Arithmetic Reasoning

1. Core Concepts & Theoretical Blueprint

Arithmetic Reasoning questions present real-world word problems requiring logical translation into mathematical operations — testing the ability to identify WHICH mathematical relationship a verbal scenario describes and set up the correct equation(s), rather than testing calculation speed or advanced mathematical technique alone. This topic sits at the intersection of reasoning and basic mathematics: the reasoning skill is in correctly decoding the problem's structure; the mathematics itself (usually basic arithmetic, simple algebra, ratios, percentages) is intentionally kept straightforward.

The underlying logical structure is verbal-to-symbolic translation: every arithmetic reasoning problem describes relationships between quantities using natural language, and the core skill is converting phrases like "twice as many," "the difference between," "increased by," or "in the ratio of" into precise mathematical expressions before any calculation begins. Misreading the verbal relationship (translating "increased BY 20%" as "increased TO 20%," for instance) produces a completely wrong setup even if all subsequent arithmetic is executed perfectly.

Five recurring problem structures are tested:

• Direct Computation Problems: A straightforward calculation is described in words (total cost, combined distance, remaining quantity) requiring simple translation to arithmetic.
• Ratio and Proportion Problems: Quantities are related by a stated or implied ratio, requiring proportional reasoning to find unknown values.
• Percentage-Based Problems: Quantities change by a stated percentage, requiring correct base-value identification (percentage of WHAT) before calculation.
• Speed-Distance-Time Problems: Motion scenarios requiring the fundamental relationship Distance = Speed × Time, applied directly or in reverse.
• Mixture/Average Problems: Combining quantities with different individual values (prices, concentrations, scores) to find a combined or average value.

Reference Table: Verbal Phrase to Mathematical Operation Translation

Verbal Phrase	Mathematical Translation
"increased by 20%"	New value = Old value × 1.20
"decreased by 20%"	New value = Old value × 0.80
"20% of X"	0.20 × X
"the ratio of A to B is 3:5"	A/B = 3/5, or A = 3k, B = 5k for some k
"A is 20% more than B"	A = B × 1.20 (NOT B = A × 1.20)
"average of a set"	Sum of all values ÷ number of values

The Universal Trap: (1) Students confuse "X% of A" with "X% more than A" — "20% of 100 is 20" but "20% more than 100 is 120" — these are fundamentally different operations, and misreading between them produces answers that are wrong by a large, systematic margin. (2) Students apply a percentage change to the wrong base value, especially in successive percentage change problems (a 10% increase followed by a 10% decrease does NOT return to the original value, since the second 10% is calculated on the NEW, already-changed base) — always identify precisely which value each percentage is "of." (3) In ratio problems, students treat the ratio numbers as the actual quantities themselves rather than as proportional MULTIPLES of an unknown common factor — always introduce a variable k (ratio parts = 3k, 5k, etc.) and solve for k using any given total or additional constraint before finding actual quantities.

2. Exhaustive Question Typology

                          ARITHMETIC REASONING
                                  |
      -------------------------------------------------------------------
      |               |                |                |               |
   Type 1          Type 2           Type 3           Type 4          Type 5
 Direct            Ratio and        Percentage-       Speed-          Mixture/
 Computation        Proportion       Based             Distance-       Average
 Word Problems                       Word Problems      Time            Problems
                                                          Problems

Type 1 — Direct Computation Word Problems

Core Scenario: "A shopkeeper buys an item for ₹250 and sells it for ₹320. What is his profit?" Governing Rule/Logic: IF the problem describes a simple direct relationship (profit = selling price − cost price) THEN translate and compute directly: Profit = 320−250 = ₹70.

Type 2 — Ratio and Proportion Problems

Core Scenario: "The ratio of boys to girls in a class is 4:5. If there are 36 students in total, how many boys are there?" Governing Rule/Logic: IF quantities are related by a ratio AND a total is given THEN let the ratio parts = 4k and 5k, set up 4k+5k=36, solve for k=4, then boys = 4×4=16.

