Practicing with Maths Mela Class 5 Solutions Chapter 5 Far and Near Question Answer NCERT Solutions improves a student’s confidence in the subject.
Class 5 Maths Chapter 5 Far and Near Question Answer Solutions
Far and Near Class 5 Maths Solutions
Class 5 Maths Chapter 5 Solutions
Let Us Find (NCERT Pg 57)
Question 1.
Identify the appropriate units for measuring each of the following.
Answer:
(i) Since, The India Gate is very tall structure, so we use metres.
Thus, the height of India Gate is 42 m.
(ii) Since, a handkerchief is a small piece of cloth, so we use centimetres.
Thus, the length of a handkerchief is 40 cm.
(iii) Since, the depth of a well is a large measurements, so it is expressed in metres.
Thus, the depth of well is 50 m.
(iv) Since, the length of a mobile phone is a smaller measurement, so it is expressed in centimetres.
Thus, the length of mobile phone is 13 cm.
(v) Since, the length of an elephant’s trunk can vary, but it’s a relatively large measurement, so it is expressed is metres. Thus, the length of an elephant’s trunk is 2 m.
(vi) The distance between two buttons on a shirt is a small measurement, so it is expressed in centimetres.
Thus, the distance between two buttons on a shirt is 5 cm.
Students can write alternate answer as per their choice.
Different Units but Same Measure (NCERT Pg 57-58)
Question 2.
Shikha and Sonu are measuring the lengths of saris and stoles in the village weaving centre. Find which measures represent the same sari or stole. You can take help of the double number line below.
Match the following
Answer:
100 cm × 2 = 200 cm
We know that 1 cm = \(\frac{1}{100}\) m
∴ 400 cm = \(\frac{400}{100}\) m = 4 m
500 cm = \(\frac{500}{100}\) m = 5 m
and 700 cm = \(\frac{700}{100}\) m = 7 m
We know that 100 cm = 1 m
(i) (c) 204 cm = 200 cm + 4 cm
= 2 × 100 cm+4 cm
= 2 m + 4 cm [∵ 1 m = 100 cm]
= 2 metre 4 cm
(ii) (a)
540 cm = 500 cm + 40 cm
= 5 × 100 cm + 40 cm
= 5 m + 40 cm [∵ 1 m = 100 cm]
= 5 metre 40 cm
(iii) (e)
750 cm = 600 cm + 150 cm
= 6 × 100 cm + 150 cm
= 6 m + 150 cm [∵ 1 m = 100 cm]
= 6 metre 150 cm
(iv) (d)
240 cm = 200 cm + 40 cm
= 2 × 100 cm + 40 cm
= 2 m+40 cm [∵ 1 m = 100 cm]
= 2 metre 40 cm
(v) (b)
404 cm = 200 cm + 204 cm
= 2 × 100 cm + 204 cm
= 2 m + 204 cm [∵ 1 m = 100 cm]
= 2 metre 204 cm
Let Us Compare (NCERT Pg 58)
Question 3.
Ritika is comparing the lengths of different rods. Compare them using < , = , > signs.
(i) 456 cm ______ 5 m
(ii) 55 cm + 200 cm ______ 200 cm + 54 cm
(iii) 6 m 5 cm ______ 6 m 50 cm
(iv) 2 m 150 cm ______ 3 m 50 cm
(v) 238 cm ______ 138 cm+1 m
Answer:
(i) We know that, 1 m = 100 cm
→ 5 m = 5 × 100 cm = 500 cm
∵ 456 cm < 500 cm
∴ 456 cm < 5 m
(ii) We have, 55 cm + 200 cm
= 255 cm and 200 cm + 54 cm = 254 cm
∵ 255 cm > 254 cm
∴ 55 cm + 200 cm > 200 cm + 54 cm
(iii) We have, 6 m 5 cm = 6 × 100 cm + 5 cm
[∵ 1 m = 100 cm]
and 6 m cm = 600 cm + 5 cm = 605 cm
= 600 + 500 cm + 50 cm
∵ 605 cm < 650 cm
∴ 6 m 5 cm < 650 cm
(iv) We have, 2 m 150 cm
= 2 × 100 cm + 150 cm and = 200 cm + 150 cm
= 350 cm 3 m 50 cm
= 3 × 100 cm + 50 cm
= 300 cm + 50 cm = 350 cm
∴ 2 m 150 cm = 3 m 50 cm
(v) We have, 138 cm + 1 m
= 138 cm + 100 cm = 238 cm
∴ 238 cm = 138 cm + 1 m
Question 4.
