Question 1

Question 1(ii)Convert the following statements into R.F.(a) 25 cm on the map = 5 km on ground.(b) 2½ inches on the map = 5 miles on ground.(c) 7 cm on the map = 63000 metres on ground.(d) 5 cm = 500 metres.(e) 15 cm = 6 km.

Accepted Answer

(a) The scale is 25 cm to 5 km = 25 cm to 5,00,000 cm(Since 1 km = 1,00,000 cm)R.F. = Distance on the map / Distance on the groundR.F. = 25 / 5,00,000R.F. = 1:20,000(b) The scale is 2½ inches to 5 miles = 2.5 inches to 3,16,800 inches(Since 1 mile = 63,360 inches)R.F. = Distance on the map / Distance on the groundR.F. = 2.5 / 3,16,800R.F. = 1:1,58,400(c) The scale is 7 cm to 63000 m = 7 cm to 63,00,000 cm(Since 1 m = 100 cm)R.F. = Distance on the map / Distance on the groundR.F. = 7 / 63,00,000R.F. = 1:9,00,000(d) The scale is 5 cm to 500 m = 5 cm to 50,000 cm(Since 1 m = 100 cm)R.F. = Distance on the map / Distance on the groundR.F. = 5 / 50,000R.F. = 1:10,000(e) The scale is 15 cm to 6 km = 15 cm to 6,00,000 cm(Since 1 km = 1,00,000 cm)R.F. = Distance on the map / Distance on the groundR.F. = 15 / 6,00,000R.F. = 1:40,000

Question 2

Question 2Convert the following representative fractions into statements:(i) 1: 63360 (to show miles).(ii) 1: 1000000 (to show kilometres).(iii) 1: 100 (to show metres).(iv) 1: 10 (to show metres).(v) 1: 200,000 (to show kilometres).

Accepted Answer

(i) R.F. = 1:63360That means 1 inch on the map = 63360 inches on the groundThe scale is 1 inch to 1 mile(Since 1 mile = 63,360 inches)(ii) R.F. = 1:10,00,000That means 1 cm on the map = 10,00,000 cm on the groundThe scale is 1 cm to 10 km(Since 1 km = 1,00,000 cm)(iii) R.F. = 1:100That means 1 cm on the map = 100 cm on the groundThe scale is 1 cm to 1 m(Since 100 cm = 1 m)(iv) R.F. = 1:10That means 1 cm on the map = 10 cm on the groundThe scale is 10 cm to 1 m(Since 1 m = 100 cm)(v) R.F. = 1:2,00,000That means 1 cm on the map = 2,00,000 cm on the groundThe scale is 1 cm to 2 km(Since 1 km = 1,00,000 cm)

Question 3

Question 3Draw linear scales and mark the following distances on them.(i) 1 cm = 1 km (9 km).(ii) 1 cm = 100 km (1100 km).(iii) 1: 500000 (13 km).(iv) 1: 20000 (250 metres).

Accepted Answer

(i) Given, 1 cm = 1 kmHence, scale is 1 cm to 1 km.(ii) Given, 1 cm = 100 kmHence, scale is 1 cm to 100 km.3-(iii) Given, R.F. = 1: 500000We know that: 1 km = 1,00,000 cm1 cm = 5 kmHence, scale is 1 cm to 5 km.(iv) Given, R.F. = 1: 20000We know that: 1 m = 100 cm1 cm = 200 mHence, scale is 1 cm to 200 m.

Question 4

Question 4The distance between New Delhi Station and Safdarjung Enclave bus stop is 20 km. On the map of Delhi, it has been shown by a line of 3.5 cm. Draw the linear scale of the map and calculate the R.F.

Accepted Answer

According to the given question,The scale is 3.5 cm to 20 km = 3.5 cm to 20,00,000 cm(Since 1 km = 1,00,000 cm)R.F. = Distance on the map / Distance on the groundR.F. = 3.5 / 20,00,000R.F. = 1 : 571428.57R.F. ≈ 1 : 5714291 cm on graph ≈ 5.7 Km on Map

Question 5

Question 5On the map of India the cartographer forgot to draw the scale of the map. The student who knows the distance between Meerut and Delhi (70km), was asked to complete the scale. How will the student draw the scale if he measures the distance between Delhi and Meerut on the map to be 5 cm? Give the procedure, draw the scale and find out the R.F.

Accepted Answer

According to the given question,The scale is 5 cm to 70 km = 5 cm to 70,00,000 cm(Since 1 km = 1,00,000 cm)R.F. = Distance on the map / Distance on the groundR.F. = 5 / 70,00,000R.F. = 1:14,00,0001 cm on graph = 14 Km on Map

Question 6

Question 6The map given below depicts the houses of four friends A, B, C and D. Find the distances using the main road and the kutcha road to answer the following questions.(a)	Which child would reach school first if all, A, B, and C, leave their houses at the same time and walk at the same speed?(b)	Which child lives closest to the school, according to the road distance?(c)	How far must D walk to reach the village well?(d)	How far is B's house from the bus depot?(e)	How much distance will C have to walk to reach the railway station?(f) In what direction is C's house from the school?

Accepted Answer

The scale is 1 cm = 10,000 cm1 cm = 10,000 / 100 m∴ 1 cm = 100 metres(a) B will reach the school first.Reason — If A, B, and C leave for school at the same time and walk at the same speed, we can conclude from the map that B will reach the school first, as B lives closer to the school than A and C.(b) D lives closest to the school according to the road distance.(c) The distance between D and the village well is 12 cm, measured with the help of the thread. If 1 cm is equal to 100 metres, then 12 cm will be equal to 1.2 Km. So D has to walk 1.2 Km to reach the village well.(d) With the help of the thread the distance between Bus depot and B's house is 12 cm. So, it is 1.2 Km far from B's house.(e) By thread, the distance is 9 cm. So the distance between C and Railway station is:9 x 100 = 900 metres.(f) C's house is in the North-West direction from the school.

Question 7

Question 1Study the given map extract and answer the following questions:(a) Find the area in sq. km. of the area enclosed within the Eastings 64 to 64 and Northings 32 to 35.(b) Find the compass direction of the following:(i) Gulabganj from Pithapura(ii) Pamera from Malgaon.(c) Find the direct distance of the metalled road from the Post Office at Pamera till it intersects with National Highway- NH 168 in grid square 6432.

Accepted Answer

(a)Number of complete squares = 3Since each grid square measures 1 km x 1 km, the area enclosed within the Eastings 64 to 64 and Northings 32 to 35 will be 3 sq. km.(b)(i) Gulabganj lies to the south-west of Pithapura.(ii) Pamera lies to the North-east of Malgaon.(c)Distance between them = 4.6 cmGiven scale: 2 cm = 1 km1 cm = 0.5 km4.6 cm = 0.5 x 4.6 = 2.3 km

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