Estimating first, so a wrong answer is visible

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What to assume they know

Multiply as whole numbers, then count the decimal digits — multiplying a decimal by a whole number, and by 100
Why multiplying by a decimal below 1 shrinks a number — which side of 1 a factor sits on, and what that does to a product
Dividing when the divisor has a decimal — quotients that exceed their dividend, and rescaling a known division
Divisibility by 4, by 100 and by 400, and counting how many numbers in a range are divisible by a given number
Writing a situation as an expression before computing it — writing a situation as an expression before evaluating it
Reading a decision tree with yes and no branches

What they should be able to do

Explain why a small decimal cannot be dismissed until you know what it is multiplied by
Compute the accumulated shortfall of a 365-day calendar over 100, 1000 and 10,000 years
Follow each of the chapter's three leap-year decision trees and classify a given year by it
Write, as an expression, how many days a stated run of calendar years holds
Compare what a scheme delivers with what is actually needed, and state whether it overshoots or undershoots
Rescale a known product or quotient to answer a new one without multiplying again
Order a set of products and quotients by size using only the position of each factor relative to 1
Judge whether an answer is plausible before checking it in detail

Where it usually goes wrong

"A small decimal can be ignored." The section exists to refute this. 0.2422 is small; 0.2422 multiplied by a thousand is 242.2 days of drift — getting on for eight months. Whether a number is negligible depends entirely on what it will be multiplied by.
"Every fourth year is a leap year." Only under the first of the three schemes. The chapter replaces that rule twice and ends with a three-level test. An explanation that stops at the first tree teaches the rule the chapter spends four pages correcting.
"The final rule is exact." It is not, and the book says so: over a thousand years the calendar still falls about two-tenths of a day short, and the calendar makers decided to live with it. Presenting the Gregorian rule as perfect throws away the section's conclusion.
"Estimating is what you do when you cannot be bothered to compute." In this section the estimate is what tells you whether the exact computation is worth doing and whether its answer is credible. Both totals are always computed; the sizing decides what to make of them.
"Checking means doing the same sum twice." The chapter never re-does a sum. It computes the same quantity by two independent routes — days the scheme delivers, days the Earth needs — and compares. That is what makes an error visible.
"Q11 needs a calculator." It needs one observation: multiplying by 7.9682 enlarges, multiplying by 0.942368 shrinks slightly, and dividing does the opposite of each. The six expressions can be ordered from that alone.

Questions to check understanding

Given a small daily or yearly discrepancy, compute its accumulated effect over a stated number of years
Classify a given year as 365 or 366 days by the three-test rule, and justify each branch
Write, as an expression, how many days a stated run of calendar years holds, then evaluate it
Derive a family of products or quotients from one given whole-number fact
Fill blanks with supplied digits to hit a stated target product
Arrange products and quotients of a common number in increasing order without computing them
Comparison shopping: which of two differently sized packets is cheaper
Decide how many objects of a given thickness fit in a given length, and how much is left over