World’s tallest statue
(i) What is the difference between the height of the tallest statue in the world and the Statue of Liberty?
(ii) Identify the statues whose heights have the least difference.
(iii) Identify the statues whose heights have the largest difference.
(iv) The height of which statue will be equal to the height of the Statue of Unity, if it is doubled?
Answer:
(i) From the figure, the tallest statue of the world is statue of Unity. Given, the height of statue of Unity = 182 m and the height of statue of Liberty = 93 m So, the required difference = 182 m – 93 m = 89 m
(ii) From the figure, the statues with the least height difference are Statue of Liberty (93 m) and The Motherland Calls (91 m) and their height difference is 2 m, which is the least difference.
(iii) From the figure, the statues with the largest height difference are the statue of Unity (182 m) and Christ the Redeemer (38 m) and their height difference is 144 m, which is the largest difference.
(iv) We have, height of statue of Unity = 182 m the height of the Motherland calls = 91 m If we double 91 metres (91 × 2); we get 182 metres, which is exactly the height of the statue of Unity. Hence, the height of the Motherland calls will be equal to the height of the statue of Unity, if it is doubled.
Let Us Do (NCERT Pg 59)
Question 5.
Measure 100 m and 200 m on your school playground or any other place in and around your school, using a Long Tape. Mark these points and draw a straight line. Walk on the lines and count the number of steps. Use this relationship between the number of steps taken and distance walked to find distances around you for at least 3 locations. Wherever possible, walk and find the number of steps. Otherwise, find the distance and estimate the number of steps.
(i) Identify and write the locations that are the nearest and the farthest from your home.
Nearest location ______.
Farthest location ______.
(ii) Write the distances obtained above in increasing order. ______.
(iii) Name a location that is equal to or more than 1,000 m from your home.
Answer:
Do yourself
Let Us Explore (NCERT Pg 59)
Question 6.
Answer:
We know that 1 km=1000 m
(ii) Given, the length of rope = 100 m
So, number of ropes needed to make 1 km
= \(\frac{1000}{100}\) = 10
(iii) Given, the length of rope = 10 m
So, number of ropes needed to make 1 km
= \(\frac{1000}{10}\) = 100
(iv) Given, the length of rope = 200 m
So, number of ropes needed to make 1 km
= \(\frac{1000}{200}\) = 5
(v) Given, the length of rope = 500 m
So, number of ropes needed to make 1 km
= \(\frac{1000}{500}\) = 2
(vi) Given, the length of rope = 250 m
So, number of ropes needed to make 1 km
= \(\frac{1000}{250}\) = 4
Kilometer Race (NCERT Pg 60)
Question 7.
Sheena and Jennifer are helping to organise a 3-km race. Help them with the arrangements for the race.
(i) Water stations are to be arranged after every 500 m. How many water stations must be set up? At what positions from the starting point will these water stations be placed?
(ii) Children need to stand at an interval of 300 m to direct the runners. How many children are needed? At what positions from the starting point will the children be standing?
(iii) Red and blue flags are to be placed alternately at every 50 m. How many red and blue flags are needed till the finish line?
Answer:
(i) Given, the race is 3 km (3000 m) long and water stations are arranged after every 500 m.
So, the number of intervals = \(\frac{3000}{500}\) = 6 Hence, total 6 water stations must be set up.
And the positions of the water stations from the staring point are 500 m, 1000 m, 1500 m, 2000 m, 2500 m and 3000 m.
