CSAT (Paper II) · MPSC Prep

CSAT · Comprehension and decision-making · Topic 1 of 11

Reading Comprehension: Main Idea, Inference and Assumption

Mark as done

Think of it like…

Listening to a gram sabha speech

At the gram sabha the sarpanch speaks for ten minutes about the new water scheme. An alert villager walking home can tell his wife three different things. First, what the speech was about: “the village needs to save water this summer” (the main idea), not the one line about the old well. Second, what follows from it even though it was not said in words: if the sarpanch said tankers will come only twice a week, then water will be scarce on the other days (an inference). Third, what the sarpanch took for granted: when he said “we will repair the pipeline before May”, he assumed the money from the zilla parishad will arrive in time (an assumption).

A careless listener adds his own gossip: “the sarpanch is doing this for the election.” That may even be true, but it was not in the speech. CSAT comprehension punishes exactly this habit. The passage is the only gram sabha you attended; your outside knowledge stays at home.

In everyday life	In the exam topic
What the whole speech was about	Main idea or central theme of the passage
What must be true if the sarpanch is right	Inference: follows necessarily from the text
What the sarpanch took for granted without saying	Assumption: unstated premise the argument needs
Village gossip added on the way home	Outside knowledge or extreme option: reject it
The one remark about the old well	A detail that is true but not the main idea

The concept

MPSC CSAT passages are short (often 100–400 words) and the questions test four skills: the main idea (what the whole passage is about), a stated detail, an inference (something that must be true given the passage) and an assumption (an unstated premise the author’s argument depends on). Because Paper II is qualifying, comprehension questions are the safest marks for candidates from a non-mathematics background, so they are worth doing first and carefully.

The main idea must cover the whole passage. A wrong option is usually either too narrow (one example or one paragraph) or too broad (a sweeping claim the author never makes). A good test: could the option serve as a title for every paragraph? An inference is stricter than “could be true”; it must be supported by the text without extra facts. Options with extreme words (always, never, only, all, completely) are rarely correct inferences, because authors usually argue for degrees.

An assumption is the hidden bridge between the author’s evidence and conclusion. Use the negation test: negate the option; if the argument collapses, it was an assumption. For example, “The district should shift farmers from paddy to millets because millets need less water” assumes that millets can actually be grown and sold profitably there. If that were false, the recommendation falls apart.

Read the question stem first, then the passage once with attention to the first and last sentences and to signal words: however, but, yet (turn), therefore, thus, hence (conclusion), for example (supporting detail). Then eliminate: an option that contradicts the passage, goes beyond it, or answers a different question is out.

Key points & formulas

Main idea = covers the whole passage; too narrow (one detail) and too broad (beyond the text) are the two classic wrong options
Inference = must be true from the passage alone; “could be true” is not enough
Assumption = unstated premise; negate it, and if the argument collapses it was assumed
Extreme words (always, never, only, all, completely) usually signal a wrong option
Signal words: however, but, yet mark a turn; therefore, thus, hence mark the conclusion
Outside knowledge, even if true, is not a valid basis for an answer
Tone words to know: critical, cautionary, optimistic, neutral, sarcastic, analytical
Order: read the question stem first, then the passage, then eliminate options

Worked example

Passage: In many villages of Marathwada, the response to every dry year has been to drill a deeper borewell. Each new bore brings water for a season, and then the water table falls further and the next bore must go deeper still. Families spend borrowed money on drilling, and when the bore fails the debt remains. A few villages have tried the opposite path. They built small check dams, stopped new borewells by agreement in the gram sabha and decided together which crops to grow based on the water measured after the monsoon. Their wells have not become full overnight, but they no longer fail every April.
Which of the following best expresses the main idea of the passage?
(a) Borewells should be banned by law in Marathwada.
(b) Collective water management works better than competitive deep drilling.
(c) Check dams fill wells overnight.
(d) Farmers in Marathwada borrow money mainly for drilling borewells.

Show solution

Read the first and last sentences: the passage begins with a problem (drilling deeper every dry year) and ends with a contrast (villages that managed water together no longer fail every April).
Option (a) is too strong: the passage describes villages agreeing in the gram sabha, not a legal ban. Extreme and beyond the text.
Option (c) contradicts the passage, which says the wells “have not become full overnight”.
Option (d) picks one detail (borrowed money) and adds “mainly”, which the passage never claims. Too narrow and too strong.
Option (b) covers both halves: the failure of competitive drilling and the success of collective management. It fits every part of the passage.

Answer: (b) Collective water management works better than competitive deep drilling.

Exam shortcut

For “main idea” questions, ask “would the author put this on the title of the passage?” For “assumption” questions, negate each option: the one whose negation destroys the argument is the answer.

Common traps

Choosing an option that is true in real life but not stated or implied in the passage.
Confusing a supporting example with the main idea: the example is there to prove the main idea, not to replace it.
Treating an inference as “anything that could be true”: it must follow from the passage.
Confusing an assumption (unstated, needed) with a conclusion (stated, the point being made).

Practice · tap an option to check

Q1. Passage: Along the Konkan coast, mangrove forests were long treated as wasteland, to be cleared for salt pans, prawn farms or housing. Scientists now describe them differently. Their tangled roots slow down storm surges and hold the soil of the creek banks in place. Their shallow waters are nurseries where young fish and crabs grow before moving out to sea, so the catch of fishing villages depends on them. They also store large amounts of carbon in their muddy soil. Protecting mangroves is therefore not a favour to nature at the cost of people; it protects the livelihoods and safety of the coastal villages themselves.
Which of the following best states the central idea of the passage?

Q2. Passage: Two years ago a gram panchayat in Satara district began painting its income and expenditure on the wall outside its office every month: how much came from the state, how much was spent on each road, drain and handpump, and who the contractor was. Villagers began stopping to read it. At the next gram sabha they asked why one drain had cost twice as much as a similar one the year before, and the work was re-measured. Complaints to the block office about the panchayat, which used to arrive every week, have since become rare.
Which of the following can be inferred from the passage?

Q3. Passage: The district agriculture officer argues that paddy farmers in the eastern talukas should be encouraged to shift to millets such as jowar and nachni. Paddy needs standing water for weeks, while millets need a fraction of that water. Farmers in these talukas now pump groundwater through the dry months to keep their paddy fields flooded, and the water level in village wells falls further every year. Millets, he points out, were grown on the same land two generations ago and are now promoted for school meals and ration shops. A shift to millets, he says, would let the taluka save water and still give farmers a living.
The officer’s argument assumes that:

Q4. Passage: Every monsoon, parts of Mumbai go under water, and every year the explanation offered is the same: the rain was unusually heavy. Heavy rain is only half the story. Over decades, open ground that once soaked up water has been paved over, and the nullahs that carry storm water to the sea have been narrowed by encroachment and choked with plastic. When high tide coincides with a downpour, the outfalls cannot drain at all. A city that wants to stay dry must treat drainage, open land and waste as one problem, all year round, not as a desilting drive in May.
Consider the following statements:
1. The author holds that heavy rainfall alone explains Mumbai’s flooding.
2. The author suggests that desilting nullahs once before the monsoon is sufficient.
Which of the statements given above is/are supported by the passage?

