Mathematics Class IX

Question 1 of 30

1. Question
1 point(s)
The range of the data:- 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19,8, 11, 20 is
- 10
- 15
- 18
- 26
Correct

Incorrect

Hint

$32-6=26$
Question 2 of 30

2. Question
1 point(s)
In a frequency distribution, the Mid value of a class is 10 and the width of the class is 6. The lower limit of the class is
- 6
- 7
- 8
- 12
Correct

Incorrect

Hint

Class $=7-13$
Lower limit $=7$
Question 3 of 30

3. Question
1 point(s)
The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is
- 15
- 25
- 35
- 40
Correct

Incorrect

Hint

$10-15\\ \\ 15-20\\ \\ 20-25\\ \\ 25-30\\ \\ 30-35$
Question 4 of 30

4. Question
1 point(s)
Let m be the midpoint and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit is
- $2m+l $
- $2m-l $
- $m- l $
- $m-2 l $
Correct

Incorrect

Hint

$m=\dfrac{\text{lower class + upper class}}{2}\\ \\ \Rightarrow m=\dfrac{\text{lower class + l}}{2}\\ \\ 2m-l=\text{lower class}$
Question 5 of 30

5. Question
1 point(s)
The class marks are 15, 20, 25……
The class corresponding to the class mark 20 is
- $12.5-17.5 $
- $17.5-22.5 $
- $18.5-21.5 $
- $19.5-20.5 $
Correct

Incorrect

Hint

Class corresponding to the class mark $=17.5-22.5$
Question 6 of 30

6. Question
1 point(s)
In the class intervals 10-20, 20-30 the number 20 in included in
- $10-20 $
- $20-30 $
- both the intervals
- none of these
Correct

Incorrect

Hint

20 is included in 20 – 30
Question 7 of 30

7. Question
1 point(s)
To draw a histogram to represent the following frequency distribution:-

The adjusted frequency for the class 25-45 is
- 6
- 5
- 3
- 2
Correct

Incorrect

Hint

Adjusted frequency = 10 – 8 = 2
Question 8 of 30

8. Question
1 point(s)
The mean of 5 numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is
- 28
- 30
- 35
- 38
Correct

Incorrect

Hint

Sum of 5 numbers $=5\times30=150$
Sum of 4 numbers $=28\times4=112$
Excluded numbers $=150-112=38$
Question 9 of 30

9. Question
1 point(s)
The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150. The resulting mean will be:-
- 50.5
- 51
- 51.5
- 52
Correct

Incorrect

Hint

$\dfrac{(100\times50)-50+150}{100}=\dfrac{5000+100}{100}\\ \\ =\dfrac{5100}{100}=51$
Question 10 of 30

10. Question
1 point(s)
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
- 45
- 49.5
- 54
- 56
Correct

Incorrect

Hint

22, 34, 39, 45, 54, 54, 56, 78, 84
Median = 54
Question 11 of 30

11. Question
1 point(s)
Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is
- 14
- 15
- 16
- 17
Correct

Incorrect

Hint

15 occur most of the times so 15 is the mode
Question 12 of 30

12. Question
1 point(s)
The arithmetic mean is 25. Determine the value of p.
- 15
- 13
- 14
- 16
Correct

Incorrect

Hint

$x=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{940+35p}{44+p}=25\\ \\ \Rightarrow 940+35p=25(44+p)\\ \\ \Rightarrow10p=160\Rightarrow p=16$
Question 13 of 30

13. Question
1 point(s)
The mean of 6 numbers is 20. If one number is deleted their mean is 15. Find the deleted number.
- 45
- 52
- 20
- 36
Correct

Incorrect

Hint

Sum of 6 numbers $=20\times6=120$
Sum of 5 number $=15\times5=75$
Deleted number $=120-75=45$
Question 14 of 30

14. Question
1 point(s)
The mean of a, b, c, d, e is 28. If the mean of a, c, e is 24. What is the mean of b and d
- 31
- 32
- 33
- 34
Correct

Incorrect

Hint

$a+b+c+d+e=140\ ……(1)\\ \\ a+c+e=72\ ……(2)\\ \\ b+d+72=140\\ \\ b+d=68\\ \\ \text{Mean}=\dfrac{b+d}{2}=\dfrac{68}{2}=34$
Question 15 of 30

15. Question
1 point(s)
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. What is the lower limit of the class?
- 6
- 7
- 8
- 12
Correct

Incorrect

Hint

Let $x_1$ = lower limit
$x_2$ = upper limit
$\dfrac{x_1+x_2}{2}=10\Rightarrow x_1+x_2=20\ …..(1)\\ \\ x_1-x_2=6\ …..(2)\\ \\ x_1=7,x_2=13$
Question 16 of 30

16. Question
1 point(s)
The average of $10 $ numbers is $12 $. What should be added as $\mathrm{11^{th}} $ number to make the average $13 $?
- 28
- 25
- 30
- 23
Correct

Incorrect

Hint

Sum of numbers $=10\times12=120$
Sum of 11 numbers $=13\times11=143$
$11^{th}$ number $= 143-120=23$
Question 17 of 30

17. Question
1 point(s)
Which of the following is the empirical relation between mean, mode and median?
- mode = 3 median – 2 mean
- mode = 2 median – 3 mean
- median = 3 mode – 2 mean
- mean = 3 median – 2 mode
Correct

