Quantitative Aptitude Test - 15

The average age of 10 students = 20 years

Average age is increased by 10%, when teacher age is included.

Sum of the ages of 10 students = 200 years

⇒ 10% increased in average age = 10% of 20 = 2

⇒ New average age = 22 years

⇒ Sum of the ages of 10 students and teacher = 22 × (11) = 242 years

⇒ Age of teacher = 242 – 200 = 42 years

Number of employees = 6,000

Ratio of males and females employees = 2 : 3

Number of male employees: \(\frac{2}{5} × 6000 = 2400\)

Number of males employees get promotion = 30% of 2400= 720

Number of female employees: \(\frac{3}{5} × 6000 = 3600\)

⇒ Number of females employees get promotion = 50% of 3600= 1800

⇒ Number of employees not get promotion = 6000 – 720 – 1800= 3480

⇒ Percentage of employees who do not get promotion \(= \frac{3480}{6000} × 100\)= 58 %

Equation I:

\(x^{2}-6 x-112=0\)

\(\Rightarrow x^{2}-14 x+8 x-112=0\)

\(\Rightarrow x(x-14)+8(x-14)=0\)

\(\Rightarrow(x-14)(x+8)=0\)

\(\Rightarrow x=14,-8\)

Equation II:

\(y^{3}=3375\)

\(\Rightarrow {y}=15\)

∴ x < y

Equation I:

\(x^{2}-25 x+154=0\)

\(\Rightarrow x^{2}-14 x-11 x+154=0\)

\(\Rightarrow x(x-14)-11(x-14)=0\)

\(\Rightarrow(x-14)(x-11)=0\)

\(\Rightarrow x=14,11\)

Equation II:

\(y^{2}-24 y+135=0\)

\(\Rightarrow y^{2}-15 y-9 y+135=0\)

\(\Rightarrow y(y-15)-9(y-15)=0\)

\(\Rightarrow(y-15)(y-9)=0\)

\(\Rightarrow y=15,9\)

∴ Relationship between x and y cannot be established.

Equation I:

\(x^{2}+3 x-130=0\)

\(\Rightarrow x^{2}+13 x-10 x-130=0\)

\(\Rightarrow x(x+13)-10(x+13)=0\)

\(\Rightarrow(x-10)(x+13)=0\)

\(\Rightarrow x=10,-13\)

Equation II:

\(2 y^{2}-10 y-208=0\)

\(\Rightarrow y^{2}-5 y-104=0\)

\(\Rightarrow y^{2}-13 y+8 y-104=0\)

\(\Rightarrow y(y-13)+8(y-13)=0\)

\(\Rightarrow(y+8)(y-13)=0\)

\(\Rightarrow y=-8,13\)

∴ x = y or the relationship between x and y cannot be established.

Equation I:

\(x^{2}+50 x+600=0\)

\(\Rightarrow x^{2}+30 x+20 x+600=0\)

\(\Rightarrow x(x+30)+20(x+30)=0\)

\(\Rightarrow(x+20)(x+30)=0\)

\(\Rightarrow x=-20,-30\)

Equation II:

\(y^{2}+5 y-50=0\)

\(\Rightarrow y^{2}+10 y-5 y-50=0\)

\(\Rightarrow y(y+10)-5(y+10)=0\)

\(\Rightarrow(y-5)(y+10)=0\)

\(\Rightarrow y=5,-10\)

∴ x < y

The series follows following pattern:

9 × 1 + 2 = 11

10 × 2 + 3 = 23

11 × 3 + 4 = 37

12 × 4 + 5 = 53

13 × 5 + 6 = 71

14 × 6 + 7 = 91

∴ The value of ? is 91.

The pattern of the number series is:

⇒ 320 ÷ 2 + 2 =162

⇒ 162 ÷ 2 + 3 = 84

⇒ 84 ÷ 2 + 4 = 46

⇒ 46 ÷ 2 + 5 = 28

⇒ 28 ÷ 2 + 6 = 20

∴ The value of ? is 20.

The pattern of the number series:

⇒ 7 × 1 = 7

⇒ 7 × 2 = 14

⇒ 7 × 3 = 21

⇒ 7 × 4 = 28

⇒ 7 × 5 = 35

⇒ 7 × 6 = 42

∴ The value of ? is 28.

A given number is divisible by 11, if the difference of the sum of the odd place digit and the sum of even place digit is 0 or divisible by 11.

Calculation:

Sum of odd place digit = 5 + 7 + 2

⇒ 14

Sum of even place digit = 6 + 3 + x

⇒ 9 + x

⇒ x + 3 + 6 = 9 + x

⇒ 2 + 7 + 5 = 14

⇒ 14 = 9 + x

⇒ 14 - 9 = x

⇒ 5 = x

6 + 3 + 5 = 14,

The sum of the digit of the odd place digit and even place digit will be equal when x = 5.

∴ x = 5