UPSC CSE · CSAT · Numeracy and data interpretation · TOPIC LESSON
Use bounds, denominators and subgroup weights to decide which comparisons the data really support.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This original common-core lesson supports its declared paper scope. Optional subjects, qualifying language papers and the interview are outside this resource.
LEARN THROUGH ORIGINAL PRACTICE
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01 / UNDERSTAND & APPLY
Absolute difference subtracts values; a ratio divides one by another; relative change divides the difference by a stated base. These are not interchangeable. An increase from 40 to 50 is 10 units or 25% of 40, while the reverse decrease is 20% of 50.
Totals and per-unit rates can rank groups differently. Thirty events in 300 opportunities is a larger event count but a lower rate than 20 in 100. Before evaluating performance, define the relevant outcome, exposure, period and comparison direction.
A zero base makes an ordinary relative-change formula undefined. A negative base can also make intuitive percentage language confusing; state the exact calculation and context rather than assuming every relative percentage has the usual positive-base interpretation.
ORIGINAL PRACTICE · 1
The second has more events, but the first has the higher rate:20% versus 10%.
Compute 20/100 and 30/300 separately. The denominators differ, so the larger numerator does not establish the larger rate.
ORIGINAL PRACTICE · 2
25% increase and 20% decrease.
The same absolute difference 10 uses different bases:10/40 and 10/50. Direction matters.
02 / UNDERSTAND & APPLY
Estimation replaces values with nearby convenient ones to check scale or make a decision robust to small errors. It is useful only relative to the precision the question needs. Rounding 198 to 200 may be harmless for a rough magnitude check but not for a threshold separated by one unit.
Use compatible rounding and ask whether errors may reinforce each other. For a positive product, rounding both factors upward makes an upper estimate; for a positive ratio, rounding the numerator upward and denominator downward increases the estimate. A rigorous bound requires knowing which direction each approximation takes.
Do not report an estimate as exact. If the task asks the nearest integer and the estimate is too rough to decide it, perform the exact calculation or tighten the bounds.
ORIGINAL WORKED EXAMPLE
Estimate 198×51/9.9, then calculate it exactly. Is the rough estimate enough if the task asks for the exact integer result?
Answer Rough estimate 1000; exact result 1020. Required precision determines whether estimation alone is sufficient.
03 / UNDERSTAND & APPLY
For positive half-up rounding to the nearest 10, a reported value x represents actual values in[x−5,x+5). The lower halfway point rounds up to x, while the upper halfway point rounds up to x+10. State the tie rule: another convention can change endpoint inclusion.
A continuous measurement can approach an excluded endpoint without attaining it. An integer quantity has discrete possibilities, so its smallest or largest possible difference can differ from the continuous case. Identify the domain before giving an interval or an attained minimum.
Subtracting reported values gives a central estimate, not necessarily the actual difference. Derive the range from both input intervals and check whether its boundary combinations are permitted.
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ORIGINAL WORKED EXAMPLE
Positive quantities A and B are reported as 80 and 70 after half-up rounding to the nearest 10. Find the possible range of A−B for continuous quantities, and for integer quantities.
Answer Continuous differences form(0,20); integer differences are 1 through 19. The reported difference 10 is not the only possible actual difference.
ORIGINAL PRACTICE · 1
45 is permitted;55 is not.
The interval is[45,55). The lower tie rounds upward to 50, while 55 rounds upward to 60.
ORIGINAL PRACTICE · 2
Because a discrete domain can have attained endpoint-near values that a continuous open interval does not.
In the 80/70 example, continuous differences approach 0 without reaching it. Integer differences cannot lie between 0 and 1, so their minimum is 1. The rounding rule alone does not specify the underlying domain.
04 / UNDERSTAND & APPLY
Suppose numerator N and denominator D vary independently within finite positive bounds 0<a≤N≤b and 0<c≤D≤d. The ratio N/D increases as N increases and decreases as D increases, so its lower bound is a/d and upper bound b/c. Positivity, especially c>0, keeps the quotient defined and the bound finite.
For closed independent intervals, the boundary pairs are feasible, so the bounds are attained minimum and maximum. With open endpoints, the same expressions may be infimum and supremum only: values can approach them without attaining them. Specify endpoint inclusion rather than calling every bound a smallest or largest possible value.
