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  • SSC CHSL Exam SSC CHSL Quantative Aptitude Study Material

Digitization help student to explore and study their academic courses online, as this gives them flexibility and scheduling their learning at their convenience. Kidsfront has prepared unique course material of SSC CHSL Quantative Aptitude Statistics for SSC CHSL Exam student. This free online SSC CHSL Quantative Aptitude study material for SSC CHSL Exam will help students in learning and doing practice on Statistics topic of SSC CHSL Exam SSC CHSL Quantative Aptitude. The study material on Statistics, help SSC CHSL Exam SSC CHSL Quantative Aptitude students to learn every aspect of Statistics and prepare themselves for exams by doing online test exercise for Statistics, as their study progresses in class. Kidsfront provide unique pattern of learning SSC CHSL Quantative Aptitude with free online comprehensive study material and loads of SSC CHSL Exam SSC CHSL Quantative Aptitude Statistics exercise prepared by the highly professionals team. Students can understand Statistics concept easily and consolidate their learning by doing practice test on Statistics regularly till they excel in SSC CHSL Quantative Aptitude Statistics.


Statistics
Given the mean of five numbers is 27. If one of them is excluded , the mean gets reduced by 2. Determine the excluded number.

a) 45
b) 55
c) 25
d) 35


✔Answer
✔Solution
Correct Answer Is : 35
Solution Is : Mean of 5 numbers = 27
Sum of 5 numbers = 27x5 = 135
Now , Mean of 4 numbers after exclusion of a number = 25
Sum of 4 numbers = 25 x 4 = 100
Excluded number = Sum of 5 numbers - sum of 4 numbers
= 135 - 100 = 35
If x is the mean of n observations x1 , x2 , ...... xn , then the mean of x1/a , x2/a , ,,,,, xn/a is

a) X
b) A x
c) X/a
d) X + a


✔Answer
✔Solution
Correct Answer Is : X/a
Solution Is : When each number is divided by`a` then mean is also divided by`a`
New mean = x/a
The mean weight of 34 students of a school is 42 kg. If the weight of the teacher be included, the mean rises by 400 g. find the weight of the teacher in kg

a) 66
b) 56
c) 55
d) 57


✔Answer
✔Solution
Correct Answer Is : 56
Solution Is : Total weight of student = number of student x avearege =34x42=1428kg
now,when teacher is included
total member = 34+1=35
Average Weight =42+0.400=42.4kg
similarly their total weight = 42.4x35=1484kg
therefore weight of teacher =(total weight including teacher -
total weight excluding teacher)
= 1484-1428 = 56 kg
Study the chart and answer the following question

Themean of the highest and the lowest sale (in rs crore )is

a) 4922
b) 4922.5
c) 4827
d) 4365


✔Answer
✔Solution
Correct Answer Is : 4827
Solution Is : Mean = Highest sale +
Lowest sale / 2
= 8730 + 924 / 2
= 4827 core
The average marks secured by 36 students was 52. but it was discovered that an item 64 was misred as 64.what is the correct mean of mark?

a) 54
b) 53.5
c) 53
d) 52.5


✔Answer
✔Solution
Correct Answer Is : 52.5
Solution Is : Correct mean=52*36-46+64/36
52*36+18/36
104+1/2=105/2=52.5
The mean high temperature of the first four days of a week is 25°C whereas the mean of the last four days is 25.5°C. If the mean of the whole week is 25.2°C, then the temperature of the 4th day is

a) 25.2°C
b) 25.5°C
c) 25.6°C
d) 25°C


✔Answer
✔Solution
Correct Answer Is : 25.6°C
Solution Is : Average higher temperature of four days = 25oC
Total higher temperature of four days = 4 x 25 = 100oC
Average temperature of four days = 25.5oC
Total temperature of these four days = 4 x 25.5 = 102oC
Average temperature of a week = 7 x 25.2 = 176.4oC
Hence, temperature of fourth day = (100 + 102) - 176.4 = 25.6oC
Find the mean of the first 6 prime numbers?

a) 15 / 3
b) 6
c) 41 / 6
d) 12 /4


✔Answer
✔Solution
Correct Answer Is : 41 / 6
Solution Is : The first 6 prime numbers are 2 , 3 , 5 , 7 ,11 , 13 Mean = ( 2 + 3 + 5 + 7 + 11 + 13 ) / 6 = 41 / 6
The mean proportion of 1.21 and 0.09 is

a) 0.33
b) 3.03
c) 3.3
d) 0.033


✔Answer
✔Solution
Correct Answer Is : 0.33
Solution Is : Mean proportion = √ab here, a=121 and b= 0.09 So, Mean proportion = √121*0.09 = √121/100 * 9/100 √11*11/10*10 * 3*3/10*10 = 11*3/100 = 33/100 = 0.33.
The arithmetic mean (average) of the first 10 whole numbers is

a) 5
b) 4
c) 5.5
d) 4.5


✔Answer
✔Solution
Correct Answer Is : 4.5
Solution Is : (4) Required average =(0+1+2+3+....+9)/10 =(9*(9+1))/(2*10) 4.5
The mean of 9 observations is 16. One more observation is included and the new mean becomes 17. The 10th observation is

a) 9
b) 16
c) 26
d) 30


✔Answer
✔Solution
Correct Answer Is : 26
Solution Is : (3) Tricky Approach Tenth observation =10*17-16*9 =170-144 =26
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