Quantitative Aptitude

Averages

1,527 Questions
All Basic average and reverse calculation 405 Adding or removing an observation 342 Weighted and combined averages 320 Replacement and correction of a recorded value 233 Ratio shortcut for two-group mixtures 111 Equal-distance average speed 59 Age, score, weight, and spending applications 32 Consecutive terms and arithmetic progressions 25

Averages Questions

Multiple choice

Directions: The following problem has a question and two statements labelled (1) and (2). You have to decide whether the statements are sufficient to answer the question. Mark your answer as

Mr. Parker earned Rs. 20,000 in 1967. What is Mr. Parker’s average yearly income for the three years 1997 to 1999? (1) He made 10% more in each year than he did in the previous year. (2) His total combined income for 1998 and 1999 was Rs. 46,200.

  1. if you can get the answer from (1) alone, but not from (2) alone

  2. if you can get the answer from (2) alone, but not from (1) alone

  3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices

  4. if (1) alone suffices and (2) alone suffices

  5. if you cannot get the answer from (1) and (2) together, and need even more data

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Correct option is (4).

Multiple choice statistics moving averages moving average and variation simple moving average uses of average in day-to-day life introduction to time series introduction to time series and forecasting

The average weight of $A,B,C$ is $45kg$ . If the average weight of $A$ and $B$ be $40kg$ and that of $B$ and $C$ be $43kg$, find the weight of $B$ ?

  1. $21kg$
  2. $20kg$
  3. $38kg$
  4. $31kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\cfrac{A+B+C}{3}=45$ $kg$

$\Rightarrow A+B+C=45\times3=135$ $kg$
$\Rightarrow \cfrac{A+B}{2}=40$ $kg$
$\Rightarrow A+B=80$ $kg$
$\therefore C=135-80=55$ $kg$
$\Rightarrow \cfrac{B+C}{2}=43$ $kg$
$\Rightarrow B+C=86$ $kg$
$\therefore B=86-55=31$ $kg$
$\therefore$ The weight of $B$ is $31$ $kg$.

Multiple choice statistics moving averages moving average and variation simple moving average uses of average in day-to-day life introduction to time series introduction to time series and forecasting

In a call of 100 students there are 70 boys whose average marks in a subject is 75. If the average marks of the complete class is 72 . then the average marks of the girls is 

  1. 73

  2. 65

  3. 68

  4. 74
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Total student = 100
For 70 students total marks = 75×7075×70
=5250=5250
Total marks of girls = & 7200-5250$
=1950=1950
Average of girls = 195030195030
=65
Multiple choice statistics moving averages uses of average in day-to-day life moving average and variation simple moving average introduction to time series introduction to time series and forecasting

David obtained 76,65,82,67,and 85 marks(out of 100) in English,Mathematics,Physics,Chemistry and Biology.What are his average marks?

  1. 75

  2. 69

  3. 72

  4. 76
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Average is equal to the sum of marks of all the subjects by the total number of subjects.

$\therefore Avg=\dfrac{76+65+82+67+85}{5}=75$
Multiple choice statistics moving averages uses of average in day-to-day life moving average and variation simple moving average introduction to time series introduction to time series and forecasting

The average weight of 10 men is decreased by 3 kg when one of them whose weight is 80 kg is replaced by a new person. The weight of the new person is

  1. 70

  2. 60

  3. 50

  4. 73
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let weight of $9$ men $=x$.

Weight of new men $=y$

According to the question,
$\dfrac{x+80}{10}-3=\dfrac{x+y}{10}$
$\dfrac{x+80-30}{10}=\dfrac{x+y}{10}$
$x+50=x+y$
$y=50\ kg$

Hence, this is the answer.

Multiple choice statistics moving averages moving average and variation simple moving average uses of average in day-to-day life introduction to time series introduction to time series and forecasting

The problem given below consists  of a question followed by three statements. You have to study the question and the statements and decide which of the statements(s) is / are necessary to answer the question.
How many marks did Tarun secure in English?
(i)The average marks obtained by Tarun in four subjects including English are 60.
(ii) The total marks obtained by him in English and Mathematics together are 170.
(iii)The total marks obtained by him in Mathematics and science together are 180.

