LESSON 13.4 — Numerical Ability

A. Standard Map

Why this lesson is longest: Each topic carries its own formula set and a distinct set of base-change traps. GATE GA numerical questions are uniformly 1–2 marks and reward formula recall + correct base selection over complex derivations. Every mark here is recovered by applying the right formula to the right number.

B. Mechanism in Words

1. Read the question and tag the topic type before computing. The problem type determines the formula. Misidentifying “two-speed journey” as a “relative speed” problem leads to the wrong equation setup. Name the type: percentage, average, work-rate, TSD, and so on.

2. Write down what is given and what is asked separately. GATE numerical problems often embed a distractor quantity — a number that is given but not used in the answer, or a rate that must be converted before use. Listing given vs asked takes five seconds and prevents 80% of errors.

3. Convert all units to a consistent system before substituting. Speed in km/h and time in seconds produce nonsense. Distance in km and speed in m/s produce nonsense. Convert first; compute second.

4. Choose the base carefully on percentage problems. Every percentage is a fraction of something. Identify the base before writing the formula. When in doubt: “what is the denominator of this percentage?” That is your base.

5. Set up an algebraic equation for problems with an unknown. Do not trial-and-error through options — it wastes time and fails on 2-mark questions. Write: [known] + [expression in x] = [total], then solve.

6. Verify with a sanity check. After computing, ask: Is the answer physically possible? (Negative time, probability > 1, and rate > 100% are impossible.) If the answer violates common sense, recheck the base or the formula.

C. Core Concept Explanations

C1. Percentages — Base Change, Successive %, % Error

Core formula: Percentage change = [(New − Old) / Old] × 100

The base is always the original (Old) value, never the new value.

Successive percentage changes: If a value changes by a% and then by b% (b is negative for a decrease):

This formula replaces the slow “calculate intermediate step, then second step” approach.

Example: Price rises 20%, then falls 15%.
Net = 20 + (−15) + (20 × −15)/100 = 20 − 15 − 3 = +2% (net increase of 2%)

Example: Price rises 25%, then falls 20%.
Net = 25 + (−20) + (25 × −20)/100 = 25 − 20 − 5 = 0% (no net change)

Percentage error: Error% = |Measured − True| / True × 100. The base is always the true value, not the measured value — a common trap.

“What % of X is Y?”: Answer = (Y / X) × 100. Identify X clearly before dividing.

“Increase X by P%”: New value = X × (1 + P/100). Do not add P to X directly — that only works when X = 100.

C2. Ratios and Proportions — Mixing, Partnership

Part : Part vs Part : Whole: If A : B = 3 : 5, then A is 3/(3+5) = 3/8 of the total. Converting part-to-part ratio to part-to-whole is required for mixing and sharing problems.

Mixing (alligation shortcut): To find the ratio in which two ingredients at prices p₁ and p₂ are mixed to get a mixture at price pₘ:

where p₁ < pₘ < p₂.

Example: Milk at Rs. 20/litre and water (Rs. 0/litre) mixed to get a mixture worth Rs. 15/litre.
Ratio = (15 − 0) : (20 − 15) = 15 : 5 = 3 : 1 (milk : water)

Partnership profit split: Partners split profit in proportion to their investment × duration, not just their investment.
If A invests Rs. 5000 for 12 months and B invests Rs. 8000 for 9 months:
A’s share : B’s share = 5000 × 12 : 8000 × 9 = 60000 : 72000 = 5 : 6

Proportionality: If A ∝ B, then A/B = constant. If A ∝ 1/B, then A × B = constant. These are the two standard forms — identify which applies before setting up the equation.

C3. Averages — Weighted Average Trap

Simple average: Sum of all values ÷ count of values.

Weighted average: Used when groups have different sizes.

The simple average trap: Adding two percentages from groups of different sizes and dividing by 2 gives the wrong combined percentage. GATE AR 2023 Q3 directly exploited this: universities (total 504) had 7% Grade III; colleges (total 25,951) had 23% Grade III. Simple average = (7 + 23)/2 = 15% — wrong. Weighted answer = (0.07 × 504 + 0.23 × 25951) / (504 + 25951) ≈ 22.7% — correct.