(ii) Since, the children are needed at every interval of 300.
So, the number of intervals = \(\frac{3000}{300}\) = 10
Hence, total 10 children are needed to stand.
And the positions of children from the starting point are 300 m, 600 m, 900 m, 1200 m, 1500 m, 1800 m, 2100 m, 2400 m, 2700 m and 3000 m.
(iii) Since, red and blue flags are placed alternately at every 50 m.
So, the number of intervals = \(\frac{3000}{50}\) = 60
Hence, 30 red flags and 30 blue flags are needed till the finish line.
Let Us Do (NCERT Pg 60-61)
Question 8.
Longest Train Journey
The longest train journey in India is by The Vivek Express which runs from Dibrugarh in Assam to Kanniyakumari in Tamil Nadu. Look at the stations on the route shown in the table below and answer the questions.
(i) The total length of the route from Dibrugarh to Kanniyakumari is ______ km.
(ii) The distance between Vijayawada and Jalpaigur1 read is ______.
(iii) Distance between Vijayawada and Visakhapatnam is ______.
(iv) Which two stations are farther apart-Guwahati and Dimapur or Bhubaneswar and Jalpaiguri Road?
(v) What is the distance between Guwahati and Coimbatore JN?
Answer:
(i) The total length of the route from Dibrugarh to Kanniyakumari is 4187 km.
(ii) Given, the distance of Vijaywada from Dibrugarh = 2800 km and the distance of Jalpaiguri Road from Dibrugarh = 983 km
So, the distance between Vijaywada and Jalpaiguri Road = 2800 km-983 km = 1817 km
(iii) Given, the distance of Vijaywada from Dibrugarh = 2800 km and the distance of Visakhapatnam from Dibrugarh = 2450 km. So, the distance between Vijaywada and Visakhapatnam = 2800 – 2450 = 350 km.
(iv) Given, the distance of Guwahati from Dibrugarh = 556 km and the distance of Dimapur from Dibrugarh = 306 km So, the distance between Guwahati and Dimapur = 556 km – 306 km = 250 km Now, the distance of Bhubaneswar from Dibrugarh = 2007 km and the distance of Jalpaiguri Road = 983 km
So, the distance between Bhubaneswar and Jalpaiguri Road = 2007 – 983 = 1024 km
∵ 1024 km > 250 km
Therefore, Bhubaneswar and Jalpaiguri Road are farther apart.
(v) Given, the distance of Guwahati from Dibrugarh = 556 km and the distance of Coimbatore JN from Dibrugarh = 3675 km
So, the distance between Guwahati and Coimbatore JN = 3675 – 556 = 3119 km
Let Us Measure (NCERT Pg 61)
Question 9.
Measure the lines in the design and write their measurements in cm and mm.
Answer:
Do Yourself
Let Us Do (NCERT Pg 62)
Question 10.
Soak some seeds of whole moong or black or white chana overnight, Next morning, take them out and wrap them in a moist cloth to sprout them. Over the next 4 days, take out one seed each day and measure the length of sprout. For ease of measurement, you can either place the seed on a paper and mark the length of the sprout or use a thread to find its length.
Answer:
Do yourself.
Let Us Draw (NCERT Pg 62)
Question 11.
(i) Draw lines of the following lengths in your notebook using a scale.
(a) 5 cm 5 mm
(b) 3 cm 6 mm
(c) 8 cm 3 mm
(d) 36 mm
(e) 67 mm
(ii) How did you draw lines of lengths 36 mm and 67 mm? Share your thoughts in class.
Answer:
(i) Do yourself
(ii) The key is to understand that 1 centimetre (cm) equals to 10 millimetres (mm).
Therefore, lengths given in millimetres can be converted to centimetres and millimetres
(e.g. 36 mm = 3 cm2 6 mm and 67 mm = 6 cm 7 mm) to easily locate the corresponding marks on a standard ruler or scale.