Type 3 — Percentage-Based Word Problems

Core Scenario: "A product's price was increased by 25% and then decreased by 20%. What is the net percentage change from the original price?" Governing Rule/Logic: IF successive percentage changes are applied THEN calculate sequentially using the actual resulting value at each step (not by simply adding/subtracting percentages): Start with 100. After +25%: 100×1.25=125. After −20% (applied to the NEW value 125): 125×0.80=100. Net change = 0%.

Type 4 — Speed-Distance-Time Problems

Core Scenario: "A train travels 240 km in 4 hours. At the same speed, how long will it take to travel 360 km?" Governing Rule/Logic: IF the fundamental relationship Distance=Speed×Time is given with sufficient information THEN first solve for the constant (Speed=240/4=60 km/h), then apply to the new scenario (Time=360/60=6 hours).

Type 5 — Mixture/Average Problems

Core Scenario: "A shopkeeper mixes 20 kg of rice costing ₹40/kg with 30 kg of rice costing ₹60/kg. What is the average price per kg of the mixture?" Governing Rule/Logic: IF quantities with different individual values are combined THEN compute total value ÷ total quantity: Total value = (20×40)+(30×60) = 800+1800=2600. Total quantity = 50 kg. Average price = 2600/50 = ₹52/kg.

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Direct Computation Word Problems

Q1. A man buys a bicycle for ₹3,000 and spends ₹500 on repairs. He then sells it for ₹4,200. What is his profit? (A) ₹500 (B) ₹700 (C) ₹900 (D) ₹1,200

Correct Answer: (B) ₹700 Solution: Total cost = Purchase price + Repair cost = 3000+500 = 3500. Profit = Selling price − Total cost = 4200−3500 = ₹700.

Q2. A tank can be filled by Pipe A in 6 hours and by Pipe B in 12 hours. If both pipes are opened together, how long will it take to fill the tank? (A) 3 hours (B) 4 hours (C) 5 hours (D) 8 hours

Correct Answer: (B) 4 hours Solution: Pipe A's rate = 1/6 tank per hour. Pipe B's rate = 1/12 tank per hour. Combined rate = 1/6+1/12 = 2/12+1/12 = 3/12 = 1/4 tank per hour. Time to fill = 1÷(1/4) = 4 hours.

Q3. A sum of ₹5,000 becomes ₹5,800 after one year of simple interest. What is the annual interest rate? (A) 12% (B) 14% (C) 16% (D) 18%

Correct Answer: (C) 16% Solution: Interest earned = 5800−5000 = 800. Rate = (Interest/Principal)×100 = (800/5000)×100 = 16%.

Type 2 — Ratio and Proportion Problems

Q1. The ratio of the ages of A and B is 3:4. If A's age is 21 years, what is B's age? (A) 24 (B) 26 (C) 28 (D) 30

Correct Answer: (C) 28 Solution: Let A=3k, B=4k. Given A=21, so 3k=21, k=7. B=4×7=28.

Q2. ₹1,540 is divided among X, Y, and Z in the ratio 3:4:7. Find Z's share. (A) ₹440 (B) ₹550 (C) ₹660 (D) ₹770

Correct Answer: (D) ₹770 Solution: Total ratio parts = 3+4+7=14. Value of each part = 1540/14=110. Z's share = 7×110 = 770.

Q3. Two numbers are in the ratio 5:7. If their sum is 96, find the larger number. (A) 40 (B) 48 (C) 52 (D) 56

Correct Answer: (D) 56 Solution: Let the numbers be 5k and 7k. Sum: 5k+7k=96, 12k=96, k=8. Larger number = 7×8=56.

Type 3 — Percentage-Based Word Problems

Q1. In an election between two candidates, the winner received 60% of the total votes and won by 4,800 votes. What was the total number of votes cast? (A) 18,000 (B) 20,000 (C) 22,000 (D) 24,000

Correct Answer: (D) 24,000 Solution: Winner got 60%, loser got 40%. Margin = 60%−40% = 20% of total votes = 4800. Total votes = 4800/0.20 = 24,000.