Overlapping groups subtraction (GATE AR 2022): When two groups share common members, subtract totals to isolate individual values.
– Average of {M, N, S} = 4000 → Total = 12,000
– Average of {N, S, P} = 5000 → Total = 15,000
– P − M = 15,000 − 12,000 = 3,000

Average with one added/removed: New average = (Old sum ± added/removed value) / new count.
If average of 5 numbers is 20, and one number (35) is removed:
New average = (5 × 20 − 35) / 4 = (100 − 35) / 4 = 65 / 4 = 16.25

C4. Profit, Loss, Discount — Marked Price vs Cost Price

Definitions:
– CP = Cost Price (what the seller paid)
– MP = Marked Price / List Price (labelled price, before discount)
– Discount = MP − SP (amount subtracted from MP)
– SP = Selling Price = MP × (1 − D%/100)
– Profit = SP − CP; Loss = CP − SP

Key formulas:

Two-step chain: A merchant marks up 40% on CP, then gives a 25% discount on MP. Net effect?
Using successive formula: a = +40%, b = −25%: Net = 40 − 25 − (40 × 25)/100 = 15 − 10 = +5% profit.

“Gain on CP, discount on MP” — these are different bases. Gain % uses CP as base; discount % uses MP as base. Never apply both percentages to the same number.

C5. Simple and Compound Interest — Formula and Half-Yearly Awareness

Simple Interest (SI):

Compound Interest (CI):

Half-yearly compounding: Rate is halved (R/2 per period), time doubles (2n periods).

Example: P = Rs. 10,000, R = 10%, T = 2 years.
– Annual CI: Amount = 10,000 × (1.1)² = 12,100 → CI = Rs. 2,100
– Half-yearly CI: Amount = 10,000 × (1.05)⁴ = 10,000 × 1.2155 = 12,155 → CI = Rs. 2,155

Half-yearly compounding always gives more interest than annual compounding at the same nominal rate.

SI vs CI difference for 2 years:

This is a direct shortcut for 2-year problems. For P = 10,000 and R = 10%:
CI − SI = 10,000 × (0.1)² = 10,000 × 0.01 = Rs. 100

C6. Time and Work — Combined Work, Alternate Days, Efficiency

Core principle: Work rate = work done per unit time.
If A completes a job in a days, A’s rate = 1/a per day.
If B completes in b days, B’s rate = 1/b per day.
Combined rate = 1/a + 1/b = (a + b) / (ab).
Time together = ab / (a + b).

Alternate day work: A works Day 1, B works Day 2, A works Day 3, …
– In every 2-day cycle: work done = 1/a + 1/b
– Count complete cycles, then check the remaining fraction.
– Do not apply the “together” formula directly — A and B alternate, not work simultaneously.

A does part of work alone, then B joins:
Let A work alone for k days. Work remaining = 1 − k/a. Then A and B work together.
Time for remainder = (1 − k/a) / (1/a + 1/b).

Efficiency ratio approach: Assign total work = LCM(a, b) units.
– A does 1 unit/day if total work = a units, so A does LCM(a,b)/a units per day.
– B does LCM(a,b)/b units per day.
– Sum gives combined daily output; divide total work by combined output.

Example: A takes 12 days, B takes 18 days. LCM = 36. A does 3 units/day, B does 2 units/day. Together: 5 units/day. Time = 36/5 = 7.2 days.

C7. Pipes and Cisterns — Inlet / Outlet Net Rate

Structurally identical to time-and-work, with one important sign convention: inlets add to the fill rate, outlets subtract from it.

Net rate = Σ (1/inlet_time) − Σ (1/outlet_time)

If net rate > 0: tank fills. If net rate < 0: tank empties.
Time to fill / empty = 1 / |net rate|.

Example: Pipe A fills in 6h, Pipe B fills in 12h, Pipe C drains in 8h.
Net rate = 1/6 + 1/12 − 1/8 = 4/24 + 2/24 − 3/24 = 3/24 = 1/8 per hour
Time to fill = 8 hours

Pipe opened / closed midway: Compute work done in the first phase, find remaining, then apply the new configuration.

