Simplification and Approximation

This is the plain BODMAS line of MTS, GD and Delhi Police papers. The expression is one line of whole numbers with +, −, ×, ÷, perhaps one bracket and perhaps an "of". You need nothing except the order, and the patience to follow it.

Step 1: brackets first. Solve whatever is inside ( ).

Step 2: "of" next. 15 of 2 = 30. Treat it like a bracket: (15 × 2).

Step 3: division and multiplication, left to right. Do each ÷ and × as you meet it from the left.

Step 4: addition and subtraction, left to right. Now only + and − are left.

Step 4 goes much faster if you collect all the positive terms together and all the negative terms together, then subtract once. For 85 − 37 + 46 − 19 + 12: positives = 85 + 46 + 12 = 143, negatives = 37 + 19 = 56, answer = 143 − 56 = 87. This also saves you from the old mistake of doing 85 − (37 + 46).

[[fig:0]]

Take 90 − 24 ÷ 6 × 3 + 7 of 2. "Of" first: 7 of 2 = 14. Then from the left: 24 ÷ 6 = 4, and 4 × 3 = 12. Now 90 − 12 + 14 = 92.

[[fig:1]]

Word form. Delhi Police papers often write the same thing in words: "He divides 56 by 7 and adds the result to the product of 3 and 5." Translate it line by line: 56 ÷ 7 + 3 × 5 = 8 + 15 = 23. "Difference between 15 and 8" means (15 − 8). "Product" means ×, and "quotient" means ÷.

In CGL, CHSL and CPO the same BODMAS comes with three or four layers of brackets. The rule does not change: open the innermost bracket first and move outwards. The usual order of brackets is:

• Vinculum (a bar over numbers): 9 − 4 means (9 − 4). It is solved before everything else.
• Small brackets ( )
• Curly brackets { }
• Square brackets [ ]

Inside every bracket the normal BODMAS order applies again: "of", then ÷ and × from left to right, then + and −.

The minus sign before a bracket. A minus sign outside a bracket changes the sign of every term inside when you open it: 20 − (8 − 3 + 2) = 20 − 8 + 3 − 2 = 13. The safest habit is to finish the bracket first (8 − 3 + 2 = 7) and then subtract (20 − 7 = 13). With letters you have no choice, and each sign must be flipped: x − (2y − x) = x − 2y + x = 2x − 2y.

[[fig:0]]

Radhika simplifies 50 − [18 + {12 − (9 − 5 − 1)}]. Vinculum: 5 − 1 = 4. Small bracket: 9 − 4 = 5. Curly: 12 − 5 = 7. Square: 18 + 7 = 25. Answer: 50 − 25 = 25.

[[fig:1]]

Finding the missing number (k, p, ? and so on). First simplify every known part on both sides, so that only one operation is left around the unknown, and then solve. For "40 ÷ 8 + 3 × k − 6 = 14": 5 + 3k − 6 = 14 ⇒ 3k = 15 ⇒ k = 5. Never carry a number across the "=" sign while the × and ÷ around it are still pending.

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Interchanged signs. Questions like "if + means ×, × means −, …" are only a job of rewriting. First rewrite the whole expression with the true signs, and only then apply BODMAS. Mistakes come when both are done in one go.

Many students slow down on fractions. The fractions are not hard; the trouble is that they turn everything into improper fractions and then fight with big numbers. There is a shorter way.

Adding and subtracting mixed numbers. Handle the whole parts and the fraction parts separately. For 634 + 212 − 314:

• Whole parts: 6 + 2 − 3 = 5
• Fraction parts: 34 + 12 − 14 = 34 + 24 − 14 = 44 = 1
• Answer: 5 + 1 = 6

[[fig:0]]

If the fraction part comes out negative, take 1 from the whole part. For 516 − 223: whole = 3, fraction = 16 − 46 = −36 = −12, so the answer is 3 − 12 = 212.

"Of" with fractions. 25 of 45 = 25 × 45 = 18. Divide by the denominator first (45 ÷ 5 = 9), then multiply by the numerator (9 × 2 = 18). The numbers stay small.

