Quantitative Aptitude PYQs

Given:

4 – 2sin2 θ – 5cos θ = 0

Identity used:

sin²θ = 1 – cos²θ 

Calculation:

4 – 2sin2 θ – 5cos θ = 0

⇒ 4 – 2 × (1 – cos2θ) – 5cosθ = 0

⇒ 4 – 2 + 2cos2θ – 5cosθ = 0

⇒ 2cos2θ – 5cosθ + 2 = 0

⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0

⇒ 2cosθ × (cosθ – 2) – 1 × (cosθ – 2) = 0

⇒ (2cosθ – 1) × (cosθ – 2) = 0

⇒ cosθ = 1/2 and cosθ = 2

Rejecting cosθ = 2 as 0° < θ < 90°

So, cosθ will be 1/2 

⇒ θ = 60° 

The value of cos θ + tan θ = cos60° + tan60° 

⇒ (1/2) + √3

∴ The value of cos θ + tan θ is \(\dfrac{1+2\sqrt{3}}{2}\)

Given:

In ΔABC, the bisectors of ∠B and ∠C meet at O. If ∠BOC = 142°

Concept Used:

The sum of three angles of a triangle is 180° 

Bisector of an angle bisects the angle in two equal parts.

Calculation:

Let, ∠B = x and ∠C = y

OB and OC are two angle bisectors and meets at O

In ΔBCO,

(x/2) + (y/2) + 142° = 180° 

⇒ (x/2) + (y/2) = 38° 

⇒ x + y = 76° 

In ΔABC,

∠B + ∠C = 76° 

∠A = 180° – 76° 

⇒ ∠A = 104° 

∴ The measure of ∠A is 104°

Given:

Percentage increase in the price of diesel = 16%

Percentage increase in expenditure = 10%

Concept used:

P × Q = E

Where P is Price, Q is Quantity and E is expenditure

Calculation:

Let initial price be P₁, quantity be Q₁, and expenditure be E₁ 

P₁ × Q₁ = E₁ 

⇒ Q1 = E₁/P₁

Let new price be P₂, new quantity consumed be Q₂, and new expenditure be E₂ 

P₂ = P₁ + 16% of P₁

⇒ P2 = 1.16P₁ 

E₂ = E₁ + 10% of E₁ 

⇒ E2 = 1.1E₁ 

P₂ × Q₂ = E₂ 

1.16P₁ × Q₂ = 1.1E₁ 

⇒ Q₂ = (1.1/1.16) × E₁/P₁ 

⇒ Q₂ = 0.948 × E1/P1 

⇒ Q2 = 0.948 × Q₁ 

Change in Quantity = Q₁ – 0.948 × Q1 

⇒ Change in Quantity = 0.052Q₁ 

Percentage change = (Change in Quantity/Initial Quantity) × 100

⇒ Percentage change = (0.052Q₁/Q₁) × 100

⇒ 5.2%

∴ Percentage change is 5.2%

Alternate MethodLet the price of diesel and quantity of diesel be Rs. 100 and 100 litres respectively

Total expenditure = Rs. 100 × 100

⇒ Rs. 10,000

Increased expenditure = Rs. 10,000 × 110%

⇒ Rs. 11,000

Increased price of diesel = Rs. 100 × 116%

⇒ Rs. 116

New consumption of diesel = 11,000/116

⇒ 94.82 ≈ 94.8 litres

Percentage decrease in consumption of diesel = [(100 – 94.8)/100] × 100

⇒ 5.2%

∴ Percentage change is 5.2%

Given:

1% = Rs. 10000

In 2005,

Strawberry = 18.90%, Butterscotch = 10.80%

In 2015,

Strawberry = 20.50%, Butterscotch = 16.60%

Calculation:

Total sales in 2005

⇒ 18.90 + 10.80

⇒ 29.70%

Total sales in 2015

⇒ 20.50 + 16.60

⇒ 37.10%

Percentage Increase = (37.10 – 29.70)

⇒ 7.4%

⇒ Rs. (7.4 × 10000)

⇒ Rs. 74,000

∴ The required increase from 2005 to 2015 is Rs. 74000

Given:

4(cosec2 57° – tan2 33°) - cos 90° – y tan2 66° tan2 24° \(=\dfrac{y}{2}\)

Formula used:

cosec²θ – cot²θ = 1

tan(90 – θ) = cotθ

Calculation:

4(cosec2 57° – tan2 33°) – cos 90° – y tan2 66° tan2 24° \(=\dfrac{y}{2}\)

⇒ 4[cosec2 57° – cot2(90° – 33°)] – cos 90° – y tan2 66° cot2(90° – 24°) \(=\dfrac{y}{2}\)

⇒ 4(cosec2 57° – cot2 57°) – cos 90° – y tan2 66° cot2 66° \(=\dfrac{y}{2}\)

⇒ 4 – 0 – y = y/2

⇒ 8 = 3y

⇒ y = 8/3

∴ The value of y is 8/3.

