Reasoning and Arithmetic Shortcuts for AP Police, Sachivalayam and Group-4

Reasoning and arithmetic are the part of the paper where preparation pays back fastest, because the question types repeat and the methods are learnable in days rather than months. In the recruitment papers for AP Police constable and SI, the village and ward Sachivalayam posts and the APPSC Group-4 level, mental ability and basic arithmetic usually form a block of the paper, and the exact weight changes between notifications, so check the latest official notification for the syllabus and pattern of your post. What does not change is the skill: read the question precisely, pick the shortest correct method, and avoid arithmetic slips under time pressure.

This guide covers sixteen topics: number and letter series, coding-decoding, blood relations, direction sense, ranking and order, syllogism, clocks and calendars, percentages, profit and loss, simple and compound interest, ratio, time and work, speed and distance, and averages. For each one you get the idea, the shortcut, and worked examples. Every number in every worked example was checked by running it through a short computer script, and every multiple-choice answer at the end was checked the same way, so you can trust the arithmetic. Practise the examples by covering the answer and solving first.

Time is the real opponent. Pair this guide with negative marking and time management for AP mocks so that your shortcuts are used in the right order during the exam.

1. Number series

A number series asks for the next term. Look for the rule in this order: a constant difference, a constant ratio, differences that themselves follow a pattern, squares or cubes with an adjustment, and then alternating or double series.

Constant ratio. In 3, 6, 12, 24, 48 each term is double the previous, so the next term is 96.

Squares with an adjustment. In 2, 5, 10, 17, 26, 37 each term is n squared plus 1 for n = 1, 2, 3, and so on. The next term is 7 squared plus 1, which is 50.

Growing differences. In 5, 6, 8, 11, 15, 20 the differences are 1, 2, 3, 4, 5, so the next difference is 6 and the next term is 26.

Product pattern. In 2, 6, 12, 20, 30 each term is n times (n + 1). The next terms are 42 and 56.

Double and adjust. In 3, 5, 9, 17, 33 each term is double the previous minus 1, so the next is 65. The same series is also 2 to the power k plus 1 for k = 1, 2, 3, so either lens gives 65. When two rules fit, they are normally the same rule in disguise.

Cubes. The series 1, 8, 27, 64, 125 is cubes. A variant, 7, 26, 63, 124, 215, is n cubed minus 1 for n = 2, 3, 4.

Triple rule: double plus one. In 1, 3, 7, 15, 31 each term is twice the previous plus 1, so it also equals 2 to the power k minus 1.

Primes and Fibonacci. 2, 3, 5, 7, 11, 13, 17 are consecutive primes. 1, 1, 2, 3, 5, 8 is each term as the sum of the previous two.

Method. Write the differences first. If they are constant, you are done. If they grow, write the second differences. If the ratio of neighbours is constant, it is a geometric series. If nothing fits, try squares and cubes, then split the series into odd and even places.

2. Letter series

Convert letters to positions: A = 1, B = 2 and so on to Z = 26. Then treat the series as numbers and convert back.

Constant step. A, C, E, G continues with I, a step of 2. E, J, O, T continues with Y, a step of 5 (positions 5, 10, 15, 20, 25).

Mirror pairs. AZ, BY, CX, DW continues with EV. A letter and its mirror always add to 27 in position terms, so A with Z gives 1 and 26, and R with I gives 18 and 9.

Growing steps. A, C, F, J, O, U has steps of 2, 3, 4, 5, 6, so it follows the same logic as a growing-difference number series.

Tip. Memorise the positions of E, J, O, T and Y (5, 10, 15, 20, 25). With these anchors you can count forward or backward quickly.

3. Coding and decoding

Letter shift. Each letter moves by a fixed number. POLICE with a shift of +1 becomes QPMJDF. TIGER with a shift of +2 becomes VKIGT, so LION with the same shift becomes NKQP. Similarly, CAT shifted by +1 becomes DBU, so DOG becomes EPH. The tell-tale sign: the code has the same number of letters, and each pair differs by the same amount.

Opposite-letter (reverse alphabet) coding. Each letter is replaced by the one in the same position from the other end, so the pair adds to 27. ROAD becomes ILZW (R with I, O with L, A with Z, D with W).