Fraction of the remaining part. This is the word problem examiners like most. Suppose Radhika reads 14 of a book on day one and 13 of the remaining part on day two:

• Left after day one = 1 − 14 = 34
• Day two she reads 13 of 34 = 14 of the book
• Left after day two = 34 × 23 = 12 of the book

[[fig:1]]

So the fractions of the remaining part are multiplied one after another. After taking 13 of something, 23 of it remains, and that 23 multiplies whatever was left before.

Mixed ↔ improper is still needed for × and ÷. 313 × 115 = 103 × 65 = 4. Do not multiply the whole parts and the fraction parts separately; the split works only for + and −.

This type looks frightening because of its height, but it is only the fraction rules of the last type, used in layers.

Complex fraction: a fraction whose numerator or denominator itself contains fractions. Simplify the top and the bottom separately, then divide. 12 + 1356 = 5656 = 1.

Continued fraction (ladder fraction): something like 2 + 13 + 14. Always start from the bottom and climb up.

• Bottom: 3 + 14 = 134
• Its reciprocal: 1134 = 413
• Top: 2 + 413 = 3013

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The one move you need again and again is 1ab = ba. Each time you meet "1 over a fraction", turn that fraction upside down.

Fraction series (telescoping). When each denominator is a product of two numbers with a fixed gap between them, every term splits into a difference:
1n(n + 1) = 1n − 1n + 1, and in general 1n(n + k) = 1k(1n − 1n + k).

When you add many such terms, the middle parts cancel, the way a telescope folds shut, and only the first and the last parts are left.

Take 13 × 4 + 14 × 5 + 15 × 6. It is (13 − 14) + (14 − 15) + (15 − 16) = 13 − 16 = 16.

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Learn to spot these denominators: 2, 6, 12, 20, 30, 42, 56, 72, 90, 110 are 1×2, 2×3, 3×4, … 10×11 (gap 1), and 3, 15, 35, 63, 99, 143 are 1×3, 3×5, 5×7, … 11×13 (gap 2).

Decimal questions test one thing: whether you can keep the decimal point in the right place. Here are the rules again, this time with the reasons, so that you never have to guess.

Adding a pattern. In sums like 0.3 + 0.03 + 0.003, write the numbers one below the other. Each digit falls in a different column, so the answer is 0.333 straight away. Look for such patterns before adding digit by digit.

Multiplying. The number of decimal places in the product = the total decimal places in all the numbers. The reason: 0.3 = 310 and 0.02 = 2100, so the product is 61000 = 0.006.

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Dividing. Shift the decimal point of both numbers by the same number of places until the divisor is a whole number. 0.36 ÷ 0.009 = 360 ÷ 9 = 40. All you have done is multiply the top and bottom of 0.360.009 by 1000.

Powers of decimals. (0.3)² has 2 decimal places: 0.09. (0.02)³ has 6 decimal places: 0.000008. Decimal places of a power = decimal places of the number × the power.

The common multiplier. Look at 0.06³ + 0.09³0.2³ + 0.3³. Each number on top is 0.3 times the matching number below (0.06 = 0.3 × 0.2 and 0.09 = 0.3 × 0.3). So the top is (0.3)³ times the bottom, and the answer is (0.3)³ = 0.027, with no big calculation at all. Whenever the top terms are a fixed multiple of the bottom terms, the answer is that multiple raised to the power.

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Decimal ↔ fraction. 0.45 = 45100 = 920. For the reverse, try to make the denominator 10, 100 or 1000: 725 = 28100 = 0.28. If that is not possible, divide: 58 = 0.625 (from the table).

Radhika has 2.4 m of ribbon and cuts pieces of 0.15 m each. Number of pieces = 2.4 ÷ 0.15 = 240 ÷ 15 = 16.

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Now the last few types come together. The expression has brackets and "of" as in Types 1–2, but the numbers are decimals and fractions as in Types 3–5. Nothing new is needed, only one good habit:

Decide the form first. If most numbers are friendly fractions (114, 38, 0.25, 0.75), work in fractions. If the decimals are not friendly (3.914 or 1.03), work in decimals. Errors come from mixing both in one step.

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Friendly decimals become fractions. 0.25 = 14, 0.4 = 25, 1.25 = 54, 0.75 = 34, 1.6 = 85. So 212 ÷ 1.25 = 52 × 45 = 2.

