Quantitative Aptitude PYQs

Given:

Other category of ice-cream is 15.70%

10% of the 'other' category is mix fruit flavor and 1570 people surveyed preferred mix fruit flavor

Concept Used:

Concept of percentage

Calculation:

10% of 15.70% is 1.570%

Let, the total number of people surveyed be x

Accordingly, 

1.570% of x = 1570

⇒ (1.570/100)x = 1570

⇒ x = 100000

∴ 100000 people were surveyed.

Given:

Principal (P) = Rs. 12,000

Rate of interest (R) = 10%

Amount (A) = Rs. 20,400

Formula used:

A = P + SI

SI = (P × R × T)/100

Where P is Principal, R is Rate of interest, T is Time and SI is Simple Interest

Calculation:

A = P + SI

20400 = 12000 + SI

⇒ SI = 20400 – 12000 = 8400

SI = (P × R × T)/100

⇒ 8400 = (12000 × 10 × T)/100

⇒ 8400 = 1200 × T

⇒ T = 8400/1200 years

⇒ 7 years

∴ Time is 7 years

Given:

Pipe A can fill in 16 hours

Pipe B can fill in 24 hours

Pipe C can empty in 40 hours

Concept used:

If a pipe can fill or empty in A hours, then it can fill or empty 1/A part in 1 hour.

Calculation:

Pipe A can fill 1/16 in 1 hour

Pipe B can fill 1/24 in 1 hour

Pipe C can empty 1/40 in 1 hour

Together they can fill 1/16 + 1/24 – 1/40 part in 1 hour

⇒ Together they can fill 19/240 parts in 1 hour

⇒ Together they can fill 190/240 = 19/24 part in 10 hours

Remaining tank = 1 – 19/24 = 5/24

B and C can do 1/24 – 1/40 = 2/120 = 1/60

B and C can fill a tank in 60 hours

B and C can fill 5/24 tank in 60 × 5/24 = 25/2 = \(12 \dfrac{1}{2}\) hours

∴ The remaining tank will be filled in \(12 \dfrac{1}{2}\) hours

Alternate Method

Calculation:

LCM of 16, 24, and 40 = 240 

Let the total work be 240 units

Efficiency of A = 240/16 = 15 units

Efficiency of B = 240/24 = 10 units

Efficiency of C = 240/40 = 6 units

When all three pipes were opened together for 10 hours,

Work done ⇒ (15 + 10 - 6) × 10 = 190 units

Remaining tank = 240 units - 190 units = 50 units

Now, Pipe A is closed and the remaining tank will be filled by B and C in = 50/(10 - 6) hours.

⇒ 50/4 = \(12 \dfrac{1}{2}\) hours.

Hence, the correct answer is \(12 \dfrac{1}{2}\) hours.

Given:

\((5√{5}x^3 - 3√{3}y^3) \div (√{5}x-√{3}y) = (Ax^2 + By^2+ Cxy)\)

Formula Used:

a³ - b³ = (a - b)(a² + ab + b²)

Calculation:

\((5√{5}x^3 - 3√{3}y^3) \div (√{5}x-√{3}y) = (Ax^2 + By^2+ Cxy)\)

\(⇒ (x√{5})^3 - (y√{3})3 \div (√{5}x-√{3}y) = (Ax^2 + By^2+ Cxy)\)

\(⇒ (√{5}x - √{3}y) (5x^2 + √15xy + 3y^2) \div (√{5}x-√{3}y) = (Ax^2 + By^2+ Cxy)\)

⇒ 5x² + √15xy + 3y² = Ax² + By² + Cxy

Comparing both sides get,

A = 5, B = 3 and C = √15

3A + B - √15C

⇒ 3 × 5 + 3 - √15 × √15

⇒ 15 + 3 - 15

⇒ 3

∴ The required value is 3.

Given:

The perimeter of square = Perimeter of rectangle

Perimeter of square = 40 m

Breadth of rectangle = Two-third of its length

Formula used:

Perimeter of rectangle = 2 × (Length + Breadth)

Area of rectangle = Length × Breadth 

Calculation:

Let the length rectangle be 3x

Breadth = 2/3 of 3x = 2x

Perimeter = 2 × (Length + Breadth)

⇒ 40 = 2 × (3x + 2x)

⇒ 20 = 5x

⇒ x = 4

Length = 3x = 12 m

Breadth = 2x = 8 m

Area = Length × Breadth 

⇒ Area = 12 × 8 = 96 m² 

∴ Area of rectangle is 96 m2

Given:

AQ = 3.5 cm

PC = 4.5 cm

BR = 7 cm

Concept used:

Tangents drawn from same point on the circle are equal.

Perimeter of triangle = Sum of all sides

Calculation:

As, tangents drawn from same point on the circle are equal.

