New Pattern and Miscellaneous

A reference address (also called the master address) is printed, followed by four options. Exactly one option is a perfect copy, and each of the others hides a small change. Sometimes the four addresses are numbered 1 to 4 and the options read "1 Only", "3 Only" and so on. The same question is also asked on a name-and-date line or a short code line.

Here is the logic that saves time. A copy must agree with the reference in every field. So, if one of the four options is the copy, every part on which all four options agree is automatically right, and you never need to check it against the reference. The paper-setter builds the options from the reference by changing one or two places, so the four options are mostly alike and differ only in a few spots. Those few spots are the only places that matter.

Step 1. Run your eye down the four options together and note the places where they are not all the same: it may be the PIN, a flat number or one word. Step 2. Only at those places, compare each option with the reference and strike the options that disagree. Step 3. If two options are left, compare them with each other; they differ at one spot, and a single look at the reference decides.

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In the figure, option (a) falls at the PIN (201310 against 201301), option (c) falls at "Rd", and option (d) falls because an extra "UP" has been printed. Only (b) is left.

Cancel what all four options share; check only the moving part against the reference.

Other forms. A date-and-place line or a form number is handled in exactly the same way. When two addresses are printed and you are asked whether they refer to the same entry, compare field by field: a changed house number, flat number or PIN decides the matter, while a difference only in a short form of a name (Rohit and R.) does not by itself prove that they are different persons.

Here four addresses (or name-and-address sets) are printed, numbered 1 to 4 or marked P, Q, R, S. The options are pairs such as "1 and 3". Only one pair is identical.

Think of an address as a chain of fields: house or flat number, building, road, area, city and PIN code. Two addresses are twins only if every field matches, character for character. The wrong addresses differ from the twin pair by a very small change, so your job is not to read each address carefully. Your job is to sieve.

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Look at the figure. Four addresses share the same house number. The PIN of S is different, and the word in Q is different. The two clean addresses, P and R, are the twins. Notice that you never had to read the whole line again and again: you compared one field at a time down a column.

Now the logic. Twins agree in every field, so every field on which all four addresses are equal can be ignored. Look only at the fields that move. For the first moving field, split the addresses into groups of equal values: the twins must be in the same group, so every option that pairs addresses from different groups is struck. Then repeat with the next moving field on the options that survive. The pair left standing is the answer. The cheapest field to compare is the PIN, then the numbers, then the words.

Compare field by field, down the column, never address by address.

Stop the moment one option is confirmed and the others are struck. Many students lose time by checking the remaining options after the answer is already clear. When all PINs are the same, which happens often, go straight to numbers and then words.

Two kinds of short-string questions appear. In the first, a code, serial number or name line is printed and you pick the option that copies it exactly. In the second, which is the newer form, several people remember the same number or ID differently, and the question tells you what they agree on. You must work out the one correct string.

Recall with agreement at each place. Four friends remember a number as PR73T, KS73T, KR78T and KR73X, and it is known that at every place at least three of them agreed. Write the recalls one under another and read down each place. The value that three or more wrote is the true one, because three out of four is a majority and a majority cannot be two different values. Place 1 is P, K, K, K, so K. Place 2 is R, S, R, R, so R. Place 4 is 3, 3, 8, 3, so 3. The number is KR73T, although no friend had the whole number right: the vote works place by place, not friend by friend.

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Recall with clues. The other form lists what the people said: "it begins with ART", "the next part is 456", "it has the letter R", "it ends with 78". Treat every clue as a filter and run it over the options, striking as you go. Start with the clue that removes the most options, usually a fixed beginning or ending. The option that passes every clue is the answer, and in these questions exactly one does. The order is part of the clue: 456 is not 465, and 78 is not 87.

Exact copy. Here the original is cut into the chunks in which it is printed, and every option is compared chunk by chunk. The first chunk that does not match puts the option out, and the eye is much better at checking a short chunk than one long run of characters.

[[fig:1]]

The figure shows the working method. The original is cut into chunks exactly as they are printed (blocks of 2 to 4 characters), and each option is compared chunk by chunk. The first chunk that does not match puts the option out. Here (d) fails in the first chunk, (a) in the second, (b) in the third, and only (c) passes all three.

For a swap check, compare the first character, then the last character, then the middle pairs. When the front looks the same, start from the end of the string. Cut into printed chunks, compare chunk by chunk, strike at the first mismatch.