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CSAT · Comprehension and decision-making · Topic 2 of 11

Decision Making and Problem Solving (administrative situations)

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Think of it like…

The talathi’s four-question sieve

An experienced talathi in a taluka office handles a dozen problems a day: a farmer whose name is missing from the 7/12 extract, a neighbour asking for a “quick favour”, a widow who cannot read the form. Over the years he has learned to run every response through four questions in his head. Is it within the law and the rules? Is it honest, with nothing I would hide from my senior? Does it treat the person in front of me with dignity? Will it actually solve the problem, not just make it go away for today?

Shouting at the farmer fails the third question. Doing the neighbour’s favour fails the first two. Telling the widow to come back next week with someone literate passes the first two but fails the third and fourth. The response he finally gives (helping her fill the form, recording it properly, and finishing the work within the rules) passes all four. CSAT decision-making questions are built exactly this way: usually only one option passes all four sieves.

In everyday life	In the exam topic
Is it within the law and the rules?	Legality: no option that breaks or bends rules
Would I hide it from my senior?	Integrity: no favours, gifts or conflict of interest
Does it respect the person in front of me?	Empathy: dignity, especially for the weak and poor
Will it solve the problem, not just postpone it?	Effectiveness: action that fixes the cause in time
The reply that passes all four	The one best option in the question

The concept

These questions put you in the chair of a public official (tahsildar, BDO, sub-divisional officer, district collector, police officer, head of an office) facing a situation: a flood, a protest, a bribe offer, a relative’s file, a subordinate’s mistake, a complaint from a citizen. Four or so responses are offered, and you must pick the best, not merely an acceptable one. The examiners reward the response a model civil servant would give: legal, honest, empathetic and effective, and proactive rather than passive.

Eliminate in this order. First drop options that are illegal or unethical (accepting a gift, doing a favour, hiding a mistake, using force first). Then drop passive or buck-passing options (“wait and watch”, “refer it to the senior and do nothing”, “tell them it is not my department”). Then drop harsh or disproportionate options (suspending a clerk for a first honest mistake, lathi-charge on a peaceful crowd). Among what remains, pick the option that solves the problem at the root and within the time it matters, usually by combining communication with action.

Typical good behaviours: talking to people before using force; following the procedure but speeding it up; recusing yourself when a relative or friend is involved; recording and reporting a bribe offer rather than quietly refusing; fixing the system after correcting an individual case; putting human life and safety first in a disaster. Typical bad behaviours: extremes in either direction, being soft and doing nothing, or being harsh and doing too much.

Marking for this block differs from the rest of the paper. In MPSC CSAT, decision-making questions are commonly said to carry no negative marking, which, if true, means you should never leave one blank. Check the current MPSC notification for the exact rule before the exam.

Key points & formulas

Four tests: legality, integrity, empathy, effectiveness; the best option passes all four
Life and safety come first in any disaster or emergency situation
Dialogue before force; force (if ever) only as a last resort and within law
Conflict of interest (relative, friend, own interest): disclose and recuse, do not decide yourself
Bribe or gift offer: refuse, record and report as per rules, then decide the file on merits
Passive options (wait and watch, pass it on, ignore) are almost never the best
Harsh options (suspend, punish first, use force first) are rarely the best for a first or honest lapse
Fix the individual case, then fix the process so it does not recur
Negative marking for decision-making is commonly said to be nil in MPSC CSAT: check the current notification

Worked example

You are the tahsildar of a riverside taluka. The irrigation department warns that water will be released from an upstream dam tonight and three low-lying villages may flood. Many families in one village refuse to move to the relief camp because they will not leave their cattle behind. What is the best course of action?
(a) Order the police to remove the families by force immediately.
(b) Record their refusal in writing and leave them, since you have done your duty.
(c) Arrange a safe place for the cattle on high ground near the camp, involve the sarpanch and elders to persuade families, and evacuate everyone from the danger zone before the release.
(d) Ask the irrigation department to postpone the release until the villagers agree.

Show solution

Life comes first, so the villagers must be out of the danger zone before the release. That rules out (b), which saves the officer’s file but not the families.
(d) is not effective: dam releases are a safety decision for the dam itself, and waiting for consent may be fatal for everyone downstream.
(a) may move people but uses force first, ignores the genuine reason (their livestock is their livelihood) and creates distrust that will hurt future evacuations.
(c) removes the real cause of refusal (cattle), uses trusted local leaders to persuade, and still completes the evacuation in time. It is legal, empathetic and effective.

Answer: (c)

Exam shortcut

Strike out every option containing ignore, wait, refuse to meet, force first, or quietly. Of the rest, prefer the one with two verbs: talk and act (for example “meet the delegation and fix a time-bound panchanama”).

Common traps

Choosing the “strict” option because it sounds like a strong officer: disproportionate punishment or force is marked as worse, not better.
Choosing “refer the matter to the senior” when the matter is within your own powers: that is buck-passing.
Choosing a kind but illegal option (relaxing a rule for a sympathetic person); empathy must work within the law.
Leaving decision-making questions blank out of fear of negative marking: they are commonly said to carry none in MPSC CSAT, but check the current notification.

Practice · tap an option to check

Q1. You are a block development officer. A contractor whose bills for a road work are pending with you visits your home on Diwali with an expensive gift hamper and says it is “only a festival courtesy”. What should you do?

Q2. You are the sub-divisional officer. After unseasonal hail, about 300 farmers block the state highway demanding immediate compensation for crop loss. Ambulances and ST buses are stuck. What is the best response?

Q3. As district collector you are to decide applications for a commercial land-use permission. You find that one of the applicants is your brother-in-law. What should you do?

Q4. At a public hearing, an elderly widow tells you, the district social welfare officer, that her name is missing from the pension beneficiary list although she is eligible. On checking, you find that a clerk in your office made a data-entry error that left out several eligible names. What is the best course of action?

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CSAT · Basic numeracy · Topic 3 of 11

Number System: Divisibility, HCF and LCM, Remainders, Unit Digits

Mark as done

Think of it like…

Packing mangoes into crates at the Ratnagiri orchard

At a Ratnagiri orchard, three heaps of Alphonso mangoes, 48, 72 and 120, must be packed into crates of one size, with no mango left over in any heap. The biggest crate size that fits every heap exactly is 24, the HCF. Pack in 24s and you get 2, 3 and 5 crates. Any larger crate leaves mangoes over in some heap.