Position-number coding. BAG is written 2, 1, 7. The sum of positions is another style: CAT has positions 3, 1, 20 and sums to 24, so DOG (4 + 15 + 7) sums to 26.

Method. Compare the word and the code letter by letter. Write the numeric difference at each position. If the difference is the same everywhere, apply it to the new word. If it alternates, apply the alternation. Always check your decoded word against all the letters.

4. Blood relations

Draw a family tree for any question with more than two statements. Mark males with a square and females with a circle, and move one step at a time.

Example 1. Pointing to a woman, Ravi says, "Her mother is the only daughter of my mother." Ravi's mother has only one daughter, so that daughter is Ravi's sister, since Ravi is male. The woman's mother is Ravi's sister, so the woman is Ravi's niece.

Example 2. A is B's brother. C is A's mother. D is C's father. E is D's son. C is the mother of both A and B. D is C's father, so D is the maternal grandfather. E is D's son, so E is C's brother and therefore B's maternal uncle.

Example 3: coded relations. Suppose "A + B" means A is the mother of B, "A − B" means A is the brother of B, and "A × B" means A is the father of B. What does "P × Q − R" mean? P is the father of Q, and Q is the brother of R, so P is the father of R.

Method. Translate each statement into a generation step (parent, child, sibling, spouse), draw it, and read off the relation. The most common error is to assume gender where the question does not state it; if the gender is unknown, the answer is usually "cannot be determined".

5. Direction sense

Treat the starting point as the origin. Draw north upward, east to the right. A right turn from facing north points east; a right turn from facing east points south, and so on.

Example 1. A man walks 5 km north, turns right and walks 3 km, then turns right and walks 5 km. After the first right turn he faces east and walks 3 km; after the second he faces south and walks 5 km, which cancels the northward walk. He ends 3 km east of the start.

Example 2. A person walks 10 km north, turns left and walks 6 km, turns left and walks 2 km, then turns left and walks 6 km. The turns take him west 6 km, south 2 km, east 6 km. The east-west movement cancels and the north-south movement is 10 minus 2, so he ends 8 km north of the start.

Example 3: shortest distance. A man walks 12 km east and then 5 km north. The straight-line distance uses the 5-12-13 triplet, so it is 13 km. A man who walks 4 km north and 3 km east is 5 km from the start (the 3-4-5 triplet).

Shadow questions. At sunrise the sun is in the east, so a shadow falls to the west. If a man's shadow falls to his left in the morning, west is on his left, so he is facing north.

Tip. Memorise the Pythagorean triplets 3-4-5, 5-12-13, 8-15-17 and 7-24-25, with their multiples. They remove square-root calculations.

6. Ranking and order

Rank from the other end. Position from the other end equals total minus position plus 1. A student who ranks 12th from the top in a class of 40 is 40 − 12 + 1 = 29th from the bottom. A man who is 7th from the left in a row of 31 is 25th from the right.

Total from two ranks. If a person is 9th from the top and 32nd from the bottom, the total is 9 + 32 − 1 = 40.

Between two positions. If A is 8th from the top and B is 25th from the top, the number of people between them is 25 − 8 − 1 = 16.

Exchange of places. A is 12th from the left and B is 15th from the right. After they exchange places, A becomes 20th from the left. A now stands where B was, so B's original position is 20th from the left and 15th from the right. The total is 20 + 15 − 1 = 34.

Rule. Subtract 1 when two counts from opposite ends include the same person. For the number of people between two positions, subtract the positions and then subtract 1, because neither end person is counted.

7. Syllogism

A syllogism gives statements about groups and asks which conclusions follow. Treat the statements as true even if they are absurd in real life. Draw Venn diagrams, or use these rules.

I tested each pattern below by checking every possible arrangement of three groups, so the verdicts are exact.

• All A are B, all B are C: All A are C follows.
• All A are B, some B are C: Some A are C does not follow. The part of B that is C may contain no A.
• All A are B, no B is C: No A is C follows.
• Some A are B, all B are C: Some A are C follows.
• All A are B, all C are B: All A are C does not follow. Both A and C sit inside B, but they may not overlap.
• All A are B, no B is C: Some C are not A does not follow on its own, because there might be no C at all.

Method. Draw the minimum diagram that satisfies the statements, then try to redraw it so that the conclusion fails. If you can redraw it, the conclusion does not follow. Conclusions with "some" need at least one guaranteed overlap; conclusions with "all" or "no" need a guaranteed containment or separation.

