Question 1 1 / -0
The mean of the numbers, 5, 6, 7, 8, 9 and x, is 8 and the median of the numbers, 4, 5, 5, 6, 6 and p, is 5. If p is a one-digit number, then what can be the highest value of x + p?
Solution
Mean of 5, 6, 7, 8, 9 and x is 8
35 + x = 48
x = 48 – 35
x = 13 --- (i)
Median of 4, 5, 5, 6, 6 and p is 5.
Since, median of 4, 5, 5, 6, 6 and p is 5, so p can't be more than 5, and as we need to find the highest value of x + p, so p can't be less than 5.
Thus, the value of p is 5.
x + p = 13 + 5 = 18 (from (i))
Question 2 1 / -0
A type of frequency polygon that shows cumulative frequencies is called ________.
Solution
An ogive is a type of frequency polygon that shows cumulative frequencies. In other words, the cumulative percents are added on the graph, from left to right. An ogive graph plots cumulative frequencies on the y-axis and class boundaries along the x-axis.
Question 3 1 / -0
The mean of 7 observations is 71. Even if two numbers are excluded, the mean stays the same. One of the excluded numbers is 72. What is the other excluded number?
Solution
When the mean stays the same after excluding some numbers, the mean of the excluded numbers is the same as that of the original set.
Let one of the excluded numbers be x and the other is given as 72.
Their mean is 71.
x + 72 = 142
x = 142 - 72 = 70
Hence, the other excluded number is 70.
Question 4 1 / -0
Find the mode for the given data (Write the answer up to two decimal places).
Money (in lakhs) Frequency 2-9 32 9-16 30 16–23 10 23-30 40 30-37 8 37-44 21 44-51 13
Solution
Since the maximum class frequency is 40, the mode class is 23–30.
Mode =
Here,
l = Lower limit of the modal class
h = Width of the modal class
f
1 = Frequency corresponding to the modal class
f
2 = Frequency corresponding to the post-modal class
f
0 = Frequency corresponding to the pre-modal class
Mode =
= 23 +
= 23 + 3.39 = 26.39
So, the mode is 26.39.
Question 5 1 / -0
The average of two observations, a and 12, will become c, when both the observations are doubled and one is added to each observation. The average of two observations, 5a and 16, will become d, when both the observations are doubled and one is added to each observation. What is the average of c and d?
Solution
c =
=
= a + 13
d =
=
= 5a + 17
Average of c and d =
=
=
= 3a + 15
Question 6 1 / -0
In an athletic event held at a school, the average score of boys was 67 and that of girls was 65. If average score of all the students in the athletics event was 65.8, then find the ratio of the number of boys who participated in the athletics event to the number of girls who participated in the athletics event.
Solution
Let number of boys in school be x.
Average score of boys = 67
Total score of boys = 67
x = 67x
Let number of girls in school be y.
Average score of girls = 65
Total score of girls = 65 × y = 65y
Now, average score of all the students in the athletics event = 65.8
So,
= 65.8
= 65.8
67x + 65y = 65.8x + 65.8y
67x - 65.8x = 65.8y - 65y
1.2x = 0.8y
=
=
x : y = 2 : 3
Therefore, the ratio of number of boys who participated in the athletics event to the number of girls who participated in athletics events is 2 : 3.
Question 7 1 / -0
Find median using an empirical relation when it is given that mean and mode are 12.5 and 9.8, respectively.
Solution
Given:
Mean = 12.5
Mode = 9.8
Empirical relation is,
3 Median = Mode + 2(Mean)
3 Median = 9.8 + 2(12.5)
3 Median = 9.8 + 25
3 Median = 34.8
Median =
Median = 11.6
Question 8 1 / -0
Out of 40 numbers, the mean of 15 observations is 18, and the mean of the remaining observations is 15. What is the mean of all 40 observations?
Solution
Total number of observations = 40
Question 9 1 / -0
The mean of n numbers of a series is
and the sum of first (n - 1) numbers is λ. The value of the last number is
Solution
Sum of the first n numbers =
Sum of the first (n - 1) number =
Required last number = Sum of n numbers - {Sum of first (n - 1) numbers) =
Question 10 1 / -0
The mean of 12 observations was 40. On rechecking, it was detected that the value of 106 was wrongly copied as 10, for computation of mean. Find the correct mean.
Solution
Mean of 12 observations = 40
Sum of the observations = 40 × 12 = 480
Correct mean =
Mean =
Mean = 48
Question 11 1 / -0
The mode of a frequency distribution with class intervals is found by using which type of graphical representation?
Solution
In a specific data set, the mode can be defined as the value that has occurred most of the times. Mode is represented by histograms. The highest peak of the histogram represents the location of the mode of the data set.
