.65)^3}}{{((53.5)^2 + (36.5)^2 - 53.5 × 36.5)}}}\)

\(⇒ \frac{{ {(5.35 : + :3.65) \times [(5.35)^2 + (3.65)^2 - 5.35 × 3.65}]}}{{100 \times [(5.35)^2 + (3.65)^2 - 5.35 × 3.65}]}\)

⇒ (5.35 + 3.65)/100

⇒ 9/100

∴ 0.09

Given:

The ladder leaning against the wall making an angle θ with the horizontal ground such that tan θ = 12/5. The height of the top of the ladder from the wall is 24 m

Concept Used:

tanθ = Height/Base

Calculation:

Accordingly,

Let the base = x that means the distance of the foot of the ladder from the wall is x

tanθ = 12/5

⇒ 24/x = 12/5

⇒ x = 10

∴ The distance of the foot of the ladder from the wall is 10 m

Given:

BD ⊥ AC at D.

E is a point on BC such that ∠BEA = x°

∠EAC = 62°

∠ EBD = 60°

Calculations:

In ΔAOD,

∠AOD + ∠ADO + ∠DAO = 180°

⇒ ∠AOD + 90° + 62° = 180°

⇒ ∠AOD = 180° - 152°

⇒ ∠AOD = 28°

∵ ∠AOD and ∠EOB are vertical opposite angles,

∠AOD = ∠EOB = 28°

In ΔBOE,

∠BEO + ∠EOB + ∠OBE = 180°

⇒ x° + 28° + 60° = 180°

⇒ x° = 180° - 88°

⇒ x° = 92°

∴ The value of x is 92°.

Given:

Radius = 14 cm

Volume = 6160 cm³

Formula used:

(i) Volume of cylinder = π × r² × h

(ii) Curved surface area = 2 × π × r × h

Calculation:

Volume = π × r² × h

⇒ 6160 = (22/7) × (14)² × h

⇒ h = (6160 × 7)/(22 × 196)

⇒ h = 10 cm

Curved surface area = 2 × π × r × h

⇒ 2 × (22/7) × 14 × 10

⇒ 2 × 22 × 2 × 10

⇒ 880 cm²

∴ The curved surface area is 880 cm²

Given:

Average of twelve numbers = 39

Average of last five numbers = 35

Average of first four numbers = 40

Fifth number = sixth number - 6

Fifth number = seventh number + 5

Formula used:

Average = (Sum of values)/(Number of values)

Calculations:

Let twelve numbers be x₁, x₂, x₃, x₄, x₅, x₆, x₇, x8, x9, x10, x11, x₁₂.

x5 = x6 - 6      ----(i)

x5 = x7 + 5      ----(ii)

Average of first four numbers,

(x1 + x2 + x3 + x₄)/4 = 40

⇒ (x1 + x2 + x3 + x4) = 160

Average of last five numbers,

(x₈ + x₉ + x₁₀ + x₁₁ + x₁₂)/5 = 35

⇒ (x8 + x9 + x10 + x11 + x12) = 175

Average of twelve numbers,

(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12)/12 = 39

⇒ (x1 + x2 + x3 + x4 + ​x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12) = 468

⇒ (160 + ​x5 + x6 + x7 + 175) = 468

⇒ ​x5 + x6 + x7 = 468 - 160 - 175

⇒ ​x5 + x6 + x7 = 133      ----(iii)

Putting value of x₅ from equation (i),

⇒ x6 - 6 + x6 + x7 = 133

⇒ 2x6 + x7 = 139

⇒ x7 = 139 - 2x6      ----(iv)

Putting value of x5 from equation (ii),

x7 + 5 + x6 + x7 = 133

⇒ 2x7 + x6 = 128

⇒ 2(139 - 2x6) + x6 = 128

⇒ 278 - 4x₆ + x₆ = 128

⇒ 3x₆ = 150

⇒ x₆ = 50

Putting the value of x₆ in equation (iv),

x7 = 139 - 2x6

⇒ x7 = 139 - 2(50)

⇒ x7 = 139 - 100

⇒ x7 = 39

Average of x₆ and x₇ = (50 + 39)/2

⇒ 89/2

⇒ 44.5

∴  The average of the sixth and seventh numbers is 44.5

Alternate MethodSum of values = Avg × Number of values

Sum of 12 numbers = 39 × 12 = 468

Sum of last 5 numbers = 35 × 5 = 175

Sum of first 4 numbers = 40 × 4 = 160

Sum of 9 numbers = 175 + 160 = 335

Sum of 5^th + 6^th + 7^th numbers = Sum of 12 numbers - Sum of 9 numbers

⇒ 468 - 335

⇒ 133

According to the question,

Let 5^th number be x.

