Question 1 - Page 49 - Tenth - Chapter 3 Class 7 - A peek beyond the Point (Ganita Prakash)

Transcript

Question 1 – Page 49 Arrange these lengths in increasing order (a) 9/10 (b) 1 7/10 (c) 130/10 (d) 13 1/10 (e) 10 5/10 (f) 7 6/10 (g) 6 7/10 (h) 4/10 We convert all lengths into normal fractions (a) 𝟗/𝟏𝟎 (b) 𝟏 𝟕/𝟏𝟎= (1 × 10 + 7)/10=𝟏𝟕/𝟏𝟎 (c) 𝟏𝟑𝟎/𝟏𝟎 (d) 𝟏𝟑 𝟏/𝟏𝟎= (13 × 10 + 1)/10=𝟏𝟑𝟏/𝟏𝟎 (e) 𝟏𝟎 𝟓/𝟏𝟎= (10 × 10 + 5)/10=𝟏𝟎𝟓/𝟏𝟎 (f) 𝟕 𝟔/𝟏𝟎= (7 × 10 + 6)/10=𝟕𝟔/𝟏𝟎 (g) 𝟔 𝟕/𝟏𝟎= (6 × 10 + 7)/10=𝟔𝟕/𝟏𝟎 (h) 𝟒/𝟏𝟎 Thus, we need to compare (a) 𝟗/𝟏𝟎 (b) 𝟏𝟕/𝟏𝟎 (c) 𝟏𝟑𝟎/𝟏𝟎 (d) 𝟏𝟑𝟏/𝟏𝟎 (e) 𝟏𝟎𝟓/𝟏𝟎 (f) 𝟕𝟔/𝟏𝟎 (g) 𝟔𝟕/𝟏𝟎 (h) 𝟒/𝟏𝟎 Since denominators are same, writing numerators in increasing order 4, 9, 17, 67, 76, 105, 130, 131 So, numbers in increasing order areThus, we need to compare (h) 𝟒/𝟏𝟎, (a) 𝟗/𝟏𝟎 , (b) 𝟏𝟕/𝟏𝟎 , (g) 𝟔𝟕/𝟏𝟎 , (f) 𝟕𝟔/𝟏𝟎 , (e) 𝟏𝟎𝟓/𝟏𝟎 , (c) 𝟏𝟑𝟎/𝟏𝟎, (d) 𝟏𝟑𝟏/𝟏𝟎