Q2. The price of an item is first increased by 10% and then decreased by 10%. What is the net percentage change from the original price? (A) No change (B) 1% decrease (C) 1% increase (D) 2% decrease

Correct Answer: (B) 1% decrease Solution: Start with 100. After +10%: 100×1.10=110. After −10% (on the new value 110): 110×0.90=99. Net change = 99−100 = −1, i.e., a 1% decrease from the original.

Q3. A student needs 40% marks to pass an exam. He scored 150 marks and failed by 30 marks. What is the maximum marks of the exam? (A) 400 (B) 450 (C) 480 (D) 500

Correct Answer: (B) 450 Solution: Passing marks = 150+30 = 180. Passing marks represent 40% of maximum marks: 0.40×Max=180, Max=180/0.40=450.

Type 4 — Speed-Distance-Time Problems

Q1. A car travels at a speed of 80 km/h. How long will it take to cover a distance of 320 km? (A) 3 hours (B) 4 hours (C) 5 hours (D) 6 hours

Correct Answer: (B) 4 hours Solution: Time = Distance/Speed = 320/80 = 4 hours.

Q2. Two trains start from the same station at the same time, traveling in opposite directions, at speeds of 60 km/h and 90 km/h respectively. After how many hours will they be 450 km apart? (A) 2 hours (B) 2.5 hours (C) 3 hours (D) 3.5 hours

Correct Answer: (C) 3 hours Solution: Since they travel in opposite directions, their speeds add: combined speed = 60+90=150 km/h. Time = Distance/Combined Speed = 450/150 = 3 hours.

Q3. A boat travels 60 km downstream in 3 hours and returns upstream in 5 hours. Find the speed of the boat in still water. (A) 12 km/h (B) 14 km/h (C) 16 km/h (D) 18 km/h

Correct Answer: (C) 16 km/h Solution: Downstream speed = 60/3=20 km/h. Upstream speed = 60/5=12 km/h. Speed in still water = (Downstream+Upstream)/2 = (20+12)/2 = 16 km/h.

Type 5 — Mixture/Average Problems

Q1. The average of 5 numbers is 24. If one number is excluded, the average of the remaining 4 numbers becomes 22. What is the excluded number? (A) 28 (B) 30 (C) 32 (D) 34

Correct Answer: (C) 32 Solution: Sum of 5 numbers = 5×24=120. Sum of remaining 4 numbers = 4×22=88. Excluded number = 120−88=32.

Q2. A shopkeeper mixes two varieties of tea costing ₹80/kg and ₹120/kg in the ratio 3:2. Find the price per kg of the mixture. (A) ₹92 (B) ₹96 (C) ₹100 (D) ₹104

Correct Answer: (B) ₹96 Solution: Using ratio parts: let the mix be 3 kg at ₹80 and 2 kg at ₹120 (5 kg total, matching the ratio). Total value = (3×80)+(2×120) = 240+240=480. Average price = 480/5=₹96/kg.

Q3. The average marks of 30 students in a class is 68. If the average marks of 18 boys is 72, find the average marks of the remaining 12 girls. (A) 60 (B) 62 (C) 64 (D) 66

Correct Answer: (B) 62 Solution: Total marks of all 30 students = 30×68=2040. Total marks of 18 boys = 18×72=1296. Total marks of 12 girls = 2040−1296=744. Average marks of girls = 744/12=62.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1: The "Of vs. More Than" Base-Value Check Application: For every percentage-related phrase, explicitly identify whether it says "X% OF [value]" (direct multiplication of that exact value) or "X% MORE/LESS THAN [value]" (multiplication by 1±X% applied to that value to get a NEW, different value) — never conflate the two. Mental Model: "Of" describes a static portion of a single value, while "more/less than" describes a transformation FROM one value TO a new value — these are grammatically similar but mathematically distinct operations, and explicitly distinguishing them prevents the single most common systematic error in percentage-based arithmetic reasoning.

Shortcut 2: The Assumed-Base-100 Technique for Successive Percentage Changes Application: For any problem involving successive percentage changes (increase then decrease, or multiple sequential changes) without a specific starting value given, assume a convenient base value of 100, apply each percentage change sequentially to the RUNNING result, and read the net percentage change directly from the final value. Mental Model: Using 100 as an assumed starting value makes every percentage calculation a direct, intuitive read (e.g., "125" directly means "25% more than the original 100") without needing to track abstract variables or fractions, and works correctly precisely because percentage relationships are scale-invariant — the final net percentage change is identical regardless of what the actual original value happens to be.

