Aptitude · Percentages · Model 5
The majority is the gap, not the winner's share
A candidate takes 57% of the votes and wins by 42,000. That 42,000 is not 57% of anything — it is the 14-point gap between winner and loser. Getting that one reading right is the entire model.
Set the winner’s share and the majority →01 The idea
Why the gap is double the lead
Two candidates contest an election. The winner takes 57% of the votes, so the loser takes 43%, and the winner’s majority — the margin of victory — is the difference between them: 57 − 43 = 14%. If that majority is 42,000 votes, then 14% is 42,000, so 1% is 3,000 and the total is 300,000.
The number to be careful with is 14, not 57 and not 7. Students reach for the winner’s 57%, or for the 7 percentage points above half. The majority counts the votes the winner has more than the loser, and every vote above 50% is worth two in that gap — one gained and one denied.
That doubling is worth stating as a rule: for two candidates, a winner on w% has a majority of (2w − 100)%. At 57% the majority is 14%; at 60% it is 20%; at 72% it is 44%. It is always twice the lead over half.
The harder variant adds invalid votes, and it adds a second base. If 10% of the votes cast are invalid, then the candidates’ percentages are shares of the valid votes while the 10% is a share of the votes cast. Two different denominators in one question, and mixing them is what the variant is set to catch.
02 Worked example
57% of the votes, a majority of 42,000
The standard form of the model. Two candidates contested an election. The winning candidate secured 57% of the total votes and won by a majority of 42,000. Find the total number of votes polled.
Check it by counting both candidates: 57% of 300,000 is 171,000 and 43% is 129,000, and the difference is exactly 42,000. Note the trap answers this question generates: matching 42,000 to the winner’s 57% gives about 73,684, and matching it to the 7-point lead over half gives 600,000. Both are in the options, and only the 14% reading is right.
03 The method
The variants, including the second base
The two-candidate case is one reading. The last two rows are where a second denominator appears.
| Winner’s share | Loser | Majority as % of valid |
|---|---|---|
| 57% | 43% | 14% |
| 60% | 40% | 20% |
| 70% | 30% | 40% |
| 72% | 28% | 44% |
| Three candidates | — | third = 100 − the other two |
| With i% invalid | — | valid = (100−i)% of cast |
| Majority = winner’s share | wrong | it is the difference |
05 Cheat sheet
Model 5 on one page
One reading, one relation, and the second base that the harder version adds.
| Case | Route | Worked |
|---|---|---|
| Loser’s share | 100 − w | 57% → 43% |
| Majority % | 2w − 100 | 57% → 14% |
| Total from the majority | maj × 100/gap | 42,000 → 3,00,000 |
| Winner’s votes | w% of the total | 57% → 1,71,000 |
| Third candidate | 100 − the other two | 40 + 36 → 24% |
| Valid from cast | (100 − i)% of cast | 10% invalid → 90% |
| Majority = w% | the standard trap | gives 73,684 not 3,00,000 |
06 Where & why
Where Model 5 shows up
A small, reliably set model whose entire difficulty is one reading.
The standard form. Both the correct total and the winner’s-share misreading are among the options every time.
The harder variant, and the two-base structure is what is being tested rather than the arithmetic.
Simpler than it looks — the third share is a hundred minus the other two, and then it is one multiplication.
Sales targets, poll leads, market share. Whenever a question gives a share and a margin, the margin is the doubled gap.
07 Interview questions
What gets asked
Nine, and the first two are the whole model.
A winner takes 57% of the votes and wins by 42,000. Find the total votes.
Why is the majority 14% rather than 57% or 7%?
Give the general relation.
A candidate gets 40% and loses by 298 votes. Find the total.
A candidate on 60% wins by 14,000. How many votes did the WINNER poll?
How do invalid votes change things?
Ten per cent of votes are invalid, the winner takes 70% of the valid and wins by 1,800. How many were cast?
How do you handle three candidates?
What is the fastest way to spot the trap in this model?
08 Practice problems
Six elections
In each one, write down the gap percentage before touching the votes. Two of these add invalid votes.