"Of" with decimals. 0.6 of 2.5 = 0.6 × 2.5 = 1.5. It is still done before ÷ and ×.

Word problems on decimals. These are one division or one multiplication dressed up as a story. "How many pieces of 0.4 m can be cut from 6 m?" is 6 ÷ 0.4 = 15. "How many full pieces, and how much is left?" means: take the whole-number part of the division, then subtract the length used. For 5 m cut into 0.75 m pieces: 5 ÷ 0.75 = 6.66…, so 6 full pieces use 4.5 m and 0.5 m is left.

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Radhika buys 112 kg of tomatoes at ₹40 per kg and 0.75 kg of beans at ₹80 per kg. Tomatoes cost 32 × 40 = ₹60 and beans cost 34 × 80 = ₹60, so the total is ₹120.

Units. Bring both quantities to one unit before you divide: 1 kg = 1000 g, 1 L = 1000 mL, 1 m = 100 cm, 1 km = 1000 m.

Percentage of a number inside BODMAS. In these questions the word of means multiply, and the percentage must first be turned into a fraction. Do that at the very beginning and the rest is plain fraction work: 25% of 480 + 12.5% of 64 = 14 × 480 + 18 × 64 = 120 + 8 = 128.

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Keep the common ones ready: 12.5% = 18, 37.5% = 38, 62.5% = 58, 1623% = 16, 3313% = 13, 8313% = 56. A small percentage like 0.5% is 1200.

Two habits save time. First, treat each "% of number" term as one block and finish the blocks before the +, −, ×, ÷ between them. Second, remember that x% of y = y% of x, so 8% of 250 and 250% of 8 are the same number, 20. Choose the form that is easier to see. When the question asks "what percent of A is B", the answer is BA × 100.

Some fractions never end as decimals: 13 = 0.333…, 411 = 0.3636…. The repeating block is shown with a bar: 0.3, 0.36. Every such recurring decimal is a fraction, and the question asks you to find it.

Why the rule works. Let x = 0.4 = 0.444…. Then 10x = 4.444…. Subtract: 10x − x = 4, so 9x = 4 and x = 49. With two repeating digits we multiply by 100 instead: x = 0.27 ⇒ 100x − x = 27 ⇒ x = 2799 = 311.

[[fig:0]]

Rule 1: pure recurring (the bar starts right after the point).
Numerator = the repeating block. Denominator = as many 9s as there are digits under the bar.
0.5 = 59, 0.45 = 4599 = 511, 0.108 = 108999 = 437.

Rule 2: mixed recurring (some digits come before the bar).
Numerator = (all the digits after the point) − (the digits not under the bar). Denominator = as many 9s as digits under the bar, followed by as many 0s as digits not under the bar.
0.16 = 16 − 190 = 1590 = 16. 0.318 = 318 − 3990 = 315990 = 722.

[[fig:1]]

With a whole-number part, keep the whole number aside: 3.6 = 3 + 69 = 323.

One fact surprises many students: 0.9 = 99 = 1 exactly.

Comparing recurring decimals. Write out a few digits of each: 0.4 = 0.4444…, while 0.43 = 0.4333…. Then compare place by place, as with ordinary decimals.

Radhika's calculator shows 0.272727… for 311. By Rule 1, 0.27 = 2799 = 311, so the calculator agrees.

"Which is the largest?", "Which lies between …?", "Arrange in ascending order": these come out quickly once you know which tool suits which situation.

Tool 1: convert to decimals. Divide each fraction out to two or three digits. 58 = 0.625, 711 = 0.636…, 914 = 0.642…. This always works, and it is the best tool when there are four or more fractions.

[[fig:0]]

Tool 2: cross-multiplication for two fractions. To compare ab and cd (b and d positive), compare a × d with b × c. For 57 and 710: 5 × 10 = 50 and 7 × 7 = 49. Since 50 > 49, 57 > 710.

[[fig:1]]

Tool 3: the same gap between numerator and denominator. In 45, 910, 1415, each fraction is 1 minus 1denominator. When the gap (denominator − numerator) is the same, the fraction with the bigger numbers is closer to 1, so it is bigger. So 1415 is the largest.