Leakage problems: A tap fills a full tank in a hours with no leakage. With a leakage, it takes b hours (b > a).
Leakage rate = 1/a − 1/b = (b − a)/(ab).
Time for leakage alone to empty = ab/(b − a).

Don’t confuse hours to fill vs rate: If a pipe fills in 4 hours, its rate is 1/4 per hour, not 4. Rate and time are reciprocals — substituting one where the other is needed gives an answer that is off by a factor of 16 or more.

C8. Time, Speed, Distance — Relative Speed, Trains, Upstream / Downstream

Core formula: Distance = Speed × Time. All three forms: D = S × T, S = D/T, T = D/S.

Unit conversion: 1 km/h = 5/18 m/s. 1 m/s = 18/5 km/h.
Always convert before substituting — mixing km/h with seconds is the single most common arithmetic error in TSD problems.

Relative speed:

Train passing problems:
– Train passes a pole / standing person: Distance = Length of train
– Train passes a platform / another stationary object: Distance = Length of train + Length of platform
– Two trains passing each other: Distance = Sum of lengths; use appropriate relative speed

Boats / streams (upstream–downstream):
– Boat speed in still water = v; stream speed = u
– Downstream speed = v + u (stream aids motion)
– Upstream speed = v − u (stream resists motion)
– Given downstream D and upstream U: v = (D + U)/2, u = (D − U)/2

Two-part journey equation (GATE AR 2022):
If a journey covers d₁ km at speed S₁ and d₂ km at speed S₂, with total distance D = d₁ + d₂ and total time T:

Substitute d₂ = D − d₁, solve for d₁.

Average speed over two legs: NOT the arithmetic mean of the two speeds.

D. Worked Examples and Practice Sets

Formula Reference Table

Worked Example 1 — Successive Percentage Change

Question: The budget for a municipal project is first increased by 30%, then reduced by 20% due to fund constraints. What is the net percentage change in the budget?

Step 1 — Identify the type: Two successive percentage changes applied to the same base. Use the net formula.

Step 2 — Apply formula: a = +30 (increase), b = −20 (decrease)
Net % change = a + b + (a × b)/100
= 30 + (−20) + (30 × −20)/100
= 30 − 20 − 6
= +4%

Step 3 — Interpret: The net effect is a 4% increase from the original budget.

Verification: Original = 100. After 30% increase → 130. After 20% decrease → 130 × 0.80 = 104. Net change = (104 − 100)/100 × 100 = 4% ✓

Why not 10%: Adding 30% and −20% = 10% ignores the cross-term −6%. This is the most common trap in successive percentage problems.

Worked Example 2 — Overlapping Averages (GATE AR 2022 Pattern)

Question: The average of monthly salaries of M, N, and S is ₹4,000. The average of monthly salaries of N, S, and P is ₹5,000. The monthly salary of P is ₹6,000. Find M’s salary as a percentage of P’s salary.

Step 1 — Convert averages to totals:
M + N + S = 3 × 4,000 = ₹12,000 … (i)
N + S + P = 3 × 5,000 = ₹15,000 … (ii)

Step 2 — Subtract (i) from (ii) to eliminate shared members (N, S):
(N + S + P) − (M + N + S) = 15,000 − 12,000
P − M = ₹3,000

Step 3 — Substitute P = ₹6,000:
6,000 − M = 3,000 → M = ₹3,000

Step 4 — Express as percentage:
M as % of P = (3,000 / 6,000) × 100 = 50%

Key insight: Never average the two averages (4,000 + 5,000)/2 = 4,500 — this is meaningless. The subtraction trick isolates the two unique members directly from the overlapping group structure.

Worked Example 3 — Combined Work Rate

Question: Pipe A fills a tank in 12 hours; Pipe B fills it in 18 hours; Pipe C (an outlet) drains it in 9 hours. All three are opened simultaneously with the tank initially empty. How many hours does it take to fill the tank?

