Aptitude · Percentages · Model 6

Add the shortfall back before you do anything else

A candidate scores 220 and fails by 20. The pass mark is 240, not 220 — and once that one line is written the rest is a single division. The two-subject version needs one more idea: anyone who passed both has been counted twice.

Choose the shape and watch the characteristic move →

Failing by b marks means the pass mark is score + b. Write that line first, every time.

01 The idea

Two shapes, and one counting rule

A candidate must secure 40% to pass. He scores 220 marks and fails by 20. The 220 is not the pass mark — he was 20 short, so the pass mark is 240. And 240 is the 40%, so the paper is out of 600.

That is the first shape, and the whole trick is writing 220 + 20 = 240 before anything else. Every version of it works the same way: recover the true pass mark, match it to the pass percentage, scale up.

The second shape gives you both figures as percentages instead. “A student needs 33% and got 25%, failing by 40 marks.” Here the two percentages share a base, so they subtract directly: the 8-point gap is 40 marks, so the paper is out of 500. No conversion needed, which makes this the quicker variant.

The third shape is different in kind. “80% passed English, 85% passed Maths, 75% passed both.” You cannot add 80 and 85, because everyone who passed both has been counted in each figure. Passing at least one is 80 + 85 − 75 = 90%, so 10% failed both. That subtraction of the overlap is inclusion-exclusion, and it is the only formula in this lesson.

For two subjects, passed at least one = A + B − both. The overlap was counted twice, so it comes out once.

Pass markThe marks actually needed. If a candidate scored s and failed by b, it is s + b — never just s.

Percentage-point gapWhen both the requirement and the score are percentages of the same maximum, their difference is a percentage of that maximum and equals the shortfall in marks.

Inclusion-exclusion|A ∪ B| = |A| + |B| − |A ∩ B|. For pass rates: passed either = passed A + passed B − passed both.

02 Worked example

Scored 220, failed by 20, pass mark 40%

The first shape, worked with the key line made explicit. In an examination a candidate must secure 40% of the marks to pass. A candidate who gets 220 marks fails by 20 marks. Find the maximum marks.

Recover the pass markFailing by 20 means he was 20 marks short of what was needed, so add them back.220 + 20 = 240 marks needed to pass

Match it to the pass percentageThe rules say 40% is required, and we now know 40% in marks.40% of the maximum = 240

Find one per centOne division.1% = 240 / 40 = 6 marks

Scale to the full paperA hundred of those.100% = 6 × 100 = 600 marks

Check it both waysForty per cent of 600 should be the pass mark, and the candidate should be 20 short.40% of 600 = 240, and 240 − 220 = 20 ✓

The trap answer here is treating the 220 as the 40%, which gives a maximum of 550 — a perfectly plausible-looking number that will be among the options. The protection is mechanical: whenever a question says “fails by” or “short by”, write the addition on its own line before you touch the percentage.

03 The method

The three shapes side by side

Deciding which shape you are in takes one read of the question. Each has one characteristic move.

Shape 1 (score and shortfall in marks): max = (s + b) × 100 / pass%. Shape 2 (both as percentages): max = b × 100 / (pass% − got%). Shape 3 (two subjects): either = A + B − both.

Shape 3 runs in either direction. Given the two pass rates and the failed-both rate, you can find the overlap: if 70% passed English, 80% passed Maths and 10% failed both, then 90% passed at least one, so both = 70 + 80 − 90 = 60%. That reverse reading is set as often as the forward one, and it is the same single equation rearranged.

The question gives	Shape	Characteristic move
Score in marks, fails by marks	1	`add the shortfall`
Needs x%, got y%, short by marks	2	`subtract the percentages`
Two subjects and the overlap	3	`A + B − both`
Two subjects and failed-both	3 reversed	`solve for the overlap`
Separate boy and girl pass rates	counting	`count failures, then divide`
Treating the score as the pass mark	`wrong`	gives 550, not 600
Adding the two pass rates	`wrong`	double-counts the overlap

05 Cheat sheet

Model 6 on one page

Three shapes, their moves, and the two errors each is set to catch.

Case	Route	Worked
Score + shortfall	`max = (s+b) × 100/pass%`	240 at 40% → 600
Two percentages	`max = b × 100/(pass−got)`	40 at 8% → 500
Two subjects, either	`A + B − both`	80+85−75 = 90%
Two subjects, failed both	`100 − either`	10%
Reverse: find the overlap	`both = A + B − either`	70+80−90 = 60%
Boys and girls separately	`count failures, then divide`	800/1800 = 44.44%
Score treated as pass mark	`wrong`	550 instead of 600

Add the shortfall firstWrite score + shortfall on its own line before touching any percentage. It is one line and it prevents the model’s main trap.