AQ = AR = 3.5 cm

CQ = CP = 4.5 cm

BP = BR = 7 cm

Perimeter = AB + BC + CA

⇒ Perimeter = AQ + CQ + CP + BP + BR + AR

⇒ Perimeter = 3.5 + 4.5 + 4.5 + 7 + 7 + 3.5

∴ Perimeter is 30 cm

Given:

Calculation:

Number of students in the interval of 170-175 = 12

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

Difference = 14 – 12 = 2

Percentage = (Difference/Number of students in the interval of 165-170) × 100

⇒ Percentage = (2/14) × 100

⇒ Percentage ≈ 14.28 ≈ 14.3

∴ Students in the class interval of 170-175 are nearly 14.3% less than the students in the class interval of 165-170

Given:

Two numbers are in the ratio 7 ∶ 11. If their HCF is 28

Concept Used:

If the ratio of two numbers is a : b and their HCF is p then those two number will be ap and bp

calculation:

Let, the numbers be 7x and 11x

HCF of those number is x

Difference between two numbers is (11x - 7x) = 4x

The difference between the numbers is 4 × 28

⇒ 112

 ∴ the difference between the two numbers is 112.

Given:

A train x running at 74 km/h crosses another train y running at 52 km/h in the opposite direction in 12 seconds. The length of y is two-thirds that of x.

Concept Used:

When two trains crossing each other, their relative speed will be the sum of their speed.

1 km/h = (5/18) m/s

Distance = Time × speed

Calculation:

Relative speed (74 + 52) = 126 km/h

⇒ 126 × (5/18)

⇒ 35 m/s

Time taken to cross each other 12 second

Total distance covered (12 × 35) = 420 m

The length of y is two-thirds that of x

Let, the length of the train x is a meter

Accordingly,

(a + (2a/3) = 420    

⇒ a = 420 × (3/5)

⇒ a = 252

∴ The length of the train X is 252 m

Given:

∠APB = 128° 

Concept used:

Angle made by the point of contact of the tangent and the centre of the circle is 90°

The Sum of all angles of a quadrilateral is 360° 

In a triangle angles opposite to equal sides are equal.

Sum of all angles of triangle is 180° 

Calculation:

As, PA and PB are tangents 

∠OAP = 90° 

∠ OBP = 90° 

As, OAPB is a quadrilateral

∠OAP + ∠APB + ∠PBO + ∠BOA = 360° 

⇒ 90° + 128° + 90° + ∠BOA = 360° 

⇒ ∠BOA = 360° – 308° 

⇒ ∠BOA = 52° 

As, OA = OB (Radii)

In ΔOAB, ∠OAB = ∠OBA (In a triangle angles opposite to equal sides are equal)

∠OAB + ∠ OBA + ∠BOA = 180° 

⇒ 2 × ∠OAB + 52° = 180° 

⇒ ∠OAB = (180° – 52°)/2

∴ ∠OAB is 64°

Given:

The length of a rectangle is decreased by 11%

The breadth of the rectangle is decreased by 11%

Formula used:

Area of rectangle = Length × Breadth

Calculation:

Let the length of the rectangle be L and breadth be B

The initial area of rectangle = LB

As the length of the rectangle is decreased by 11%

Decreased length = L – 11% of L

⇒ Decreased length = 0.89L

As, the breadth of the rectangle is decreased by 11%

Decreased breadth = B – 11% of B

⇒ Decreased Breadth = 0.89B

New area of rectangle = 0.89L × 0.89B

⇒ New area of the rectangle = 0.7921LB

Total decrease in area = LB – 0.7921LB = 0.2079LB

Percentage decrease in area = (Total decrease/Initial area) × 100

⇒ Percentage decrease in area = (0.2079LB/LB) × 100 = 20.79%

∴ The percentage decrease in the area is 20.79%

Shortcut Trick

Decrease in area = -x + (-y) + (-x)(-y)/100

∴ Decrease in area = -11 - 11 + 121/100 = -20.79%

Concept used:

(a3 + b3) = (a + b) × (a2 + b2 – ab)

Calculation:

\(\frac{{6.35 × 6.35 × 6.35 + 3.65 × 3.65 × 3.65}}{{63.5 × 63.5 + 36.5 × 36.5 - 63.5 × 36.5}}\)

\(⇒ \frac{{(6.35 )^3 + (3.65)^3}}{{(63.5)^2 + (36.5)^2 - 63.5 × 36.5}}\)

\(⇒ \frac{{(6.35 )^3 + (3.65)^3}}{{(6.35 \times 10)^2 + (3.65 \times 10)^2 - 6.35 \times 10 × 3.65 \times 10}}\)

\(⇒ \frac{{ {(6.35 : + :3.65) \times [(6.35)^2 + (3.65)^2 - 6.35 × 3.65}]}}{100{[(6.35)^2 + (3.65)^2 - 6.35 × 3.65}]}\)

⇒ (6.35 + 3.65)/100

⇒ 10/100

⇒ 0.1

Given:

A sum amounts to Rs. 7,562 in 4 years and to Rs. 8,469.44 in 5 years at a certain rate percent per annum.