Let Us Do (NCERT Pg 63-64)
Question 12.
Fill in the blanks appropriately in the double number lines given below.
Answer:
(i) 7 cm = 7 × 10= 70 mm [∵ 1 cm = 10 mm]
10 mm × 7 = 70 mm
15 cm = 15 × 10 = 150 mm
250 mm=\(\frac{250}{10}\) = 25 cm [∵ 1 mm=\(\frac{1}{10}\) cm]
320 mm=\(\frac{320}{10}\) = 32 cm
450 mm=\(\frac{450}{10}\) = 45 cm
50 cm = 50 × 10 = 500 mm
(ii) 4 m = 4 × 100 = 400 cm [∵ 1 m=100 cm]
1 m = 1 × 100 = 100 cm
100 cm × 4 = 400 cm
10 m = 10 × 100 = 1000 cm
1200 cm = \(\frac{1200}{100}\) = 12 m [∵ 1 cm = \(\frac{1}{100}\) m]
1400 cm = \(\frac{1400}{100}\) = 14 m
1700 cm = \(\frac{1700}{100}\) = 17 m
21 m = 21 × 100=2100 cm
[∵ 1 m = 100 cm]
(iii) 9 km = 9 × 1000 = 9000 m
[∵ 1 km = 1000 m]
1000 m × 9 = 9000 m
25 km = 25 × 1000 = 25000 km
41000 m = \(\frac{41000}{1000}\) = 41 km
[∵ 1 m=\(\frac{1}{1000}\) km]
55000 m = \(\frac{55000}{1000}\) = 55 km
67000 m = \(\frac{67000}{1000}\) = 67 km
82 km = 82 × 1000 = 82000 m.
Question 13.
Use your understanding from above to fill in the blanks appropriately.
(i) 4 cm 5 mm = ______ mm
(ii) 89 mm = ______ cm ______ mm
89 mm = 80 mm + 9 mm = 8 cm 9 mm
(iii) 234 cm = ______ mm
(iv) 514 mm = ______ cm ______ mm
(v) 6 m 34 cm = ______ cm
(vi) 20 m 12 cm = ______ cm
(vii) 397 m= ______ cm
(viii) 5,792 cm = _____ m _______ cm
5,792 cm = 5,700 cm + 90 cm = 57 m 92 cm
(ix) 9,108 cm = ______ m ______ cm
(x) 34 km = ______ m
(xi) 6,870 m = ______ km ______ m
(xii) 10,552 m = ______ km ______ m
(xiii) 29 km 30 m = ______ m
(xiv) 32 km 359 m = ______ m
Answer:
(i) 4 cm 5 mm = 4 × 10 mm + 5 mm [∵ 1 cm = 10 mm]
= 40 mm + 5 mm = 45 mm
(iii) 234 cm = 234 × 10 mm = 2340 mm
[∵ 1 cm=10 mm]
(iv) 514 mm=510 mm+4 mm
= 51 cm 4 mm [∵ 1 mm = \(\frac{1}{10}\) cm]
(v) 6 m 34 cm = 6 × 100 cm+34 cm
[∵ 1 m = 100 cm]
= 600 cm + 34 cm = 634 cm
(iv) 20 m 12 cm = 20 × 100 cm + 12 cm
[∵ 1 m = 100 cm]
= 2000 cm + 12 cm
= 2012 cm
(vii) 397 m = 397 × 100 cm = 39700 cm
[∵ 1 m =100 cm]
(ix) 9108 cm=9100 cm+8 cm
= 91 m 8 cm [∵ 1 cm = \(\frac{1}{100}\) m]
(x) 34 km = 34 × 1000 m = 34000 m
[∵ 1 km = 1000 m]
(xi) 6870 m = 6000 m + 870 m
= 6 km 870 m [∵ 1 m = \(\frac{1}{1000}\) km]
(xii) 10552 m = 10000 m + 552 m
= 10 km 552 m [∵ 1 m = \(\frac{1}{1000}\) km]
(xiii) 29 km 30 m = 29 × 1000 m + 30 m
[∵ 1 km = 1000 m]
= 29000 m + 30 m = 29030 m
(xiv) 32 km 359 m = 32 × 1000 m + 359 m
= 32359 m [∵ 1 km = 1000 m]
Let Us Do (NCERT Pg 66)
Question 14.