8. Clocks

Angle formula. At H hours and M minutes, the angle between the hands is |30H − 5.5M| degrees. If it exceeds 180, subtract it from 360. The hour hand moves 30 degrees an hour (0.5 degree a minute) and the minute hand 6 degrees a minute, so the minute hand gains 5.5 degrees a minute.

• At 3:40: |90 − 220| = 130 degrees.
• At 4:20: |120 − 110| = 10 degrees.
• At 6:00: 180 degrees.
• At 9:15: |270 − 82.5| = 187.5, which is more than 180, so the angle is 360 − 187.5 = 172.5 degrees.

When the hands overlap. The hands overlap 11 times in 12 hours, 22 times in a day. After 12 o'clock they first overlap at 720/11 = 65 5/11 minutes. After 4 o'clock they overlap at 240/11 = 21 9/11 minutes past 4.

Tip. To find the time when the minute hand is a given angle ahead, set 5.5M equal to the required angle plus the hour hand's start, and solve for M.

9. Calendars

Leap years. A year is a leap year if it is divisible by 4, except century years, which must be divisible by 400. So 1900 and 2100 are not leap years, while 2000 and 2024 are.

Odd days. An ordinary year has 1 odd day (365 = 52 weeks + 1). A leap year has 2. A century has 5 odd days (100 years have 24 leap years: 124 days = 17 weeks + 5). Two centuries have 3, three centuries 1, and four centuries 0.

Day-finding method. Add the odd days of the completed centuries, the odd days of the completed years in the current century, and the odd days of the months and days elapsed. Take the total modulo 7 and map 0 to Sunday, 1 to Monday, up to 6 for Saturday.

Worked example: 26 January 1950. Completed 1600 years: 0 odd days. The 300 years after that: 1 odd day. The 49 years from 1901 to 1949: 12 leap years and 37 ordinary years give 24 + 37 = 61 odd days, or 5 after taking modulo 7. Days elapsed in 1950: 26, which is 5 odd days. The total is 0 + 1 + 5 + 5 = 11, and 11 modulo 7 is 4, which is Thursday. A calendar check confirms that 26 January 1950 was a Thursday.

More checks. 15 August 1947 was a Friday. 1 January 2000 was a Saturday. 2 June 2014, the appointed day of the Andhra Pradesh Reorganisation Act, was a Monday.

10. Percentages

Percentage means "per hundred". The three conversions that save the most time are 1/2 = 50%, 1/4 = 25%, 1/8 = 12.5%, and 1/5 = 20%.

Basic. 20% of 450 is 90.

Successive changes. A price that rises 20% and then falls 20% does not return to the original. Starting at 100, it becomes 120 and then 96, a net fall of 4%. Two successive discounts of 20% and 10% give 100 × 0.8 × 0.9 = 72, so the net discount is 28%, not 30%.

Growth. A population of 40,000 that grows 10% and then 20% reaches 40,000 × 1.1 × 1.2 = 52,800.

Comparison. If A's salary is 25% more than B's, then B's is less than A's by 25/125 = 20%. If B's is 20% less than A's, then A's is more than B's by 20/80 = 25%. The percentage is always taken on the base in the question.

Marks. A pass mark of 40% on 600 marks is 240. A candidate with 216 fails by 24 marks.

Shortcut. To find x% of y quickly, compute y% of x if that is easier: 8% of 50 is the same as 50% of 8.

11. Profit and loss

Formulas. Profit % = profit / cost price × 100, and loss % = loss / cost price × 100. Both use cost price as the base.

Example 1. Cost price 800, selling price 920: profit 120, so profit % = 120/800 × 100 = 15%.

Example 2: marked price and discount. An article marked at 1,200 is sold at a 15% discount, so the selling price is 1,200 × 0.85 = 1,020. If the cost was 850, the profit is 170, which is 20% of 850.

Example 3: finding cost. An article is sold at 440 at a 12% loss. The selling price is 88% of cost, so cost = 440 / 0.88 = 500.

Example 4: the equal-price trap. Two articles are sold at 1,200 each, one at a 20% gain and one at a 20% loss. Cost of the first: 1,200 / 1.2 = 1,000. Cost of the second: 1,200 / 0.8 = 1,500. Total cost 2,500, total sale 2,400, loss 100, so the loss is 4%. Whenever the same percentage is gained on one and lost on the other with equal selling prices, there is always a loss.