Now the trucks: one truck returns to the orchard every 4 days, another every 6 days, a third every 10 days. They all leave together today. The first day all three are back together is day 60, the LCM. Mangoes left over after filling crates are remainders, and the last digit on the crate count is the unit digit, which repeats in a fixed cycle like the days of the week.

In everyday life	In the exam topic
Biggest crate that empties every heap exactly	HCF: greatest common divisor
First day all trucks return together	LCM: least common multiple
Mangoes left after the last full crate	Remainder
Last digit repeating like days of the week	Cyclicity of unit digits (period 4 for 2, 3, 7, 8)
Number of crates × crate size = heap	Dividend = divisor × quotient + remainder

The concept

Divisibility rules save time. By 2: last digit even. By 3 or 9: digit sum divisible by 3 or 9. By 4: last two digits divisible by 4. By 8: last three digits divisible by 8. By 5: last digit 0 or 5. By 6: divisible by both 2 and 3. By 11: the difference between the sum of digits in odd places and the sum in even places is 0 or a multiple of 11. For a composite divisor, split it into co-prime factors (12 = 3 × 4, not 2 × 6).

HCF and LCM: HCF is found by prime factors (lowest powers) or by repeated division; LCM by highest powers. For two numbers, HCF × LCM = product of the numbers, which does not hold for three numbers. Word clues: “largest”, “greatest”, “maximum size of a tile or crate” point to HCF; “least”, “smallest”, “first time together”, “bells ringing together” point to LCM. Remainder variants: the largest number that divides a, b, c leaving remainders p, q, r is HCF(a−p, b−q, c−r); the least number that leaves the same remainder r when divided by x, y, z is LCM(x, y, z) + r.

Remainders and unit digits: the unit digit of a power depends only on the unit digit of the base. 0, 1, 5, 6 repeat themselves; 4 and 9 have cycle 2 (4, 6 and 9, 1); 2, 3, 7, 8 have cycle 4 (2, 4, 8, 6; 3, 9, 7, 1; 7, 9, 3, 1; 8, 4, 2, 6). Divide the exponent by 4: remainder 1, 2, 3 picks that position, remainder 0 picks the fourth. For remainders, replace numbers by their remainders before multiplying or adding, since (a × b) mod n = (a mod n × b mod n) mod n.

Key points & formulas

Divisible by 8: last three digits divisible by 8; by 11: (odd-place sum − even-place sum) is 0 or a multiple of 11
HCF × LCM = product of the two numbers (only for two numbers)
HCF of fractions = HCF of numerators ÷ LCM of denominators; LCM of fractions = LCM of numerators ÷ HCF of denominators
Largest number dividing a, b, c leaving remainders p, q, r = HCF(a−p, b−q, c−r)
Least number leaving the same remainder r with x, y, z = LCM(x, y, z) + r
Unit digit cycles: 2 → 2,4,8,6; 3 → 3,9,7,1; 7 → 7,9,3,1; 8 → 8,4,2,6 (period 4); 4 and 9 period 2; 0,1,5,6 fixed
Exponent remainder 0 (when divided by 4) means take the 4th term of the cycle
Dividend = divisor × quotient + remainder; remainder is always less than the divisor

Worked example

Find the least number which, when divided by 12, 15 and 20, leaves a remainder of 5 in each case, and is exactly divisible by 7.

Show solution

Numbers leaving remainder 5 with 12, 15 and 20 are of the form LCM(12, 15, 20) × k + 5.
LCM: 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5, so LCM = 2² × 3 × 5 = 60. The numbers are 60k + 5.
Divisibility by 7: 60 leaves remainder 4 with 7, so 60k + 5 leaves remainder 4k + 5. Try k = 1, 2, 3, 4: 4k + 5 = 9, 13, 17, 21. Only 21 is divisible by 7, at k = 4.
Number = 60 × 4 + 5 = 245. Check: 245 ÷ 7 = 35, and 245 leaves remainder 5 when divided by 12, 15 and 20.

Answer: 245

Exam shortcut

For unit digits of a big power, keep only the last digit of the base and the remainder of the exponent when divided by 4 (use 4 if the remainder is 0). For 7⁹⁵: 95 = 4 × 23 + 3, so take the third term of 7, 9, 3, 1: the unit digit is 3.

Common traps

Testing divisibility by 12 with 2 and 6: they are not co-prime. Use 3 and 4.
Applying HCF × LCM = product to three numbers; it holds only for two.
Exponent remainder 0 means the 4th term of the cycle, not the first.
Mixing up the two remainder patterns: different remainders → HCF of differences; same remainder → LCM plus remainder.

Practice · tap an option to check

Q1. What is the unit digit of 7⁹⁵ × 3⁵⁸?

Q2. The HCF and LCM of two numbers are 12 and 360 respectively. If one of the numbers is 72, the other number is:

Q3. Find the largest number which divides 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.

Q4. Consider the following statements:
1. 7,35,814 is divisible by 8.
2. 9,18,082 is divisible by 11.
Which of the statements given above is/are correct?

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CSAT · Basic numeracy · Topic 4 of 11

Percentages, Profit and Loss, Simple and Compound Interest

Mark as done

Think of it like…

The kirana shop’s base price

Sunita runs a kirana shop in Latur. She buys a sack of sugar for ₹1,000 (the cost price). She writes ₹1,400 on the board (the marked price), a 40% mark-up on cost. A regular customer haggles and gets 15% off the marked price, so he pays ₹1,190 (the selling price). Her profit is ₹190, and profit percent is always counted on the cost price: 19%. Notice that 40% up and 15% down did not give 25%: the 15% was taken on a bigger base.

When she deposits her savings in the cooperative bank, the same idea of the base appears. With simple interest the bank pays interest only on the original deposit, every year the same. With compound interest the interest is added to the deposit, and next year’s interest is counted on the new, bigger base. Every percentage question is really the question: percent of what?

In everyday life	In the exam topic
Price Sunita paid the wholesaler	Cost price (CP): base for profit and loss %
Price on the board	Marked price (MP): base for discount %
Price the customer paid	Selling price (SP)
Interest only on the original deposit	Simple interest = P × R × T ÷ 100
Interest added to the deposit each year	Compound interest: A = P(1 + R/100)^T

The concept

A percentage is a fraction with denominator 100, and memorising a few fraction equivalents (1/8 = 12.5%, 1/6 ≈ 16.67%, 1/3 ≈ 33.33%, 3/8 = 37.5%) speeds up CSAT arithmetic enormously. Successive changes of a% and b% give a net change of a + b + ab/100 (use negative signs for decreases). So a 10% rise followed by a 10% fall gives 10 − 10 − 1 = −1%, a net fall, because the fall is taken on a bigger base. If a price rises by r%, consumption must fall by r/(100 + r) × 100% to keep spending the same.