[[fig:2]]

Tool 4: benchmarks. Compare with 12 or 1 first. 511 is less than 12 (5 is less than half of 11), while 611 is more. Many options fall at this first check.

Tool 5: same numerators or same denominators. With the same denominator, the bigger numerator wins. With the same numerator, the smaller denominator wins: 37 > 38.

For mixed numbers, compare the whole parts first; look at the fraction parts only if the whole parts are equal.

Radhika wants the largest of 34, 0.7, 45 and 0.78. As decimals they are 0.75, 0.70, 0.80 and 0.78, so the largest is 45.

This type is about squaring exact numbers and taking square roots of perfect squares quickly. (Surds like √50 or √3 + √2, which are not perfect squares, belong to the next chapter.)

Squaring with (a ± b)². 47² = (50 − 3)² = 2500 − 300 + 9 = 2209. 104² = (100 + 4)² = 10000 + 800 + 16 = 10816.

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Squares ending in 5. For n5², multiply n by (n + 1) and write 25 at the end. 65² → 6 × 7 = 42 → 4225. 115² → 11 × 12 = 132 → 13225.

Near 50 and 100 (base method). For a number near 50: n² = (25 + (n − 50)) hundreds + (n − 50)². 53² = (25 + 3) | 3² = 28|09 = 2809. Near 100: 96² = (96 − 4) | 4² = 92|16 = 9216.

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Patterns. 99² = 9801, 999² = 998001; each extra 9 adds one more 9 in front and one more 0 in the middle, so 99999² = 9999800001. Similarly 101² = 10201 and 1001² = 1002001.

Square root of a perfect square (up to 5 digits), by unit digit and range.

• Step 1: the unit digit of the root comes from the unit digit of the number (1 → 1 or 9, 4 → 2 or 8, 6 → 4 or 6, 9 → 3 or 7, 5 → 5, 0 → 0).
• Step 2: drop the last two digits and see between which two squares the rest lies. That gives the tens digit.
• Step 3: if the unit digit has two choices, compare the number with the square of the middle (…5) number.

Radhika finds √5329. Unit digit 9 → the root ends in 3 or 7. Drop "29": 53 lies between 7² = 49 and 8² = 64, so the root is 7_. Since 75² = 5625 > 5329, the root is below 75, so it is 73. Check: 73² = 5329.

[[fig:2]]

Square roots of decimals. Count the decimal places; the count must be even. Take the root of the digits and give the answer half the decimal places. √0.0049 = 0.07 (4 places → 2 places), √1.44 = 1.2.

Perfect square check. A perfect square never ends in 2, 3, 7 or 8, and its digit sum is always 1, 4, 7 or 9.

Cube roots. For a perfect cube, split off the last three digits. The unit digit of the root comes from the unit digit of the cube (2 ↔ 8 and 3 ↔ 7 swap, all other digits stay the same), and the tens digit comes from the part that is left, by looking between which two cubes it lies.

[[fig:3]]

In ∛10648, the last three digits 648 end in 8, so the root ends in 2. The left part 10 lies between 2³ = 8 and 3³ = 27, so the tens digit is 2. The root is 22.

For decimals, count the places: the number of decimal places must be a multiple of 3, and the answer gets one third of them. ∛0.027 = 0.3 and ∛0.000216 = 0.06. A cube root of a fraction is the cube root of the top over the cube root of the bottom.

From here the questions are less about calculation and more about recognition. The examiner takes an identity from algebra, puts numbers (often awkward decimals) in place of a and b, and waits to see whether you spot it. Spot it, and the answer comes in one line. Miss it, and you face a page of multiplication.

The square identities you should recognise on sight:

• a² − b² = (a + b)(a − b)
• (a + b)² = a² + 2ab + b²
• (a − b)² = a² − 2ab + b²
• (a + b)² − (a − b)² = 4ab and (a + b)² + (a − b)² = 2(a² + b²)
• (a + b)² − 4ab = (a − b)² and (a − b)² + 4ab = (a + b)²
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• x² + (a + b)x + ab = (x + a)(x + b)

How to spot them.

