NEET Test Series from KOTA - 10 Papers In MS WORD
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DATA HANDLING
300106
The weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. Their mean weight is:
1 60.5 gram
2 60 gram
3 59 gram
4 62 gram
Explanation:
59 gram Given weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. then total weight of 9 apples = 50 + 60 + 65 + 62 + 67 + 70 + 64 + 45 + 48 = 531 then mean \( = \frac{531}{9} = {59}{\text{ gram}}\)
DATA HANDLING
300107
Let x be the mean of x\(_{\1}\), x\(_{\1}\), ... ,xn and y the mean of . If z is the mean of x\(_{\1}\), x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\), y\(_{\1}\), ... ,y\(_{\1}\) then z is equal to:
1 \(\text{x + y}\)
2 \(\frac{\text{x + y}}{2}\)
3 \(\frac{\text{x + y}}{\text{n}}\)
4 \(\frac{\text{x + y}}{\text{2n}}\)
Explanation:
\(\frac{\text{x + y}}{2}\) x is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)then \(\text{x} = \frac{\text{x}_{1}+\text{x}_{2} + ......+\text{x}_{\text{n}}}{\text{n}}\) y is the mean of y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \( \text{y} = \frac{\text{y}_{1}+\text{y}_{2} + ......+\text{y}_{\text{n}}}{\text{n}}\) z is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \(_{\1}\) \(_{\1}\)
DATA HANDLING
300108
The average (arithmetic mean) of a particular set of seven numbers is 12. When one of the numbers is replaced by the number 6, the average of the set increases to 15. What is the number that was replaced?
1 -10
2 -15
3 -12
4 0
5 12
Explanation:
-15 The average of a set of numbers is the sum of the numbers divided by the number of numbers i.e. average \( = \frac{\text{Sum of all 7 number}}{\text{No. of number(N)}}\) Sum of all 7 numbers = average × N = 12 × 7 = 84 Similarly, the sum of the new set of numbers is = 15 × 7 = 105 Now, suppose that the seven numbers are a, b, c, d, e, f, g and the g gets replaced by 6. Then we have, a + b + c + d + e + f + g = 84 ....... (1) a + b + c + d + e + f + 6 = 105 a + b + c + d + e + f = 99 ....... (2) Plugging the vlaue from (2) in (1), we get 99 + g = 84 g = -15 \(\therefore\) the number that was replaced was -15.
DATA HANDLING
300109
The mean of five numbers is 4. If 1 is added to each other, then the new mean is:
1 4
2 5
3 3
4 5.5
Explanation:
5 Mean of five numbers = 4 Sum of five numbers = 5 × 4 = 20 \(\text{New mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}\) \(=\frac{20+1+1+1+1+1}{5}\) \(=\frac{25}{5}\) \(=5\) Thus, the new mean is 5 Hence, the correct option is (b).
300106
The weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. Their mean weight is:
1 60.5 gram
2 60 gram
3 59 gram
4 62 gram
Explanation:
59 gram Given weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. then total weight of 9 apples = 50 + 60 + 65 + 62 + 67 + 70 + 64 + 45 + 48 = 531 then mean \( = \frac{531}{9} = {59}{\text{ gram}}\)
DATA HANDLING
300107
Let x be the mean of x\(_{\1}\), x\(_{\1}\), ... ,xn and y the mean of . If z is the mean of x\(_{\1}\), x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\), y\(_{\1}\), ... ,y\(_{\1}\) then z is equal to:
1 \(\text{x + y}\)
2 \(\frac{\text{x + y}}{2}\)
3 \(\frac{\text{x + y}}{\text{n}}\)
4 \(\frac{\text{x + y}}{\text{2n}}\)
Explanation:
\(\frac{\text{x + y}}{2}\) x is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)then \(\text{x} = \frac{\text{x}_{1}+\text{x}_{2} + ......+\text{x}_{\text{n}}}{\text{n}}\) y is the mean of y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \( \text{y} = \frac{\text{y}_{1}+\text{y}_{2} + ......+\text{y}_{\text{n}}}{\text{n}}\) z is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \(_{\1}\) \(_{\1}\)
DATA HANDLING
300108
The average (arithmetic mean) of a particular set of seven numbers is 12. When one of the numbers is replaced by the number 6, the average of the set increases to 15. What is the number that was replaced?
1 -10
2 -15
3 -12
4 0
5 12
Explanation:
-15 The average of a set of numbers is the sum of the numbers divided by the number of numbers i.e. average \( = \frac{\text{Sum of all 7 number}}{\text{No. of number(N)}}\) Sum of all 7 numbers = average × N = 12 × 7 = 84 Similarly, the sum of the new set of numbers is = 15 × 7 = 105 Now, suppose that the seven numbers are a, b, c, d, e, f, g and the g gets replaced by 6. Then we have, a + b + c + d + e + f + g = 84 ....... (1) a + b + c + d + e + f + 6 = 105 a + b + c + d + e + f = 99 ....... (2) Plugging the vlaue from (2) in (1), we get 99 + g = 84 g = -15 \(\therefore\) the number that was replaced was -15.
DATA HANDLING
300109
The mean of five numbers is 4. If 1 is added to each other, then the new mean is:
1 4
2 5
3 3
4 5.5
Explanation:
5 Mean of five numbers = 4 Sum of five numbers = 5 × 4 = 20 \(\text{New mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}\) \(=\frac{20+1+1+1+1+1}{5}\) \(=\frac{25}{5}\) \(=5\) Thus, the new mean is 5 Hence, the correct option is (b).