Question 12 1 / -0
There are 3 sections viz. A, B and C with 25, 40 and 35 students, respectively in grade X. The average marks of section A are 70%, of section B are 65% and of section C are 50%. Find the average marks (in percentage) of the entire grade X.
Solution
Here, n
1 = 25, n
2 = 40, n
3 = 35
= 70,
= 65,
= 50
Let
denote the average marks of grade X.
=
=
=
=
= 61%
Question 13 1 / -0
Find the value of the missing frequency if the mean of the given data is 16.8
x f 12 8 13 + t 12 17 10 21 6 25 2t
Solution
x f fx 12 8 96 13 + t 12 156 + 12t 17 10 170 21 6 126 25 2t 50t Total 36 + 2t 548 + 62t
Mean =
16.8 =
604.8 + 33.6
t = 548 + 62
t 604.8 - 548 = 62
t - 33.6
t 56.8 = 28.4
t t = 2
Missing frequency = 2
t = 4
Question 14 1 / -0
The average amount of money saved by Jasmine in 11 weeks is Rs. 1,650, out of which Rs. 1,500 is the average amount that she saved in 7 weeks and Rs. 1,780 is the average amount that she saved in the last 5 weeks. How much did she save in week 6 if she saved Rs. 1,000 more in week 6 than her savings of 7th week?
Solution
Average (Mean) = Total amount saved ÷ Number of weeksth week = Total money saved in last 5 weeks - Money saved in last 4 weeks = 8,900 – 7,650 = Rs. 1,250
Question 15 1 / -0
If the frequency of median class is 20, the number of observations are 68, the cumulative frequency of the class preceding the median class is 26, and the upper limit of the median class is 145, then find median with class size 20.
Solution
Median = l is the lower class boundary of the group containing the median.
n is the total number of values.
cf is the cumulative frequency of the groups before the median group.
f is the frequency of the median group.
h is the group width.
h = Class Size = Upper limit – Lower limit
20 = 145 – l
l = 145 – 20 = 125
Median = Median =
Median =
Median = 125 + 8 = 133
Question 16 1 / -0
The mean weight of 150 students in 6th class is 60 kg. The mean weight of boys in the class is 70 kg and that of girls is 55 kg. The number of boys and girls in the class are respectively
Solution
Let total number of boys = x
Question 17 1 / -0
The time taken by a train to travel from station A to B is 35 minutes. If the speed of the train had been 10 km/hr lesser, it would have taken 42 minutes to complete the journey. If the train covers 30% of the distance at 30 km/hr and the remaining at 35 km/hr, then find the approximate average speed of the train during the journey.
Solution
Let the distance between the stations be x km and the speed of the train be y kmph.
A.T.Q,
60x = 35y --- (i)
Also,
60x = 42y - 420 --- (ii)
From (i) and (ii),
42y - 420 = 35y
42y - 35y = 420
7y = 420
y = 420 ÷ 7
y = 60
Putting the value of y in (i), we get
60x = 35y
60x = 2100
x = 35
Time taken by the train to complete the journey =
= 21 + 42
= 63 minutes
OR
63 ÷ 60 = 1.05 hours
Required average speed = 35 ÷ 1.05
= 33.33 kmph
Question 18 1 / -0
The average marks scored by 30 students in a test are 78. The marks of student A were marked 62 instead of 78, and the marks of student B were marked 98 instead of 55. Find the correct average of marks scored by the 30 students.
Solution
Average marks scored by the students = 78
Question 19 1 / -0
The number of music CDs sold by four shopkeepers, on different days of the week, are as follows:
Shopkeeper Number of CDs Average price A 50 Rs. 35 B 70 Rs. 38 C 60 Rs. x D 30 Rs. 45
If the total cost of the CDs sold by all the shopkeepers is Rs. 8,250, find the value of x.
Solution
Total cost of CDs sold by all the shopkeepers = Rs. 8,250
Number of CDs sold by C = 60
Average price of CDs sold by C =
= Rs. 41.5
Question 20 1 / -0
The average marks scored by Harpreet and Gurpreet in three tests are 57 and 61, respectively. If the ratio of the marks scored by them in English is 7 : 9, find the average marks of Gurpreet in the other two tests, given that the average marks scored by Harpreet in the other two subjects were 54.
Solution
Total marks scored by Harpreet in all tests = 57 × 3 = 171
Question 21 1 / -0
Fill in the blanks. P the frequencies of all the classes preceding the given class. Q if each observation is decreased by 8. R .
Solution
(i) The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
(ii) Sum of 80 numbers = 80 × 36 = 2,880
If each number is decreased by 8, then the new sum = 2,880 - (8 × 80)
= 2,880 - 640 = 2,240
New mean =
= 28
(iii) Let the 17 consecutive numbers be p, p + 1, p + 2, ..., p + 16
Mean = 80 ( given )
So,
i.e.
