Telangana SCERT Solution Class VII (7) Math Chapter 7 Data Handling.
Exercise – 1
1.) Maximum day time temperatures of Hyderabad in a week (from 26th February to 4th March, 2011) are recorded as 26oC, 27oC, 30oC, 30oC, 32oC, 33oC and 32oC.
(i) What is the maximum temperature of the week?
Answer:
From above data the maximum temperature of the week is 33oC.
(ii) What is the average temperatures of the week?
Answer:
We have to find the average temperature of the week.
Average temperature of the week = (Sum of all temperatures of week / Total days in week)
Average temperature of the week =26oC + 27oC + 30oC + 30oC + 32oC + 33oC + 32oC / 7
Average temperature of the week = 210oC / 7
Average temperature of the week =30oC
2.) Rice consumed in a school under the mid-day meal program for 5 consecutive days is 15.750 kg, 14.850 kg, 16.500 kg, 14.700kg, and 17.700 kg. Find the average rice consumption for the 5days.
Answer:
Given that,
Rice consumed in a school under the mid-day meal program for 5 consecutive days.
We have to find average rice consumption for the 5days.
Average rice consumption = Sum of All Rice consumption / Total number of days
Average rice consumption =15.750 kg + 14.850 kg +16.500 kg + 14.700kg + 17.700 kg / 5 days
Average rice consumption =15.900 kg
3.) In a village three different crops are cultivated in four successive years. The profit (in rupees) on the crops, per acre is shown in the table below.
|
year |
2005 | 2006 | 2007 | 2008 |
|
Crop |
||||
| Ground nuts | 7000 | 8000 | 7500 |
7500 |
|
Jawar |
6000 | 1000 | 8000 | 1000 |
| Millets | 9000 | 5000 | 3000 |
4000 |
(i) Calculate the mean profit for each crop over the 4 years
Answer:
Here we have to calculatemean profit for each crop over the 4 years.
Mean profit forGround nuts = Sum of profit / Total number of years
Mean profit forGround nuts =7000 + 8000 + 7500 + 7500 / 4
Mean profit for Ground nuts = 30,000/4 = 7500
Now,
Mean profit forJawar = Sum of profit / Total number of years
Mean profit forJawar =6000 + 1000 + 8000 + 1000 / 4
Mean profit for Jawar =16000/4 = 4000
Now,
Mean profit forMillets = Sum of profit / Total number of years
Mean profit forMillets =9000 + 5000 + 3000 + 4000 / 4
Mean profit for Millets = 21000/4 = 5250
(ii) Based on your answers, which crop should be cultivated in the next year?
Answer:
From result Ground nutscrop should be cultivated in the next year.
4.) The number of passengers who travelled in APSRTC bus from Adilabad to Nirmal in 4 trips in a day are 39, 30, 45 and 54. What is the occupancy ratio (average number of passengers travelling per trip) of the bus for the day?
Answer:
Given that,
The number of passengers who travelled in APSRTC bus from Adilabad to Nirmal in 4 trips in a day.
We have to findthe occupancy ratio (average number of passengers travelling per trip)
Average number of passengers travelling per trip = Sum of passengers in day/ Number of trip in day
Average number of passengers travelling per trip = 39 + 30 + 45 + 54 / 4
Average number of passengers travelling per trip = 168/4
Average number of passengers travelling per trip = 42 passengers
6.) Three friends went to a hotel and had breakfast to their taste, paying Rs.16, Rs.17 and Rs.21 respectively
(i) Find their mean expenditure.
Answer:
Given, 3 friends expenditure.
We have to find their mean expenditure.
Mean expenditure = Rs.16 + Rs.17 + Rs.21 / 3
Mean expenditure = 54 / 3
Mean expenditure = Rs.18
(ii) If they have spent 3 times the amount that they have already spent, what would their mean expenditure be?
Answer:
If they have spent 3 times the amount that they have already spent.
I.e. Rs.16 x 3 = 48
Rs.17 x 3 = 51
Rs.21 x 3 = 63
Mean expenditure = Rs.48 + Rs.51 + Rs.63 / 3
Mean expenditure = Rs.54
(ii) If the hotel manager offers 50% discount, what would their mean expenditure be?
Answer:
Given that,the hotel manager offers 50% discount
Then their mean expenditure is
Each one pay half amount.