Step 1 — Assign rates (fraction of tank per hour):
Rate of A = 1/12; Rate of B = 1/18; Rate of C = −1/9 (drain, so negative)

Step 2 — Net rate:
Net = 1/12 + 1/18 − 1/9

LCM of 12, 18, 9 = 36:
= 3/36 + 2/36 − 4/36
= 1/36 per hour

Step 3 — Time to fill:
T = 1 / (1/36) = 36 hours

Step 4 — Verify sign: Net rate = +1/36 > 0, so the tank fills (net inlet dominates). ✓

Trap avoided: Outlet rate was 1/9 = 4/36, which is larger than A’s 3/36 alone, but A + B = 5/36 > C’s 4/36, so net is positive. Students who only compare one inlet to the outlet conclude wrongly that the tank never fills.

Worked Example 4 — Two-Speed Journey (GATE AR 2022 Pattern)

Question: A person travels 80 km in 6 hours. The first part of the journey is at 10 km/h; the remaining part is at 18 km/h. What percentage of the total distance is covered at 10 km/h?

Step 1 — Set up: Let distance at 10 km/h = d km. Then distance at 18 km/h = (80 − d) km.

Step 2 — Time equation: Total time = 6 hours.
d/10 + (80 − d)/18 = 6

Step 3 — Clear fractions (multiply by LCM = 90):
9d + 5(80 − d) = 540
9d + 400 − 5d = 540
4d = 140 → d = 35 km

Step 4 — Percentage:
= (35/80) × 100 = 43.75%

Answer: 43.75%

Check: Time at 10 km/h = 35/10 = 3.5h; time at 18 km/h = 45/18 = 2.5h; total = 6h ✓

Distractor trap: The option 50% corresponds to d = 40 km, which gives time = 40/10 + 40/18 = 4 + 2.22 = 6.22h ≠ 6h. Students who guess equal split choose this and are wrong.

E. Common Confusions

• Percentage base direction: “A is 25% more than B” means A = B × 1.25; base is B. “A is 25% more” does not mean B is 25% less than A — B is 20% less than A (base is now A). Increasing by 25% and decreasing by 25% are not symmetric.

• Profit% and discount% use different bases: Profit percentage always uses CP as base. Discount percentage always uses MP as base. A 20% markup on CP followed by a 20% discount on MP does not break even — apply the successive formula to confirm the net effect.

• Average speed ≠ (S₁ + S₂)/2: Arithmetic average of speeds is only correct when equal time is spent at each speed. When equal distance is covered at each speed, the correct average is the harmonic mean: 2S₁S₂/(S₁ + S₂).

• Work-rate unit mismatch: Rate must be “work per unit time” in consistent units. “A takes 5 days” → rate = 1/5 job/day. If B’s rate is given in hours, convert to days (or vice versa) before adding rates.

• Compound interest half-yearly: When compounding is half-yearly, both the rate and the period change. R becomes R/2 per period and T years becomes 2T periods. Students often apply R and 2T, or R/2 and T — both wrong.

• Relative speed direction error: When two objects move in the same direction, the closing speed is the difference of their speeds. Students often add them (the rule for opposite directions). Check the direction before applying the formula.

F. Exam Traps

G. Answer-Writing Cues

H. PYQ Linkage Note

Forecast for 2027: Percentages and averages (especially weighted average traps) are near-certain. One TSD question in the 1-mark slot is highly probable based on 2022 and 2025 recurrence. Work/pipes or profit/loss may appear in the 2-mark GA slot as the exam cycles through syllabus topics.

I. Mini-Check — Lesson 13.4

Instructions: Q1 and Q2 are MCQ with one correct answer. Q3 is MSQ — select all correct options. Q4 and Q5 are MCQ. No NAT questions.

Q1. (MCQ) A project’s estimated cost is first revised upward by 25% and then revised downward by 20%. The net percentage change in the estimated cost from the original estimate is:

(A) +5%
(B) 0%
(C) −5%
(D) +4%

Answer: (B) 0%

Solution:
Net % change = a + b + (a × b)/100 = 25 + (−20) + (25 × −20)/100
= 25 − 20 − 5 = 0%

The two changes cancel exactly. (A) adds without the cross-term (25 − 20 = 5). (C) uses −5 which is the cross-term in isolation. (D) would be the answer if a = 30, b = −20. Apply the formula: the cross-term here is exactly −5, making the net 0.