Examples worth working on the board

The opening figure (Part II, §4.4, p.88). Checked against the printed page. A dark space panel: the Sun at the top, a chequered finish flag with a torn calendar page reading DEC 31 pinned to it, and a cartoon Earth just short of the line saying it has not finished yet. The whole argument of the section is in that picture, so give it real attention.
The two numbers everything rests on (Part II, §4.4, p.88). One circuit of the Sun takes 365.2422 days; the calendar counts 365. The leftover is 0.2422. The book then multiplies that by 100.
First scheme, over four years and over a century (Part II, §4.4, p.89). Checked against the printed page. A two-row table covering eight consecutive years: a header row labelled Year 1 to Year 8, a row of day counts underneath reading 365 or 366 with the 366s falling on years 4 and 8, and a trailing dots column carrying both rows onward. Inputs: 4 × 365 + 1 against 4 × 365.2422; then 100 × 365 plus one day for each of the years divisible by 4, against 100 × 365.2422.
The three decision trees (Part II, §4.4, pp.89, 90, 91). Checked against the printed page. Each is a hand-set branching diagram with yes and no on the arms and a day count at every leaf. The first asks only about 4 — one decision node, two leaves. The second asks about 100 first and then about 4 — two nodes, three leaves. The third asks about 400, then 100, then 4 — three nodes and four leaves, counted on the printed page. A cartoon of four calendar makers round a table, one of them announcing the new rule in a speech bubble, goes with each scheme, but not on the same page as every tree: the first sits at the foot of Part II p.88 under "Making an Adjustment", with its tree overleaf at the top of p.89, while the second and third stand directly above their trees on pp.90 and 91. The same four people turn up a fourth time on p.92, above the ten-thousand-year Try This, three of them walking out; their expressions change each time.
Counting the leap years in a century (Part II, §4.4, p.90). Twenty-five years in a hundred are divisible by 4, and the hundredth itself must be taken out under the second scheme, leaving twenty-four long years and seventy-six short ones. The book writes this as a single expression built from 100 over 4 and 100 over 100 before evaluating it.
A thousand years under the final scheme (Part II, §4.4, pp.91–92). The book sorts the thousand years into four groups — divisible by 400; divisible by 100 but not by 400; divisible by 4 but not by 100; and all the rest — and gives the sizes of the first three as 2, 8 and 240, deducing the fourth.
The Try This on p.92 (Part II, §4.4). The same calculation over 10,000 years, with the difference to be found and a fix to be proposed. No answer is printed anywhere in the chapter. This is the best single exercise for the topic's thesis.
"Figure it Out" Q2, the two number lines (Part II, §4.4, p.93). Checked against the printed page, including a zoom. Two horizontal lines, each with eleven ticks — a long tick at each end, a long tick in the middle, and short ticks between — so each line is cut into ten equal intervals. The first runs from 3.1 to 3.2, the second from 2.15 to 2.17, and on each an arrow points down at the tick one step past the middle. Both endpoints and the interval count are the inputs; the value at the arrow is what the student works out.
"Figure it Out" Q8 and Q9 (Part II, §4.4, p.94). Q8 supplies 756 ÷ 36 = 21 and asks for rescaled quotients. Q9 is a five-by-five grid of divisions with three cells filled — 41, 4.1 and 4100 — and everything else blank. Both are answerable by watching powers of ten rather than by dividing.
"Figure it Out" Q10 (Part II, §4.4, p.94). Checked against the printed page. Five boxes to be filled with the digits 2, 4, 5, 8 and 0, laid out as a two-digit number with one decimal place multiplied by a one-digit number with one decimal place. Five separate targets are asked for, including a product nearest to 100 and a product nearest to 5. Those two cannot be attacked except by sizing candidates first.
"Figure it Out" Q11 (Part II, §4.4, p.95). Six expressions built from 245.05, 7.9682 and 0.942368 — two products, two quotients and the two bare numbers — to be put in increasing order. Every one can be placed by asking which side of 1 the second number sits on.
Two other exercise inputs worth using (Part II, §4.4, pp.93–94). A 210 g packet at ₹70.5 against a 110 g packet at ₹33.25, asking which is cheaper (Q1); and books 2.5 cm thick, 80 of them wanted, on a shelf 160 cm long (Q4). Both reward a rough size check before any exact arithmetic.

Figures to have open

A calendar strip that can be marked in days, so 0.2422 of a day and 24.22 days can be shown at the same scale.
An orbit diagram with a fixed calendar marker and a drifting Earth. Standard schematic; the book's own space panel is drawn artwork and should be redrawn.
A branching decision tree that can grow from one level to three, with yes and no arms and a day count at every leaf. This is the section's backbone and must be reusable in sections 4, 6 and 8.
A two-row table of consecutive years: the year labels along the top, the day counts in the row underneath.
Two number lines cut into ten equal intervals, with labelled endpoints and a movable arrow, for the Q2 figure.
A five-by-five grid for Q9 and six sortable cards for Q11.
No photograph and no data set from the textbook is required for this topic.

Where this sits in the book

NCERT Ganita Prakash, Class 7, Part II, printed Chapter 4 "Another Peek Beyond the Point", §4.4 "Look Before You Leap!", pp.88–93 — the opening panel and the 0.2422 leftover (p.88), the subheading "Making an Adjustment" and the first decision tree with its century check (pp.88–89), "Making Another Adjustment" with the second tree and the expression (pp.90–91), the third tree and the thousand-year count (pp.91–92), and "Making Yet Another Adjustment" with the ten-thousand-year Try This (pp.92–93)
Same part, same chapter, §4.4, "Figure it Out", pp.93–95 — questions 1, 2, 4, 8, 9, 10 and 11
Same part, same chapter, §4.3, p.87 — questions 3 and 4, which are the same rescaling move on a smaller scale
Backward pointers: Writing a situation as an expression before computing it, Why multiplying by a decimal below 1 shrinks a number and Dividing when the divisor has a decimal