# Averages aptitude Questions

## Average

 Question 1
Distance between two stations A and B is 778km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of train during the whole journey.
 A 60 km/hr B 57 km/hr C 30.5 km/hr D 67.2 km/hr
Question 1 Explanation:
\begin{align} & {\text{The}}\,{\text{required}}\,{\text{average}}\,{\text{speed}}\,{\text{given}}\,{\text{by}}\,{\text{the}}\,{\text{formula}}, \cr & {\text{Average}}\,{\text{speed}} = \left[ {\frac{{2xy}}{{\left( {x + y} \right)}}} \right]km/hr \cr & {\text{Where}}, \cr & x = 84\,kmph \cr & y = 56\,kmph \cr & {\text{Average}}\,{\text{speed}} \cr & = \left[ {\frac{{\left( {2 \times 84 \times 56} \right)}}{{\left( {84 + 56} \right)}}} \right] \cr & = 67.2\,kmph \cr\end{align}

 Question 2
Distance between two stations A and B is 778km. A train covers the journey from A to B at 84km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of train during the whole journey.
 A 57 km/hr B 67.2 km/hr C 60 km/hr D 30.5 km/hr
Question 2 Explanation:
\begin{align} & {\text{Average}}\,{\text{speed}} \cr & = \left[ {\frac{{2xy}}{{\left( {x + y} \right)}}} \right]\,km/hr \cr & = \left[ {\frac{{\left( {2 \times 84 \times 56} \right)}}{{\left( {84 + 56} \right)}}} \right]\,km/hr \cr & = \left[ {\frac{{\left( {2 \times 84 \times 56} \right)}}{{140}}} \right]\,km/hr \cr & = 67.2\,km/hr \cr\end{align}
 Question 3
A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
 A 280 B 250 C 285 D 276
Question 3 Explanation:
Since the month begins with a Sunday, to there will be five Sundays in the month. \begin{align} & \therefore {\text{Required}}\,{\text{average}} \cr & = \left( {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \right) \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr\end{align}

 Question 4
Which one of the following numbers can be removed from the set S = {0, 2, 4, 5, 9} without changing the average of set S?
 A 2 B 0 C 5 D 4
Question 4 Explanation:

The average of the elements in the original set S is: (0+2+4+5+9) /5 =20 /5 =4.

If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}.

The average of the elements is,

(0+2+5+9) /4=16 /4 =4.

 Question 5
Find the average of first 97 natural numbers.
 A 37 B 48 C 49.5 D 49
Question 5 Explanation:
\begin{align} & {1^{st}}\,{\text{Method}}: \cr & {\text{Average}}\,{\text{of}}\,{1^{st}}\,n\,{\text{natural}}\,{\text{number}}\,{\text{is}}\,{\text{given}}\,{\text{by}} \cr & = \frac{{\left( {\frac{{\left[ {n \times \left( {n + 1} \right)} \right]}}{2}} \right)}}{n} \cr & {\text{Average}}\,{\text{of}}\,{1^{st}}\,97\,{\text{natural}}\,{\text{number}}\,{\text{is}}\,{\text{given}}\,{\text{by}} \cr & = \left\{ {\frac{{\left( {\frac{{\left[ {97 \times \left( {97 + 1} \right)} \right]}}{2}} \right)}}{{97}}} \right\} \cr & = 49 \cr & \cr & {2^{nd\,}}{\text{Method}}: \cr & {\text{These}}\,{\text{numbers}}\,{\text{are}}\,{\text{in}}\,AP\,{\text{series,}}\,{\text{so}}\,{\text{average}}. \cr & = \frac{{\left( {{\text{sum}}\,{\text{of}}\,{\text{corresponding}}\,{\text{term}}} \right)}}{2} \cr & = \frac{{\left( {1 + 97} \right)}}{2} = 49 \cr & Or,\,\frac{{\left( {2 + 96} \right)}}{2} = 49 \cr & Or,\frac{{\left( {3 + 95} \right)}}{2} = 49\,{\text{And}}\,{\text{so}}\,{\text{on}}. \cr\end{align}

