Quantitative Aptitude PYQs

Given:

Total number of girls = 500 + 550 + 650 + 600 + 650 = 2950

Total number of boys = 600 + 450 + 750 + 700 + 800 = 3300

Concept used:

Using the concept of ratio.

Calculation:

Required ratio = Total number of girls ∶ Total number of boys

⇒ Required ratio = 2950 ∶ 3300

⇒ Required ratio = 59 ∶ 66

∴ The ratio of girls and boys from all the schools taken together is 59 ∶ 66.

Given:

Ratio of two numbers = 7 : 13

HCF = 8

Concept used:

(i) If a and b are two numbers, then a × b = HCF × LCM

(ii) HCF of two numbers is the highest common factor of both of them. 

Calculations:

Let the numbers be 7x and 13x.

Then, H.C.F. = x

As we have given, x = 8

First number is,

⇒ 7x

⇒ 7 × 8 = 56

Second number is,

⇒ 13x

⇒ 13 × 8 = 104

a × b = HCF × LCM

⇒ 56 × 104 = 8 × LCM

⇒ LCM = 728

∴ The LCM is 728.

Given:

Total sum donated by class = Rs. 2209

Total number of student in class = Money donated by each student

Concept used:

Total sum = Total number of students in class × Money donated by each student.

Calculation:

Let each student donated Rs.

x. Total number of student in class = Money donated by each student

⇒ Total number of student in class = x

According to the question,

Total sum = Total number of student in class × Money donated by each student

⇒ 2209 = x × x

⇒ x² = 2209

⇒ x = √2209

⇒ x = 47

∴ The number of students in the class is 47.

Given:

ABCD is a cyclic quadrilateral.

∠ADC = 148°

Concept used:

(i) Angle formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle.

(ii) Sum of opposite angles of cyclic quadrilateral is 180°

Calculations:

 

In quadrilateral ABCD,

∠ADC + ∠CBA = 180°

⇒ 148° + ∠CBA = 180°

⇒ ∠CBA = 180° - 148° 

⇒ ∠CBA = 32°

In ΔABC,

∠BAC + 90° + ∠CBA = 180°

⇒ ∠BAC + 90° + 32° = 180° 

⇒ ∠BAC + 122° = 180° 

⇒ ∠BAC = 180° - 122°

⇒ ∠BAC = 58°

∴ The measure of the ∠BAC is 58°

Given:

ΔABC ~ ΔRPQ and \(\frac {ar (\Delta ABC)}{ar(\Delta RPQ)} = \frac 4 9\)

AB = 3 cm, BC = 4 cm and AC = 5 cm

Concept used:

When ΔABC is similar to ΔRPQ

\(\frac{{{\rm{area;of;\Delta ABC}}}}{{{\rm{area;of;\Delta RPQ}}}} = ;\frac{{{\rm{A}}{{\rm{B}}^2}}}{{{\rm{R}}{{\rm{P}}^2}}} = ;\frac{{{\rm{B}}{{\rm{C}}^2}}}{{{\rm{P}}{{\rm{Q}}^2}}} = ;\frac{{{\rm{A}}{{\rm{C}}^2}}}{{{\rm{R}}{{\rm{Q}}^2}}}\)

Calculation:

\(\frac{{{\rm{area;of;\Delta ABC}}}}{{{\rm{area;of;\Delta RPQ}}}} = ;\frac{{{\rm{A}}{{\rm{B}}^2}}}{{{\rm{R}}{{\rm{P}}^2}}} = ;\frac{{{\rm{B}}{{\rm{C}}^2}}}{{{\rm{P}}{{\rm{Q}}^2}}} = ;\frac{{{\rm{A}}{{\rm{C}}^2}}}{{{\rm{R}}{{\rm{Q}}^2}}}\)

⇒ 4/9 = BC²/PQ²

⇒ PQ² = (4² × 9)/4

⇒ PQ² = 36

⇒ PQ = 6 cm

∴ The value of PQ is 6 cm.

Given:

The area of the square is decreased by 19%.

Formula used:

Area of the square = side² = (diagonal)²/2

Diagonal of square = √2 × side

Percentage = (part value/original value) × 100

Calculation:

Let the side of the square is 10 cm.

