Download App

Maths M.C.Q

Tabular Representation of Statistical Data · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 11 questions.

Questions

M.C.Q

🎯

Test yourself on this topic

11 questions · timed · auto-graded

Question 11 Mark
The width of each of nine classes in a frequency distribution is 2.5 and the lower class boundary of the lowest class 10.6. Then the upper class boundary of the highest class is:
  1. 35.6
  2. 33.1
  3. 30.6
  4. 28.1
Answer
  1. 33.1

Solution:

The number of classes is 9 and the uniform class size is 2.5. The lower limit of the lower class (first class) is 10.6.

Therefore, the upper limit of the last class is

10.6 + (9 × 2.5)

= 10.6 + 22.5

= 33.1

Hence, the correct choice is (b).

View full question & answer→
Question 21 Mark
The number of times a particular item occurs in a given data is called its:
  1. Variation.
  2. Frequency.
  3. Cumulative frequency.
  4. Class-size.
Answer
  1. Frequency.

Solution:

The number of times a particular item occurs in a given data is called the frequency of the item. Hence, the correct choice is (b).

View full question & answer→
Question 31 Mark
The mid-value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are:
  1. 47 and 37
  2. 37 and 47
  3. 37.5 and 47.5
  4. 47.5 and 37.5
Answer
  1. 47 and 37

Solution:

Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is $\frac{\text{l+m}}{2}$ and the class-size is (l - m).

Given that the mid-value of the class is 42 and the class-size is 10. Therefore, we have two equations

$\frac{\text{l+m}}{2}=42$

⇒ l + m = 84,

m - l = 10

Adding the above two equations, we have

(l + m) + (m - l) = 84 + 10

⇒ l + m + m - l = 94

⇒ 2m = 94

⇒ l = 37

Substituting the value of m in the first equation, we have

l + 47 = 84

⇒ l = 84 - 47

⇒ l = 37

Hence, the upper and lower limits of the class are 47 and 37 respectively. Thus, the correct choice is (a).

View full question & answer→
Question 41 Mark
The following marks were obtained by the students in a test:
81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
The range of the marks is:
  1. 9
  2. 17
  3. 27
  4. 33
Answer
  1. 33

Solution:

The marks obtained by the students are 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79 and 62.

The highest and lowest marks are 95 and 62 respectively. Therefore, the range of marks is

95 - 62

= 33

Hence, the correct option is (d).

View full question & answer→
Question 51 Mark
The difference between the upper and the lower class limits is called:
  1. Mid-points.
  2. Class size.
  3. Frequency.
  4. Mean.
Answer
  1. Class size.

    Solution:

    The difference between the upper and the lower class limits is called the class size.

Hence, the correct choice is (b).

View full question & answer→
Question 61 Mark
The difference between the highest and lowest values of the observations is called:
  1. Frequency.
  2. Mean.
  3. Range.
  4. Class-intervals.
Answer
  1. Range.

    Solution:

    The difference between the highest and lowest values of the observations is called the range.

Hence, the correct choice is (c)

View full question & answer→
Question 71 Mark
Tallys are usually marked in a bunch of:
  1. 3
  2. 4
  3. 5
  4. 6
Answer
  1. 4

Solution:

Tallies are usually marked in a bunch of 4.

Hence, the correct option is (b). 

View full question & answer→
Question 81 Mark
Tally marks are used to find:
  1. Class intervals.
  2. Range.
  3. Frequency.
  4. Upper limits.
Answer
  1. Frequency.

    Solution:

    Tally marks are used to find the frequencies.

    Hence, the correct choice is (c).

View full question & answer→
Question 91 Mark
Let l be the lower class limit of a class-interval in a frequency distribution and m be the mid point of the class. Then, the upper class limit of the class is:
  1. $\text{m}+\frac{\text{l+m}}{2}$
  2. $\text{l}+\frac{\text{m+l}}{2}$
  3. $2\text{m}-1$
  4. $\text{m}-2\text{l}$
Answer
  1. $2\text{m}-1$

Solution:

Given that, the lower class limit of a class-interval is l and the mid-point of the class is m. Let u be the upper class limit of the class-interval.

Therefore, we have

$\text{m}=\frac{\text{l+u}}{2}$

⇒ l + u = 2m

⇒ u = 2m - l

Thus the upper class limit of the class is (2m - l).

Hence, the correct choice is (c).

View full question & answer→
Question 101 Mark
In the class intervals 10-20, 20-30, 20 is taken in:
  1. The interval 10-20.
  2. The interval 20-30.
  3. Both intervals 10-20, 20-30.
  4. None of the intervals.
Answer
  1. The interval 20-30.

Solution:

The given class intervals are 10-20, 20-30. In these class intervals the value 20 is lies in the class interval 20-30.Hence, the correct choice is (b).

View full question & answer→
Question 111 Mark
In a frequency distribution, the mid-value of a class is 15 and the class intervals is 4. The lower limit of the class is:
  1. 10
  2. 12
  3. 13
  4. 14
Answer
  1. 13

Solution:

Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is $\frac{\text{l+m}}{2}$ and the class-size is (m - l).

Therefore, we have two equations

$\frac{\text{l+m}}{2}=15$

⇒ l + m = 30,

m - l = 4

Subtracting the second equation from the first equation, we have

(l + m) - (m - l) = 30 - 4

⇒ l + m - m + l = 26

⇒ 2l = 26

⇒ l = 13

Hence, the lower limit of the class is 13. Thus, the correct choice is (c).

View full question & answer→
Generate Paper FreeAll Tabular Representation of Statistical Data topics
M.C.Q - Maths STD 9 Questions - Vidyadip