Dependencies can narrow the range. If N=D within[5,10], the ratio is always 1 even though independent endpoint calculations would suggest 1/2 to 2. Before combining the smallest numerator with largest denominator, ensure that pair can actually occur together.
ORIGINAL WORKED EXAMPLE
First let N∈[40,60] and D∈[5,10] vary independently. Find the minimum and maximum of N/D. Then keep N∈[40,60] but change D to(5,10]. What changes?
Answer Closed independent ranges give minimum 4 and maximum 12. Excluding D=5 removes attainment of the upper bound, giving[4,12).
ORIGINAL PRACTICE · 1
[2,6].
The positive closed endpoints allow 12/6=2 and 18/3=6. Independence permits those combinations, and all intermediate ratios are feasible.
05 / UNDERSTAND & APPLY
A pooled success rate is total successes divided by total attempts. Subgroup rates must be weighted by their attempt counts. If methods serve different mixes of easy and hard cases, their pooled rates may compare that mix as well as their performance within each subgroup.
A reversal does not contradict arithmetic. It indicates that the groups being pooled have different weights. Nor does the reversal by itself establish a causal effect: case difficulty, selection and other conditions need examining before declaring one method intrinsically superior.
Always supply the full numerators and denominators. Percentages alone can conceal the composition needed to understand the reversal.
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ORIGINAL WORKED EXAMPLE
Hypothetical method A succeeds in 9 of 10 easy cases and 30 of 100 hard cases. Method B succeeds in 80 of 100 easy cases and 2 of 10 hard cases. Compare subgroup and pooled success rates.
Answer A has higher observed rates within both subgroups, while B has the higher pooled rate. The case mix explains how both statements can hold.
ORIGINAL PRACTICE · 1
54%.
Successes are 9 and 45, total 54 among 100 cases. The unweighted mean 70% gives equal influence to groups that differ ninefold in size.
06 / UNDERSTAND & APPLY
An index of 100 represents a chosen baseline. An index of 150 means the measured quantity is 1.5 times that baseline, or 50% higher; it does not tell you the absolute quantity unless the baseline amount is known.
Successive percentage growth multiplies factors. The arithmetic average of two percentage rates is not their combined change over both periods. If an annualized constant rate is required, a different calculation involving the overall factor and number of periods is needed; do not label a simple mean as equivalent growth.
Index comparisons also require the same base definition. Two series both labelled 100 at different dates or for different quantities cannot be interpreted as equal absolute levels.
ORIGINAL WORKED EXAMPLE
An index starts at 100, grows 20% in one period and falls 10% in the next. Find the final index and overall change.
Answer Final index 108 and overall increase 8%.
ORIGINAL PRACTICE · 1
The quantity is 50% above its baseline, or 1.5 times the baseline amount.
It does not necessarily mean 150 physical units. If the baseline were 200 units, index 150 would represent 300 units.
07 / UNDERSTAND & APPLY
Rounded inputs limit justified output precision. Computing many decimal places from coarse measurements does not create new information. Report an appropriate approximation or an interval that reflects the input uncertainty.
Absolute error is the nonnegative magnitude |estimate − reference|, measured in the original units. The signed error estimate − reference also records direction. Relative error divides the absolute error by the nonzero magnitude of the reference value. A percentage-point difference instead compares percentage levels directly. For a reference rate 20% estimated as 22%, the absolute difference is 2 percentage points and the relative error is 2/20=10%.
Near a threshold, a good scale estimate can still be inadequate. If a quantity is near a pass/fail cutoff, calculate exactly or bound it tightly enough that the whole permitted interval lies on one side.
ORIGINAL PRACTICE · 1
No; the exact product is 2005.98, which does exceed 2003.
Compute 99.8×20+99.8×0.1=1996+9.98=2005.98. The rough estimate is useful for scale but falls on the other side of a nearby threshold.
08 / UNDERSTAND & APPLY
A robust conclusion holds throughout the permitted data range. In the rounded 80/70 case, A>B is robust even though the exact difference is unknown. In a ratio problem, an upper bound may rule out a proposed value without determining the actual ratio.
Check all assumptions behind a bound: positivity, finiteness, independent variability, rounding convention and endpoint inclusion. If a dependency restricts feasible pairs, use that restriction rather than an impossible endpoint combination.
For aggregated data, test whether changed weights explain a ranking before asserting a mechanism. A sound final sentence names what was compared and what remains unresolved: observed subgroup rates, pooled rates and causal effectiveness are separate claims.
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