Area of the square

⇒ 10 × 10

⇒ 100 cm²

Diagonal of square 

⇒ √2 × 10

The area is decreased by 19%, the new area is

⇒ 100 × (81/100)

⇒ 81

Diagonal of the new square

⇒ 81 = (diagonal)2/2

⇒ diagonal = 9√2

Diagonal decreased by

⇒ 10√2 - 9√2

⇒ √2 

Percentage decrease

⇒ (√2/10√2) × 100

⇒ 10%

∴ The percentage decreased in diagonal is 10%.

Given:

Total employees = 450

Females in department E = 60% of the total employees in department E

Calculations:

Total employees in department E = 450 × 20%

⇒ 90

Females in department E = 60% × 90

⇒ 54

Males in department E = 90 - 54

⇒ 36

∴ There are 36 male employees in department E.

Given:

Average weight of A, B, and C = 70 kg

Average weight of A, B, C, and D = 60 kg

Weight of E = Weight of D + 5 kg

Average weight of B, C, D and E = 59 kg

Concept used:

Sum of weight = Average × Number of observation

Calculation:

A + B + C = 70 × 3

⇒ A + B + C = 210 kg               .............(1)

A + B + C + D = 60 × 4

⇒ A + B + C + D = 240 kg        .............(2)

B + C + D + E = 59 × 4

⇒ B + C + D + E = 236 kg         ............ (3) Subtracrt (1) in (2),

(A + B + C + D) – (A + B + C) = 240 – 210

⇒ D = 30 kg

Weight of E = Weight of D + 5 kg

⇒ E = 30 + 5

⇒ E = 35 kg

B + C + D + E = 236                 [Using eqn (3)]

⇒ B + C + 30 + 35 = 236

⇒ B + C = 236 – 65

⇒ B + C = 171 kg                ------------(4)

Subtract (4) in (1)

(A + B + C) – (B + C) = 210 – 171

⇒ A = 39 kg

∴ The weight of A is 39 kg

Given:

Time to fill the tank by two pipes = 15 hours and 4 hours

Time to empty the tank by third pipe = 12 hours

Concept used:

If Pipe A take ‘x’ hours and Pipe B takes ‘y’ hours to fill a tank then assume the total capacity of the tank is equal to LCM of ‘x’ and ‘y’ or a multiple of them.

Formula used:

Efficiency = Total work/total time

Calculations:

Let the total capacity of the tank be LCM of 15, 4, and 12 i.e. 60 units.

The efficiency of the first pipe = 60/15

⇒ 4 units/hour

The efficiency of the second pipe = 60/4

⇒ 15 units/hour

The efficiency of the third pipe = 60/12

⇒ 5 units/hour

Combined efficiency of three pipes = 4 + 15 - 5

⇒ 14 units/hour

Time to empty the tank by all the pipes together = 60/14

⇒ 30/7 hours

∴ It takes 30/7 hours to fill the empty tank if all three pipes are opened simultaneously.

Shortcut Trick Total work = 60 unit

 

Total efficiency = 4 + 15 - 5 = 14

∴ Time is taken to fill the tank = 60/14 = 30/7 hours

Given:

Reduction on amount of bill before due date(in %) = 12%

Reduction on amount of bill before due date(in Rs.) = Rs. 2,100

Calculations:

Let the house tax be x.

x × 12% = 2,100

⇒ 0.12x = 2,100

⇒ x = 2,100/0.12

⇒ x = 17,500

∴ The amount of house tax was Rs. 17,500

Given:

The number is 75 × 73 × 78 × 76.

Concept used:

Divide each number individually by 34 and break it into a small number and find a number smaller than 34.

Calculation:

75 × 73 × 78 × 76 is divided by 34.

⇒ (75 × 73 × 78 × 76)/34

⇒ (75/34) × (73/34) × (78/34) × (76/34)

⇒ 7 × 5 × 10 × 8

⇒ (35 × 80)/34

⇒ (35/34) × (80/34)

⇒ 1 × 12

⇒ 12

∴ The remainder is 12.

Given:

Profit = 10%

New profit% = 40%

Concept used:

SP = (100 + Profit%)/100 × CP

Calculation:

Let the original CP of the article be x.