Profit and loss percentages are always on the cost price; discount is always on the marked price. Profit% = (SP − CP)/CP × 100. With a mark-up of m% and a discount of d%, SP = CP × (1 + m/100) × (1 − d/100). A dishonest dealer using a false weight of 900 g for 1 kg while selling at cost price gains 100/900 = 11.11%, because the gain is counted on what he actually gives.

Simple interest SI = P × R × T/100: the same interest every year. Compound interest: amount A = P(1 + R/100)^T, CI = A − P. For 2 years, CI − SI = P(R/100)², and for 3 years CI − SI = P(R/100)²(3 + R/100). When interest is compounded half-yearly, halve the rate and double the number of periods.

Key points & formulas

Successive change of a% and b%: net = a + b + ab/100
Price up r%: cut consumption by r/(100 + r) × 100% to keep expenditure unchanged (25% up → 20% cut)
Profit or loss % is on CP; discount % is on MP
SP = CP × (100 + profit%)/100 = MP × (100 − discount%)/100
False weight: gain % = (error ÷ true value given) × 100, e.g. 900 g for 1 kg → 11.11%
SI = PRT/100; A = P(1 + R/100)^T for compound interest
CI − SI for 2 years = P(R/100)²
Two successive discounts of a% and b% = single discount of a + b − ab/100

Worked example

A shopkeeper marks his goods 40% above the cost price and allows a discount of 15% on the marked price. Find his profit percent.

Show solution

Take CP = ₹100, so that the answer comes out directly as a percent.
Marked price = 100 + 40% of 100 = ₹140.
Discount = 15% of 140 = ₹21, so SP = 140 − 21 = ₹119.
Profit = 119 − 100 = ₹19 on a CP of ₹100, i.e. 19%. By the formula: 40 − 15 − (40 × 15)/100 = 40 − 15 − 6 = 19%.

Answer: 19% profit

Exam shortcut

Assume the base is 100 (₹100 cost, 100 people, ₹100 principal). Every percentage then becomes a plain number, and the final answer is already a percent.

Common traps

Adding percentages directly: 40% mark-up and 15% discount is 19% profit, not 25%.
Calculating profit% on the selling price instead of the cost price.
Assuming a 10% rise and then a 10% fall brings the value back; it leaves a 1% net fall.
Using the simple-interest formula for compound interest, or forgetting to halve the rate for half-yearly compounding.

Practice · tap an option to check

Q1. The population of a village is 12,000. It increases by 10% in the first year and decreases by 10% in the second year. What is the population after two years?

Q2. What is the difference between the compound interest and the simple interest on ₹15,000 for 2 years at 8% per annum (interest compounded annually)?

Q3. By selling an article for ₹1,140, a trader loses 5%. At what price should he sell it to gain 5%?

Q4. Consider the following statements:
1. If the price of sugar rises by 25%, a family must reduce its consumption by 20% to keep its expenditure on sugar unchanged.
2. Two successive discounts of 20% and 10% are equal to a single discount of 30%.
Which of the statements given above is/are correct?

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CSAT · Basic numeracy · Topic 5 of 11

Ratio, Proportion, Averages and Mixtures

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Think of it like…

Sharing the sugar cooperative’s bonus

At the end of the crushing season, a cooperative sugar factory near Baramati shares a bonus among three member groups in proportion to the cane they supplied, say 3 : 2 : 1. Nobody cares about the actual tonnes to split the money: the pot is cut into 3 + 2 + 1 = 6 equal parts, and each group takes its parts. That is a ratio.

The chairman then announces the average yield: if all members’ cane were levelled out evenly, how much would each have? If one member’s weight was wrongly recorded, the total is wrong by the error, and the average is wrong by the error divided by the number of members. Finally, the factory crushes cane of two sugar contents together and wants a fixed average recovery: how much of each? That is a mixture, solved by alligation, where each grade’s share is the distance of the other grade from the target.

In everyday life	In the exam topic
Bonus pot cut into 3 + 2 + 1 parts	Ratio a : b : c, share = total × part ÷ sum of parts
Levelling all members to one yield	Average = total ÷ number of items
One member’s weight wrongly recorded	Corrected average = old average ± error ÷ n
Crushing two grades of cane to a target recovery	Mixture by alligation
Share of each grade = other grade’s distance from target	Alligation rule: (d − m) : (m − c)

The concept

A ratio a : b compares two quantities of the same kind; multiplying both terms by the same number does not change it. To divide an amount in the ratio a : b : c, give each share = total × part ÷ (a + b + c). When the ratio is given in fractions, such as 1/2 : 1/3 : 1/4, multiply by the LCM of the denominators (12) to get whole numbers 6 : 4 : 3. To combine A : B = 2 : 3 and B : C = 4 : 5, make B equal (8 and 8): A : B : C = 8 : 12 : 15. Proportion means two ratios are equal, a : b = c : d, so ad = bc.

An average is total ÷ number. Most CSAT average questions are really total questions: write the total before and after. If one item is removed or replaced, the change in total is the change in average × number of items. A misread mark changes the average by (wrong − correct) ÷ n. For equal distances at two speeds, the average speed is the harmonic mean 2xy/(x + y), not the simple average.

For mixtures, the rule of alligation: if a cheaper item at price c and a dearer item at price d are mixed to get a mean price m, then quantity of cheaper : quantity of dearer = (d − m) : (m − c). For “add water to change the milk–water ratio” problems, the milk stays fixed: compute the milk first, then the water needed for the new ratio.

Key points & formulas

Share in ratio a : b : c = total × part ÷ (a + b + c)
Fraction ratios: multiply all terms by the LCM of denominators (1/2 : 1/3 : 1/4 = 6 : 4 : 3)
Proportion a : b = c : d ⇔ ad = bc; mean proportional of a and b = √(ab)
Average = sum ÷ number; removed item = old total − new total
Corrected average = old average − (wrong value − right value) ÷ n
Alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper)
In “add water” problems, the quantity of milk does not change
Average speed over equal distances = 2xy ÷ (x + y)

Worked example

In what ratio must rice costing ₹32 per kg be mixed with rice costing ₹42 per kg so that the mixture is worth ₹36 per kg?

Show solution

Cheaper price c = 32, dearer price d = 42, mean price m = 36.
By alligation, cheaper : dearer = (d − m) : (m − c) = (42 − 36) : (36 − 32) = 6 : 4.
Simplify: 6 : 4 = 3 : 2.
Check: 3 kg at ₹32 + 2 kg at ₹42 = 96 + 84 = ₹180 for 5 kg, i.e. ₹36 per kg.

Answer: 3 : 2

Exam shortcut

Alligation cross: write the two prices at the top, the mean in the middle, and subtract diagonally. The difference under each price belongs to the other price’s quantity.

