Collins Maths Solutions Class 6 Chapter 17 Data Handling
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 17, Data Handling. Here students can easily find Exercise wise solution for chapter 17, Data Handling. Students will find proper solutions for Exercise 17.1, 17.2, 17.3, 17.4 and 17.5 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Data Handling Exercise 17.1 Solution :
Question no – (1)
Solution :
Required frequency distribution table :
| Wages | No of workers | Tally marks |
| 2216 | 4 | |||| |
| 2217 | 4 | |||| |
| 2315 | 5 | |||| |
| 2332 | 6 | |||| | |
| 2347 | 6 | |||| | |
For more better understanding :
Question no – (2)
Solution :
The frequency distribution table for the given data on heights (in feet) of some of the tallest monuments in the world
| Height | No of mountains | Tally mark |
| 132 | 6 | |||| | |
| 135 | 2 | || |
| 138 | 1 | | |
| 148 | 2 | || |
| 153 | 1 | | |
| 200 | 2 | || |
| 203 | 3 | ||| |
| 240 | 1 | | |
| 260 | 2 | || |
| 289 | 2 | || |
| 157 | 3 | ||| |
| 159 | 1 | | |
| 164 | 2 | || |
| 184 | 1 | | |
| 187 | 1 | | |
| 325 | 1 | | |
| 330 | 1 | | |
| 354 | 1 | | |
| 360 | 1 | | |
| 381 | 1 | | |
| 420 | 1 | | |
Question no – (3)
Solution :
Frequency distribution table for the given data on the number of items sold in a grocery store over a month :
| No of items sold | No of days | Tally mark |
| 30 | 10 | |||| |||| |
| 33 | 2 | || |
| 34 | 3 | ||| |
| 35 | 3 | ||| |
| 42 | 2 | || |
| 45 | 4 | |||| |
| 50 | 6 | |||| | |
For more better understanding :
Data Handling Exercise 17.2 Solution :
Question no – (2)
Solution :
Pictograph for the given data :
| Factory | Amount of coal in tones [ 🚂 = 50 tones ] |
| 200 | 🚂🚂🚂🚂 |
| 550 | 🚂🚂🚂🚂🚂🚂🚂🚂🚂🚂🚂 |
| 250 | 🚂🚂🚂🚂🚂 |
| 300 | 🚂🚂🚂🚂🚂🚂 |
| 500 | 🚂🚂🚂🚂🚂🚂🚂🚂🚂🚂 |
For more better understanding :
Question no – (3)
Solution :
(a) How many chapatis were made on Monday?
= (4 × 15)
= 200 chapatis mode on Monday?
(b) How many fewer chapatis were made on Thursday than on Tuesday?
= 1500 chapatis mode on thus day them on Thursday
(c) How many chapatis were made on Thursday and Friday altogether?
= 600 chapatis were made on Thursday and Friday togethers.
(d) On which day was the maximum number of chapatis made? How many?
= Friday, and 600 chapatis
Data Handling Exercise 17.3 Solution :
Question no – (1)
Solution :
(a) What is the information shown in the bar graph?
= The bar graph gives information about the animals liked by the students of class 6
(b) How many students like the lion?
= 6 students like lion
(c) Name the animal that is liked by most of the students.
= Tiger is liked by most of the student
(d) What is the ratio of the students who like zebra to the students who like giraffe?
= Ratio
= 10 : 12
= 5 : 6
Question no – (2)
Solution :
(a) Name the tallest mountain peak. What is its height?
= Mt. Everest is the tallest peak.
= Its, height 880 meter.
(b) Which mountain peak is 7500 m tall?
= Nanda Devi 7500 m tall
(c) How much taller is Mt. Everest than Himadri?
= 8800 – 6500
= 2300 m taller is Mt. Everest than Himadri.
(d) Arrange these mountain peaks in the decreasing order of their heights.
= Mount Everest > Kanchanjunga > Namgapabat > Nanda Devi > Himadri
Question no – (3)
Solution :
(a) On which day did Balbir earn the least amount? How much?
= Friday Balbir earn the least amount and Rs 250
(b) How much did Balbir earn on Monday?
= Rs 500 Balbir earn on the Monday
(c) The total amount he earned on Tuesday and Friday is equal to the amount he earned on Thursday.