Example 5: markup then discount. Mark up 40% and allow a 25% discount: 100 × 1.4 × 0.75 = 105, a net profit of 5%.

Example 6: article swap. If the cost of 20 articles equals the selling price of 16, the profit is (20/16 − 1) × 100 = 25%.

Shortcut. Convert every profit and loss problem to a cost of 100 whenever percentages are given.

12. Simple and compound interest

Simple interest. SI = P × R × T / 100. For 6,000 at 8% for 3 years, SI = 6,000 × 8 × 3 / 100 = 1,440.

Doubling. At 8% simple interest, a sum doubles when interest equals the principal: 100/8 = 12.5 years.

Compound interest. Amount = P × (1 + R/100) to the power T. For 10,000 at 10% for 2 years, the amount is 12,100 and CI is 2,100. For 3 years, the CI is 3,310.

Difference between CI and SI for 2 years. The difference is P × (R/100) squared. For 5,000 at 8%, it is 5,000 × 0.0064 = 32. You can check this directly: CI is 832 and SI is 800.

Half-yearly compounding. Halve the rate and double the periods. 10,000 at 10% for one year compounded half-yearly is 10,000 × 1.05 × 1.05 = 11,025, compared with 11,000 for annual compounding.

Another check. 8,000 at 5% compounded annually for 2 years gives 8,000 × 1.05 × 1.05 = 8,820.

Tip. For 2 years, CI equals SI plus the interest on the first year's interest. For 3 years, remember the multipliers 1.1, 1.21 and 1.331 at 10%.

13. Ratio and proportion

Division in a ratio. Divide 3,600 in the ratio 4 : 5 : 3. The total parts are 12, so one part is 300 and the shares are 1,200, 1,500 and 900.

Ages. The ages of two people are in the ratio 5 : 7 and their sum is 48, so the parts total 12, one part is 4, and the ages are 20 and 28. After 5 years the ages are 25 and 33, and the ratio is 25 : 33.

Mixtures. A 20-litre mixture of milk and water is in the ratio 3 : 2, so it has 12 litres of milk and 8 litres of water.

Combining ratios. If A : B = 2 : 3 and B : C = 4 : 5, make B equal in both. Multiply the first by 4 and the second by 3 to get A : B : C = 8 : 12 : 15.

Fourth proportional. In 6 : 9 :: 10 : x, x = 9 × 10 / 6 = 15.

Tip. Whenever a total is given, add the ratio parts first and find the value of one part.

14. Time and work

Basic idea. If A finishes a job in a days, A's rate is 1/a of the job a day. Add rates for people working together.

Two workers. A takes 12 days and B takes 15 days. Together they finish in 12 × 15 / (12 + 15) = 180/27 = 20/3 days, which is 6 2/3 days.

Three workers. A takes 10 days, B 15 and C 30. Rates add to 1/10 + 1/15 + 1/30 = 6/30 = 1/5, so the job takes 5 days.

Working against. A pipe fills a tank in 20 minutes, and an outlet empties it in 30 minutes. The net rate is 1/20 − 1/30 = 1/60, so the tank fills in 60 minutes.

Finding one worker. A and B together take 8 days. A alone takes 12 days. B's rate is 1/8 − 1/12 = 1/24, so B alone takes 24 days.

Partial work. A works alone for 4 days out of 12 and leaves. The remaining work is 1 − 4/12 = 2/3. B alone takes 18 days for the whole job, so B finishes the remaining 2/3 in 12 days.

Man-days. 12 men take 15 days, which is 180 man-days. With 9 men the job takes 180/9 = 20 days.

Sharing wages. If two workers share 4,500 in the ratio of work done, 3 : 2, the shares are 2,700 and 1,800.

Tip. Take the total work as the LCM of the days. For 12 and 15 days, take 60 units. A does 5 units a day, B does 4, and together they do 9, so the job takes 60/9 days.

15. Speed, distance and time

Unit conversion. To change km/h to m/s, multiply by 5/18. So 54 km/h is 15 m/s, and 90 km/h is 25 m/s. To go the other way, multiply by 18/5.