Recent papers include a few short items that look unlike any other reasoning chapter. There is no formula for them, but each kind has a fixed way of thinking, and four kinds cover almost all of them.

1. Behaviour and emotional intelligence. Emotional intelligence has five parts: self-awareness (knowing your own feelings), self-regulation (controlling them), motivation, empathy (understanding other people) and social skill. Intelligence-test skills such as logic, spatial reasoning, memory and vocabulary are not parts of it. A "most appropriate action" item asks what a sensible and honest person would do. Choose the option that is honest, lawful and calm, and that goes through the proper person or procedure. The usual wrong options give in to a friend, use force, act in a hurry or ignore the problem. A critical thinker checks evidence, reviews and comparisons before deciding.

2. Logical order. You must arrange items in a meaningful order: stages of a plant (seed, sapling, plant, flower, fruit), steps of a process (application, admission, college, degree), periods of history, planets by distance, people by year. First fix the item that must come first and the item that must come last, and strike the options that break them. Then settle the middle with one pair at a time, such as "sapling before plant".

3. Roll numbers and number properties. A list of roll numbers is asked about a property: perfect cube, perfect square, prime, palindrome, or only composite digits. Keep these ready. Cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Prime digits: 2, 3, 5, 7. Composite digits: 4, 6, 8, 9 (0 and 1 are neither). A palindrome reads the same from both ends. One number can have two properties: 64 is both a square and a cube.

4. Records and tables. A short list or table is printed with a key, such as "the first two digits show the centre". Decode the key first, then tally in one pass, then read the exact condition of the question ("more than", "at least", "which", "how many").

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Take the roll numbers in the figure. The key says the first digit is the centre code. So the first job is to decode the code key: what does each part of the code mean? Here digit 1 is Patna and digit 2 is Ranchi.

When an entry is added, work out how the total and the average change, then re-read the condition. An added entry can move the average so that a record that was above it is now below it.

Decode the key, tally once, then read the question's exact condition.

Arguments with a flaw (slippery slope, a correlation taken as a cause, hasty generalisation) and "which assumption is implied" items also reach this chapter; their full method is in the Statement-Based Reasoning chapter.

Age questions come in three shapes. A sentence connects a person's age now with an age some years ago or later ("present age is three times the age 12 years ago"). A group's total or average is given and moves with time. Or children are born at equal gaps and only the total is known. All three rest on one fact: everyone grows older at the same speed.

An age equation is a sentence turned into algebra. The skill is not in the solving; it is in the translation. Call the present age x and translate the sentence word by word.

The key phrase to read carefully is "its age n years ago". That is a past age, so it is x − n. The present age is still x. Many wrong answers come from using x − n as the answer.

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Solve in three short moves: open the bracket, collect the x terms on one side, divide. Then always substitute back into the original sentence. This single check removes most mistakes, because the wrong options in these questions are usually the past age, the future age, or a half-solved value.

Past age is x − n. Present age is x. Check the answer in the sentence.

When two persons are involved (father and son), take two unknowns, write one equation for each sentence, and subtract the equations to remove one unknown.

Think of a group of people standing in a row. As time passes everyone grows older by the same number of years. So when n years go by, each person adds n, and the group total rises by (number of people × n). The average, which is the total divided by the number of people, rises by exactly n.

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• Average n years later = present average + n; average n years ago = present average − n.
• Total n years later = present total + (members × n); n years ago = present total − (members × n).

This works only while nobody joins or leaves the group between the two moments. If a person joins (a teacher joins a class, a baby is born), the number of members changes. Then do not shift averages. Find each total separately, add or remove the person, and divide by the new count.

Average shifts by n. Total shifts by members × n. A new member means a new count.

Children born at equal gaps. The ages rise by the same amount each time, so the middle age is the average. For an odd number of children the middle child has exactly the average age; for an even number the average lies halfway between the two middle ages. Hence sum = number × average, and youngest + oldest = 2 × average. Five children, gap 4, sum 45: the middle child is 45 ÷ 5 = 9, so the oldest is 9 + 2 × 4 = 17. For any equal-gap list, terms = (last − first) ÷ gap + 1 and sum = terms × (first + last) ÷ 2. Remember that n children have n − 1 gaps. To add such a list quickly, pair the first with the last: every pair has the same total, as the figure shows for 5, 8, 11, ..., 32.

[[fig:2]]

In the figure, 5 + 32 = 37, 8 + 29 = 37, 11 + 26 = 37, and so on, five pairs of 37. So the whole sum is 5 × 37 = 185.

