Figure it Out — Proportional Reasoning in Context — Proportional Reasoning

1. 13 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 170

   The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?

   Hint. How many weeks are there in a year? Set up a proportion between distance and time.

   Step 1 — Write the given information in full. 940 million = 940 × 1,000,000 = 940,000,000 km in one year. There are 52 weeks in a year.

   Step 2 — Set up the proportion. Distance is proportional to time, so

   distance in a year : 52 weeks :: distance in a week : 1 week

   940,000,000 / 52 = x / 1

   Step 3 — Solve. x = 940,000,000 ÷ 52 = 18,076,923 km (to the nearest kilometre)

   Step 4 — Check the size of the answer. A rough check: 940 million ÷ 50 would be about 18.8 million, and dividing by 52 rather than 50 gives slightly less. Our answer of about 18.1 million sits just below that ✓

   Multiplying back: 52 × 18,076,923 = 939,999,996 km, which is 940 million to the accuracy given ✓

   Note on precision. The starting figure is stated as approximately 940 million, so quoting the answer as about 1.8 × 10⁷ km — roughly 18 million kilometres a week — is at least as honest as the full digit string. An answer can never be more precise than the data it came from.

   ✦ About 18,076,923 km, or roughly 1.8 × 10⁷ km per week, since 940,000,000 ÷ 52 = 18,076,923.

2. 24 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 170

   A mason is building a house in the shape shown. He needs both the outer walls and the inner wall separating two rooms. To build 10 feet of wall he needs approximately 1450 bricks. How many bricks would he need for the house? Assume all walls have the same height and thickness.

   Hint. Find the total length of wall first — do not forget the inner dividing wall.

   Step 1 — Add up every wall length. From the plan, the wall segments are 12 + 12 + 12 + 15 + 9 + 15 + 9 + 9 + 9 + 6

   Adding in order: 12, 24, 36, 51, 60, 75, 84, 93, 102, 108.

   Total wall length = 108 feet

   The trap in this question is forgetting the inner dividing wall. It is not part of the outer boundary, but the mason still has to build it and it still needs bricks.

   Step 2 — Set up the proportion. Bricks are proportional to wall length, since the height and thickness are the same throughout:

   10 feet : 1450 bricks :: 108 feet : x bricks

   10 / 1450 = 108 / x

   Step 3 — Solve. x = (1450 × 108) ÷ 10 = 156,600 ÷ 10 = 15,660 bricks

   Step 4 — Check by a simpler route. 1450 bricks per 10 feet means 145 bricks per foot. 145 × 108 = 15,660 ✓

   Reducing to a unit rate like this is usually the fastest way through a proportion question, and it gives an independent check on the cross-multiplication.

   A sensible note for the mason: since 1450 is stated as approximate, he would order somewhat more than 15,660 — bricks break, and running short mid-wall is costly.

   ✦ The walls total 108 feet, and at 145 bricks per foot that needs 15,660 bricks.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp107.pdf). Four 'Figure it Out' blocks sit on pages 165, 170, 175 and 176-178. Every numeric answer here was independently recomputed before comparison with the book's printed answer key — including the acre-to-square-feet conversions (1 acre = 43,560 sq ft), the Lilavati saffron problem, the population densities of Delhi and Mumbai, and the cupro-nickel coin costings. THREE NOTES ARE FLAGGED IN PLACE: (1) question 6 on page 165 depends on counting bricks in a printed pattern, so its counts are cited from the key rather than derived; (2) questions 4, 5 and 7 on page 165 are measurement and drawing activities with no single numeric answer — the method is given instead; (3) the printed key's working for question 2 on page 176 contains a garbled fraction ('x = 34/9'), so the solution is re-derived cleanly here while reaching the same final answer of 4 buses.. Questions are referenced from the NCERT textbook for identification.