Practice Statistics - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Read the following information carefully to answer the questions that follow. The arithmetic mean, geometric mean and median of 6 positive numbers a, a, b, b, c, c, where a < b < c are $7/3$, 2, 2, respectively.What is the value of c?
(a)
(b)
(c)
(d)
a < b < c
Total numbers = 6
Increasing order a, a, b, b, c, c
∴ Median = ${\text"(6/2)th term + (6/2 + 1)th term"}/2$
= ${\text"3rd term + 4th term"}/2$
2 = ${b + b}/2$ = b
Arithmetic mean = ${a + a + b + b + c + c}/6$
⇒ $7/3 = {a + b + c}/3$
⇒ a + b + c = 7
⇒ a + c = 7 – 2 = 5 ... (i)
Geometric mean = $(a^2 × b^2 × c^2)^{1/6}$
⇒ 2 = $(abc)^{1/3}$
⇒ abc = 8
⇒ ac = $8/2$ = 4 ... (ii)
⇒ c = $4/a$
From equation (i),
a + $4/a$ = 5
⇒ ${a^2 + 4}/a$ = 5
⇒ $a^2$ – 5a + 4 = 0
⇒ $a^2$ – 4a – a + 4 = 0
⇒ a(a – 4) – 1(a – 4) = 0
⇒ (a – 4) (a – 1) = 0
if a = 1 then c = 4
a = 4 then c = 1
a = 1, c = 4 and b = 2
The value of c is 4.
Q-2) Consider the following data: x 1 2 3 4 5 f 3 5 9 - 2
If the arithmetic mean of the above distribution is 2.96, then what is the missing frequency?
| x | 1 | 2 | 3 | 4 | 5 |
| f | 3 | 5 | 9 | - | 2 |
(a)
(b)
(c)
(d)
| x | f | xf |
| 1 | 3 | 3 |
| 2 | 5 | 10 |
| 3 | 9 | 27 |
| 4 | $f_1$ | $4f_1$ |
| 5 | 2 | 10 |
| Total | 19 + $f_1$ | 50 + $4f_1$ |
∴ Mean = ${Σx_if_i}/{Σ_i}$
⇒ 2.96 = ${50 + 4f_1}/{19 + f_1}$ [given]
⇒ 56.24 + 2.96 $f_1 = 50 + 4f_1$
⇒ 6.24 = 1.04 $f_1$
⇒ $f_1$ = 6
Q-3) Read the following information carefully to answer the questions that follow. The arithmetic mean, geometric mean and median of 6 positive numbers a, a, b, b, c, c, where a < b < c are $7/3$, 2, 2, respectively.What is the mode?
(a)
(b)
(c)
(d)
a < b < c
Total numbers = 6
Increasing order a, a, b, b, c, c
∴ Median = ${\text"(6/2)th term + (6/2 + 1)th term"}/2$
= ${\text"3rd term + 4th term"}/2$
2 = ${b + b}/2$ = b
Arithmetic mean = ${a + a + b + b + c + c}/6$
⇒ $7/3 = {a + b + c}/3$
⇒ a + b + c = 7
⇒ a + c = 7 – 2 = 5 ... (i)
Geometric mean = $(a^2 × b^2 × c^2)^{1/6}$
⇒ 2 = $(abc)^{1/3}$
⇒ abc = 8
⇒ ac = $8/2$ = 4 ... (ii)
⇒ c = $4/a$
From equation (i),
a + $4/a$ = 5
⇒ ${a^2 + 4}/a$ = 5
⇒ $a^2$ – 5a + 4 = 0
⇒ $a^2$ – 4a – a + 4 = 0
⇒ a(a – 4) – 1(a – 4) = 0
⇒ (a – 4) (a – 1) = 0
if a = 1 then c = 4
a = 4 then c = 1
a = 1, c = 4 and b = 2
Mode = 3 (Median) – 2 (Mean)
= $3(2) - 2(7/3) = {18 - 14}/3 = 4/3$
Q-4) The following table gives the frequency distribution of life length in hours of 100 electric bulbs having median life 20 h.Life of bulbs(in hours) Number of bulbs 8-13 7 13-18 x 18-23 40 23-28 y 28-33 10 33-38 2
What is the missing frequency 'y'?
| Life of bulbs(in hours) | Number of bulbs |
| 8-13 | 7 |
| 13-18 | x |
| 18-23 | 40 |
| 23-28 | y |
| 28-33 | 10 |
| 33-38 | 2 |
(a)
(b)
(c)
(d)
Number of total bulbs = 100
∴ 7 + x + 40 + y + 10 + 2 = 100
⇒ x + y = 41 ... (i)
| Life of bulbs (in hours) | Number of bulbs | Cumulative Frequency |
| 8 - 13 | 7 | 7 |
| 13 - 18 | x | 7 + x |
| 18 - 23 | 40 | 47 + x |
| 23 - 28 | y | 47 + x + y |
| 28 - 33 | 10 | 57 + x + y |
| 33 - 38 | 2 | 59 + x + y |
| N = 100 |
The median life is 20 h, so median interval will be (18-23).
