Mathematical Aptitude formulas for UGC NET Paper 1

Most Important Mathematical Aptitude formulas for UGC NET Paper 1

1. Percentages – Important Formulas

Formula 1 – Percentage of a Quantity

Formula:$$\text{Value} = \frac{P}{100} \times X$$

Variables:

• P: percentage value
• X: original quantity
• Value: required part of the quantity

Example MCQ:
Q1. What is 20% of 450?

A) 80
B) 85
C) 90
D) 95

Answer: C) 90

Explanation:
Use Value = (P/100) × X.
Here P = 20, X = 450$$\text{Value} = \frac{20}{100} \times 450 = 0.2 \times 450 = 90$$

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Formula 2 – Percentage Increase / Decrease

Formula (increase):$$\text{\% Increase} = \frac{\text{Increase}}{\text{Original}} \times 100$$

Formula (decrease):$$\text{\% Decrease} = \frac{\text{Decrease}}{\text{Original}} \times 100$$

Variables:

• Original: initial value
• Increase: new − original
• Decrease: original − new

Example MCQ:
Q2. A number decreases from 640 to 544. What is the percentage decrease?

A) 10%
B) 12.5%
C) 15%
D) 20%

Answer: C) 15%

Explanation:
Decrease = 640 − 544 = 96$$\text{\% Decrease} = \frac{96}{640} \times 100 = 0.15 \times 100 = 15\%$$

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Formula 3 – Successive Percentage Change

If we apply a% change and then b% change (on the new value):

Formula:$$\text{Net \% change} = a + b + \frac{ab}{100}$$

Variables:

• a: first percentage change (take negative if decrease)
• b: second percentage change
• Net % change: overall percentage change

Example MCQ:
Q3. A price is increased by 10% and then again by 20%. What is the net percentage increase?

A) 22%
B) 28%
C) 30%
D) 32%

Answer: D) 32%

Explanation:$$\text{Net \% change} = 10 + 20 + \frac{10 \times 20}{100} = 30 + 2 = 32\%$$

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Formula 4 – Reverse Percentage (Finding Original)

If a value becomes New after an increase of p%:

Formula:$$\text{Original} = \frac{\text{New}}{1 + \frac{p}{100}}$$

If it becomes New after a decrease of p%:$$\text{Original} = \frac{\text{New}}{1 – \frac{p}{100}}$$

Variables:

• Original: initial value
• New: value after percentage change
• p: percentage increase/decrease

Example MCQ:
Q4. The price of a book after a 25% increase is ₹750. What was the original price?

A) ₹500
B) ₹550
C) ₹600
D) ₹650

Answer: C) ₹600

Explanation:$$\text{Original} = \frac{\text{New}}{1 + \frac{p}{100}} = \frac{750}{1.25} = 600$$

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2. Ratio & Proportion – Important Formulas

Formula 5 – Division of Amount in a Given Ratio

Formula:
For total amount S and ratio $a:b$a:b:$$\text{Share of first} = \frac{a}{a+b} \times S$$$$\text{Share of second} = \frac{b}{a+b} \times S$$

Variables:

• S: total amount
• a, b: parts of the ratio
• Share: actual amount received

Example MCQ:
Q5. Divide ₹840 in the ratio 2 : 5. What is the share of the larger part?

A) ₹240
B) ₹360
C) ₹480
D) ₹600

Answer: D) ₹600

Explanation:
Total parts = 2 + 5 = 7
Larger part (5):$$\text{Share} = \frac{5}{7} \times 840 = 5 \times 120 = 600$$

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Formula 6 – Direct Proportion

If x is directly proportional to y:

Formula:$$\frac{x_1}{x_2} = \frac{y_1}{y_2}$$

Variables:

• x₁, x₂: two values of x
• y₁, y₂: corresponding values of y

Example MCQ:
Q6. If 5 workers produce 120 units, how many units will 8 workers produce (same rate)?