Given:

A's salary is 60% more than B's salary

Calculation:

Let the salary of B be 100

A's salary = 100 + 60% of 100

A's salary = 160

The salary of B is (160 – 100) = 60 less than A

Salary of B is {[60/(Salary of A)] × 100} percent less than A

⇒ Salary of B is {[60/160] × 100} percent less than A

⇒ 37.5%

∴ The salary of B is 37.5% less than the salary of A

Given:

A can do a job in 10 days

B can do a job in 5 days

Formula used:

Efficiency = Total work/Time taken

Calculation:

LCM of 10 and 5 = 10 = Total work

Efficiency of A = 10/10 = 1 unit/day

Efficiency of B = 10/5 = 2 units/day

A and B work for 2 days

⇒ (1 + 2) × 2 units

⇒ 6 units

Remaining work  = (10 – 6) units

⇒ 4 units

A and C can complete the remaining work in 3 days

A and C one day work

⇒ 4/3

C one day work

⇒ (4/3) – 1

⇒ 1/3

C can complete 60% of work in

⇒ 6 ÷ (1/3)

⇒ 18 days

∴ C can complete 60% of work in 18 days.

Given:

The number of students from school A in the year 2009, 2011 and 2013 is 800, 700, 750 respectively

The number of students from school B in the year 2008, 2012 and 2013 is 550, 500, 480 respectively

Concept Used:

Concept of ratio and proportion

Calculation:

The number of students from school A in the year 2009, 2011, and 2013 together is (800 + 700 + 750) = 2250

The number of students from school B in the year 2008, 2012, and 2013 together is (550 + 500 + 480) = 1530

The required ratio is 2250 : 1530

⇒ 25 : 17

∴ The ratio of the students taken for the years 2009, 2011, 2013 together from school A to the students taken for the years 2008, 2012, 2013 together from school B is 25 : 17

Concept used:

LCM of numbers

Calculation:

Take LCM of 15, 18, 36

    5, 1, 2

⇒ LCM = 2 × 3 × 3 × 5 × 1 × 2 = 180

⇒ LCM = 180

The required number which is exactly divisible by 11

⇒ 180x + 9

Put x = 6

⇒ 180 × 6 + 9

⇒ 1089 is divisible by 11.

∴ The required number is 1089.

Given:

Decreasing 110 by x% gives the same result as increasing 50 by x%

Concept Used:

Concept of percentage, it is calculated on the basis of 100

For example, x% means x out of 100

Calculation:

(100 - x) × (110/100) = (100 + x) × (50/100)

⇒ 1100 - 11x = 500 + 5x

⇒ 16x = 600

⇒ x = 75/2

Now, (75/2)% of 650 = 243.75

(75/2 + 20)% of 180 = 103.5

% more than {(243.75 - 103.5)/103.5} × 100%

⇒ 135.5%

⇒ 136% (correct to the nearest integer)

∴ The required percentage is 136%

Given:

1433 × 1433 × 1422 × 1425

Concept used:

Divide individually to get a smaller number than 12.

Calculation:

(1433 × 1433 × 1422 × 1425) ÷ 12

⇒ (1433/12) × (1433/12) × (1422/12) × (1425/12)

⇒ (5 × 5 × 6 × 9)/12

⇒ (1350/12) 

⇒ 6

∴ The remainder is 6.

Given:

The sides of the triangle are 24 cm, 26 cm, and 10 cm.

Concept Used:

The area made by three circles in three vertices of a triangle of the same radius will be half the area of the circle.

Calculation:

26² = 24² + 10²

The given triangle is a right angle triangle with the length of the base and height 24 cm and 10 cm and the length of the hypotenuse is 26 cm

Area of the triangle is (1/2) × 24 × 10

⇒ 120 cm²

The radius of the circle in three vertices is 4.2 cm

The total angle created in three circles in the vertices is 180°  [∵ Sum of three angles of a triangle is 180°]

The area of the triangle which is inside the common portion of the triangle and circle is (1/2) π (4.2)²

⇒ 27.72 cm²

The area of the triangle excluding the circle is (120 - 27.72) = 92.28 cm²

∴ The area excluding the portion covered by the section of the circles is 92.28 cm².

Given:

Investment of A = Rs. 13,750

Investment of B = Rs. 16,250

Investment of C = Rs. 18,750

B's share in profit = Rs. 5200

Calculation:

Ratio of investment 

A ∶ B ∶ C = 13,750 ∶ 16,250 ∶ 18,750

⇒ A ∶ B ∶ C = 11 ∶ 13 ∶ 15

Ratio of their profit will be same as ratio of their investment

Let profit of A be 11x, B be 13x and C be 15x

As, B's share = 5200

⇒ 13x = 5200

⇒ x = 5200/13 = 400

Total profit = 11x + 13x + 15x = 39x

⇒ Total profit = Rs. (39 × 400)

⇒ Rs. 15,600

∴ Total profit is Rs. 15,600

Given:

Given number is 785x3678y

The number is divisible by 72

Concept used:

If a number is divisible by 72 then the number will be divisible by 8 and 9

A number is divisible by 8 if the last three digits are divisible by 8.

A number is divisible by 9 if the sum of its digits is divisible by 9.