Common traps

In alligation, putting each price’s own distance against it instead of the other price’s distance.
Dividing in the ratio 1/2 : 1/3 : 1/4 as if it were 2 : 3 : 4; convert with the LCM first.
Taking the simple average of two speeds for equal distances; use the harmonic mean.
In add-water problems, changing the milk quantity along with the water.

Practice · tap an option to check

Q1. The average of five numbers is 28. If one number is removed, the average of the remaining four numbers becomes 25. What is the number removed?

Q2. ₹6,500 is divided among A, B and C in the ratio 1/2 : 1/3 : 1/4. What is C’s share?

Q3. A 60-litre mixture contains milk and water in the ratio 7 : 3. How much water must be added to make the ratio of milk to water 3 : 2?

Q4. The average marks of 40 students in a test were found to be 62. Later it was discovered that one student’s mark had been read as 84 instead of 48. What is the correct average?

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CSAT · Basic numeracy · Topic 6 of 11

Time and Work, Pipes and Cisterns

Mark as done

Think of it like…

Tankers filling the village water tank

In summer, a village in Beed gets water through tankers. The gram panchayat’s storage tank holds a fixed amount. One tanker alone can fill it in 12 hours, so it pours 1/12 of the tank every hour. A second, bigger tanker fills it in 15 hours, pouring 1/15 per hour. Run them together and the tank gets 1/12 + 1/15 of itself every hour. Nobody needs to know how many litres the tank holds.

But the outlet pipe to the village is also open, draining the full tank in 20 hours, i.e. taking away 1/20 every hour. The net hourly fill is 1/12 + 1/15 − 1/20 = 1/10, so the tank is full in 10 hours. Work problems are the same tank: each worker is a tanker, the job is the tank, and a worker who spoils work (or a leak) is the outlet.

In everyday life	In the exam topic
Village tank (one full tank)	Total work = 1 unit (or the LCM of the days)
Tanker fills the tank in 12 hours	Rate = 1/12 of the work per hour
Two tankers pouring together	Rates add: 1/a + 1/b
Outlet pipe draining the tank	Negative rate: subtract 1/c
Hours to fill = 1 ÷ net hourly share	Time = work ÷ combined rate

The concept

The key idea is rate. If A finishes a job in a days, A does 1/a of it per day. Rates add when people work together, so A and B together take 1/(1/a + 1/b) = ab/(a + b) days. Times never add. A faster way is the LCM method: take total work as the LCM of the given days. If A takes 10 days and B 15, let the work be 30 units: A does 3 units a day, B 2, together 5, so 30 ÷ 5 = 6 days.

Efficiency is the same as rate. If A is twice as efficient as B, A does 2 units for every 1 of B, and takes half the time. For men, days and hours, use the chain rule: M₁ × D₁ × H₁ ÷ W₁ = M₂ × D₂ × H₂ ÷ W₂, since the work done is proportional to man-hours. For problems where someone leaves partway, work out what was done before the change and divide the remainder by the new rate.

Pipes and cisterns are the same with signs: inlets have positive rates, outlets or leaks negative. If a pipe alone fills a tank in x hours but takes y hours with a leak, the leak alone empties it in xy/(y − x) hours. Wages for a joint job are shared in the ratio of work done, which (for equal days) is the ratio of efficiencies.

Key points & formulas

A does a job in a days ⇒ A’s rate = 1/a per day
A and B together: ab/(a + b) days
LCM method: total work = LCM of the days; rate = LCM ÷ days
Chain rule: M₁D₁H₁/W₁ = M₂D₂H₂/W₂
If A is k times as efficient as B, A takes 1/k of B’s time
Pipe fills in x h, with leak in y h ⇒ leak empties in xy/(y − x) h
Outlet or leak = negative rate; net rate = sum of inlets − sum of outlets
Wages are shared in the ratio of work done

Worked example

Pipes A and B can fill a tank in 12 hours and 15 hours respectively. An outlet pipe C can empty the full tank in 20 hours. If all three are opened together when the tank is empty, in how many hours will it be full?

Show solution

Take the tank = LCM(12, 15, 20) = 60 units.
A fills 60 ÷ 12 = 5 units per hour; B fills 60 ÷ 15 = 4 units per hour; C empties 60 ÷ 20 = 3 units per hour.
Net fill per hour = 5 + 4 − 3 = 6 units.
Time = 60 ÷ 6 = 10 hours. (By fractions: 1/12 + 1/15 − 1/20 = 6/60 = 1/10.)

Answer: 10 hours

Exam shortcut

Two workers: product over sum. A in 10 days, B in 15 days: together 10 × 15 ÷ (10 + 15) = 150 ÷ 25 = 6 days.

Common traps

Adding times instead of rates: 10 days and 15 days together is 6 days, not 25 or 12.5.
Forgetting the minus sign for an outlet or leak.
In “A leaves after some days”, forgetting that both worked during those days.
Ignoring hours per day when the number of hours changes between the two situations.

Practice · tap an option to check

Q1. A can do a piece of work in 20 days and B in 30 days. They start together, but A leaves after 4 days. In how many more days will B finish the remaining work?

Q2. 12 men working 8 hours a day complete a work in 18 days. In how many days will 16 men working 6 hours a day complete the same work?

Q3. A pipe can fill a tank in 6 hours. Because of a leak at the bottom, it takes 8 hours to fill the tank. In how many hours can the leak alone empty the full tank?

Q4. A is twice as efficient as B, and together they finish a work in 12 days. How many days will A alone take to finish it?

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CSAT · Basic numeracy · Topic 7 of 11

Time, Speed and Distance (trains, boats, relative speed)

Mark as done

Think of it like…

Two ST buses on the Pune–Nashik highway

Two red ST buses leave Pune and Nashik at the same time and drive towards each other. From the point of view of a passenger in one bus, the other bus seems to come at both speeds added together, because both are eating up the same gap. If one bus does 45 km/h and the other 60 km/h on a 210 km road, the gap shrinks by 105 km every hour, and they meet in 2 hours.

Now a faster bus chases a slower one going the same way. The gap shrinks only by the difference of their speeds. A train passing a platform is the same idea: the train has to cover its own length plus the platform’s. And a boat on the Krishna river near Sangli goes faster downstream (the current adds to it) and slower upstream (the current subtracts).

In everyday life	In the exam topic
Buses approaching each other	Relative speed = sum of speeds (opposite directions)
One bus chasing another	Relative speed = difference of speeds (same direction)
Train passing a platform	Distance = train length + platform length
Boat with the current	Downstream speed = boat + stream
Boat against the current	Upstream speed = boat − stream

The concept

Distance = speed × time. Convert units carefully: 1 km/h = 5/18 m/s and 1 m/s = 18/5 km/h, so 54 km/h = 15 m/s and 72 km/h = 20 m/s. If the distance is fixed, speed and time are inversely proportional: going at 4/5 of the usual speed means taking 5/4 of the usual time. For equal distances at speeds x and y, the average speed is 2xy/(x + y), always less than the simple average.

