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MATHS M.C.Q

Measures Of Central Tendency · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 15 questions.

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M.C.Q

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MCQ 11 Mark
The algebraic sum of the deviations of a set of $n$ values from their mean is:
  • ✓
    $0$
  • B
    $n - 1$
  • C
    $n$
  • D
    $n + 1$
Answer
Correct option: A.
$0$
if is the mean of $n$ observations $x_1, x_2, x_3, x_4 ...x_n.$
then algebraic sum of deviations $=\sum\limits^\text{n}_{\text{i}=0}\Big(\text{x}_\text{i}-{\overline{\text{X}}}\Big)$
$=\sum\limits^\text{n}_{\text{i}=0}\text{x}_\text{i}-\text{n}{\overline{\text{X}}}$
$=\text{n}\bigg(\frac{\sum^\text{n}_{\text{i}=0}\text{x}_\text{i}}{\text{n}}\bigg)-\text{n}\overline{\text{X}}$
$=\text{n}\overline{\text{X}}-\text{n}\overline{\text{X}}$
$=0$
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Question 21 Mark
For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?
  1. Mean = Median.
  2. Mean > Mode.
  3. Mean > Mode.
  4. Mode = Median.
Answer
  1. Mean > Mode.
Solution:
Median = 4
Mode = 2
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, (Mean = 5) > (Mode = 2)
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Question 31 Mark
The empirical relation between mean, mode and median is:
  1. Mode = 3 Median - 2 Mean.
  2. Mode = 2 Median - 3 Mean.
  3. Median = 3 Mode - 2 Mean.
  4. Mean = 3 Median - 2 Mode.
Answer
  1. Mode = 3 Median - 2 Mean.
Solution:
The empirical Relation between mean, median and mode is:
Mode = 3 Median - 2 mean.
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Question 41 Mark
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is:
  1. 9
  2. 11
  3. 13
  4. None of these.
Answer
  1. 9
Solution:
Mean of first five observations $=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}=11$
⇒ 5x + 20 = 55
⇒ x = 7
⇒ First three numbers are 7, 9, 11
$\text{Mean}=\frac{7+9+11}{3}=\frac{27}{3}=9$
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Question 51 Mark
The following is the data of wages per day : 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is:
  1. 7
  2. 5
  3. 8
  4. 10
Answer
  1. 5
  2. 8
Solution:
In data 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8,
We observe that values 5 and 8 both have maximum frequency i.e. 4
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Question 61 Mark
The mean of n observations is $\overline{\text{X}}.$ If each observation is multiplied by k, the mean of new observations is:
  1. $\text{k}\overline{\text{X}}$
  2. $\frac{\overline{\text{X}}}{\text{k}}$
  3. $\overline{\text{X}}+\text{k}$
  4. $\overline{\text{X}}-\text{k}$
Answer
  1. $\text{k}\overline{\text{X}}$
Solution:
$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
if each observation is multiplied by k, then
$\text{New Mean},\overline{\text{X'}}=\frac{(\text{Sum of all observations})\text{k}}{\text{n}}$
$\Rightarrow\overline{\text{X}'}=\text{k}\overline{\text{X}}$
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Question 71 Mark
If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value x is:
  1. 10
  2. 12
  3. 14
  4. 16
Answer
  1. 16
Solution:
$\text{Mean}=\frac{7+5+13+\text{x}+9}{5}=10$
$\Rightarrow34+\text{x}=50$
$\Rightarrow\text{x}=16$
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Question 81 Mark
The median of the following data : 0, 2, 2, 2, -3, 5, -1, 5, −3, 6, 6, 5, 6 is:
  1. 0
  2. -1.5
  3. 2
  4. 3.5
Answer
  1. 3.5
Solution:
Data: 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6
Rearranging data in increasing order, we have
-3, -3, -1, 0, 2, 2, 5, 5, 5, 5, 6, 6, 6
Number of observations = n = 14 (even)
Now, 
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{7^{\text{th}}\text{observation}+8^{\text{th}}\text{observation}}{2}$