• Two squares with a minus between them → a² − b². 63² − 37² = (100)(26) = 2600.
• Two squares plus a "middle term" → check whether the middle term equals 2 × a × b. In 3.6² + 1.4² + 10.08, 2 × 3.6 × 1.4 = 10.08, so it is (3.6 + 1.4)² = 25.
• Three squares plus three products → (a + b + c)². Check that each product is 2 × (that pair).
• a² ± 2ab + b² divided by (a ± b) → the answer is (a ± b).

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Whenever you see a middle term, multiply the two base numbers, double the result and compare. This one check opens most of these questions.

[[fig:1]]

Radhika sees 7.3² + 2.7² + 2 × 7.3 × 2.7. It is (7.3 + 2.7)² = 10² = 100, with no multiplication of decimals at all.

Alternating squares. 20² − 19² + 18² − 17² + … + 2² − 1²: each pair n² − (n − 1)² = n + (n − 1). So the whole sum is 20 + 19 + 18 + … + 1 = 210.

[[fig:2]]

The numbers in these questions look terrible (14.37³, or 3.321 × 2.681 × 1.245), but once the identity is seen the answer is always a small addition or subtraction.

The cube identities:

• a³ + b³ = (a + b)(a² − ab + b²)
• a³ − b³ = (a − b)(a² + ab + b²)
• (a + b)³ = a³ + b³ + 3ab(a + b) = a³ + 3a²b + 3ab² + b³
• (a − b)³ = a³ − b³ − 3ab(a − b) = a³ − 3a²b + 3ab² − b³
• a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
• If a + b + c = 0, then a³ + b³ + c³ = 3abc.
• (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)
• x⁴ + x² + 1 = (x² + x + 1)(x² − x + 1)

Three shapes the questions take:
1. a³ + b³a² − ab + b² = a + b, and a³ − b³a² + ab + b² = a − b. Look at the sign of ab in the denominator: minus goes with a³ + b³, plus goes with a³ − b³.
2. a³ + b³ + c³ − 3abca² + b² + c² − ab − bc − ca = a + b + c. Add the three numbers and you are done.
3. a + b + c = 0 hidden inside. For example, in 13³ + 7³ − 20³, 13 + 7 − 20 = 0, so the value is 3 × 13 × 7 × (−20) = −5460.

[[fig:0]]

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Hidden middle numbers. The examiner often writes ab as a decimal so that you do not notice it. In 0.7³ − 0.0080.49 + 0.14 + 0.04: 0.008 = 0.2³, 0.49 = 0.7², 0.14 = 0.7 × 0.2 and 0.04 = 0.2². So it is a³ − b³a² + ab + b² = 0.7 − 0.2 = 0.5.

[[fig:2]]

Radhika works out (0.3³ + 0.4³ − 0.7³) ÷ (3 × 0.3 × 0.4 × 0.7). Here 0.3 + 0.4 + (−0.7) = 0, so the top is 3 × 0.3 × 0.4 × (−0.7). The bottom has +0.7, so the answer is −1, not 1. Watch the sign of c.

The last type covers questions where an honest calculation is long: big cubes, products of three- and four-digit numbers, or awkward decimals. The aim is not to avoid calculation altogether. It is to do only as much as the options demand.

1. Approximation (≈). When the question says approximately, or when the options are far apart, round every number to a friendly value and work out the rough answer. Round sensibly:

• Round to numbers that make the work easy: 49.97 → 50, 0.248 → 0.25, 1199 → 1200.
• In a product, if you round one number up, try to round the other down, so that the errors cancel.
• Square roots: use the nearest perfect square. √50 ≈ 7.07 because 7² = 49; √120 ≈ 11 because 11² = 121.
• Percentages: 19.9% of 601 ≈ 20% of 600 = 120.

Radhika estimates 49.8 × 20.3 ÷ 9.9. Rounded, it is 50 × 20 ÷ 10 = 100. The exact value is about 102.1, so if the options are 82, 102, 125 and 150, she can mark 102 without doubt.

[[fig:0]]

2. Cancel before you multiply. In expressions like A × BC, look for common factors between the numbers before multiplying anything. Big numbers in SSC questions are seldom random; they are usually built from small factors that cancel.

3. Check with digit sum and unit digit. You met these checks in the foundation section. Use them when all the options are close in size but differ in their digits. For a negative answer −N, the digit sum of the expression must equal 9 − DS(N) (taking 9 as 0).