 Question 6
Of the three numbers, the first is twice the second and the second is twice the third. The average of the reciprocal of the numbers is 7/72. The numbers are:
 A 36, 18, 9 B 24, 12, 6 C 20, 10, 5 D 16, 8, 4
Question 6 Explanation:
\begin{align} & {\text{Let}}\,{\text{three}}\,{\text{numbers}}\,{\text{be}}\,x,\,y,\,z. \cr & {\text{Given}}, \cr & x = 2y \cr & \Rightarrow x = 4z \cr & \Rightarrow y = 2z \cr & \Rightarrow z = z \cr & {\text{The}}\,{\text{average}}\,{\text{of}}\,{\text{reciprocal}}\,{\text{numbers}}\,{\text{is}}\,\frac{7}{{72}} \cr & \frac{{\left[ {\left( {\frac{1}{x}} \right) + \left( {\frac{1}{y}} \right) + \left( {\frac{1}{z}} \right)} \right]}}{3} = \frac{7}{{72}} \cr & \Rightarrow \frac{{\left( {yz + xz + xy} \right)}}{{3xyz}} = \frac{7}{{72}} \cr & \Rightarrow \left( {14{z^2}24{z^3}} \right) = \frac{7}{{72}} \cr & \Rightarrow 504 = 84z \cr & z = 6 \cr & {\text{So}},\,x = 4z = 4 \times 6 = 24, \cr & \Rightarrow y = 2z = 2 \times 6 = 12 \cr & {\text{Thus}}\,{\text{the}}\,{\text{numbers}}\,{\text{are}}\,24,\,12,\,6 \cr\end{align}
 Question 7
Ajay working in a Cellular company as a salesman. His monthly salary is Rs. 200. But he gets bonus as per given rule. If he sells simcards of Rs. X then his bonus will be [(x /100)2 +10]. In the first quarter of the year his average sale was Rs. 3000 per month. In the next 5 five month his average sale was Rs. 5000 per month and for next four month his average sale was Rs. 8000 per month. What is the average earning per month for the whole year?
 A Rs. 3750 B Rs. 3610 C Rs. 3350 D Rs. 3560
Question 7 Explanation:
\begin{align} & {\text{Bonus}}\,{\text{for}}\,{\text{the}}\,{\text{first}}\,{\text{three}}\,{\text{month}}, \cr & = \left[ {{{\left( {\frac{{3000}}{{100}}} \right)}^2} + 10} \right] \times 3 \cr & = Rs.\,2710 \cr & {\text{Bonus}}\,{\text{for}}\,{\text{the}}\,{\text{next}}\,{\text{five}}\,{\text{month}}, \cr & = \left[ {{{\left( {\frac{{5000}}{{100}}} \right)}^2} + 10} \right] \times 5 \cr & = Rs.\,12550 \cr & {\text{Bonus}}\,{\text{for}}\,{\text{the}}\,{\text{next}}\,{\text{four}}\,{\text{month}}, \cr & = \left[ {{{\left( {\frac{{8000}}{{100}}} \right)}^2} + 10} \right] \times 4 \cr & = Rs.\,25640 \cr & {\text{Total}}\,{\text{earning}}\,{\text{as}}\,{\text{bonus}}\,{\text{for}}\,{\text{whole}}\,{\text{year}}, \cr & = 2710 + 12550 + 25640 \cr & = Rs.\,40900 \cr & {\text{His}}\,{\text{average}}\,{\text{bonus}} \cr & = \frac{{40900}}{{12}} = Rs.\,3410 \cr & {\text{Thus}}\,{\text{his}}\,{\text{average}}\,{\text{earning}}\,{\text{for}}\,{\text{whole}}\,{\text{year}}, \cr & = 3410 + 200 \cr & = Rs.\,3610 \cr\end{align}

 Question 8
A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
 A Rs. 8 B Rs. 7.98 C Rs. 9 D Rs. 8.50
Question 8 Explanation:
Total quantity of petrol consumed in 3 years \begin{align} & = \left( {\frac{{4000}}{{7.50}} + \frac{{4000}}{8} + \frac{{4000}}{{8.50}}} \right){\text{litres}} \cr & = 4000\left( {\frac{2}{{15}} + \frac{1}{8} + \frac{2}{{17}}} \right){\text{litres}} \cr & = \left( {\frac{{76700}}{{51}}} \right){\text{litres}} \cr & {\text{Total}}\,{\text{amount}}\,{\text{spent}} \cr & = Rs.\,\left( {3 \times 4000} \right) \cr & = Rs.\,12000. \cr & \therefore {\text{Average}}\,{\text{Cost}} \cr & = Rs.\,\left( {\frac{{12000 \times 51}}{{76700}}} \right) \cr & = Rs.\,\frac{{6120}}{{767}} \cr & = Rs.\,7.98 \cr\end{align}
 Question 9
David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, mathematics, physics, chemistry and biology. What are his average marks?
 A 75 B 69 C 65 D None of these
Question 9 Explanation:
\begin{align} & {\text{Average}} \cr & = \frac{{\left( {76 + 65 + 82 + 67 + 85} \right)}}{5} \cr & = \frac{{375}}{5} \cr & = 75 \cr & {\text{Hence,}}\,{\text{average}} = 75 \cr\end{align}

 Question 10
The average of runs of a cricket player of 10 innings was 32. How many runes must be made in his next innings so as to increase his average of runs by 4?
 A 76 B 74 C 72 D 70
Question 10 Explanation:

Average after 11 innings = 36.

Required number of runs = ( 36 * 11) - (32 * 10).

=396 -320.

=76.

There are 10 questions to complete.