Relative speed: two bodies moving in opposite directions approach (or separate) at the sum of their speeds; in the same direction, at the difference. For trains, a train passing a pole, a person or a signal covers its own length; passing a platform, bridge or another train it covers the sum of the two lengths. Two trains crossing each other: time = (L₁ + L₂) ÷ relative speed.

Boats and streams: with boat speed b in still water and stream speed s, downstream = b + s and upstream = b − s. Conversely, b = (downstream + upstream)/2 and s = (downstream − upstream)/2. For a round trip of distance d each way, total time = d/(b + s) + d/(b − s).

Key points & formulas

Distance = speed × time; 1 km/h = 5/18 m/s; 1 m/s = 18/5 km/h
Average speed (equal distances at x and y) = 2xy ÷ (x + y)
Opposite directions: relative speed = sum; same direction: difference
Train passing a pole or person: distance = train length
Train passing a platform or another train: distance = sum of lengths
Downstream = b + s; upstream = b − s
b = (down + up) ÷ 2; s = (down − up) ÷ 2
Meeting time of two bodies starting together towards each other = gap ÷ (sum of speeds)

Worked example

Two trains 150 m and 100 m long run on parallel tracks at 54 km/h and 36 km/h. How long do they take to cross each other when moving (a) in opposite directions and (b) in the same direction?

Show solution

Distance to cover in both cases = 150 + 100 = 250 m.
(a) Opposite directions: relative speed = 54 + 36 = 90 km/h = 90 × 5/18 = 25 m/s. Time = 250 ÷ 25 = 10 s.
(b) Same direction: relative speed = 54 − 36 = 18 km/h = 18 × 5/18 = 5 m/s. Time = 250 ÷ 5 = 50 s.

Answer: (a) 10 seconds; (b) 50 seconds

Exam shortcut

Remember the 18s: 18 km/h = 5 m/s, 36 = 10, 54 = 15, 72 = 20, 90 = 25. Convert first, then divide.

Common traps

Forgetting to add the platform length (or the other train’s length) to the train’s own length.
Mixing km/h with metres and seconds; convert before dividing.
Taking the simple average of two speeds for a round trip.
Using the sum of speeds for vehicles moving in the same direction.

Practice · tap an option to check

Q1. A train 240 m long passes a signal pole in 16 seconds. How long will it take to pass a platform 360 m long?

Q2. The speed of a boat in still water is 15 km/h and the speed of the stream is 3 km/h. How long will the boat take to go 36 km downstream and come back?

Q3. An ST bus goes from Pune to Nashik at 40 km/h and returns along the same road at 60 km/h. What is its average speed for the whole journey?

Q4. Two ST buses start at 8:00 a.m. from two towns 210 km apart and travel towards each other at 45 km/h and 60 km/h. At what time will they meet?

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CSAT · Reasoning · Topic 8 of 11

Number and Letter Series, Coding–Decoding

Mark as done

Think of it like…

The dabbawala’s tiffin code

Mumbai’s dabbawalas deliver thousands of tiffins a day with a simple painted code on each lid: a few letters and numbers for the station of origin, the destination station, the building and the floor. A new dabbawala does not memorise every tiffin. He learns the rule once, and then any lid can be read, and any new address can be written in code.

A series question gives you a row of lids and hides the rule: what has to happen to each term to get the next one? Add a fixed amount, add a growing amount, multiply and add, or alternate two rules. A coding question gives you one word and its coded form and asks you to find the rule (shift each letter forward by one, reverse the word, replace letters by their positions) and then write a new word in the same code.

In everyday life	In the exam topic
Painted marks on a tiffin lid	Terms of the series or letters of the code
The rule the dabbawala learns once	Pattern: differences, ratios, letter shifts
Reading a new tiffin with the same rule	Finding the next term or decoding a word
Station number written as a code	Letter positions: A = 1 … Z = 26

The concept

Number series: first look at the differences between terms. Constant differences mean an arithmetic series; differences that themselves form a series (2, 3, 4, 5…) mean a second-level pattern. If the terms grow fast, try ratios or “× a + b” rules (5, 11, 23, 47 is × 2 + 1). Watch for squares and cubes and their neighbours (n² − 1, n² + n = n(n + 1), n³ + 1), primes, and two alternating series interleaved in one row.

Letter series: convert letters to positions (A = 1, … , Z = 26) and treat the problem as a number series. Memorise the anchors E = 5, J = 10, O = 15, T = 20, Y = 25 (remember “EJOTY”), and the reverse position = 27 − position (A ↔ Z, B ↔ Y). When a shift passes Z, wrap around to A.

Coding–decoding: compare the word with its code letter by letter. Common rules: each letter shifted by a fixed number (+1, −1, +2), alternating shifts, reversal of the word, replacing letters with their reverse-alphabet partners, or numerical coding where each letter becomes its position (and sometimes the code is their sum). Check the rule on every letter before applying it to the new word.

Key points & formulas

Check differences first, then second differences, then ratios
Common forms: n² ± 1, n(n + 1), n³ ± 1, × 2 + 1, primes
Alternating series: split odd- and even-position terms and solve each
Letter positions: E = 5, J = 10, O = 15, T = 20, Y = 25 (EJOTY)
Reverse position = 27 − position (A ↔ Z, M ↔ N)
Shift past Z wraps around to A
Coding: verify the rule on every letter before applying it

Worked example

Find the next term: 3, 8, 15, 24, 35, ?

Show solution

Differences: 8 − 3 = 5, 15 − 8 = 7, 24 − 15 = 9, 35 − 24 = 11. The differences are consecutive odd numbers.
The next difference is 13, so the next term is 35 + 13 = 48.
Check with a formula: the terms are 2² − 1, 3² − 1, 4² − 1, 5² − 1, 6² − 1, so the next is 7² − 1 = 48.

Answer: 48

Exam shortcut

Write the differences under the series in one line. Two lines of differences solve most CSAT number series in under 30 seconds.

Common traps

Stopping at the first pattern that fits three terms without checking all terms.
Miscounting letter positions in the middle of the alphabet; use EJOTY as anchors.
Forgetting to wrap around past Z in letter shifts.
Assuming the same shift for every letter when the code alternates (+1, −1, +1…).

Practice · tap an option to check

Q1. Find the next number in the series: 5, 11, 23, 47, 95, ?

Q2. Find the next letter in the series: A, C, F, J, O, ?

Q3. In a certain code, SOLAPUR is written as RNKZOTQ. How is KOLHAPUR written in that code?

Q4. In a certain code, each letter is replaced by its position in the English alphabet and the code of a word is the sum of these numbers. If WARI is coded as 51, what is the code for DINDI?