Here, l = 18, $N/2$ = 50
c = 7 + x, f = 40, h = 5
∴ Median = l + ${(N/2 - C)}/f$ × h
⇒ 20 = 18 + ${(50 – 7 – x)}/{40}$ × 5
⇒ 2 = ${50 – 7 – x}/8$
⇒ 16 = 50 – 7 – x
⇒ x = 43 – 16
⇒ x = 27
Missing frequency 'y' is 14.
Q-5) What is the difference between the number of boys studying Mathematics and the number of girls studying Physics?
(a)
(b)
(c)
(d)
Difference in the number of boys studying Mathematics and Physics = 180 – 150 = 30
Q-6) Read the following information and answer the four items that follow .Let the distribution of number of scooters of companies X and Y sold by 5 showrooms (A, B, C, D and E) in a certain year be denoted by S1 and the distribution of number of scooters of only company X sold by the five showrooms in the same year be denoted by S2Showroom A B C D E Total number of scooters sold S1 (in%) 19 21 15 33 12 6400 S2(in%) 24 18 20 30 8 3000
Number of scooters of company Y sold by showroom E is what per cent of the number of scooters of both companies sold by showroom C ?
| Showroom | A | B | C | D | E | Total number of scooters sold |
| S1 (in%) | 19 | 21 | 15 | 33 | 12 | 6400 |
| S2(in%) | 24 | 18 | 20 | 30 | 8 | 3000 |
(a)
(b)
(c)
(d)
No. of scooters of Y sold by E
= 12% × 6400 – 8% × 3600
= 528
No. of total scooters by showroom
c = 15% × 6400 = 960
Reqd. % = ${528}/{960}$ × 100 = 55%
Q-7) Consider the following grouped frequency distribution : x f 0-10 8 10-20 12 20-30 10 30-40 p 40-50 9
If the mean of the above data is 25. 2, then what is the value of p ?
| x | f |
| 0-10 | 8 |
| 10-20 | 12 |
| 20-30 | 10 |
| 30-40 | p |
| 40-50 | 9 |
(a)
(b)
(c)
(d)
Mean = (sum of ∞) / (sum of f) = (5*5 + 12*15 + 10*25 + p*35 + 9*45) / (8 + 12 + 10 + P + 9) = 25.2 (875 + 35P) / (39 + P) = 25.2 ⇒ P = 11
Q-8) What is the mode of the frequency distribution of SeriesII?
(a)
(b)
(c)
(d)
Mode of frequency distribution of series II is 46.
Q-9) What is the median of the data 3, 5, 9, 4, 6, 11, 18?
(a)
(b)
(c)
(d)
Arrange the number ascending order
3, 4, 5, 6, 9, 11, 18
Therefore $4^{th}$ term out of 7 is Median.
∴ Median = 6
Q-10) The arithmetic mean of 10 numbers was computed as 7.6. It was later discovered that a number 8 was wrongly read as 3 during the computation. What should be the correct mean?
(a)
(b)
(c)
(d)
Correct A.M. = ${n \ov{x} - \text"(Sum of wrong observations) + (Sum of correct observations)"}/n$
= ${10 × 7.6 - 3 + 8}/{10} = {81}/{10}$ = 8.1
Q-11) The number of girls studying Statistics is what percent (approximate) of the total number of students studying Chemistry?
(a)
(b)
(c)
(d)
According to the question,
250 = x% of 340 ⇒ x = ${250 × 100}/{340}$ = 73.5%
Q-12) In a pi-diagram there are three sectors. If the ratio of the angles of the sectors is 1 : 2 : 3, then what is the angle of the largest sector?
(a)
(b)
(c)
(d)
Sum of angles of a pie diagram = 360°
∴ θ + 2θ + 3θ = 360°
6θ = 360°
θ = 60°
Hence, angle of Largest sector
6θ = 3 × 60°
= 180°
Q-13) The arithmetic mean of two numbers is 10 and their geometric mean is 8. What are the two numbers?
(a)
(b)
(c)
(d)
Q-14) Consider the following frequency distribution :Class Frequency 0-10 4 10-20 5 20-30 7 30-40 10 40-50 12 50-60 8 60-70 4
What is the mode of the distribution?