Case 1:

SP = (100 + Profit%)/100 × CP

⇒ SP = (100 + 10%)/100 × x

⇒ SP = 11x/10

Case2:

New CP is 20% less than the original CP and sold for Rs. 1000 more

New CP = x – 20% of x

⇒ New CP = x – 20/100 × x

⇒ New CP = 80/100 × x

⇒ New CP = 4/5 × x

According to the question,

(SP + 1000) – New CP = 40% of New CP

⇒ (11x/10 + 1000) – 4x/5 = 40% of 4x/5

⇒ 11x/10  – 4x/5 + 1000 = 40/100 × 4x/5

⇒ (11x – 8x)/10 – 8x/25 = -1000

⇒ 3x/10 – 8x/25 = -1000

⇒ (15x – 16x)/50 = -1000

⇒ -x/50 = -1000

⇒ x = Rs. 50000

∴ The Cost Price of the article is Rs. 50000.

Shortcut Trick 

Let the CP of the article = X

man sold it at 10% profit which means he sold it at 1.1X

⇒ Had he bought it at 0.8X and sold at 1.1X + 1000 

⇒ Means 1.1X+1000- 0.8X = 40 % of 0.8X

⇒ 0.3X+1000 = 0.32X

⇒ 1000 = 0.02X

⇒ X = 50000

∴ The Cost Price of the article is Rs. 50000.

Shortcut Trick

When the distance is constant, speed is inversely proportional to the time taken.

Time taken by the bus = 1 hr 15 min = 75 min

Time taken by the car = 45 min

Ratio of time taken (Bus : Car) = 75 : 45 = 5 : 3

Ratio of their speeds (Bus : Car) = 3 : 5

∴ The correct answer is 3 : 5.

Alternate Method

Distance = 60 km (Constant)

Bus

Time = 1 hr 15 min = 75 min

Car

Time = 45 min

Given:

Total distance = 60 km

Time taken by bus (T_B) = 1 hour 15 minutes = 75 minutes = 75/60 hours = 5/4 hours

Time taken by car (T_C) = 45 minutes = 45/60 hours = 3/4 hours

Formula Used:

Speed = Distance ÷ Time

Calculations:

⇒ Speed of bus (S_B) = 60 ÷ (5/4)

⇒ S_B = 60 × (4/5) = 48 km/hr

⇒ Speed of car (S_C) = 60 ÷ (3/4)

⇒ S_C = 60 × (4/3) = 80 km/hr

⇒ Ratio of speed of bus to speed of car = S_B : S_C

⇒ Ratio = 48 : 80

⇒ Ratio = 3 : 5

∴ The correct answer is 3 : 5.

Additional Information

Proportionality in Speed, Time, and Distance

If distance is constant, Speed ∝ 1 / Time. If speed is constant, Distance ∝ Time. If time is constant, Distance ∝ Speed.

Average Speed

Average Speed = Total Distance ÷ Total Time. If a body covers equal distances at speeds S₁ and S₂, then Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Relative Speed

When two objects move in the same direction, their relative speed is the difference of their individual speeds (S₁ − S₂). In opposite directions, it is the sum of their speeds (S₁ + S₂).

Given:

The numbers are 5, 6, 8, 10 and 12

Concept used:

The least number which is exactly divisible by some numbers equals the LCM of these numbers.

Calculation:

LCM of 5, 6, 8, 10 and 12 = 120

∴ The least number which is exactly divisible by 5, 6, 8, 10, and 12 is 120.

Given:

Side of cube = 9 cm

Cuboid Dimensions

Length = 5 cm, breadth = 13 cm, height = 31 cm

Formula used:

Volume of cube = (side)³

Volume of cuboid = length × breadth × height

Total surface area of cube = 6 × side²

Calculation:

Volume of Solid metallic cube

⇒ 9³

⇒ 729 cm³

Volume of Solid metallic cuboid

⇒ 5 × 13 × 31

⇒ 2015 cm³

Volume of Recast single cube

⇒ Volume of Solid metallic cube + Volume of Solid metallic cuboid

⇒ 729 + 2015

⇒ 2744 cm3

Side of Recast cube

⇒ (Side)³ = 2744

⇒ side = 14 cm

Total surface area of cube

⇒ 6 × 14 × 14

⇒ 1176 cm²

∴ The total surface area of recast cube is 1176 cm².

Given:

A = C/6

B = 2A

Average of A, B and C = 30

Formula used:

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

Calculations:

A = C/6

⇒ A : C = 1 : 6

⇒ C : A = 6 : 1

B = 2A

⇒ A : B = 1 : 2

C : A : B = 6 : 1 : 2

Let C, A and B be 6x, x, 2x respectively.