Mean expenditure = Rs.8 + Rs.8.5 + Rs.10.5 / 3
Mean expenditure = Rs. 27/3
Mean expenditure = Rs.9
(iii) Do you notice any relationship between the change in expenditure and the change in mean expenditure.
Answer:
From above result, there is relationship between the change in expenditure and the change in mean expenditure.
Change in expenditure is directly proportional to the change in mean expenditure.
7.) Find the mean of the first ten natural numbers.
Answer:
We have to find the mean of the first ten natural numbers.
1st 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10
Mean of the first ten natural numbers = Sum of first ten natural numbers/ Total number of natural numbers
Mean of the first ten natural numbers = 1+2+3+4+5+6+7+8+9+10 / 10
Mean of the first ten natural numbers = 55
8.) Find the mean of the first five prime numbers.
Answer:
We have to find the mean of the firstfive prime numbers.
Firstfive prime numbers are 2,3,5,7 and 11.
Meanof the first five prime numbers = 2 + 3 + 5 + 7 + 11 / 5
Mean of the first five prime numbers = 28/5
Mean of the first five prime numbers = 5.6
9.) In a set of four integers, the average of the two smallest integers is 102, the average of the three smallest integers is 103, the average of all four is 104. Which is the greatest of these integers?
Answer:
Given that,
The average of the two smallest integers is 102
The average of the three smallest integers is 103
The average of all four is 104.
We have to find greatest of these integers.
We know,
Average of the two smallest integers = Sum of two smallest integers / 2
102 = Sum of two smallest integers / 2
Sum of two smallest integers = 204
Now,
We know,
Average of the 3 smallest integers = Sum of 3 smallest integers / 3
103 = Sum of 3 smallest integers / 3
Sum of 3 smallest integers = 309
But sum of two smallest integers = 204
Sum of two smallest integers + 3rd integer = 309
3rd integer = 309 – 204
3rd integer = 105
Now,
We know,
Average of the 4 smallest integers = Sum of 4 smallest integers / 4
104 = Sum of 4 smallest integers / 4
Sum of 4 smallest integers = 416
But sum of 3 smallest integers = 309
Sum of 3 smallest integers + 4th integer = 416
4th integer = 416 – 309
4th integer = 107
From above,
Greatest integer is 107.
10.) Write question to find the mean, giving suitable data.
Answer:
Students in class 1 to 5 are 34,35,45,34 and 35.
We have to mean.
Mean = Sum of students in class 1 to 5 / total number of class
Mean = 34 + 35 + 45 + 36 + 35 / 5
Mean = 185/5
Mean =37
Exercise – 2
1.) Long jumps by 7 students of a team are 98cm, 125cm, 140cm, 155cm, 174cm, 140cm and 155cm. Find the mode of the data.
Answer:
Given that,
Long jumps by 7 students of a team are 98cm, 125cm, 140cm, 155cm, 174cm, 140cm and 155cm.
We have to find the mode of the data.
The mode = Observations which occurs more number of times.
The mode =155cm Long jump occurs 2 times.
The mode = 155cm
2.) Ages of players in a cricket team are 25, 26, 25, 27, 28, 30, 31, 27, 33, 27, 29.
(i) Find the mean and mode of the data.
Answer:
Given that,
Ages of players in a cricket team are 25, 26, 25, 27, 28, 30, 31, 27, 33, 27, 29.
We have to findthe mean and mode of the data.
The mode = Observations which occurs more number of times.
The mode = 27 age occurs 3 times.
The mode = 27 age
Now,
Mean = Sum of all observations / Number of observations
Mean = 25 + 26 + 25 + 27 + 28 + 30 + 31 + 27 + 33 + 27 + 29 / 11
Mean = 28 age
3.) Find the mode of the following data.
12, 24, 36, 46, 25, 38, 72, 36, 25, 38, 12, 24, 46, 25, 12, 24, 46, 25, 72, 12, 24, 36, 25, 38 and 36.
Answer:
Here, we have to find the mode of the given data.
The mode = Observations which occurs more number of times.
From above data,
Observation 25 occurs 5 times.
The mode = 25
4.) Decide whether mean or mode is a better representative value in the following situations
(i) A shop keeper, who sells tooth paste tubes of different sizes, wants to decide which size is to be ordered more
Answer:
Here, we have to decide thatmean or mode is a better representative value of given condition.
A shop keeper, who sells tooth paste tubes of different sizes, wants to decide which size is to be ordered more.
The size of tooth paste tubes large, that tooth paste tubesordered more.
