Average Questions and answers

Given that, number of boys \( \Large n_{1} \) = 15

Number of girls \( \Large n_{2} \) = 10

Average weight of boys and girls = 38.4 kg

Average of boys = 40 kg

Let average of girls be \( \Large a_{2} \)

By formula, average = \( \Large \frac{n_{1}a_{1} + n_{2}a_{2}}{n_{1} + n_{2}} \)

=> \( \Large 38.4 = \frac{15 \left(40\right)+ 10 \times a_{2}}{15+10} \)

=> \( \Large 38.4 \times 25 = 600 + 10a_{2} \)

=> \( \Large a_{2} = \frac{900-600}{10} = \frac{360}{10} = 36 kg \)

Given that, total number of students = 150 = \( \Large \left(n_{1}+n_{2}\right) \) and mean weight of 150 students = 60 kg. The mean weight of boys a1 = 70 kg and the mean weight of girls \( \Large a_{2} \) = 55 kg Let \( \Large n_{1} \) and \( \Large n_{2} \) be the number of boys and girl, respectively.

Therefore, \( \Large n_{1} + n_{2} \) = 150 ...(i)

\( \Large11 n_{1} + 11n_{2} \) = 1650 ...(ii)

Average weight = \( \Large \frac{n_{1}a_{1} + n_{2}a_{2}} {n_{1} + n_{2}} \)

=> \( \Large 60 = \frac{70n_{1} + 55n_{2}}{150} \)

=> \( \Large 70n_{1} + 55n_{2} \) = 9000

=> \( \Large 14n_{1} + 11n_{2} = 1800 \) .....(iii)

On subtracting Eq. (ii) from Eq. (iii). we get

\( \Large 3n_{1} = 150 => n_{1} = 50 \)

Hence, required number of boys is 50

Monthly income of 4 persons = \( \Large 7350 \times 4 = Rs.29400 \)

Monthly income of 3 persons (excluding the dead person) = \( \Large 6500 \times 3 = Rs.19500 \)

Therefore, Monthly income of dead person = 29400 - 19500 = Rs.9900

Let the 5 consecutive odd positive integersbex+1,x+ 3,x+ 5,x+ 7,x+ 9

Average = Sum of all terms / Number of terms

\( \Large 9 = \frac{x+1+x+3+x+5+x+7+x+9}{5} = 9 : 5 = 5 \)

\( \Large 9 =  \frac{5x +25}{5} \)

=> 45 = 5x + 25 

5x = 45 - 25 = 20

5x = 20 => x = 4

The least one is x+ 1 = 4 + 1 = 5

\( \Large 47 \left(a+b\right) = 5452 \)

\( \Large a + b = \frac{5452}{47} = 116 \)

Therefore, Average value = \( \Large \frac{a+b}{2} = \frac{116}{2} = 58 \)

Here, n = 120, a = 35, x = 39, y =15

Number of students passed = \( \Large \frac{n \left(a-y\right)}{x-y} \)

= \( \Large \frac{120 \left(35-15\right) }{39-15} = \frac{120 \times 20}{24} \)

= 100 students passed

Here, P = 9 km, Q = 25 km, R = 30 km

x = 3 km/h, y = 5 km/h and z = 10km/h

Therefore, Required average speed

\( \Large \frac{P + Q + R}{ \LARGE \frac{P}{x}+\frac{Q}{y}+\frac{R}{z}} \)

\( \Large \frac{9 + 25 + 30}{ \LARGE \frac{9}{3}+\frac{25}{5}+\frac{30}{10}} \)

\( \Large \frac{64}{3+5+3} = \frac{64}{11} \)

\( \Large 5\frac{9}{11} km/h \)