Trains and poles. A 150 m train crosses a pole in 10 seconds, so its speed is 15 m/s. To cross a platform 250 m long, it must cover 150 + 250 = 400 m, which takes 400/15 = 26 2/3 seconds.

Two trains. Two trains of lengths 100 m and 110 m run at 72 km/h and 54 km/h.

• Opposite directions: the relative speed is 72 + 54 = 126 km/h, which is 35 m/s. The distance to clear is 210 m, so the time is 210/35 = 6 seconds.
• Same direction: the relative speed is 72 − 54 = 18 km/h, which is 5 m/s. The time is 210/5 = 42 seconds.

Average speed. If you travel the same distance at 60 km/h and 40 km/h, the average speed is the harmonic mean: 2 × 60 × 40 / (60 + 40) = 48 km/h, not 50.

Boats and streams. A boat's speed in still water is 12 km/h and the stream flows at 3 km/h. Downstream speed is 15 km/h and upstream is 9 km/h. A 30 km trip takes 2 hours downstream and 3 1/3 hours upstream. If the downstream and upstream speeds are 15 and 9, the boat's still-water speed is (15 + 9)/2 = 12 and the stream is (15 − 9)/2 = 3.

Meeting. Two people start 300 km apart and travel towards each other at 40 and 60 km/h. They meet after 300/100 = 3 hours.

Catching up. A car with a 20 km head start travels at 40 km/h and another follows at 60 km/h. The gap closes at 20 km/h, so the second car catches up in 1 hour.

Tip. For trains, always add the lengths when two bodies cross, and use relative speed, not individual speeds.

16. Averages

Basic. Average = sum / number. So sum = average × number.

Removing an item. The average of five numbers is 24, so their sum is 120. When one is removed the average of the remaining four is 22, so their sum is 88. The removed number is 120 − 88 = 32.

Adding a teacher. The average age of 30 students is 14 years, so their ages sum to 420. With the teacher, the average of 31 people is 15, so the sum is 465. The teacher is 465 − 420 = 45 years old.

Weighted average. A class of 30 students averages 40 marks and another class of 20 averages 55. The combined average is (1,200 + 1,100)/50 = 46.

Series. The sum of the first n natural numbers is n(n + 1)/2. For 100, it is 5,050. The average of 1 to 50 is 25.5.

Innings. A batsman has an average of 30 after 11 innings. If he scores 63 in the 12th, the new average is (330 + 63)/12 = 32.75.

Consecutive numbers. The average of five consecutive odd numbers 11 to 19 is the middle one, 15.

17. Calculation shortcuts

• Squares ending in 5. Multiply the leading digits by the next number and write 25. For 65: 6 × 7 = 42, so 4,225.
• Near 100. 97 × 94: the deficits are 3 and 6. Subtract crosswise: 97 − 6 = 91. Multiply the deficits: 3 × 6 = 18. The answer is 9,118.
• Near 50. 48 squared is (50 − 2) squared = 2,500 − 200 + 4 = 2,304.
• Squares ending in 5 above 100. 105 squared is 11,025.
• Multiplying by 11. For 63 × 11, write 6, (6 + 3), 3 to get 693. For 87 × 11, 8, (8 + 7) = 15, 7 with the carry gives 957.
• Multiplying by 5. Halve and multiply by 10: 48 × 5 = 240.
• Multiplying by 25. Divide by 4 and multiply by 100: 36 × 25 = 900.
• Fractions to percentages. 1/8 = 12.5%, 1/6 = 16.67%, 1/7 = 14.29%, 1/9 = 11.11%.

18. A fourteen-day plan

Study two topics a day for the first seven days, writing out three worked examples each from memory. On days eight to ten, take mixed sets of twenty questions on the topics, with a stopwatch. On days eleven to thirteen, take a full timed paper and review every wrong answer, writing the correct method beside it. On day fourteen, revise only the shortcut lists. Skip a question after one minute and come back to it; speed on easy questions is what gains marks.

Key facts for exams

• Series: check difference, ratio, squares, cubes, then second differences.
• Letter positions: A = 1 to Z = 26; mirror letters add to 27; anchors E, J, O, T, Y = 5, 10, 15, 20, 25.
• Clock angle = |30H − 5.5M|; hands overlap 22 times a day; 1 century = 5 odd days.
• Equal percentage gain and loss on equal sale prices always gives a loss.
• Successive changes: the net change is not the sum.
• CI minus SI for 2 years = P × (R/100) squared.
• Two workers: ab/(a + b) days. Average speed over equal distances: 2xy/(x + y).
• km/h to m/s: multiply by 5/18.