Here we find mode is a better representative value of given condition.
(ii) An invigilator wants to bring sufficient number of additional papers to the examination hall.
Answer:
Here, we have to decide thatmean or mode is a better representative value of given condition.
An invigilator wants to bring sufficient number of additional papers to the examination hall.
Here, we find mean of all students in examination hall who requires additional papers.
Here we find mean is a better representative value of given condition.
(iii) Preparation of the number of laddus for a marriage.
Answer:
Here, we have to decide thatmean or mode is a better representative value of given condition.
Preparation of the number of laddus for a marriage.
Here, we find mean of all people who present to marriage for eat laddus.
Here we find mean is a better representative value of given condition.
(iv) For finding the favourite cricketer in a class
Answer:
Here, we have to decide thatmean or mode is a better representative value of given condition.
For finding the favourite cricketer in a class
Here, we find Mode for finding the favourite cricketer in a class.
Here we find mode is a better representative value of given condition.
Exercise – 3
1.) Say true or false and why?
(i) The difference between the largest and smallest observations in a data set is called the mean.
Answer:False.
(ii) In a bar graph, the bar which has greater length may contains mode.
Answer:True.
In a bar graph, the bar which has greater length may contains mode.
(iii) Value of every observation in the data set is taken into account when median is calculated.
Answer:False.
Value of every observation in the data set isnot taken into account when median is calculated.
(iv) The median of a set of numbers is always one of the numbers
Answer:False.
The median of a set of numbers isnot always one of the numbers.
2.) The monthly income (in rupees) of 7 households in a village are 1200, 1500, 1400, 1000,
1000, 1600, 10000.
(i) Find the median income of the households.
Answer:
Given that,
The monthly income (in rupees) of 7 households in a village are 1200, 1500, 1400, 1000, 1000, 1600, 10000.
We have to findthe median income of the households.
The median income of the households =
Here, Number of observations are 7.
We 1st arrange given data in ascending order.
1000, 1000,1200,1400,1500,1600,10000
Given data has odd number of observations. Hence our median is middle number.i.e 4th number.
The median income of the households =Rs.1400
(ii) If one more household with monthly income of Rs.1500 is added, what will the median income be?
Answer:
Now,
One more household with monthly income of Rs.1500 is added.
We have to findthe new median income of the households.
Here, Number of observations are 8.
We 1st arrange given data in ascending order.
1000, 1000,1200,1400,1500,1500,1600,10000
Given data has even number of observations. Hence our median is average of middle numbers.i.e 4th and 5th number.
The median income of the households = 1400 + 1500 / 2
The median income of the households = 2900 / 2
The median income of the households = 1450
3.) Observations of a data are 16, 72, 0, 55, 65, 55, 10, and 41. Chaitanya calculated the mode and median without taking the zero into consideration. Did Chaitanya do the right thing?
Answer:
Given data,
Observations of a data are16, 72, 0, 55, 65, 55, 10, and 41
Chaitanya calculated the mode and median without taking the zero into consideration.
We have to find whether Chaitanya do the right thing or not.
We 1st find mode.
Mode = Observations which occurs more number of times.
From above data,
Observations 55 occurs 2 times.
Mode = 55
Chaitanya do the right thing when he find Mode.
Now,
We find median.
Here, Number of observations are 8.
We 1st arrange given data in ascending order.
0, 10,16,41,55,55,65,72
Given data has even number of observations. Hence our median is average of middle numbers.i.e 4th and 5th number.
The median = 41 + 55 / 2
The median = 48
When Chaitanya find median without taking the zero into consideration, median is wrong.
4.) How many distinct sets of three positive integers have a mean of 6, a median of 7, and no mode?
Answer:
Here we have to find,many distinct sets of three positive integers have a mean of 6, a median of 7, and no mode.
For no mode, there is only 1 observations. Not repeat any observations.
For median 7, middle number is always 7 because observations are odd numbers (3)
For mean 6, Sum of all observations are 18.
Distinct sets of three positive integers are,
1, 7, 10; 2, 7, 9; 3, 7, 8
Exercise – 4
1.) Draw a bar graph for the following data.
Population of India in successive census years.
| Year | 1941 | 1951 | 1961 | 1971 | 1981 | 1991 | 2001 |
| Population | 320 | 360 | 440 | 550 | 680 | 850 | 1000 |
Answer:
We have to draw bar graph for the given data of Population of India in successive census years.