Calculation:

As, the number 785x3678y is divisible by 8

⇒ 78y is divisible by 8

For y = 4, the number is divisible by 8

As, the number 785x3678y is divisible by 9

⇒ 7 + 8 + 5 + x + 3 + 6 + 7 + 8 + y is divisible by 9

⇒ 44 + x + y is divisible by 9

⇒ 44 + x + 4 is divisible by 9

⇒ 48 + x is divisible by 9

Nearest integer to 48 is 54 which is divisible by 9

48 + x = 54

⇒ x = 6

x – y = 6 – 4

⇒ 2

∴ The value of (x – y) is 2

Given:

Seven years ago the ratio of ages of A and B = 4 : 5

After 8 years from present the ratio of ages of A and B = 9 : 10

Calculation:

Seven years ago the ratio of ages of A and B

⇒ 4x : 5x

Ratio of Present ages of A and B

⇒ 4x + 7 : 5x + 7

After 8 years the ratio of ages of A and B

⇒ (4x + 7 + 8) : (5x + 7 + 8) = 9 : 10

⇒ 40x + 150 = 45x + 135

⇒ 5x = 15

⇒ x = 3

Sum of present ages of A and B

⇒ (4 × 3 + 7) + (5 × 3 + 7)

⇒ 41 years

∴ The sum of present ages of A and B is 41 years.

Given:

A takes 2 hours 30 minutes more than B to walk 40 km. If A doubles his speed, then he can make it in 1 hour less than B.

Concept Used:

Time = Distance/ Speed

Calculation:

Let, Speed of A is x km/h

Time taken by A to cover the distance 40/x hour

Time taken by B (40/x - 2.5) hours

When A doubles his speed time taken by A 40/2x hour

Accordingly,

 [(40/x) – 2.5)] – (40/2x) = 1

⇒ (40/x) – (40/2x) = 3.5

⇒ 40/2x = 7/2

⇒ x = 40/7

Time taken by A to cover the distance 40 ÷ (40/7) 

⇒ 7 hours

∴ A needs 7 hours to walk a 40 km distance.

Given:

Discount 1 percent = 22%

Discount 2 percent = 17%

Discount 3 percent = 11%

Calculation:

Let the initial price be 100

Discount 1 = 22% of 100 

Discount 1 = 22

Price after discount 1 = 100 – 22 = 78

Discount 2 = 17% of 78

Discount 2 = 13.26

Price after discount 2 = 78 – 13.26 = 64.74

Discount 3 = 11% of 64.74

Discount 3 = 7.1214

Price after discount 3 = 64.74 – 7.1214 

Final price after third discount = 57.6186

Total discount = Initial price – Final price = 100 – 57.6186 = 42.3814

Total discount ≈ 42

Discount% ≈ (Total discount/Initial price) × 100

⇒ Discount% ≈ (42/100) × 100

⇒ 42%

∴ Discount percentage is approximately 42%

Given:

x4 + x-4 = 194

Identity used:

(a + b)² = a² + b² + 2ab

Calculation:

x4 + x-4 = 194

Adding 2 on both sides

x4 + x-4 + 2 = 194 + 2

⇒ (x² + 1/x²)² = 196 = 14²

⇒ x2 + 1/x² = 14

Again adding 2 on both sides

x2 + 1/x2 + 2 = 14 + 2

⇒ (x + 1/x)² = 16 = 4² 

∴ x + 1/x is 4

Given:

The average of twelve numbers = 39.

The average of the last five numbers = 35,

The average of the first four numbers = 40.

The fifth number is 6 less than the sixth number and 5 more than the seventh number. 

Concept Used:

Sum = Average × total number

Solution:

The Sum of the twelve numbers = 12 × 39 = 468

The sum of the last five number = 5 × 35 = 175

The sum of the first four number = 4 × 40 = 160

The sum of the fifth, sixth and seventh number = (468 – 175 – 160) = 133

Let, the fifth number is x

Then, the sixth number is (x + 6) and the seventh number is (x – 5)

x + (x + 6) + (x – 5) = 133

⇒ 3x + 1 = 133

⇒ 3x = 132

⇒ x = 44

The fifth number is 44 and the sixth number is (44 + 6) = 50

Average of fifth and sixth number is (44 + 50)/2 = 47

∴ The average of fifth and sixth number is 47

Given:

The single discount is Rs. 331.20 and the ed price is Rs. 920

Concept Used:

Selling price = Marked price - discount

Calculation:

1^st discount = x% of 920 = Rs. 9.2x

Have to pay amount Rs. (920 – 9.2x)

2^nd discount x% of Rs (920 - 9.2x) = (920 – 9.2x)x/100

Accordingly, 

9.2x + x(920 – 9.2x)/100 = 331.20

⇒ 9200x + 9200x – 92x² = 331200

⇒ 92x² – 18400x + 331200 = 0

⇒ x² – 200x + 3600 = 0

⇒ x² – 180x - 20x + 3600 = 0

⇒ (x – 180)(x – 20) = 0

x = 180 or x = 20

180% invalid as per the question

∴ The value of x is 20