| Class | Frequency |
| 0-10 | 4 |
| 10-20 | 5 |
| 20-30 | 7 |
| 30-40 | 10 |
| 40-50 | 12 |
| 50-60 | 8 |
| 60-70 | 4 |
(a)
(b)
(c)
(d)
| Class | Mid values | Frequency $F_i$ | $d_I = x_i$ - 35 | $U_i = {x_1 - 35}/{10}$ | $f_I U_I$ | Cumulative Frequency |
| 0 - 10 | 5 | 4 | -30 | -3 | -12 | 4 |
| 10 - 20 | 15 | 5 | -20 | -2 | -10 | 9 |
| 20 - 30 | 25 | 7 | -10 | -1 | -7 | 16 |
| 30-40 | 35 | 10 | 0 | 0 | 0 | 26 |
| 40 - 50 | 45 | 12 | 10 | 1 | 12 | 38 |
| 50 - 60 | 55 | 8 | 20 | 2 | 16 | 46 |
| 60 - 70 | 65 | 4 | 30 | 3 | 12 | 50 |
$ΣF_iU_i = 11$
N = $Σf_i = 50$
$N/2 = {50}/2 = 25$
Mode = $l + {f - f_1}/{2f - f_1 - f_2} × h$
Here, the maximum frequency is 12 their class is
40 – 50, then l = 40 f = 12, $f_1 = 10, f_2$ = 8
mode = 40 + ${12 - 10}/{2 × 12 - 10 - 8}$ × 10
= 40 + $2/6$ × 10 = 40 + 3.33 = 43.33
Q-15) An individual purchases three qualities of pencils. The relevant data is given below :Quality Price per Pencil (in Rs.) Money spent (in Rs.) A 1.00 50 B 1.50 x C 2.00 20
It is known that the average price per pencil is Rs.1.25. What is the value of x ?
| Quality | Price per Pencil (in Rs.) | Money spent (in Rs.) |
| A | 1.00 | 50 |
| B | 1.50 | x |
| C | 2.00 | 20 |
(a)
(b)
(c)
(d)
Number of Type A pencil = ${50}/1$ = 50
Number of Type B pencil = $x/{1.50}$
Number of Type C pencil = ${20}/2$ = 10
Average = ${\text"Total money spent"}/{\text"totalno.of pencil"}$ = 1.25
= ${x + 50 + 20}/{50 + 10 + {x/{1.50}}}$ = 1.25
= 70 + x = 1.25 $(60 + x/{1.50})$
70 + x = 75.00 + ${1.25}/{1.50}$x
x - ${125}/{150}$ x = 5
= ${25}/{150}$ x = 5
x = 30
Q-16) Consider the following statements related to cumulative frequency polygon of a frequency distribution, the frequencies being cumulated from the lower end of the range :
1. The cumulative frequency polygon gives an equivalent representation of frequency distribution table.
2. The cumulative frequency polygon is a closed polygon with one horizontal and one vertical side. The other sides have non–negative slope.
Which of the above statements is / are correct ?
(a)
(b)
(c)
(d)
Here, Statement 1 is correct but Statement 2 is not correct.
Q-17) Square diagrams are drawn to represent the following data: Country Pakistan India Myanmar China Labour 36 81 25 100 Production(inRs.)
Using the scale 1 $cm^2$ = Rs.25, what is the length of the representative square for India?
| Country | Pakistan | India | Myanmar | China |
| Labour | 36 | 81 | 25 | 100 |
| Production(inRs.) |
(a)
(b)
(c)
(d)
Rs.25 = 1 $cm^2$
∴ Rs.1 = $1/{25} cm^2$
∴ Rs.81 = ${81}/{25} cm^2$ = Area of square
Side of square = $√{{81}/{25}} = 9/5$ = 1.8 cm
Q-18) Read the following information carefully to answer the questions that follow. The average age of 6 persons living in a house is 23.5 years. Three of them are majors and their average age is 42 years. The difference in ages of the three minor children is same.What is the median of the ages of minor children?
(a)
(b)
(c)
(d)
Total age of six persons = 23.5 × 6 = 141 years
Total age of three major persons = 42 × 3 = 126 years
∴ Total age of three minor children = 141 – 126 = 15 years
The difference in ages of the three minor children is same.
Therefore, we take ages may be:
5, 5, 5; 3, 5, 7; 2, 5, 8 and 1, 5, 9
In all the cases, median will be 5 years.
Median age of minor children = 5 years.
Q-19) Consider the following statements in respect of a histogram:
I. The histogram consists of vertical rectangular bars with a common base such that there is no gap between consecutive bars.
II. The height of the rectangle is determined by the frequency of the class it represents.
Which of the statements given above is/are correct?
(a)
(b)
(c)
(d)
Statement I :
A graph which displays the data by using vertical bars of various heights in rectangular shapes to represent frequencies. Such that there is no gap between consecutive bars and also the height of the rectangle.
Statement II :
The height of the rectangle is determined by the frequency of the class it represents.
So, both the statements are correct.
Q-20) Consider the following statements in respect of the set S = {1, 2, 3, ..., n}.
I. ${(n + 1)}/2$ is the median of the numbers in S.
II. n is the mode of the numbers in S.
Which of the above statements is/are correct?
(a)
(b)
(c)
(d)
S = {1, 2, 3, ..., n}
Statement I
Median : It is the middle term. So, ${n + 1}/2$ here it is not define n is even or odd.
Hence, we cannot say that ${n + 1}/2$ is median. It is not correct.
Mode : It is the value that appears most of ten in a set of data.
S = {1, 2, 3, ..., n}
Here, all elements in S have same frequency.
So, both the statements are not correct.