300106
The weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. Their mean weight is:
1 60.5 gram
2 60 gram
3 59 gram
4 62 gram
Explanation:
59 gram Given weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. then total weight of 9 apples = 50 + 60 + 65 + 62 + 67 + 70 + 64 + 45 + 48 = 531 then mean \( = \frac{531}{9} = {59}{\text{ gram}}\)
DATA HANDLING
300107
Let x be the mean of x\(_{\1}\), x\(_{\1}\), ... ,xn and y the mean of . If z is the mean of x\(_{\1}\), x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\), y\(_{\1}\), ... ,y\(_{\1}\) then z is equal to:
1 \(\text{x + y}\)
2 \(\frac{\text{x + y}}{2}\)
3 \(\frac{\text{x + y}}{\text{n}}\)
4 \(\frac{\text{x + y}}{\text{2n}}\)
Explanation:
\(\frac{\text{x + y}}{2}\) x is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)then \(\text{x} = \frac{\text{x}_{1}+\text{x}_{2} + ......+\text{x}_{\text{n}}}{\text{n}}\) y is the mean of y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \( \text{y} = \frac{\text{y}_{1}+\text{y}_{2} + ......+\text{y}_{\text{n}}}{\text{n}}\) z is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \(_{\1}\) \(_{\1}\)
DATA HANDLING
300108
The average (arithmetic mean) of a particular set of seven numbers is 12. When one of the numbers is replaced by the number 6, the average of the set increases to 15. What is the number that was replaced?
1 -10
2 -15
3 -12
4 0
5 12
Explanation:
-15 The average of a set of numbers is the sum of the numbers divided by the number of numbers i.e. average \( = \frac{\text{Sum of all 7 number}}{\text{No. of number(N)}}\) Sum of all 7 numbers = average × N = 12 × 7 = 84 Similarly, the sum of the new set of numbers is = 15 × 7 = 105 Now, suppose that the seven numbers are a, b, c, d, e, f, g and the g gets replaced by 6. Then we have, a + b + c + d + e + f + g = 84 ....... (1) a + b + c + d + e + f + 6 = 105 a + b + c + d + e + f = 99 ....... (2) Plugging the vlaue from (2) in (1), we get 99 + g = 84 g = -15 \(\therefore\) the number that was replaced was -15.
DATA HANDLING
300109
The mean of five numbers is 4. If 1 is added to each other, then the new mean is:
1 4
2 5
3 3
4 5.5
Explanation:
5 Mean of five numbers = 4 Sum of five numbers = 5 × 4 = 20 \(\text{New mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}\) \(=\frac{20+1+1+1+1+1}{5}\) \(=\frac{25}{5}\) \(=5\) Thus, the new mean is 5 Hence, the correct option is (b).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
DATA HANDLING
300106
The weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. Their mean weight is:
1 60.5 gram
2 60 gram
3 59 gram
4 62 gram
Explanation:
59 gram Given weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. then total weight of 9 apples = 50 + 60 + 65 + 62 + 67 + 70 + 64 + 45 + 48 = 531 then mean \( = \frac{531}{9} = {59}{\text{ gram}}\)
DATA HANDLING
300107
Let x be the mean of x\(_{\1}\), x\(_{\1}\), ... ,xn and y the mean of . If z is the mean of x\(_{\1}\), x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\), y\(_{\1}\), ... ,y\(_{\1}\) then z is equal to:
1 \(\text{x + y}\)
2 \(\frac{\text{x + y}}{2}\)
3 \(\frac{\text{x + y}}{\text{n}}\)
4 \(\frac{\text{x + y}}{\text{2n}}\)
Explanation:
\(\frac{\text{x + y}}{2}\) x is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)then \(\text{x} = \frac{\text{x}_{1}+\text{x}_{2} + ......+\text{x}_{\text{n}}}{\text{n}}\) y is the mean of y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \( \text{y} = \frac{\text{y}_{1}+\text{y}_{2} + ......+\text{y}_{\text{n}}}{\text{n}}\) z is the mean of x\(_{\1}\)?, x\(_{\1}\), ... ,x\(_{\1}\)y\(_{\1}\)?, y\(_{\1}\)?, .... ,y\(_{\1}\) \(_{\1}\) \(_{\1}\)
DATA HANDLING
300108
The average (arithmetic mean) of a particular set of seven numbers is 12. When one of the numbers is replaced by the number 6, the average of the set increases to 15. What is the number that was replaced?
1 -10
2 -15
3 -12
4 0
5 12
Explanation:
-15 The average of a set of numbers is the sum of the numbers divided by the number of numbers i.e. average \( = \frac{\text{Sum of all 7 number}}{\text{No. of number(N)}}\) Sum of all 7 numbers = average × N = 12 × 7 = 84 Similarly, the sum of the new set of numbers is = 15 × 7 = 105 Now, suppose that the seven numbers are a, b, c, d, e, f, g and the g gets replaced by 6. Then we have, a + b + c + d + e + f + g = 84 ....... (1) a + b + c + d + e + f + 6 = 105 a + b + c + d + e + f = 99 ....... (2) Plugging the vlaue from (2) in (1), we get 99 + g = 84 g = -15 \(\therefore\) the number that was replaced was -15.
DATA HANDLING
300109
The mean of five numbers is 4. If 1 is added to each other, then the new mean is:
1 4
2 5
3 3
4 5.5
Explanation:
5 Mean of five numbers = 4 Sum of five numbers = 5 × 4 = 20 \(\text{New mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}\) \(=\frac{20+1+1+1+1+1}{5}\) \(=\frac{25}{5}\) \(=5\) Thus, the new mean is 5 Hence, the correct option is (b).