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CSAT · Reasoning · Topic 9 of 11

Blood Relations and Direction Sense

Mark as done

Think of it like…

Drawing the family tree at a village wedding

At a big joint-family wedding in a village near Kolhapur, a guest asks the bride’s aunt, “Who is that boy?” She does not answer in one jump. She says, “His father is my husband’s younger brother,” and in her head she draws the tree: generations stacked one above another, brothers and sisters side by side, husband and wife joined by a line. Once the tree is drawn, any relation can be read off it.

A farmer describing his walk is the same: “I went 4 fields north, turned right for 9, right again for 12…” You cannot hold all that in your head, but on paper it becomes a few lines on a map, and the distance from the start is just the straight line from the first point to the last, found with Pythagoras.

In everyday life	In the exam topic
Generations stacked one above another	Rows of the family tree (grandparents, parents, children)
Brothers and sisters side by side	Same generation; male (+) and female (−) marked
“My husband’s younger brother”	Read the chain from the end, one link at a time
Farmer’s walk drawn on paper	Direction diagram with N up and E right
Straight line from start to finish	Shortest distance = √(x² + y²)

The concept

Blood relations: draw a tree. Put one generation per row; join spouses with a double line and siblings with a single horizontal line; mark gender with + (male) and − (female) or by shapes. Break a long sentence (“the son of my grandfather’s only son”) from the inner end outwards: grandfather’s only son = my father (if the speaker’s father is on that side); his son = me or my brother. Do not assume gender from a name or a word like “child”, “sibling” or “parent”: if gender is not stated, the answer may be “cannot be determined”.

Direction sense: draw with north up, east right. A right turn when facing north means facing east; when facing south it means facing west. Turning clockwise means turning right. Angles: 45° turns take you to the corner directions (NE, SE, SW, NW). The final distance from the start is found by adding all north–south moves and all east–west moves separately, then applying Pythagoras.

Shadows: in the morning the sun is in the east, so shadows fall to the west; in the evening shadows fall to the east. At noon in India the shadow is very short. If a person’s shadow falls to his right in the morning, his right is west, so he faces south.

Key points & formulas

Draw a family tree: one generation per row, spouses joined, gender marked
Read long relation chains from the inner end outwards
If gender is not given, the answer may be “cannot be determined”
Right turn = clockwise 90°; left turn = anticlockwise 90°
Facing north: right is east; facing south: right is west
Net displacement: add N–S moves and E–W moves separately, then √(x² + y²)
Morning shadow falls west; evening shadow falls east
Clockwise order of directions: N, NE, E, SE, S, SW, W, NW (45° apart)

Worked example

From his house, Ramesh walks 4 km north, turns right and walks 9 km, turns right and walks 12 km, then turns right again and walks 3 km. How far, and in which direction, is he from his house?

Show solution

Place the house at (0, 0), with north as +y and east as +x.
4 km north → (0, 4). Turning right from north means facing east: 9 km → (9, 4).
Turning right from east means facing south: 12 km → (9, −8). Turning right from south means facing west: 3 km → (6, −8).
Net displacement: 6 km east and 8 km south. Distance = √(6² + 8²) = √100 = 10 km, towards the south-east.

Answer: 10 km, south-east of his house

Exam shortcut

For distance, ignore the path: net east–west and net north–south are all you need. The 3-4-5 and 6-8-10 triangles appear very often.

Common traps

Assuming the gender of a person called “child”, “sibling” or by a gender-neutral word.
Turning right from a south-facing position as if you were facing north.
Adding all the distances walked instead of finding the straight-line distance.
Forgetting that in the morning shadows fall to the west, not the east.

Practice · tap an option to check

Q1. Pointing to a girl in a photograph, Anil says, “She is the daughter of the only son of my paternal grandfather.” How is the girl related to Anil?

Q2. Seema is facing north. She turns 90° clockwise, then 135° anticlockwise, and then 180° clockwise. Which direction is she facing now?

Q3. A is the brother of B. B is the sister of C. C is the son of D. How is D related to A?

Q4. One morning after sunrise, Raju and Sonu were standing in a field facing each other. Raju’s shadow fell exactly to his right. Which direction was Sonu facing?

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CSAT · Reasoning · Topic 10 of 11

Syllogisms and Statement–Conclusion

Mark as done

Think of it like…

Venn circles of a village’s farmers and voters

The gram sevak draws circles on the back of the voter register. “All farmers are voters”: the farmers’ circle sits entirely inside the voters’ circle. “Some voters are teachers”: the teachers’ circle overlaps the voters’ circle somewhere. Now someone asks, “So some farmers are teachers?” The gram sevak tries drawing it another way: the teachers’ circle overlaps the voters’ circle, but in the part away from the farmers. Both drawings fit the statements, and in the second, no farmer is a teacher. So the conclusion does not follow.

That is the whole method. A conclusion follows only if it is true in every possible drawing that fits the statements. If you can draw even one picture that obeys the statements and breaks the conclusion, the conclusion fails, however reasonable it sounds in real life.

In everyday life	In the exam topic
Farmers’ circle inside the voters’ circle	“All A are B”
Teachers’ circle overlapping voters	“Some A are B”
Two circles kept apart	“No A is B”
Trying every way of drawing the circles	Checking all possible Venn diagrams
True in every drawing	Conclusion definitely follows

The concept

A syllogism gives two or more statements, taken to be true even if they sound odd, and asks which conclusions definitely follow. The four statement types are: All A are B (A inside B), Some A are B (A and B overlap), No A is B (A and B apart), and Some A are not B (part of A outside B). Draw the minimum diagram that satisfies the statements, then try to break each conclusion with an alternative diagram.

Useful rules: “All A are B” gives “Some A are B” and “Some B are A”. “Some A are B” gives “Some B are A”. “No A is B” gives “No B is A”. Two “some” statements never give a definite conclusion between the end terms. All + All gives All; All + No gives No (All A are B, No B is C ⇒ No A is C). Either–or case: when two conclusions have the same subject and predicate, neither follows on its own, and one is a “some” and the other a “no” (or “all” and “some … not”), the answer is “either I or II follows”, because one of them must be true in every diagram.

In statement–conclusion and statement–assumption questions with ordinary sentences (a notice, a government decision), a conclusion follows if it is a direct, reasonable implication of the statement without extra facts. Reject conclusions with “only”, “all”, “never” or with information not present. These are logic questions, not opinion questions.

Key points & formulas

All A are B ⇒ Some A are B and Some B are A (but not All B are A)
Some A are B ⇒ Some B are A; No A is B ⇒ No B is A
All A are B + All B are C ⇒ All A are C
All A are B + No B is C ⇒ No A is C
Some + Some ⇒ no definite conclusion between the end terms
Either–or: same terms, neither follows alone, one is “some” and the other “no” (or “all” and “some not”)
A conclusion must hold in every possible Venn diagram; one counter-diagram kills it
Statements are taken as true even if they contradict common knowledge

Worked example

Statements: All farmers are voters. Some voters are teachers.
Conclusions: I. Some farmers are teachers. II. Some teachers are voters.
Which conclusion(s) follow?