Rani has two red-coloured ribbon rolls, one of length 3 m 75 cm and another 2 m 25 cm long. How much ribbon does she have?
Answer:
Given, length of the first ribbon = 3 m 75 cm
= 3 × 100 cm + 75 cm [∵ 1 m = 100 cm]
= 300 cm + 75 cm = 375 cm
and the length of the second ribbon
= 2 m 25 cm
= 2 × 100 cm + 25 cm
= 200 cm + 25 cm = 225 cm
So, the total length of ribbon
= 375 cm + 225 cm = 600 cm
= 6 m
[∵ 1 cm = \(\frac{1}{100}\) m]
Hence, Rani have total 6 m ribbon.
Question 15.
The distance from Bhopal to Sanchi is 48 km 700 m. Bhadbhada Ghat waterfall is on the way and 17 km 900 m away from Bhopal. How far is Sanchi from the waterfall?
Answer:
Given, distance from Bhopal to Sanchi
= 48 km 700 m
= 48 × 1000 m + 700 m
= 48700 m [∵ 1 km = 1000 m]
and the distance from Bhopal to the waterfall
= 17 km 900 m
= 17 × 1000 m + 900 m
= 17900 m
Since, waterfall is on the way.
So, the distance from Sanchi to waterfall
= Distance from Bhopal to Sanchi – Distance from Bhopal to the waterfall
= 48700 m-17900 m
= 30800 m
= 30000 m + 800 m
= 30 km 800 m [∵ 1 m = \(\frac{1}{1000}\) km]
Question 16.
Gulmarg Gondola in Gulmarg, Kashmir is the second longest and second highest cable car in the world. It is divided into two sections. The first section covers 2 km 300 m and the second section covers 2 km 650 m. What is the total distance covered by the cable car?
Answer:
Given, the distance covered by the first section = 2 km 300 m and the distance covered by the second section = 2 km 650 m
So, the total distance covered by the cable car
= Distance covered by the first section + Distance covered by the second section
= 2 km 300 m + 2 km 650 m
= 4 km 950 m
Question 17.
Circle the bigger length and find the difference.
(i) 11 mm and 1 cm
Difference ______.
(ii) 26 mm and 2 cm
Difference ______.
(iii) 20 cm and 201 mm
Difference ______
(iv) 1,020 mm and 1 m
Difference ______.
(v) 2 m and 245 cm
Difference ______.
(vi) 5,678 m and 6 km
Difference ______.
(vii) 6 km 1,480 m and 7 km 479 m
Difference ______.
Answer:
(i) We have, 11 mm and 1 cm
∵ 1 cm = 10 mm
∴ 11 mm > 10 mm → 11 mm > 1 cm
Now, difference = 11 mm – 10 mm = 1 mm
(ii) We have, 26 mm and 2 cm
∵ 1 cm = 10 mm
∴ 2 cm = 20 mm
So, 26 mm > 20 mm
→ 26 mm > 2 cm
Now, difference = 26 mm – 20 mm = 6 mm
(iii) We have 20 cm and 201 mm
∵ 1 cm = 10 mm
∴ 20 cm = 200 mm
So, 200 mm < 201 mm
→ 20 cm < 201 mm
Now, difference = 201 mm – 200 mm = 1 mm
(iv) We have, 1020 mm and 1 m
∵ 1 m = 1000 mm
∴ 1020 mm > 1000 mm → 1020 mm > 1 m
Now, difference = 1020 mm – 1000 mm
= 20 mm
(v) We have, 2 m and 245 cm
∵ 1 m = 100 cm
∴ 2 m = 200 cm
So, 200 cm < 245 cm
→ 2 m < 245 cm
Now, difference = 245 cm – 200 cm = 45 cm
(vi) We have, 5678 m and 6 km
∵ 1 km = 1000 m
∴ 6 km = 6000 m
So, 5678 m < 6000 m
→ 5678 m < 6 km
Now, difference = 6000 m – 5678 m = 322 m
(vii) We have, 6 km 1480 m and 7 km 479 m
∵ 1 km = 1000 m
∴ 6 km 1480 m = 6 × 1000 m + 1480 m
= 6000 m + 1480 m
= 7480 m and 7 km 479 m
= 7 × 1000 m + 479 m
= 7000 m + 479 m
= 7479 m
So, 7480 m > 7479 m
→ 6 km 1480 m > 7 km 479 m
Now, difference = 7480 m – 7479 m = 1 m
Multiplying and Dividing Lengths (NCERT Pg 66-67)
Question 18.