(A + B + C)/3 = 30

⇒ (x + 2x + 6x)/3 = 30

⇒ 9x/3 = 30

⇒ 3x = 30

⇒ x = 10

A = x = 10

C = 6x

⇒ 6 × 10

⇒ 60

C - A = 60 - 10

⇒ 50

∴  The difference between A and C is 50.

Given:

Number of girls in school:

A = 500, B = 550, C = 650, D = 600, E = 650

Total number of student in school:

A = 1100, B = 1000, C = 1400, D = 1300, E = 1450

Concept used:

Percentage of girls = Number of girls/Total number of student × 100

Calculation:

Percentage of girls in school:

A = 500/1100 × 100

⇒ A = 45.4%

B = 550/1000 × 100

⇒ B = 55%

C = 650/1400 × 100

⇒ C = 46.4%

D = 600/1300 × 100

⇒ D = 46.1%

E = 650/1450 × 100

⇒ E = 44.8%

∴ The percentage of girls the lowest in school E.

Given:

New selling price = Rs. 29 per kg

Old selling price = Rs. 25 per kg

Expenditure = constant

Formula used:

(i) Expenditure = Price × Quantity

(ii) Percentage difference = {(Old price - New price)/Old price} × 100

Calculations:

Let expenditure, initial selling price and initial quantity be C, P₁ and Q₁ respectively.

C = P₁ × Q₁

⇒ C = 25 × Q₁

⇒ Q₁ = C/25

Let final selling price and final quantity be P₂ and Q₂ respectively.

C = P₂ × Q₂

⇒ C = 29 × Q₂

⇒ Q₂ = C/29

Percentage difference = (Q₂ - Q₁)/Q₁ ×100

⇒ {(C/29 - C/25)/(C/25)} ×100

⇒ [{(25C - 29C)/(25 × 29)}/(C/25)] × 100

⇒ {–(4C/725)/(C/25)} × 100

⇒ {–(4C × 25)/725C} × 100

⇒ –13.79% ≈ –14%

∴ By 14% should a family reduce its consumption, so as to keep the expenditure the same as before.

Shortcut Trick

                         Price           Consumption

Old                   25                29

New                 29                25

Expenditure     25 × 29        29 × 25

Thus, family should reduce their consumption by 4 kg

% Reduction = (4/29) × 100 = 13.79 ≈ 14%

∴ The family should reduce the consumption by 14%.

Given:

\(\frac {\sin^2 30^\circ + \cos^2 60^\circ - \sec 35^\circ.\sin 55^\circ}{\sec 60^\circ + \rm cosec;30^\circ}\) 

Calculations:

\(\Rightarrow\frac {\sin^2 30^\circ + \cos^2 60^\circ - \sec 35^\circ.\sin 55^\circ}{\sec 60^\circ + \rm cosec;30^\circ}\)

\(\Rightarrow\frac {(1/2)^2 + (1/2)^2 - \sec 35^\circ.\sin (90 - 55)^\circ}{2 + 2}\)

\(\Rightarrow\frac {(1/4) + (1/4) - \sec 35^\circ.\cos 35^\circ}{2 + 2}\)

\(\Rightarrow\frac {(2/4) - (1/\cos 35^\circ).\cos 35^\circ}{4}\)

\(\Rightarrow\frac {(1/2) - 1}{4}\)

\(\Rightarrow\frac {-1}{8}\)

∴ The answer is -1/8

Given:

∠APB = 100°

O is the centre of the circle.

PA and PB are two tangents drawn from P outside the circle

Concept used:

A tangent to a circle forms a right angle with the circle's radius.

The Sum of all angles of a quadrilateral is 360°.

Calculation:

A tangent to a circle forms a right angle with the circle's radius

⇒ ∠OAP = ∠OBP = 90°

The Sum of all angles of a quadrilateral is 360°

⇒ ∠OAP + ∠OBP + ∠APB + ∠AOB = 360°

⇒ 90° + 90° + 100° + ∠AOB = 360°

⇒ ∠AOB = 360° - 280°

⇒ ∠AOB = 80°

In ΔAOB

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ 80° + x + x = 180°

⇒ 2x = 100°

⇒ x = 50°

∴ The value of ∠OAB is 50°.