Percentages of the same base subtract“Needs 33%, got 25%” gives an 8-point gap directly, because both are shares of the same maximum. No conversion is needed.

Never add two pass ratesAnyone passing both is counted twice, so the sum overcounts. Subtract the overlap once — that is inclusion-exclusion and it is the only formula here.

06 Where & why

Where Model 6 shows up

Two of the three shapes are pure reading; the third brings in a genuine counting principle that recurs well beyond this chapter.

SSC CGL · RRB · CHSL

“Fails by n marks”

The most-set version. The trap answer from treating the score as the pass mark is always in the options.

Bank PO

Two-subject pass rates

Inclusion-exclusion, forwards and backwards. Adding the two rates gives over 100%, which is itself the signal that an overlap must be removed.

Counting versions

Separate boy and girl pass rates

“1,000 boys and 800 girls, 60% and 50% pass.” Count the failures in each group, add, then divide by the combined total — not an average of the two rates.

Set theory and probability

The same inclusion-exclusion

|A ∪ B| = |A| + |B| − |A ∩ B| is the identical rule. Meeting it here makes it familiar when it reappears with Venn diagrams.

Two habits cover this whole model: write the shortfall addition on its own line, and never add two pass rates without removing the overlap.

07 Interview questions

What gets asked

Ten, covering all three shapes and the reverse reading of the third.

A candidate needs 40% to pass, scores 220 and fails by 20. Find the maximum marks.

Six hundred. The pass mark is 220 + 20 = 240, and that is the 40%, so 1% is 6 and the maximum is 600. Treating the 220 as the 40% gives 550, which is the trap answer.

Why add the shortfall?

Because failing by 20 means the candidate was 20 marks below what was needed. The score is not the requirement; the requirement is the score plus the shortfall. That one line is the whole first shape.

A student needs 33%, gets 25% and fails by 40 marks. Find the total marks.

Five hundred. Both figures are percentages of the same maximum, so they subtract directly: an 8-point gap. That 8% is 40 marks, so 1% is 5 and the total is 500. No conversion is needed, which makes this variant quicker.

80% passed English, 85% passed Maths and 75% passed both. What percentage failed both?

Ten per cent. Passing at least one is 80 + 85 − 75 = 90%, so 10% failed both. The overlap has to be subtracted because everyone who passed both was counted once in each of the first two figures.

Why can’t you just add 80 and 85?

Because that totals 165%, which is impossible as a share of the students. The excess is exactly the double-counted overlap. Subtracting the 75% who passed both removes one of the two copies and brings it back to a sensible 90%.

70% passed English, 80% passed Maths, and 10% failed both. What percentage passed both?

Sixty per cent. If 10% failed both then 90% passed at least one, so 90 = 70 + 80 − both, giving both = 60%. This is the same equation rearranged, and it is set as often as the forward version.

There are 1,000 boys and 800 girls. 60% of boys and 50% of girls pass. What percentage failed?

About 44.44%. Count the failures: 40% of 1,000 is 400 boys and 50% of 800 is 400 girls, so 800 failed out of 1,800, which is 44.44%. Averaging the two pass rates would give 45%, which is wrong because the groups are different sizes.

Why is averaging the two rates wrong there?

Because it weights the two groups equally when they are not equal in size. A weighted average would work, but counting the actual failures is simpler and harder to get wrong. The two coincide only when the groups are the same size.

65% passed Maths, 48% passed Physics and 30% passed both. What percentage failed both?

Hard

In an examination 75% passed in Maths, 60% passed in Physics and 55% passed in Chemistry. Of these, 45% passed in Maths and Physics, 35% in Physics and Chemistry, and 40% in Maths and Chemistry, while 25% passed in all three. (a) Find the percentage who passed in at least one subject. (b) Find the percentage who failed in all three. (c) State the general three-set inclusion-exclusion formula you used, and explain why the triple overlap is ADDED back rather than subtracted.

Follow-up

Part (c) is the real content. The three pairwise subtractions remove the triple overlap three times having added it three times, so it must be added back once to be counted at all. Getting that reasoning — rather than memorising the alternating signs — is what makes the formula safe to use.

Show the hint

For three sets: add the singles, subtract the pairs, add the triple. Track what happens to a student who passed all three at each stage.