5 GMAT Statistics Practice Questions & Explanations
Question 1:
The average weight of n students in a certain class was 24 kg. A new student named Jack joined the class and one existing student named Carol left the class. What is the new average?
I. Carol’s weight was 1 kg more than that of Jack
II. Carol’s weight was 24kg.
Answer:
Let’s say the weight of Jack is J, weight of Carol is C and the sum of weights of all the students in the class is S
It is given that the average weight of n students in a certain class was 24 kg.
Since,
Average=Sum of the terms/Total number of terms
Thus, Average=S/n = 24
Or, S = 24n…..(i)
Since, Carol has left and Jack has joined the class, thus one person left and one person joined due to which the total number of students in the class will still be n.
Thus, the new average will be;
New Average = S - weight of Carol + weight of Jack/n..(ii)
1. Carol’s weight was 1 kg more than that of Jack
Or, C = J + 1
Using equation (ii), we get,
New Average = S - C + J/n
Or, New Average = S - ( J + 1 ) + J / n = S - J - 1 + J/n = S - 1 /n
Or, New Average = 24n - 1/n {as S = 24n}
=24- 1n
But since we don’t know the value of n, hence we cannot find a definite answer; NOT sufficient.
2. Carol’s weight was 24 kg
Using equation (ii), we get,
New Average = S - C + J/n = S - 24 + J/n = 24n - 24 + Jn {as S = 24n}
But since we don’t know the value of n and J, hence we cannot find a definite answer
The correct answer is E; both statements together are still not sufficient.
Question 2:
The ratio of boys to girls in a class is 4 : 3. The average weight of all the students in the class is 45kg. If the average weight of girls is 7kg less than that of boys, what is the ratio of the average weight of boys to the average weight of girls?
47 : 40
48 : 41
49 : 42
50 : 43
4 : 3
Answer:
Ratio of Boys to Girls = 4:3, So let’s consider there are “4x” boys and “3x” girls in the class.
Let’s also consider the average weight of the boys = “y” kgs. Therefore, average weight of girls will be “y-7” kgs.
The average weight of Class = (Average weight of Boys * Number of Boys + (Average weight of Girls * Number of Girls)/(Total Number of Students)
=> 45 = y× 4x + (y-7)×3x/4x+3x
=> 45 = 7y - 21/7
y = 48, and also, y-7 = 41
Therefore, ratio of Average weight of Boys to Average weight of Girls = 48:41
The correct answer is B.
Question 3:
Each of the following linear equations defines y as a function of x for all integers x from 1 to 20. For which of the following equations is the standard deviation of the y – values corresponding to all the x – values, the greatest?
Y = -(x/5)
Y = -(x/5)
Y = x
Y = -2x + 7
Y = -4x - 9
Answer :
We are given some linear equations which defines y as a function of x for all integers x from 1 to 20.
1 ≤ x ≤ 20
We have to find, in which case the standard deviation of y will be the greatest.
As y is a function of x, so we will have 20 different values of y for 20 values of integer x varying from 1 to 20
=> POINTS TO REMEMBER:-
Whenever we add or subtract a constant in all the observations, the standard deviation of the data remains unchanged.
Whenever we multiply or divide all the observations by a constant, the standard deviation will be respectively multiplied or divided by the absolute value of the constant.
(A) y = - x/5
In this option, we will get the values of ‘y’ by dividing the values of x by (-5). So, every observation in set x will be divided by (-5) to get the corresponding values in set y.
(Standard Deviation)Y = ((Standard Deviation of x))/5
(B) y = - x/2 + 2
In this option, we will first divide every value of x by (-2) and then we will add 2 in all the observations.
(Standard Deviation)Y = ((Standard Deviation of x))/2
(C) y = x
In this option, set y will have the same values as set x.
(Standard Deviation)Y = [(Standard Deviation)X]
(D) y = -2x + 7
In this option, we will first multiply every observation in x by (-2) and then we will add 7 in every observation.
So, from all the options above, we will have the greatest standard deviation of ‘y’ in option E.
The correct answer is E.
Question 4:
An alumni meet is attended by 150 people. Are there any two people of the same age (rounded to the nearest year) who live in the same city?
(1) The attendees’ age ranges from 30 to 33 years, inclusive.
(2) The attendees live in 30 different cities.
Answer :
Number of people attending the alumni meet = 150
We have to find – Are there any two people of the same age (rounded to the nearest year) who lives in the same city?
STATEMENT 1: The attendees’ age ranges from 30 to 33, inclusive.
From this statement, we are only getting the Information about the age but nothing is mentioned about the cities.
So, this statement is not alone sufficient to answer the question; NOT sufficient.
STATEMENT 2: The attendees’ live in 30 different cities.
In this statement we are mentioned about the cities but nothing is mentioned about the ages of the attendees.
So, this statement is also not alone sufficient to answer this question; NOT sufficient
Taking statement 1 and 2 together,
Now, the ages of attendees can be 30, 31, 32 and 33 only.
And, there are 30 different cities.
Now, If we assume there is one person each of age 30, 31, 32 & 33 in each city, so there will be 4 persons of different ages in each city.
So, total number of persons in 30 different cities will be = 30 × 4 = 120.
But, total number of attendees are 150.
So, there will be 30 persons left who will also be from one of the 30 cities with age either 30, 31, 32 or 33.
So, there will be at least two people of the same age living in the same city.
The correct answer is C; both statements together are sufficient.
Question 5:
S is a set of n integers, where 0 < n < 11. If the arithmetic mean of set S is a positive integer b, which of the following could NOT be the median of set S?
(A) 0
(B) b
(C)(-b)
(D) n/5
(E) 5n/11
Answer:
S is a set of n integers, where 0 < n < 11.
Arithmetic mean of Set S is a positive integer ‘b’.
We have to find which of the following could not be the median of Set S.
Now, We know that if the number of terms are odd, then there will be only one middle term and that will be the median, which will be an Integer value.
(A) And, If number of terms are even, then there will be two middle terms, and the median will be the average of those two middle terms, in the form of (a+b)/2, where a & b will be those two middle terms.
(F) So, the median can be either an integer or can be in the form of I/2, where I will be an integer.
So, option (A), (B) & (C) can be the median because all three are integers.
(A) Option (D), (n/5) can also be the median because 0
But, option (E), 5n/11 cannot be either an integer or in the form of I/2, as n is not a multiple of 11. Hence, this cannot be the median.
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