$=\frac{2+5}{2}$
$\Rightarrow\text{Median}=3.5$
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Question 91 Mark
A, B, C are three sets of values of x:
A: 2, 3, 7, 1, 3, 2, 3
B: 7, 5, 9, 12, 5, 3, 8
C: 4, 4, 11, 7, 2, 3, 4
Which one of the following statements is correct?
  1. Mean of A = Mode of C
  2. Mean of C = Median of B
  3. Median of B = Mode of A
  4. Mean, Median and Mode of A are equal.
Answer
  1. Mean, Median and Mode of A are equal.
Solution:
A: 1, 2, 2, 3, 3, 3, 7
B: 3, 5, 5, 7, 8, 9, 12
C: 2, 3, 4, 4, 4, 7, 11
$\text{Mean of A}=\frac{1+2+2+3+3+3+7}{7}$
$=\frac{21}{7}=3$
$\text{Mean of B}=\frac{3+5+5+7+8+9+12}{7}$
$=\frac{49}{7}=7$
$\text{Mean of C}=\frac{2+3+4+4+4+7+11}{7}$
$=\frac{35}{7}=5$
Mode of A = 3; Median of A = 3
Mode of B = 5; Median of B = 7
Mode of C = 4; Median of C = 4
(Mean of A = 3) $\neq$ (Mode of C = 4)
(Mean of C = 5) $\neq$ (Median of B = 4)
(Median of B = 7) $\neq$ (Mode of A = 3)
Mean of A = 3, Mode of A = 3, Median of A = 3
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Question 101 Mark
Mode is:
  1. Least frequent value.
  2. Middle most value.
  3. Most frequent value.
  4. None of these.
Answer
  1. Most frequent value.
Solution:
Most Frequent value is called mode.
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Question 111 Mark
The mean of a, b, c, d and e is 28. If the mean of a, c, and e is 24, What is the mean of b and d?
  1. 31
  2. 32
  3. 33
  4. 34
Answer
  1. 34
Solution:
$\text{Mean}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
⇒ a + b + c + d + e = 140 ...(1)
Also, $\text{Mean}=\frac{\text{a}+\text{c}+\text{e}}{3}=24$
⇒ a+ c + e = 72 ...(2)
Subtracting equation (2) from (1), we have
b + d = 68
$\text{Mean}=\frac{\text{b}+\text{d}}{2}=\frac{68}{2}=34$
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Question 121 Mark
The mean of n observations is $\overline{\text{X}}.$ If k is added to each observation, then the new mean is:
  1. $\overline{\text{X}}$
  2. $\overline{\text{X}}+\text{k}$
  3. $\overline{\text{X}}-\text{k}$
  4. $\text{k}\overline{\text{X}}$
Answer
  1. $\overline{\text{X}}+\text{k}$
Solution:
$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
Now if k is aded to each observation
$\text{New Mean},\overline{\text{X'}}=\frac{\text{Sum of all observations}+\text{nk}}{\text{n}}$
$=\frac{\text{Sum of all observations}}{\text{n}}+\text{k}$
$\Rightarrow\overline{\text{X}'}=\overline{\text{X}}+\text{k}$
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Question 131 Mark
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is:
  1. 98
  2. 99
  3. 100
  4. 101
Answer
  1. 99
Solution:
$\text{Mean}=81=\frac{\text{Sum of seven numbers}}{7}$
⇒ Sum of seven numbers = 81 × 7 = 567
Let the discared number be x.
⇒ Sum of 6 numbers = 567 - x
Now, mean of remaining 6 numbers $=\frac{567-\text{x}}{6}=78$
⇒ 567 - x = 468
⇒ x = 99
So, discarded number is 99
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Question 141 Mark
For which set of numbers do the mean, median and mode all have the same value?
  1. 2, 2, 2, 2, 4
  2. 1, 3, 3, 3, 5
  3. 1, 1, 2, 5, 6
  4. 1, 1, 1, 2, 5
Answer
  1. 1, 3, 3, 3, 5
Solution:
 
Mean
Median
Mode
2, 2, 2, 2, 4
$\frac{12}{5}=2.4$
2
2
1, 3, 3, 3, 5
$\frac{15}{5}=3$
3
3
1, 1, 2, 5, 6
$\frac{15}{5}=3$
2
1
1, 1, 1, 2, 5
$\frac{10}{5}=2$
1
1
From above table, data 1, 3, 3, 3, 5 has mean, median, mode all have same value, i.e. 3
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Question 151 Mark
Which one of the following is not a measure of central value?
  1. Mean.
  2. Range.
  3. Median.
  4. Mode.
Answer
  1. Range.
Solution:
The difference between the highest value and the lowest value in the data set is called Range.
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