4. Use a given value. "If 12.5 = 0.4, find 10.0025" only asks you to shift the decimal point: 10.0025 = 12.5 × 1000 = 400. Count how many places the decimal point moved in the divisor, and move the answer the same number of places the other way.

[[fig:1]]

5. Repeating-digit numbers. 7 + 77 + 777 = 7 × (1 + 11 + 111) = 7 × 123 = 861. Taking the digit out as a common factor turns a long addition into one multiplication.

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Square root of a number that is not a perfect square. Approximation questions often have a root such as √623 or √80 in them. Find the nearest perfect square, call its root n, and use

√N ≈ n + N − n²2n

[[fig:3]]

For √87, the nearest square is 81 = 9², so √87 ≈ 9 + 618 = 9.33 (the true value is 9.327). The formula is very accurate when N is close to n², which is how these questions are set. If N is just below a square, the correction is negative: √80 ≈ 9 − 118 = 8.94.

When the options are far apart you do not even need the correction. √623 is a little less than 25 (25² = 625), and √401 is a little more than 20, so rounding to 25 and 20 is enough.

A number series is a row of numbers made by one hidden rule. Your job is to find that rule and use it once more. Do not guess. Check the possible rules in the same order every time, and the answer comes out quickly.

1. Differences first. Subtract each term from the next one.

• Same difference every time: 12, 19, 26, 33 has +7, so the next term is 40.
• Differences that grow by a fixed amount: 6, 11, 21, 36, 56 has +5, +10, +15, +20, so the next difference is +25 and the term is 81.
• Differences that double: 2, 3, 5, 9, 17 has +1, +2, +4, +8, so the next one is +16.

[[fig:0]]

2. Multiply and adjust (× k ± c). If the terms roughly double or treble, the rule is of the form × k ± c. In 5, 11, 23, 47, each term is 2 × the previous one + 1, so the next is 95. To find k and c, compare two neighbouring terms and test.

3. Squares and cubes with a small adjustment. Terms like 2, 5, 10, 17, 26 are 1² + 1, 2² + 1, 3² + 1, … and terms like 3, 10, 29, 66 are n³ + 2. Know the squares up to 25 and the cubes up to 12 by heart, because this rule is only recognised if you see the squares behind the numbers. Other common shapes are n(n + 1) (2, 6, 12, 20, 30) and n³ − n (0, 6, 24, 60, 120).

4. Products with 1, 2, 3, 4 … The series 1, 2, 6, 24, 120 multiplies by 1, 2, 3, 4, 5 in turn.

5. Two series woven together. If nothing above works, split the terms into odd places and even places. In 4, 3, 8, 6, 16, 12, the odd places give 4, 8, 16 and the even places give 3, 6, 12, and each doubles.

[[fig:1]]

Wrong term. Find the rule from the terms that agree with each other, and the one that breaks it is the wrong term. Radhika checks 3, 10, 29, 66, 129, 218: the rule n³ + 2 gives 127 in the fifth place, so 129 is the wrong term.

In this type every question gives two quantities, Quantity I and Quantity II, and asks how they are related. The two quantities may be plain values (a percentage of a number, a root, a fraction of a fraction) or unknowns defined by an equation, and you must pick one of the relations: I > II, I < II, I = II, or no relation.

Values. Work out both and compare. Convert percentages to fractions first, as you do in BODMAS, so that the two numbers can be compared in your head.

Equations with two roots. A quadratic gives two values of the unknown, so a quantity may be a pair of numbers. Solve each equation by factorising, then compare every root of I with every root of II.

• If every pair gives the same sign (all >, all <, all =), that relation holds.
• If some pairs give > or = and none gives <, the relation is I ≥ II; in the same way, no > and some <, with or without =, gives I ≤ II.
• If the pairs give both > and <, no relation can be established.

[[fig:0]]

[[fig:1]]

For x² − x − 12 = 0 and y² + 7y + 12 = 0, Radhika gets x = 4, −3 and y = −3, −4. The pairs give 4 > −3, 4 > −4, −3 = −3 and −3 > −4. Nothing is smaller and one pair is equal, so I ≥ II. Pay attention to negative roots: −3 is greater than −4, because it lies to the right on the number line.