Practice questions

1. What comes next: 2, 6, 12, 20, 30, ? a) 36 b) 40 c) 42 d) 56 Answer: c. Each term is n × (n + 1), so the next is 6 × 7 = 42.

2. If CAT is coded DBU, then DOG is coded: a) EPH b) ENF c) CNF d) EOH Answer: a. Each letter moves one step forward.

3. A man walks 12 km east and then 5 km north. How far is he from the start? a) 7 km b) 13 km c) 15 km d) 17 km Answer: b. The 5-12-13 triplet.

4. A student is 12th from the top in a class of 40. What is the rank from the bottom? a) 27 b) 28 c) 29 d) 30 Answer: c. 40 − 12 + 1 = 29.

5. All A are B and no B is C. Which conclusion follows? a) All A are C b) No A is C c) Some A are C d) Some C are A Answer: b. A lies inside B, and B has no overlap with C.

6. What is the angle between the hands of a clock at 3:40? a) 120 degrees b) 130 degrees c) 140 degrees d) 150 degrees Answer: b. |30 × 3 − 5.5 × 40| = |90 − 220| = 130.

7. A price rises by 20% and then falls by 20%. The net change is: a) No change b) 2% fall c) 4% fall d) 4% rise Answer: c. 100 becomes 120 and then 96.

8. The compound interest on 10,000 at 10% a year for 2 years is: a) 2,000 b) 2,100 c) 2,200 d) 2,310 Answer: b. The amount is 12,100.

9. A can do a job in 12 days and B in 15 days. Together they take: a) 6 2/3 days b) 7 days c) 13 1/2 days d) 27 days Answer: a. 12 × 15/27 = 20/3.

10. A car covers the same distance at 60 km/h and 40 km/h. The average speed is: a) 45 km/h b) 48 km/h c) 50 km/h d) 52 km/h Answer: b. 2 × 60 × 40/100 = 48.

Frequently asked questions

Which topics should I do first? Start with percentages, ratio, averages and series, because their methods feed into profit and loss, interest, and time and work.

How many questions can I expect from reasoning and arithmetic? It depends on the post and the notification. Check the official notification for the current pattern and marks.

Should I learn tables and squares? Yes. Squares to 30 and cubes to 10, with the multiplication tables to 20, save more time than any single shortcut.

Why does a gain and a loss of the same percentage give a loss? Because the loss percentage applies to a larger cost price when the selling price is the same.

How do I avoid silly mistakes? Write the unit next to each number, re-read the question's last line before marking, and estimate the answer first so that you can reject an absurd option.

Is guessing sensible? Only after eliminating options. Read the marking scheme in your notification and see the time management guide for a policy.

Do the same methods work for Group-4 and Police papers? The methods are the same; the difficulty and the count of questions differ by post and year. Check the latest notification.

Related reading

• AP Police Constable: exam and selection guide
• AP Police SI guide
• AP Sachivalayam Secretary guide
• APPSC Group-4 guide
• Negative marking, time management and mock strategy
• Telugu version of this post

Practise on pareeksha.in

Shortcuts only help when they become automatic. Practise them in timed sets and mock papers on pareeksha.in, where the platform offers 10 Lakh+ MCQs, 100+ Exams and 5,000+ Full-Length Mocks. Open a mock test and start with reasoning and arithmetic sections. Confirm the pattern and syllabus for your post in the latest official notification.

Sources and verification

Pages opened for this revision: none. This post is based on standard reasoning and arithmetic methods, and no website was used as a source.

How the numbers were checked: every worked example and every practice answer in this post was recomputed with a Python script that uses exact fractions, direct simulation for walks and calendars, and an exhaustive check of all arrangements of three groups for the syllogism verdicts. The script passed all checks before the text was written, and the calendar days (26 January 1950, 15 August 1947, 1 January 2000, 2 June 2014) were checked against the Python date library.

Not verified: the exact number of reasoning and arithmetic questions, their marks, and the syllabus for each AP post. These change with each notification, so the text only says to check the latest official notification.