A ratio such as 7 : 2 tells you the shape of the ages, not the real ages. Two numbers that make the same ratio can be 7 and 2, 14 and 4, or 42 and 12. To find the real ages you need one more fact: the difference, the sum, or one actual age. That is the clue the question always gives.

Ratio a : b with a difference D. The difference in parts is (a − b), and that equals D years, so one part = D ÷ (a − b). Then the ages are a × part and b × part. If the sum S is given, one part = S ÷ (a + b).

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In the figure the ratio is 7 : 2 and the father is 30 years older. Five parts are 30 years, so one part is 6; father 42, son 12. After 10 years both move by 10 and become 52 and 22. The gap is 30 at both times, but the ratio has changed from 7 : 2 to 52 : 22, which is 26 : 11.

A ratio moves with time; the gap does not. Find the real ages first, then move both.

Notice the usual wrong step: adding the 10 years to the ratio numbers (7 + 10 : 2 + 10). The numbers 7 and 2 are not ages, they are parts, so that step has no meaning.

When both ratios are given (a : b now, c : d after n years). Write the ages as ak and bk. After n years, d(ak + n) = c(bk + n). Cross-multiplying gives k(ad − bc) = n(c − d), so k = n(c − d) ÷ (ad − bc). Use a negative n for "years ago". For 4 : 7 now and 5 : 8 after 6 years, k = 6 × (5 − 8) ÷ (4 × 8 − 7 × 5) = (−18) ÷ (−3) = 6. If the gap is the same number of parts in both ratios (7 − 4 = 3 and 8 − 5 = 3), the formula shrinks to one idea: one part = years ÷ (change in one person's share). Here the share goes from 4 to 5, so one part = 6 years.

These are short stories, one or two lines long, that the paper uses as fillers. They look like arithmetic, but each is solved by one idea. Learn to spot the idea in the first sentence.

1. Sum and difference. If two numbers have sum S and difference D, then larger = (S + D) ÷ 2 and smaller = (S − D) ÷ 2. Reason: adding the two facts doubles the larger number.

2. "Thrice as many". Turn the sentence into parts. "Thrice as many wrong as right" means right : wrong = 1 : 3, so the total is 4 parts and one part is the number right.

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In the figure, 60 questions make 4 parts of 15, so 15 are right and 45 are wrong.

3. Working backwards. When something is done to a quantity stage by stage (half is taken away, one-third is drunk, an amount is spent) and only the end is known, start from the end and undo every stage in reverse order. "Took half" becomes "double", and "gave back 2" becomes "take away 2". A boy ends with 10 apples after three checkpoints, where each guard takes half and returns 2. Undo the last checkpoint: 10 − 2 = 8, doubled 16. Then 16 − 2 = 14, doubled 28. Then 28 − 2 = 26, doubled 52. He started with 52. A person who spends "half of what he has plus 1" in each of 6 shops and is left with nothing undoes each shop by (amount + 1) × 2: 2, 6, 14, 30, 62, 126.

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4. Transfers: "after taking ₹x, A equals B". If A takes ₹x from C and then has as much as B, then B has ₹x more than A at the start. With the total of A and B you now know a sum and a difference, so use (S ± D) ÷ 2. If A gives away ₹x to a third person and then equals B, A had ₹x more than B. But if B gives ₹x to A and they become equal, B had ₹2x more than A, because B loses x and A gains x, which closes a gap of 2x.

5. Positions and gaps. The k-th odd number is 2k − 1. For houses 5, 7, 9, ..., the "sixth odd house after 5" is 5 + 2 × 6 = 17. Equally spaced posts or houses: n of them have n − 1 gaps, so 15 houses from the first to the last have 14 gaps.

6. Shares and comparisons. For weights, write every object as a multiple of the one that is known, then compare. For "every 2 people share a cake, every 6 a soup, every 8 a platter", each person uses 1 ÷ 2 + 1 ÷ 6 + 1 ÷ 8 = 19 ÷ 24 of an item, so 38 items serve 38 × 24 ÷ 19 = 48 people. If 5 machines make 5 items in 5 minutes, one machine makes one item in 5 minutes, so 100 machines make 100 items in 5 minutes.

7. Sitting together. People who must sit together are glued into one block. For n people in a row with k particular people together, the count is (n − k + 1)! × k!. For k people sitting side by side in a row of n seats, it is (n − k + 1) × k!, so 3 people in 8 seats give 6 × 6 = 36. "Not together" is the total minus "together".