Show solution

Draw the farmers’ circle inside the voters’ circle (All farmers are voters).
Draw the teachers’ circle overlapping the voters’ circle (Some voters are teachers).
Conclusion I: draw the teachers’ circle overlapping only the part of the voters’ circle outside the farmers. The statements still hold, but no farmer is a teacher. I does not follow.
Conclusion II: “Some voters are teachers” reverses directly to “Some teachers are voters”. True in every diagram, so II follows.

Answer: Only conclusion II follows.

Exam shortcut

Before drawing, check for a “some + some” pair (no conclusion between end terms) and for the either–or pair (same two terms, “some” vs “no”). These two checks settle many questions without a diagram.

Common traps

Reversing “All A are B” into “All B are A”: the correct reverse is “Some B are A”.
Accepting a conclusion because it is true in real life even though the statements do not force it.
Missing the either–or pair and marking “neither follows”.
Using common knowledge to reject a statement: in syllogisms the statements are always taken as true.

Practice · tap an option to check

Q1. Statements: All sugarcane growers are cooperative members. No cooperative member is a defaulter.
Conclusions:
I. No sugarcane grower is a defaulter.
II. Some defaulters are sugarcane growers.

Q2. Statements: Some teachers are writers. Some writers are poets.
Conclusions:
I. Some teachers are poets.
II. No teacher is a poet.

Q3. Statement: The district collector has ordered all schools in the taluka to remain closed tomorrow because the India Meteorological Department has issued a red alert for extremely heavy rain.
Conclusions:
I. The administration expects that heavy rain could make it unsafe for children to travel to school.
II. Schools in the taluka are never closed for rain unless there is a red alert.

Q4. Statements: All gram sevaks are graduates. Some graduates are women. No woman is a contractor.
Conclusions:
I. Some gram sevaks are women.
II. Some graduates are not contractors.

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CSAT · Data interpretation · Topic 11 of 11

Data Interpretation: Tables, Bar and Pie Charts

Mark as done

Think of it like…

Reading the APMC market’s daily arrivals board

Every morning the APMC market in Lasalgaon writes up the arrivals and prices of onions on a board. A trader who glances at it does three things before reacting. He checks the units: quintals or tonnes, rupees per quintal or per kg. He compares like with like: today’s price against yesterday’s, not against last year’s total. And he thinks in percentages and ratios, because a ₹200 rise means a lot on a ₹1,000 price and little on a ₹4,000 one.

Data interpretation in CSAT is that board. A table, bar chart or pie chart is given, and the questions ask for totals, averages, percentage changes and ratios. The arithmetic is simple; the marks are lost by misreading the board.

In everyday life	In the exam topic
Quintals vs tonnes on the board	Units and scale (thousands, lakhs, %)
Today’s price vs yesterday’s	Percentage change = (new − old) ÷ old × 100
Share of each variety in total arrivals	Pie chart: share % and angle = % × 3.6°
Arrivals over the week	Bar or line chart: trend over time
Rough mental total before calculating	Approximation to eliminate options

The concept

A DI set gives data and asks four kinds of question: totals and averages, percentage change (growth over a previous year), percentage share (one part of a total), and ratios. Percentage change is always on the old (base) value: (new − old) ÷ old × 100. The largest absolute increase need not be the largest percentage increase; a small base makes a modest increase look large in percentage terms.

In a pie chart the whole circle is 100% or 360°, so 1% = 3.6° and the central angle of a share = percentage × 3.6. Amounts are percentage × total. In bar charts, read the scale carefully (does the axis start at zero?) and the units. In tables, read the row and column headings twice before calculating.

Speed comes from approximation: if options are far apart, round the numbers and compare. Use fraction equivalents (1/8 = 12.5%, 1/6 ≈ 16.7%, 1/9 ≈ 11.1%) to estimate percentage changes quickly. Calculate exactly only when two options are close.

Key points & formulas

Percentage change = (new − old) ÷ old × 100, always on the base (old) value
Percentage share = part ÷ total × 100
Pie chart: 100% = 360°, so 1% = 3.6°; angle = share % × 3.6
Amount in a pie sector = share % × total
Highest absolute increase is not necessarily the highest percentage increase
Average = total ÷ number of items; check units (thousand tonnes, ₹ lakh)
Approximate first; calculate exactly only when options are close

Worked example

Production of three crops in a Maharashtra district (thousand tonnes):
Year | Soybean | Cotton | Tur
2019 | 120 | 80 | 45
2020 | 135 | 72 | 50
2021 | 150 | 90 | 48
2022 | 140 | 96 | 60
2023 | 165 | 84 | 66
(a) In which year was the total production of the three crops the highest? (b) By what percentage did tur production increase from 2019 to 2023?

Show solution

Totals by year: 2019: 120 + 80 + 45 = 245; 2020: 257; 2021: 288; 2022: 296; 2023: 165 + 84 + 66 = 315.
(a) The highest total is 315 thousand tonnes, in 2023.
(b) Tur rose from 45 to 66, an increase of 21. Percentage increase = 21 ÷ 45 × 100 = 46.67% (about 46.7%).
Note the base: dividing by 66 (the new value) would give about 31.8%, which is wrong.

Answer: (a) 2023; (b) about 46.7%

Exam shortcut

For “highest percentage growth” questions, compare increase ÷ base as fractions by eye: 15/120 = 1/8, 15/135 = 1/9, 25/140 ≈ 1/5.6. The smallest denominator wins.

Common traps

Calculating percentage change on the new value instead of the old value.
Choosing the year with the largest absolute increase when the question asks for percentage increase.
Ignoring units such as thousand tonnes or ₹ lakh when an option is in different units.
In pie charts, using percentages as degrees or forgetting to multiply by 3.6.

Practice · tap an option to check

Q1. Soybean production in a district (thousand tonnes):
Year | 2019 | 2020 | 2021 | 2022 | 2023
Production | 120 | 135 | 150 | 140 | 165
In which year was the percentage increase in production over the previous year the highest?

Q2. Tap-water connections in four talukas:
Taluka | Households | Households with tap connection
A | 12,000 | 9,000
B | 8,000 | 6,400
C | 15,000 | 10,500
D | 10,000 | 8,500
Which taluka has the highest percentage of households with a tap connection?

Q3. A pie chart shows how a gram panchayat spends its annual budget of ₹48 lakh:
Roads 30% | Water supply 25% | Education 20% | Health 15% | Others 10%
How much more is spent on roads than on health?

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