Simplification for CDS

Simplification questions look like arithmetic, but they are really tests of order and recognition. The candidate who spots that 7.5² − 2.5² is a difference of squares finishes in ten seconds. The one who squares both numbers takes a minute and risks a slip.

In CDS, these questions are usually quick marks, and they often hide an algebraic identity in decimal clothing. Treat them as the place where you bank time for harder questions later in the paper. This guide covers the order of operations, decimal handling, continued fractions, the identities UPSC builds questions around, and when to approximate.

The order of operations

Work in this order, which many textbooks call VBODMAS:

1. Vinculum (a bar over an expression, which acts like a bracket)
2. Brackets, innermost first: ( ), then { }, then [ ]
3. Of and orders (powers and roots)
4. Division and Multiplication, left to right
5. Addition and Subtraction, left to right

Two points cause most errors.

Division and multiplication have equal rank. Work left to right. 18 ÷ 3 × 2 is 6 × 2 = 12, not 18 ÷ 6 = 3. So 18 ÷ 3 × 2 + 4 − 6 = 12 + 4 − 6 = 10.

"Of" comes before division. In 20 ÷ 4 of 5, do "4 of 5" first: 20 ÷ 20 = 1. Compare 20 ÷ 4 × 5, which is 25.

Decimals without errors

• Multiplying: ignore the points, multiply, then count the decimal places. 0.25 × 0.4 → 25 × 4 = 100, and 2 + 1 = 3 decimal places gives 0.100 = 0.1.
• Dividing: move the point in both numbers by the same number of places until the divisor is whole. 0.1 ÷ 0.05 → 10 ÷ 5 = 2. This works because multiplying top and bottom by 100 does not change a fraction.
• Recurring decimals: convert them to fractions first. 0.333… is 1/3, and 0.4545… is 5/11. The fractions and roots guide explains the conversion.

Continued fractions

Work from the bottom up, one level at a time. Write each step; the error rate on these comes almost entirely from skipping steps.

For 1 + 1/(1 + 1/2): the bottom is 1 + 1/2 = 3/2. Its reciprocal is 2/3. So the value is 1 + 2/3 = 5/3.

Identities that save time

The last identity has a special case worth memorising: if a + b + c = 0, then a³ + b³ + c³ = 3abc. UPSC-style questions often hide it: 13³ − 8³ − 5³ has a = 13, b = −8 and c = −5, which add to 0.

Approximation

When the options are far apart, round first. For 49.8 × 20.1 ÷ 9.97, think 50 × 20 ÷ 10 = 100. If the options are 80, 100, 120 and 140, the answer is clear without exact work. (The exact value is about 100.4.)

Use approximation only when the options are well spread. If two options are within a few per cent of each other, calculate properly.

Worked questions

Question 1: Simplify 1 + 1/(1 + 1/(1 + 1/2)).

• Innermost: 1 + 1/2 = 3/2, and its reciprocal is 2/3.
• Next level: 1 + 2/3 = 5/3, and its reciprocal is 3/5.
• Top level: 1 + 3/5 = 8/5.

Question 2: Simplify (0.75 × 0.75 × 0.75 + 0.25 × 0.25 × 0.25) ÷ (0.75 × 0.75 − 0.75 × 0.25 + 0.25 × 0.25).

• The top is a³ + b³ and the bottom is a² − ab + b², with a = 0.75 and b = 0.25.
• Since a³ + b³ = (a + b)(a² − ab + b²), the quotient is a + b = 1.

Question 3: Find the value of 13³ − 8³ − 5³.

• Take a = 13, b = −8, c = −5. Then a + b + c = 0.
• So a³ + b³ + c³ = 3abc = 3 × 13 × (−8) × (−5) = 1,560.
• Check: 2,197 − 512 − 125 = 1,560.

Question 4: Find (5.6 + 2.4)² − (5.6 − 2.4)².

• This is (a + b)² − (a − b)² = 4ab.
• 4 × 5.6 × 2.4 = 4 × 13.44 = 53.76.
• Check: 8² − 3.2² = 64 − 10.24 = 53.76.

Question 5: Simplify 12 + 20 ÷ 4 of 5 − 3.

• "Of" first: 4 of 5 = 20.
• Division: 20 ÷ 20 = 1.
• Then 12 + 1 − 3 = 10.

Question 6: Simplify (3/4 of 64) ÷ (2/3 of 18).

• 3/4 of 64 = 48, and 2/3 of 18 = 12.
• 48 ÷ 12 = 4.

Common slips

• Doing a multiplication before a division that comes first on the left.
• Treating "of" as ordinary multiplication when it sits next to a division sign.
• Losing a decimal place. Count places before and after every step.
• Opening brackets with a minus sign in front and forgetting to change every sign inside.
• Calculating cubes in full when an identity was waiting in the denominator.

Practice set

1. 36 ÷ 4 × 3 − 7
2. 1/2 of 3/5 of 200
3. 0.6 × 0.5 ÷ 0.03
4. 8.5² − 1.5²
5. (1.2³ − 0.8³) ÷ (1.2² + 1.2 × 0.8 + 0.8²)
6. 2 + 1/(2 + 1/(2 + 1/2))
7. 25³ − 15³ − 10³
8. 48 ÷ [16 − {4 + (18 − 6 × 2)}]

Answers

1. 20. Left to right: 36 ÷ 4 = 9, 9 × 3 = 27, 27 − 7 = 20.
2. 60. 3/5 of 200 = 120, and half of 120 is 60.
3. 10. 0.6 × 0.5 = 0.3, and 0.3 ÷ 0.03 = 30 ÷ 3 = 10.
4. 70. (8.5 − 1.5)(8.5 + 1.5) = 7 × 10.
5. 0.4. This is (a³ − b³) ÷ (a² + ab + b²) = a − b = 1.2 − 0.8.
6. 29/12. 2 + 1/2 = 5/2, reciprocal 2/5; 2 + 2/5 = 12/5, reciprocal 5/12; 2 + 5/12 = 29/12.
7. 11,250. 25 − 15 − 10 = 0, so the value is 3 × 25 × (−15) × (−10) = 11,250. Check: 15,625 − 3,375 − 1,000 = 11,250.
8. 8. 6 × 2 = 12; 18 − 12 = 6; 4 + 6 = 10; 16 − 10 = 6; 48 ÷ 6 = 8.

What to do next

• Copy the identity table onto your formula sheet, with one "disguise" example for each
• Solve the simplification questions from the last five CDS papers, aiming for under 45 seconds each
• For each question, write down whether it needed an identity, BODMAS or approximation
• Deepen the algebra behind these shortcuts with the algebraic identities guide