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The mean age of 5 students of a class is 10 years. If one student is excluded from the class, the new mean age becomes 9 years. The age of student who is excluded from the class is :
Age of the excluded student
years
The ratio between upper limit and lower limit of a class is 3 : 2. If the class mark is 15, then lower limit of the classinterval is :
class mark
lower limit
For \(x = 1\), the mode of data \(x + 1, 0, x + 2, x + 3\) and \(x + 4\) is :
0, 0, 1, 2, 3
A cricket player has a mean score of 50 runs in the four innings. How many runs to be scored by the player in the fifth innings to raise the mean score by five ?
The mean of the prime factors of 24 is :
Prime factors of
The mode of the following data 44, 40, 40, 41, 45, 43, 45, 40, 41 is :
As 40 occur most of the times
If lower limit and upper limit of a class are in the ratio 3 : 5 and its class size is 6, then the class mark of this class is :
lower limit
upper limit
class mark
\(x + 8,\ x\) and \(x – 3\) are the frequencies of three numbers 5, 8 and 10 respectively. If their arithmetic mean is 6, then the value of \(x \) is :
The median of first six whole numbers is :
Median
Two students scored following marks (out of 10 marks) in five different subjects :
Student I : 2, 3, 4, 5, 6
Student II : 3, 3, 3, 4, 8
Which student scored better ?
Mean of student
Mean of student
The mean of perimeters of two squares having sides \(x \) units and \(y \) units is :
Perimeter of square
Perimeter of square
Mean
The mean weight of 10 students in a class is 50 kg. If the sum of the weights of 9 students is 453 kg, the weight of 10th student is :
The mean of three numbers is 4. If two of the numbers are 1 and 2, then the third number is :
In the class intervals 1020, 2030, the number 20 is included in :
20 is included in 20 – 30
If each observation of a data is increased by 5, then their mean :
original mean
= original mean + 5
The class mark of the class 90120 is :
The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is :
In a given data, maximum value = 32 and minimum value = 6
We know, range of the data = maximum value – minimum value
Range 32 – 6 = 26
The marks obtained by 17 students in a mathematics test (out of 100) are given below : 91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49. The range of the data is :
Range = R = Highest value – Lowest value
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
Median
Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is :
15 occur most times so mode is 15
The mean temperature of \(30 \) days was taken as \(15^oC \), but later it was found that temperature of one day was noted as \(16^oC \) instead of \(13^oC \). The correct mean is :
Numbers \(7,\ 8, 3x + 2, 5x – 4, 40, 41\) are given in ascending order. If the median of the data is \(15\), then the value of \(x \) is :
A/Q
The mean marks obtained by 15 students in a class is 73. If the marks of each student are doubled, then the new mean marks is :
Sum
= Mean
New mean
The median of first 49 whole numbers is :
The mean of class–marks of (20 – 30) and (30 – 40) is :
Class mark of interval
Class mark of
Mean
The mean of 15 numbers is 18 and the mean of 10 numbers is 13. The mean of all the 25 numbers is :
The sum of first n natural numbers is given by \(\dfrac{n(n+1)}{2} \). The mean of first 100 natural numbers is:
Sum of first 100 natural numbers
Mean
If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5, then the class corresponding to the class mark 33.5 is
class mark
The mean of 10 numbers is 55. If one number is excluded, their mean becomes 50, the excluded number is :
Excluded number
Class mark of a particular class is 6.5 and class size is 3, then class interval is :
upper limit – lower limit = 3
lower limit + 3 + lower limit = 13
lower limit = 5
upper limit = 8
The range of the data 25.7, 16.3, 2.8, 21.7, 24.3, 22.7, 24.9 is :
The range is the difference between the smallest value and largest value of the data set.
Given data set is
The smallest value in the given data is 2.8 and the largest value is 25.7
Therefore the range is