Then, 6^th number = x + 6, and 7^th number = x - 5

Sum of 5^th + 6^th + 7^th numbers = 133

⇒ x + x + 6 + x - 5 = 133

⇒ 3x = 133 - 1

⇒ x = 132/3 = 44

Average of 6^th number + 7^th number = (x + 6 + x - 5)/2 

⇒ (2x + 1)/2

⇒ (2 × 44 + 1)/2

⇒ 89/2

⇒ 44.5

∴  The average of the sixth and seventh numbers is 44.5

Given:

4 - 2sin2θ - 5cosθ = 0

Formula used:

sin²θ + cos²θ = 1

Calculation:

4 - 2sin2θ - 5cosθ = 0

⇒ 4 - 2(1 - cos2θ ) - 5cosθ = 0

⇒ 4 - 2 + 2cos2θ - 5cosθ = 0

⇒ 2cos2θ - 5cosθ + 2 = 0

⇒ cosθ = [5 ± √(25 - 16)]/4

⇒ cosθ = (5 ± 3)/4

⇒ cosθ = 2 or 1/2

cosθ can't more than 1, cosθ = 1/2

tanθ = √3

cosθ - tanθ

⇒ (1/2) - √3

⇒ \(\frac{{1 - 2\surd 3}}{2}\)

∴ The value is \(\frac{{1 - 2\surd 3}}{2}\).

Given:

Total student in 2008, 2012, 2013 together from school A = 640 + 900 + 750

Total student in 2009, 2010, 2011 together from school B = 820 + 600 + 750

Concept used:

Using the concept ratio

Calculation:

Total student in 2008, 2012, 2013 together from school A = 640 + 900 + 750

⇒ Required sum of A = 2290

Total student in 2009, 2010, 2011 together from school B = 820 + 600 + 750

⇒ Required sum of B = 2170

Required ratio = 2290 ∶ 2170

⇒ Required ratio = 229 ∶ 217

∴ The required ratio is 229 ∶ 217.

Given:

Calculation:

Number of students in the class interval 165-170 = 14

Number of students in the class interval 150-155 = 15

Difference = 15 – 14 = 1

Required percentage = (1/15) × 100

⇒ Required percentage ≈ 6.7%

∴  The number of students in the class interval 165-170 is 100/15 = 6.7% less than the number of students in class interval of 150-155

Given:

Each discount = 20%

Single discount = Rs. 331.20

Concept used:

Using the concept of percentage.

Calculation:

Let the MP is Rs.

x. 1^st discount at MP = (100 – 20)% of x

⇒ 1^st discount = 80% of x

⇒ 1^st dicount = 80/100 × x

⇒ 1^st dicount = 0.8x

2^nd discount = (100 – 20)% of 0.8x

⇒ 2^nd discount = 80% of 0.8x

⇒ 2^nd dicount = 80/100 × 0.8x

⇒ 2^nd dicount = 0.8 × 0.8x

⇒ 2^nd dicount = 0.64x

Equivalent discount = x – 0.64x

⇒ 331.20 = 0.36x

⇒ x = 331.20/0.36

⇒ x = Rs. 920

∴ The marked price of an article is Rs. 920.

Given:

Distance between mall and person = 1215 m

Distance between tree or person = 60 m 

Height of tree = 20 m

Concept used:

Using the concept of trigonometric ratios

tanθ = (Perpendicular)/(Base)

Calculation:

Let the height of mall be AB

In ΔDEC,

tanθ = DE/CE

⇒ tanθ = 20/60

⇒ tanθ = 1/3

In ΔABC,

tanθ = AB/BC

⇒ 1/3 = AB/1215

⇒ AB = 405 m

∴ The height of mall is 405 m.