Concept Used:

If P be the principal, r be the rate of compound interest and t be time then amount after t year p(1 + r/100)^t

Calculation:

Let, P be the principal

P(1 + r/100)⁴ = 7562      ----(1)

P(1 + r/100)⁵ = 8469.44      ----(2)

Divided (2) by (1) get,

(1 + r/100) = 846944/756200

⇒ r/100 = 1.12 - 1

⇒ r = 12

∴ The rate of interest is 12%

Calculation:

Total number of students from school A

⇒ 640 + 800 + 500 + 700 + 900 + 750

⇒ 4290

Total number of students from school B

⇒ 550 + 820 + 600 + 750 + 500 + 480

⇒ 3700

Ratio of average of total students from school A to school B

⇒ (4290/6) : (3700/6)

⇒ 4290 : 3700

⇒ 429 : 370

∴ The required ratio is 429 : 370

Given:

The person is 1425 m away from the mall. The height of the tree is 10 m and it is 30 m away from him.

Concept Used:

 \(tanθ = {{Height} \over {Base}}\)

Calculation:

 

Let, the height of the mall (AB) = x m

From the figure,

CD = 10 m

DM = 30 m

BM = 1425 m

Let, ∠AMB = ∠CMD = θ 

From ΔAMB and ΔCMD get,

\({{AB} \over {BM}} = {{CD} \over {DM}}\)

⇒ x/1425 = 10/30

⇒ x/1425 = 1/3

⇒ x = 475

∴ The mall is 475 m tall.

Given:

A is twice as good a workman as B and together they finish a piece of work in 13 days.

Concept Used:

Efficiency ∝ 1/Time

Calculation:

The efficiency ratio of A and B is 2 : 1

The time ratio of A and B is 1 : 2

Let, A took time x days and B took time 2x days to complete the work alone

In 1 day A can do 1/x part of the work

In 1 day B can do 1/2x part of the work

Together in 1 day they can do 1/13 part of the work

(1/x) + (1/2x) = 1/13

⇒ 3/2x = 1/13

⇒ x = 39/2

B alone will finish the work in 2 × (39/2)

= 39 days

∴ In 39 days B alone will finish the work.

Given:

Radius of cylinder = 14 cm

The curved surface area of the cylinder = 880 cm²

Formula used:

The curved surface area of cylinder = 2πrh

The volume of cylinder = πr²h

Where,

r = radius of the cylinder

h = height of the cylinder

Calculation:

Height of the cylinder

⇒ 2πrh = 880

⇒ 2 × (22/7) × 14 × h = 880

⇒ h = 10 cm

The volume of the cylinder

⇒ (22/7) × (14)² × 10

⇒ 6160 cm³

∴ The volume of the cylinder is 6160 cm³.

Given:

Side of rhombus = 13 cm

D₁ (Diagonal) = 10 cm

Concept and Formula used:

Diagonals of rhombus bisect each other at 90° 

Area of rhombus = 1/2 × (D₁ × D₂)

Where D₁ and D₂ are diagonals of the rhombus

Calculation:

Let the length of other diagonal (D₂) be 2x cm

As, diagonals of rhombus bisect each other at 90° 

AB = 10/2 = 5 cm

BC = 2x/2 = x cm

In ΔABC, ∠ABC = 90° 

AC² = AB² + BC² 

⇒ 13² = 5² + x² 

⇒ 169 – 25 = x2 

⇒ x = 12 cm

D₂ = 2x = 24 cm

Area of rhombus = 1/2 × (D1 × D2) = 1/2 × (10 × 24)

∴ Area of rhombus is 120 cm²

Given:

cosθ = 5/13

The height of the top of the ladder from the foot of the wall is 18 m.

Concept Used:

Cosθ = Base/ Hypotenuse

Calculation:

cosθ = 5/13 = Base/ Hypotenuse

Let, Base (BC) = 5x and Hypotenuse (AC) = 13x where x is a constant

Accordingly, AB = 18 m

From ΔABC, 

18² + (5x)² = (13x)²

⇒ 18² = 169x² - 25x²

⇒ 324 = 144x²

⇒ x² = 324/144

⇒ x = 3/2

Base (BC) = 5 × 3/2

⇒ 7.5

∴ The distance of the foot of the ladder from the wall is 7.5 m.

Given:

Calculation:

Total number of students = 15 + 13 + 10 + 14 + 12 + 6

⇒ Total number of students = 70

Students in class interval 160-170 = 10 + 14 = 24

Percentage of students in class interval 160-170 = (24/Total students) × 100

⇒ Percentage of students in class interval 160-170 = (24/70) × 100 = 34.28

∴ The percentage of students whose height is in the class interval of 160-170 is 34% (correct to the nearest integer)