= False
(d) How much less money did Balbir earn on Friday than on Thursday?
= (600 – 250)
= 350 On Friday than on Thurs day.
Question no – (4)
Solution :
Bar graph according to the given data,
Question no – (5)
Solution :
Required Bar graph on the given data :
Data Handling Exercise 17.4 Solution :
Question no – (1)
Solution :
As we know that, Mean = sum of the given number/number of observations
(a) 3, 8, 13, 18, 23, 28, 33
∴ Mean = 3 + 8 + 13 + 18 + 23 + 28 + 33/7
= 126/7
= 18
(b) 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20
∴ Mean = 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 + 20/11
= 132/11
= 12
(c) 21, 28, 35, 42, 49, 56, 70, 77
∴ Mean = 21 + 28 + 35 + 42 + 49 + 56 + 70 + 77/8
= 378/8
= 47.25
(d) 101, 103, 107, 109, 113
∴ Mean = 101 + 103 + 107 + 109 + 113/5
= 533/5
= 106.6
(e) 82, 76, 70, 64, 58, 52, 46, 40
∴ Mean = 82 + 76 + 70 + 64 + 58 + 52 + 46 + 40/8
= 488/8
= 61
(f) 14, 32, 59, 67, 89, 105, 200, 350
∴ Mean = 14 + 32 + 59 + 67 + 89 + 105 + 200 + 350/8
= 916/8
= 114.5
Question no – (2)
Solution :
According to the question,
The wages of a worker from Monday to Saturday are,
750, 600, 550, 1000, 850, 900
∴ Mean wage of the worker per day,
= 750 + 600 + 550 + 1000 + 850 + 900/6
= 46050/6
= 775
Therefore, the mean wage of the worker per day is Rs 775
Question no – (3)
Solution :
As per the question,
Rohan played outdoor for 30 min, 1 h, 45 min, 50 min and 40 min
∴ Rahul’s mean play time,
= 30 + 60 + 45 + 50 + 40/5
= 225/5
= 45
Therefore, Mean play time per day 45 min.
Data Handling Exercise 17.5 Solution :
Question no – (1)
Solution :
(a) 3, 6, 7, 4, 5, 9, 8, 9,7
= n = 9 is odd
Therefore median,
= value of [1/2 (n + 1)] the observation
= 1/2 (9 + 1)
= 5th observation
= 5
Hence, the median will be 5
(b) 20, 22, 30, 32, 40, 42, 50, 62,70
= n = 9
median = value [1/2 (n + 1)]th observation
= 1/2 (9 + 1) = 1/2 × 10
= 5th observation
= 40
Thus, the median will be 40
(c) 75, 35, 48, 47, 50, 55, 58, 60, 70
= n = 9
median = value of [1/2 (n + 1)]th observation
= 50
So, the median will be 50
(d) 51, 53, 55, 57, 59, 61, 63, 65
= n = 8 is even
median = 1/2 [n/2)th observation + (n/2 + 1)th observation
= 1/2 [8/2 = 4th observation + (4 × 1) = 5th observation]
= 1/2 × [57 + 59]
= 1/2 × 116 = 58
Therefore, the median will be 58
(e) 80, 82, 84, 86, 88, 90, 92
= n = 7 odd
median = value of [1/2 (7 + 1)th observation
= 4th observation
= 86
Thus, the median will be 86
Question no – (2)
Solution :
Here, n = 10 even
∴ Median = 1/2 910/2th observation + 910/2 + 1)th observation)
= 1 1/2 (5th + 6th)
= 1/2 (4th + 45)
= 85/2
= 42.5
Therefore, the median cost will be Rs 42.5
Question no – (3)
Solution :
Here, n = 10 even
∴ Median = 1/2 (n/2th observation + (n/2 + 1) th observation)
= 1/2[10/2th + (5 + 1)th]
= 1/2 (5th + 6th)
= 1/2 × (29 + 28)
= 1/2 × 57
= 28.5
Hence, the median mark will be 28.5
Question no – (4)
Solution :
Here, n = 5 odd
Median = value of 1/2 (n + 1)th observation
= 1/2 (5 + 1) th observation
= 1/2 × 6 th observation
= 3rd observation
= 40
Therefore, the median cost of the books is Rs 40
Previous Chapter Solution :
👉 Chapter 1 👈