Statistics Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Statistics topic of quantitative aptitude
(a) 5 years
(b) 7 years
(c) 3 years
(d) Cannot be determined
The correct answers to the above question in:
Answer: (a)
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.
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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) Only 2
b) Both 1 and 2
c) Only 1
d) Neither 1 nor 2
Answer »Answer: (c)
Here, Statement 1 is correct but Statement 2 is not correct.
Question : 2
Let 1 $\ov{x_1}$ and $\ov{x_2}$ (where $\ov{x_2} > \ov{x_1}$) be the means of two sets comprising $n_1 and n_2$ (where $n_2 < n_1$) observations respectively. If $\ov{x}$ is the mean when they are pooled, then which one of the following is correct?
a) $\ov{x} > \ov{x_2}$
b) $\ov{x} < \ov{x_1}$
c) $\ov{x_1} < \ov{x} < \ov{x_2}$
d) $(\ov{x_1} - \ov{x}) + (\ov{x_2} - \ov{x}) = 0$
Answer »Answer: (c)
Let if possible $\ov{x}_1 < \ov{x} < \ov{x}_2$
Consider
$\ov{x}_1 < \ov{x}$
$\ov{x}_1 < {n_1 \ov{x}_1 + n_2 \ov{x}_2}/{n_1 + n_2}$
$n_1 \ov{x}_1 + n_2\ov{x}_1 < n_1\ov{x}_1 + n_2 \ov{x}_2$
⇒ $\ov{x}_1 < \ov{x}_2$
[it is true as given in Question]
Hence $\ov{x}_1 < \ov{x}$ ...(1)
Consider $\ov{x} < \ov{x}_2$
${n_1\ov{x}_1 + n_2 \ov{x}_2}/{n_1 + n_2} < \ov{x}_2$
⇒ $\ov{x}_1 < \ov{x}_2$
[True as given in Question]
Hence $\ov{x} < \ov{x}_2$
From (1) & (2)
$\ov{x}_1 < \ov{x} < \ov{x}_2$
Question : 3
The number of girls studying Statistics is what percent (approximate) of the total number of students studying Chemistry?
a) 73.5
b) 78.7
c) 58.8
d) 80.6
Answer »Answer: (a)
According to the question,
250 = x% of 340 ⇒ x = ${250 × 100}/{340}$ = 73.5%
Question : 4
What is the difference between the number of boys studying Mathematics and the number of girls studying Physics?
a) 30
b) 60
c) 20
d) 80
Answer »Answer: (a)
Difference in the number of boys studying Mathematics and Physics = 180 – 150 = 30
Question : 5
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 |
a) 24
b) 14
c) 27
d) 11
Answer »Answer: (b)
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.
Question : 6
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 |
a) Rs. 30
b) Rs.40
c) Rs.10
d) Rs.60
Answer »Answer: (a)
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
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