2p + 16 = 80 × 2
2p = 160 - 16
2p = 144
p = 72
The 9
th term is the centre-most term which is in the form (p + 8).
p + 8 = 72 + 8 = 80
Question 22 1 / -0
The median of the following data is 425. Find the missing frequencies if there are 150 observations in the data.
Solution
Class Interval xi (Mid point of class interval) fi Cumulative Frequency (cf) 0 - 100 50 3 3 100-200 150 5 8 200-300 250 7 15 300-400 350 f1 15 + f1 400-500 450 12 27 + f1 500-600 550 16 43 + f1 600-700 650 20 63 + f1 700-800 750 f2 63 + f1 + f2 800–900 850 8 71 + f1 + f2 900- 1000 950 3 74 + f1 + f2
Given,
n = 150
74 + f
1 + f
2 = 150
f
1 + f
2 = 76
Since median is 425, median class is 400–500.
Median = l +
× h
l is the lower class boundary of the group containing the median.
n is the total number of values.
cf is the cumulative frequency of the groups before the median group.
f is the frequency of the median group.
h is the group width.
425 = 400 +
× 100
25 =
× 100
= 60 – f
1 f
1 = 57
Therefore, f
2 = 76 – 57 = 19
Question 23 1 / -0
Find the mean, median and mode of the following data.
Class Interval cf 0-30 4 30-60 10 60-90 18 90-120 28 120-150 33 150-180 37 180-210 40
Solution
Class (xi ) Frequency (fi ) xi fi Cumulative frequency (cf) 0-30 15 4 60 4 30-60 45 6 270 10 60-90 75 8 600 18 90-120 105 10 1050 28 120-150 135 5 675 33 150-180 165 4 660 37 180-210 195 3 585 40
Σ f
i = 40
Σ f
i x
i = 3900
(I) Mean =
=
= 97.5
n = 40,
=
= 20
(ii) Median class is 90–120.
Median = l +
× h
l is the lower class boundary of the group containing the median.
n is the total number of values.
cf is the cumulative frequency of the groups before the median group.
f is the frequency of the median group.
h is the group width.
= 90 +
× 30
90 + 6 = 96
(iii) Maximum frequency is 10, so modal class is 90–120.
Mode = l +
× h
Here,
l = Lower limit of the modal class
h = Width of the modal class
f
1 = Frequency corresponding to the modal class
f
2 = Frequency corresponding to the post-modal class
f
0 = Frequency corresponding to the pre-modal class
= 90 +
× 30
= 90 +
= 98.57
Therefore, mean = 97.5
Median = 96
Mode = 98.57
Question 24 1 / -0
The table shows the distribution of marks obtained by a group of 165 students in Science examination (maximum marks 100).
Marks Obtained Number of students More than 0% 165 More than 10% 155 More than 20% 140 More than 30% 122 More than 40% 94 More than 50% 78 More than 60% 53 More than 70% 27
Calculate the median and mode of the percentage marks obtained. (Up to two decimal places.)
Solution
Marks Cumulative Frequency (cf) Frequency (f) cf(Ne w) 0-10 165 10 10 10-20 155 15 25 20-30 140 18 43 30-40 122 28 71 40-50 94 16 87 50-60 78 25 112 60-70 53 26 138 70-80 27 27 165
n = 165,
=
= 82.5
(i) Median class is 40-50.
Median = l +
× h
l is the lower class boundary of the group containing the median.
n is the total number of values.
cf is the cumulative frequency of the groups before the median group.
f is the frequency of the median group.
h is the group width.
= 40 +
× 10
= 40 + 7.1875 = 47.1875 or 47.19
(ii) Maximum frequency is 28.
So, the modal class is 30–40.
Mode = l +
× h
Here,
l = Lower limit of the modal class
h = Width of modal class
f
1 = Frequency corresponding to the modal class
f
2 = Frequency corresponding to the post-modal class
f
0 = Frequency corresponding to the pre-modal class
= 30 +
× 10
= 30 + 4.545
= 34.55
Therefore, median = 47.19
Mode = 34.55
Question 25 1 / -0
Find the median for the following distribution of employees.
Solution
Daily payment Number of employeesi ) Cumulative frequency (cf) 1-4 6 6 4-8 8 14 8-12 14 28 12-16 36 64 16-20 18 82 20-24 14 96 24-28 9 105 28-32 5 110
Here,
n = 110,
=
= 55
Median class is 12–16.
Median = L +
× h
L is the lower class boundary of the group containing the median.
n is the total number of values.
cf is the cumulative frequency of the groups before the median group.
f is the frequency of the median group.
h is the group width.
= 12 +
× 4
= 12 + 3
= 15
Therefore, median of the given data is 15.
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