  1. I and II only

  2. II and III only

  3. I and III only

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

These statements are not enough to find out his English marks. 
So Ans = D

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

In a study on a certain population, the following data was given.

Population (X) 2000-2001 2001-2002 2002-2003 2003-2004 2004-2005 2005-2006 2006-2007
Number of people 10 20 30 40 50 60 70


Find the average number of population using step deviation method.

  1. $2002$
  2. $2003$
  3. $2004$
  4. $2005$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 X Midpoint (X)  Frequency (F)  i=class interval width  A=2003.5 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 2000-2001 2000.5  10  1  2003.5  -3  -30 
2001-2002  2001.5  20  1  2003.5  -2  -40 
2002-2003  2002.5  30  1  2003.5  -1  -30 
2003-2004  2003.5=A  40  1  2003.5  0   0
2004-2005  2004.5  50  1  2003.5  1  50 
2005-2006  2005.5  60  1  2003.5  2  120 
2006-2007  2006.5  70  1  2003.5  3  210 
    $\Sigma f=280$        $\Sigma fd'=280$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =2003.5 +\dfrac{280}{280}\times 1$
$= 2003.5 + 1$
$= 2004.5$ $\approx$ $2005$

Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

A boy takes 20 minutes to reach the school at an average speed of 12 km/hour. If he wants to reach the school in 15 minutes, his average speed (in km/hour) will be

  1. 16

  2. 17

  3. 18

  4. 19
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A boy takes $20\ min$ to reach the school at an average speed $12\ km/hr$.


Let he takes $15\ min$ to reach the school at an average speed $x\ km/hr$.

Therefore,
$20\times 12=15\times x$
$x=16\ km/hr$

Hence, this is the answer.

Multiple choice maths how much does it weigh? define weight and units of weight using decimals in weight conversion of length measurement (length) basic operations with same units operations involving units of length

The average weight of three boys $P$, $Q$ and $R$ is $\displaystyle 54\frac {1}{3}$ $Kg$, while the average weight of three boys $Q$, $S$ and $T$ is $53$ $kg$. What is the average weight of $P$, $Q$, $R$, $S$ and $T$?

  1. $52.4$ $kg$
  2. $53.2$ $kg$
  3. $53.8$ $kg$

  4. Data inadequate
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As per question, 
$P+Q+R=54\frac { 1 }{ 3 } $---------eq 1
$Q+S+T=53 $-----------eq2
Here is 5 variable and 2 euqation.Data inadequate to solve the question.
Answer (D) Data inadequate

Multiple choice business mathematics and statistics index numbers weighted methods to calculate index numbers construction of index numbers applied statistics

Taking 1975 as the base year with an index number 100 , calculate an index number for 1985 based on weighted average of price relatives.

Commodity A B C D
weight 20 30 10 40
Price per unit in 1975 10 20 5 40
Price per unit in 1985 30 35 10 80

  1. 212.5

  2. 217.5

  3. 219.5

  4. 345.65
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$Commodity$ $Weight$$w$  $Price\,in\,1975$$P _0$  $Price\,in\,1985$ $P _1$ $Price\,relative $$I=\dfrac{P _1}{P _0}\times 100$ $I.w$ 
 $A$ $20$  $10$  $30$  $300$  $6000$ 
$B$  $30$  $20$  $35$  $175$  $5250$
$C$ $10$  $5$  $10$  $200$  $2000$ 
$D$ $40$  $40$  $80$  $200$  $8000$ 
 $Total$ $\sum w=100$        $\sum I.w=21250$

$\Rightarrow$  By using weighted average of price relative method,

$\Rightarrow$  $P _{01}=\dfrac{\sum I.w}{\sum w}=\dfrac{21250}{100}=212.5$