5. Deep-Dive: Most Frequently Asked Questions (Exam-Style Walkthroughs)

Problem 1 (SSC/RRB Level): A shopkeeper marks up the price of an item by 40% above cost price, then offers a 25% discount on the marked price during a sale. If the item's cost price is ₹200, what is the final selling price, and what is the shopkeeper's actual profit percentage?

Traditional Method (Slow) — approx. 40-50 seconds: A slow solver calculates the marked price (200×1.40=280) correctly, but then mistakenly applies the 25% discount to the ORIGINAL cost price (200) instead of the marked price (280), computing 200×0.75=150 as a wrong intermediate step, before realizing the error and having to restart the discount calculation correctly on 280.

Exam Shortcut (Fast) — approx. 15-20 seconds: Apply the "Of vs. More Than" Base-Value Check rigorously: the discount is explicitly "on the marked price," so it must be applied to 280, NOT to the original 200. Marked price = 200×1.40=280. Selling price after discount = 280×(1−0.25) = 280×0.75=210. Profit = Selling price − Cost price = 210−200=10. Profit percentage = (10/200)×100=5%. Answer: Final selling price = ₹210; profit percentage = 5%. Correctly anchoring each percentage operation to its stated base value (cost price for markup, marked price for discount) avoids the compounding error of using the wrong base.

Problem 2 (UPSC/Banking Advanced Level): A container is filled with a mixture of milk and water in the ratio 5:3. 24 liters of the mixture is removed and replaced with pure water. If the new ratio of milk to water becomes 5:7, find the original total quantity of the mixture in the container.

Step-by-step derivation:

1. Let the total original quantity of the mixture = T liters. Given the original ratio milk:water = 5:3, so original milk = (5/8)T and original water = (3/8)T.
2. When 24 liters of the mixture is removed, since the mixture is uniform, the removed portion maintains the same 5:3 ratio internally: milk removed = (5/8)×24=15 liters, water removed = (3/8)×24=9 liters.
3. After removal, remaining milk = (5/8)T − 15, and remaining water = (3/8)T − 9.
4. 24 liters of pure water is then added back: new water quantity = [(3/8)T − 9] + 24 = (3/8)T + 15. Milk quantity remains unchanged after this step (only water was added): new milk quantity = (5/8)T − 15.
5. Set up the new ratio equation using the given final ratio 5:7: [(5/8)T − 15] / [(3/8)T + 15] = 5/7.
6. Cross-multiply: 7×[(5/8)T − 15] = 5×[(3/8)T + 15].
7. Expand: 7×(5/8)T − 7×15 = 5×(3/8)T + 5×15 → (35/8)T − 105 = (15/8)T + 75.
8. Collect T terms: (35/8)T − (15/8)T = 75+105 → (20/8)T = 180 → (5/2)T = 180.
9. Solve for T: T = 180×(2/5) = 72.

Final Answer: The original total quantity of the mixture in the container was 72 liters. Verification: original milk = (5/8)×72=45, original water=(3/8)×72=27. Removing 24 liters (ratio 5:3) removes 15 milk and 9 water: remaining milk=45−15=30, remaining water=27−9=18. Adding 24 liters pure water: new water=18+24=42, milk stays 30. New ratio = 30:42 = 5:7 ✓, confirming the answer.

6. Chapter Checklist for Students

• I apply the "Of vs. More Than" Base-Value Check on every percentage phrase, explicitly identifying which value each percentage is calculated FROM.
• I use the Assumed-Base-100 Technique for successive percentage change problems without a specific given starting value.
• I introduce a variable k for ratio parts (3k, 5k, etc.) rather than treating raw ratio numbers as actual quantities, solving for k using any given total first.
• I correctly identify which base value a discount or markup percentage applies to (cost price vs. marked/selling price), especially in multi-step markup-then-discount problems.
• I verify my final numeric answer by substituting it back into all the original given conditions, confirming full internal consistency before finalizing.