Incorrect

Hint

mode = 3 median – 2 mean
Question 18 of 30

18. Question
1 point(s)
If the mean of 9 observation p, p + 2, p + 4, p + 6, p + 8, p – 2, p – 4, p – 6 and p – 8’ is 10. Find the mean of the least 5 observations.
- 6
- 25
- 10
- 9
Correct

Incorrect

Hint

$\text{P + P + 2 + P + 4 + P + 6 + P + 8 + P - 2 + P - 4 + P - 6 + P - 8 = 90}\\ \\ 8=90\\ \\ \text{9P + 0 = 90}\\ \\ \text{P=10}$
Least observation are 10, 8, 6, 4, 2
Mean $=\dfrac{10+8+6+4+2}{5}=\dfrac{30}{5}=6$
Question 19 of 30

19. Question
1 point(s)
Find the median of 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6
- 0
- -1.5
- 2
- 3.5
Correct

Incorrect

Hint

$-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6$ No. of observation (n) $= 14$
Median = $\dfrac{1}{2}[7^{th}\ \text{item}+8^{th}\ \text{item}]$
$=\dfrac{1}{2}[2+5]=3.5$
Question 20 of 30

20. Question
1 point(s)
For what value of n, the mode of the following data is 18?
2, 5, 3, 18, 5, 18, 6, 5, n, 7, 18
- 18
- 5
- 3
- None of these
Correct

Incorrect

Hint

There are three 5 and three 18. For mode to be 18, n = 18
Question 21 of 30

21. Question
1 point(s)
There are 150 numbers. Each number is subtracted from 60 and the mean of the numbers so obtained is found to be -4.5.
The mean of the given numbers is
- 400
- 34.5
- 64.5
- 55.5
Correct

Incorrect

Hint

Mean of number = 60 + (-4.5) = 55.5
Question 22 of 30

22. Question
1 point(s)
The median and mean of the first 10 natural numbers are
- 5.5, 5.5
- 5.5, 6
- 5, 6
- 5, 3
Correct

Incorrect

Hint

Mean $=\dfrac{1+2+3+4+5+6+7+8+9+10}{10}=5.5$
median $=\dfrac{5+6}{2}=\dfrac{11}{2}=5.5$
Question 23 of 30

23. Question
1 point(s)
The following observations are arranged in ascending order.
20, 23, 42, 53, x, x+2, 70, 75, 82, 96 If the median is 63 Find the value of x
- 62
- 64
- 63
- 60
Correct

Incorrect

Hint

Median is mean of $5^{th}$ and $6^{th}$ term.
$x+1=63\\ \\ \Rightarrow x=2$
Question 24 of 30

24. Question
1 point(s)
The mean of 20 observations was 60. It was detected on rechecking that the value of 125 was wrongly copied as 25 for computation of mean. Find the correct mean.
- 67
- 66
- 65
- 64
Correct

Incorrect

Hint

Let x be the sum of observation of 19 numbers leaving 125.
$x+25=20\times60=1200\ …..(1)\\ \\ x+125=20\times y=20y\ ……(2)$
subtracting $\mathrm{eq^n}$ (1) form $\mathrm{eq^n}$ (2)
$125-25=20y-1200\\ \\ y=65$
Question 25 of 30

25. Question
1 point(s)
Find x and y so that the ordered data set has a mean of 42 and a median of 35.
17, 22, 26, 29, 34, x, 42, 67, 70, y
- 34, 77
- 36, 77
- 77, 34
- 77, 36
Correct

Incorrect

Hint

Mean $=42$
$\Rightarrow \dfrac{17+22+26+29+34+x+42+67+70+y}{10}=420\\ \\ \Rightarrow307+x+y=420\\ \\ \Rightarrow x+y=113\\ \\ \dfrac{34+x}{2}=35\\ \\ x=70-34=36\\ \\ y=113-36=77$
Question 26 of 30

26. Question
1 point(s)
What do you call the value in a data around which the values of all the other observations tend to concentrate?
- common value
- range
- measure of central tendency
- mid value
Correct

Incorrect

Hint

The value in a data around which the values of all the other observations tend to concentrate is measure of central tendency.
Question 27 of 30

27. Question
1 point(s)
The following is a chart showing the temperature of a patient.

What is the temperature of the patient at 21 hours?
- $100^o\ F $
- $101^o\ F $
- $102^o\ F $
- $103^o\ F $
Correct

Incorrect

Hint

$100^o\ F$
Question 28 of 30

28. Question
1 point(s)
The following is a chart showing the temperature of a patient.

What is the percentage increase in temperature between 9 hours and 15 hours?
- 0.04
- 0.03
- 0.02
- 0.01
Correct

Incorrect

Hint

$\dfrac{104-100}{100}\times100=\dfrac{4}{100}\times100=4\%$
Question 29 of 30

29. Question
1 point(s)
Kavita obtained 16, 14, 18 and 20 marks out of 25 in Maths in weekly tests in the month of Jan 2000 ; then mean marks of Kavita is:
- 16
- 16.5
- 17
- 17.5
Correct

Incorrect

Hint

Mean marks $=\dfrac{16+14+18+20}{4}$
$=\dfrac{63}{4}=17$
Question 30 of 30

30. Question
1 point(s)
The median of first 12 prime numbers is :
- 19
- 12
- 15
- 17
Correct

Incorrect

Hint

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
17 is median

Mathematics Class IX

Olympiad Questions-Statistics-Class IX

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