Q2. (MCQ) Pipe A fills a tank in 8 hours. Pipe B fills the same tank in 12 hours. Both pipes are opened simultaneously. How many hours will it take to fill the tank?

(A) 4 hours
(B) 4.8 hours
(C) 10 hours
(D) 20 hours

Answer: (B) 4.8 hours

Solution:
Combined rate = 1/8 + 1/12 = 3/24 + 2/24 = 5/24 tank per hour.
Time = 24/5 = 4.8 hours

(A) is the arithmetic mean of 8 and 12, divided by 2 (= 5) — wrong. (C) is the arithmetic mean (8+12)/2 — wrong. (D) is 8+12 — wrong. The correct formula is ab/(a+b) = (8×12)/(8+12) = 96/20 = 4.8h.

Q3. (MSQ) Which of the following statements about combined rates are correct? Select all that apply.

(A) If Pipe A fills a tank in 6 hours and Pipe B fills it in 9 hours, together they fill 5/18 of the tank per hour.
(B) If Pipe A fills a tank in 8 hours and Pipe C drains it in 12 hours, with both open simultaneously from an empty tank, the tank fills completely in 24 hours.
(C) If Pipe A fills in x hours and Pipe C drains in y hours where y > x (drain is slower than fill), opening both from an empty tank will eventually fill it.
(D) Three pipes filling a tank in 6, 8, and 24 hours respectively together fill it in 4 hours.

Answer: (A), (B), (C)

Explanation:
– (A) Rate = 1/6 + 1/9 = 3/18 + 2/18 = 5/18 per hour ✓
– (B) Net rate = 1/8 − 1/12 = 3/24 − 2/24 = 1/24 per hour. Time = 24 hours ✓
– (C) y > x means drain rate 1/y < fill rate 1/x, so net rate = 1/x − 1/y > 0. Tank fills. ✓
– (D) Wrong: Net rate = 1/6 + 1/8 + 1/24 = 4/24 + 3/24 + 1/24 = 8/24 = 1/3 per hour. Time = 3 hours, not 4 hours as stated. (D) is a distractor designed to test whether you verify the formula result rather than accept the “intuitive” answer.) ✗

Q4. (MCQ) A train 200 metres long passes a platform 300 metres long at a uniform speed. The train takes 25 seconds to completely pass the platform. What is the speed of the train in km/h?

(A) 36 km/h
(B) 72 km/h
(C) 54 km/h
(D) 90 km/h

Answer: (B) 72 km/h

Solution:
Total distance = length of train + length of platform = 200 + 300 = 500 metres
Time = 25 seconds
Speed = 500/25 = 20 m/s
Convert: 20 × 18/5 = 72 km/h

(A) 36 km/h = 10 m/s → would cover 250m in 25s — only the train length, ignoring platform. (C) 54 km/h = 15 m/s → 375m in 25s — wrong. (D) 90 km/h = 25 m/s → 625m in 25s — too fast.

Q5. (MCQ) Rs. 8,000 is invested at 10% per annum compound interest, compounded annually. The compound interest earned after 2 years is:

(A) Rs. 1,600
(B) Rs. 1,680
(C) Rs. 1,600
(D) Rs. 2,000

Answer: (B) Rs. 1,680

Solution:
CI = P[(1 + r/100)ⁿ − 1] = 8,000[(1.1)² − 1] = 8,000[1.21 − 1] = 8,000 × 0.21 = Rs. 1,680

Shortcut check using SI vs CI difference formula:
SI for 2 years = 8,000 × 10 × 2 / 100 = Rs. 1,600
CI − SI = P(r/100)² = 8,000 × (0.1)² = 8,000 × 0.01 = Rs. 80
CI = 1,600 + 80 = Rs. 1,680 ✓

(A) is the Simple Interest — ignores the compounding effect. (D) is 25% of principal — wrong formula.