When the numbers are small, put the options into the story; the wrong ones fail at once.

In a round-robin league every player meets every other player exactly once. Handshake questions are the same idea: everyone shakes hands with everyone else, once.

Count from one player's point of view. The first player has n − 1 opponents. The second player has already met the first, so there are n − 2 new opponents. The third has n − 3, and so on down to the last player, who has none left to meet.

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In the figure there are 6 players. A plays 5 matches (red lines). B has 4 new matches (blue lines), because A–B was already counted. Then 3, 2 and 1. The total is 5 + 4 + 3 + 2 + 1 = 15.

There is a faster way. Pick any one player and pair him with the remaining n − 1, which gives n × (n − 1). But the match A–B has been counted twice, once from A and once from B. So halve it: Matches = n(n − 1) ÷ 2. For 6 players, 6 × 5 ÷ 2 = 15.

Three situations to keep apart. Single league: n(n − 1) ÷ 2. Double league (each pair plays twice, home and away): n(n − 1). Knock-out: n − 1 matches, because every match removes exactly one team and only one team stays at the end.

Variants in recent papers. If each person shakes hands with exactly k others (people in a circle greeting their 3 neighbours), the count is n × k ÷ 2: 20 people with 3 neighbours each give 20 × 3 ÷ 2 = 30. If some people refuse, count only those who take part: of 28 persons, 7 refuse, so 21 take part and 21 × 20 ÷ 2 = 210.

These are small timetable and cycle puzzles. The cure is always the same: draw a time line, or a distance-time picture, and mark what you know.

Every k minutes. If a bus leaves every 30 minutes, the previous bus and the next bus are exactly 30 minutes apart, and any moment lies between them. Suppose a passenger says "the last bus left 12 minutes ago, and the next is at 4:10 pm."

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The previous bus left at 4:10 − 30 min = 3:40 pm. He spoke 12 minutes after that, so it was 3:52 pm. Check: 12 minutes have passed since the last bus and 18 remain to the next, and 12 + 18 = 30. Always run that check, because the commonest slip is to read "left 12 minutes ago" as "next bus in 12 minutes".

Overlapping shifts. When several shifts run at once, stack them on one strip. The busiest stretch is where the most shifts overlap, and the overlap of intervals runs from the latest start to the earliest end.

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Shifts 7 am to 1 pm (4 people), 9 am to 3 pm (6 people) and 11 am to 5 pm (3 people): all three are on from 11 am to 1 pm, so that stretch has 4 + 6 + 3 = 13 people. Compare with 9 to 11 (4 + 6 = 10) and 1 to 3 (6 + 3 = 9).

Catch-up. A faster vehicle starts later. When it catches the first, both have covered the same distance, but the first has been travelling for longer by the delay. So speed × time is equal on both sides, with the delay added to the first one's time. Train A leaves at 8:00 at 90 km/h and Train B leaves at 8:10 at 120 km/h. B catches A at 8:40, when A has travelled 40 minutes and B only 30. A: 90 × 40 ÷ 60 = 60 km. B: 120 × 30 ÷ 60 = 60 km. The lines in the figure meet at the point where the distances are equal.

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Teams working together. When groups work at the same time, the total time is the time of the biggest load, not the sum. Three groups search 2 floors each, 20 rooms per floor, at 1 minute a room: every group has 40 rooms, so the search takes 40 minutes. Repeating turns. A shape that turns 45° each step has turned 8 × 45° = 360° after 8 steps, so it is back in its original position. Whenever the total turn is a multiple of 360°, the shape is back where it began.

Climb and slip. A snail climbs 5 m each day and slips back 3 m each night. How long does it take to climb a 20 m well? The tempting answer is to divide 20 by the net gain of 2 m and say 10 days. It is wrong, because the last climb has no slip after it. Once the snail reaches the top, it is out.

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1. Take the final climb out first: H − c = 20 − 5 = 15 m must be covered by full days (climb and slip).
2. Divide by the net gain per cycle: 15 ÷ 2 = 7.5, and round up to 8 full cycles.
3. Add one more cycle for the final climb: 8 + 1 = 9.

Cycles = (H − c) ÷ (c − s), rounded up, plus 1.

Before using the formula, state the order of events: climb first, slip afterwards, in every cycle. If a question counts time (a climb takes 4 minutes, a slip 1 minute), find the number of full cycles, multiply by the cycle time, and then add only the time of the final climb.