Either head or tail. When we toss a coin, two outcomes are possible, i.e. head or tail.
DATA HANDLING
300175
The mean weight of 21 students is 21kg. If a student weighing 21kg is removed from the group, then the mean of of the remaining students is:
1 20kg
2 21kg
3 19kg
4 18kg
Explanation:
21kg Mean weight = 21kg Number of students = 21 Sum of weights of 21 students = 21 × 21 = 441 Sum of weights of 20 students left = 441 - 21 = 420 Mean of remaining students \(=\frac{420}{20}=21\text{kg}\) Hence, the correct option is (b).
DATA HANDLING
300176
If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x:
1 8
2 5
3 6
4 7
Explanation:
7 Mean of observations x, x + 2, x + 4, x + 6, x + 8 is 11 Mean \( =\frac{ \text{Sum}}{\text{Number of observation}}\) Mean \( = \frac{\text{x+x+2+x+4+x+6+x+8}}{5} = {11}\) \(\frac{\text{5x+20}}{5} = {11}\) x + 4 = 11 x = 7
DATA HANDLING
300177
2, 10, m, 12, 4 A group of 5 integers is shown above. If the average (arithmetic mean) of the numbers is equal to m, find the value of m:
1 7
2 8
3 9
4 10
5 11
Explanation:
7 We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation: \(\Rightarrow \frac{{2+10+}{\text{m}}{+12+4}}{5} = {\text{m}}\) \(\Rightarrow{28}+\text{m} = {5}{\text{m}}\) \(\Rightarrow {\text{m}} - {5}{\text{m}} = -{28}\) \(\Rightarrow -{4}{\text{m}} = -{28}\) \(\Rightarrow{\text{m}} = {7}\)
Either head or tail. When we toss a coin, two outcomes are possible, i.e. head or tail.
DATA HANDLING
300175
The mean weight of 21 students is 21kg. If a student weighing 21kg is removed from the group, then the mean of of the remaining students is:
1 20kg
2 21kg
3 19kg
4 18kg
Explanation:
21kg Mean weight = 21kg Number of students = 21 Sum of weights of 21 students = 21 × 21 = 441 Sum of weights of 20 students left = 441 - 21 = 420 Mean of remaining students \(=\frac{420}{20}=21\text{kg}\) Hence, the correct option is (b).
DATA HANDLING
300176
If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x:
1 8
2 5
3 6
4 7
Explanation:
7 Mean of observations x, x + 2, x + 4, x + 6, x + 8 is 11 Mean \( =\frac{ \text{Sum}}{\text{Number of observation}}\) Mean \( = \frac{\text{x+x+2+x+4+x+6+x+8}}{5} = {11}\) \(\frac{\text{5x+20}}{5} = {11}\) x + 4 = 11 x = 7
DATA HANDLING
300177
2, 10, m, 12, 4 A group of 5 integers is shown above. If the average (arithmetic mean) of the numbers is equal to m, find the value of m:
1 7
2 8
3 9
4 10
5 11
Explanation:
7 We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation: \(\Rightarrow \frac{{2+10+}{\text{m}}{+12+4}}{5} = {\text{m}}\) \(\Rightarrow{28}+\text{m} = {5}{\text{m}}\) \(\Rightarrow {\text{m}} - {5}{\text{m}} = -{28}\) \(\Rightarrow -{4}{\text{m}} = -{28}\) \(\Rightarrow{\text{m}} = {7}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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DATA HANDLING
300174
On tossing a coin, the outcome is:
1 Only head.
2 Only tail.
3 Neither head nor tail.
4 Either head or tail.
Explanation:
Either head or tail. When we toss a coin, two outcomes are possible, i.e. head or tail.
DATA HANDLING
300175
The mean weight of 21 students is 21kg. If a student weighing 21kg is removed from the group, then the mean of of the remaining students is:
1 20kg
2 21kg
3 19kg
4 18kg
Explanation:
21kg Mean weight = 21kg Number of students = 21 Sum of weights of 21 students = 21 × 21 = 441 Sum of weights of 20 students left = 441 - 21 = 420 Mean of remaining students \(=\frac{420}{20}=21\text{kg}\) Hence, the correct option is (b).
DATA HANDLING
300176
If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x:
1 8
2 5
3 6
4 7
Explanation:
7 Mean of observations x, x + 2, x + 4, x + 6, x + 8 is 11 Mean \( =\frac{ \text{Sum}}{\text{Number of observation}}\) Mean \( = \frac{\text{x+x+2+x+4+x+6+x+8}}{5} = {11}\) \(\frac{\text{5x+20}}{5} = {11}\) x + 4 = 11 x = 7
DATA HANDLING
300177
2, 10, m, 12, 4 A group of 5 integers is shown above. If the average (arithmetic mean) of the numbers is equal to m, find the value of m:
1 7
2 8
3 9
4 10
5 11
Explanation:
7 We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation: \(\Rightarrow \frac{{2+10+}{\text{m}}{+12+4}}{5} = {\text{m}}\) \(\Rightarrow{28}+\text{m} = {5}{\text{m}}\) \(\Rightarrow {\text{m}} - {5}{\text{m}} = -{28}\) \(\Rightarrow -{4}{\text{m}} = -{28}\) \(\Rightarrow{\text{m}} = {7}\)
Either head or tail. When we toss a coin, two outcomes are possible, i.e. head or tail.
DATA HANDLING
300175
The mean weight of 21 students is 21kg. If a student weighing 21kg is removed from the group, then the mean of of the remaining students is:
1 20kg
2 21kg
3 19kg
4 18kg
Explanation:
21kg Mean weight = 21kg Number of students = 21 Sum of weights of 21 students = 21 × 21 = 441 Sum of weights of 20 students left = 441 - 21 = 420 Mean of remaining students \(=\frac{420}{20}=21\text{kg}\) Hence, the correct option is (b).
DATA HANDLING
300176
If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x:
1 8
2 5
3 6
4 7
Explanation:
7 Mean of observations x, x + 2, x + 4, x + 6, x + 8 is 11 Mean \( =\frac{ \text{Sum}}{\text{Number of observation}}\) Mean \( = \frac{\text{x+x+2+x+4+x+6+x+8}}{5} = {11}\) \(\frac{\text{5x+20}}{5} = {11}\) x + 4 = 11 x = 7
DATA HANDLING
300177
2, 10, m, 12, 4 A group of 5 integers is shown above. If the average (arithmetic mean) of the numbers is equal to m, find the value of m:
1 7
2 8
3 9
4 10
5 11
Explanation:
7 We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation: \(\Rightarrow \frac{{2+10+}{\text{m}}{+12+4}}{5} = {\text{m}}\) \(\Rightarrow{28}+\text{m} = {5}{\text{m}}\) \(\Rightarrow {\text{m}} - {5}{\text{m}} = -{28}\) \(\Rightarrow -{4}{\text{m}} = -{28}\) \(\Rightarrow{\text{m}} = {7}\)