A shop sells cloth for making bags at ₹ 100 for 5 m. How much money is needed to buy a 1 m cloth?
If 5 m cloth costs ₹ 100, then a 1 m cloth costs 100 ÷ 5 = ₹ 20.
Now, use the double number line to find the cost of the cloth or the length of cloth that we can buy at a particular cost.
Answer:
We know that the cost of 1 m cloth = ₹ 20
5 m × 2 = 10 m
5 m × ₹ 20 = ₹ 100
10 m × ₹ 20 = ₹ 200
₹ 100 × 2 = ₹ 200
20 m × ₹ 20 = ₹ 400
40 m × ₹ 20 = ₹ 800
₹ 2000 ÷ ₹ 20 = 100
Question 19.
Anita is making an embroidery on the border of a sari. She needs a 1 m long thread to embroider a 50 cm sari. How much thread would she need for a 5 m sari border?
A 1 m long thread costs ₹50. How much money will be needed to buy the thread?
Answer:
Given, thread needed to embroider a 50 cm sari = 1 m = 100 cm
So, thread needed for 1 cm sari = \(\frac{100}{50}\) = 2 cm
Hence, the thread needed for 5 m sari
= 5 m × 2 cm
= 5 × 100 cm × 2 cm [∵ 1 m=100 cm]
= 500 cm × 2 cm
= 1000 cm
= 10 m [∵ 1 cm = \(\frac{1}{100}\) m]
Now, given the cost of 1 m long thread = ₹ 50 Since, we need total 10 m thread.
So, the cost of 10 m thread = 10 ×(₹ 50)
= ₹ 500
Hence, ₹ 500 will be needed to buy the thread.
Question 20.
A road 12 km 600 m long is being laid in a town. The workers lay an equal length of road each day and complete the work in 6 days.
How much road-laying work is done on each day?
Answer:
Given, the length of the road
= 12 km 600 m
= 12 × 1000 m + 600 m [∵ 1 km = 1000 m]
= 12000 m + 600 m = 12600 m
Since, the workers lay an equal length of road each day and complete the work in 6 days.
So, length of road laid each day
= Total length of road + Number of days.
= 12600 m ÷ 6 = 2100
= 2000 m + 100 m [∵ 1 m = \(\frac{1}{1000}\) km]
= 2 km 100 m
Hence, 2 km 100 m of road-laying work is done on each day.
Let Us Estimate (NCERT Pg 68)
Question 21.
Estimate the following. Share your reasoning in class.
(i) The height of the tallest building in your neighbourhood. What did you use as a reference to estimate the height?
(ii) The height of the tallest tree in your neighbourhood. What did you use as a reference?
(iii) The depth of a well or swimming pool in your neighbourhood. How did you find out?
Answer:
Do yourself.
Measure your Height (NCERT Pg 69)
Question 22.
Stand against a wall and mark your height. Measure the distance between the floor and the marked point in feet and inches.
Similarly, other students in the class can also measure their heights.
Find out who is the tallest student in your class. What is his or her height in feet and inches?
Answer:
Do yourself.