A) 150
B) 160
C) 180
D) 192

Answer: D) 192

Explanation:$$\frac{5}{8} = \frac{120}{\text{Units}} \Rightarrow \text{Units} = \frac{8 \times 120}{5} = 8 \times 24 = 192$$

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Formula 7 – Inverse Proportion

If x is inversely proportional to y:

Formula:$$x_1 y_1 = x_2 y_2$$

Variables:

• x₁, x₂: one quantity
• y₁, y₂: another quantity inversely related

Example MCQ:
Q7. If 6 men can complete a work in 15 days, in how many days will 10 men complete the same work (same speed)?

A) 8 days
B) 9 days
C) 10 days
D) 12 days

Answer: B) 9 days

Explanation:
Men × Days = constant$$6 \times 15 = 10 \times D \Rightarrow D = \frac{90}{10} = 9$$

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3. Fractions & Decimals – Important Formulas

Formula 8 – Fraction to Percentage

Formula:$$\text{Percentage} = \frac{a}{b} \times 100\%$$

Variables:

• a: numerator
• b: denominator
• Percentage: equivalent % value

Example MCQ:
Q8. What is $\dfrac{3}{8}$​ as a percentage?

A) 35%
B) 37.5%
C) 40%
D) 42.5%

Answer: B) 37.5%

Explanation:$$\frac{3}{8} \times 100 = 37.5\%$$

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Formula 9 – Addition of Fractions

Formula:$$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$

Variables:

• a, c: numerators
• b, d: denominators
• Result: sum of the two fractions

Example MCQ:
Q9. What is $\dfrac{1}{4} + \dfrac{1}{3}$​ ?

A) $\dfrac{5}{12}$ ; B) $\dfrac{7}{12}$ ; C) $\dfrac{2}{7}$ ; D) $\dfrac{3}{7}$ ;

Answer: B) $\dfrac{7}{12}$​

Explanation:
LCM of 4 and 3 is 12$$\frac{1}{4} = \frac{3}{12}, \quad \frac{1}{3} = \frac{4}{12}$$$$\frac{3}{12} + \frac{4}{12} = \frac{7}{12}$$

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4. Averages – Important Formulas

Formula 10 – Basic Average

Formula:$$\text{Average} = \frac{\text{Sum of observations}}{\text{Number of observations}}$$

Variables:

• Sum of observations: total of all values
• Number of observations: count of values

Example MCQ:
Q10. The sum of 5 numbers is 260. What is their average?

A) 50
B) 51
C) 52
D) 53

Answer: C) 52

Explanation:$$\text{Average} = \frac{260}{5} = 52$$

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Formula 11 – Combined Average

Formula:$$\text{Combined avg} = \frac{A_1 n_1 + A_2 n_2}{n_1 + n_2}$$

Variables:

• A₁, A₂: averages of two groups
• n₁, n₂: number of items in each group

Example MCQ:
Q11. Group A has 20 students with average marks 60. Group B has 30 students with average marks 70. What is the combined average?

A) 64
B) 65
C) 66
D) 68

Answer: C) 66

Explanation:$$\text{Combined avg} = \frac{60 \times 20 + 70 \times 30}{20 + 30} = \frac{1200 + 2100}{50} = \frac{3300}{50} = 66$$

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Formula 12 – Average Speed (Equal Distances)

Formula:
If equal distances at speeds $v_1$v1​ and $v_2$v2​:$$v_{\text{avg}} = \frac{2 v_1 v_2}{v_1 + v_2}$$

Variables:

• v₁, v₂: speeds in two parts
• v_avg: average speed for whole journey

Example MCQ:
Q12. A person travels a distance at 40 km/h and returns over the same distance at 60 km/h. What is the average speed?

A) 45 km/h
B) 48 km/h
C) 50 km/h
D) 52 km/h

Answer: B) 48 km/h

Explanation:$$v_{\text{avg}} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h}$$

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5. Profit, Loss & Discount – Important Formulas

Formula 13 – Profit % and Loss %

Formula (Profit %):$$\text{Profit \%} = \frac{\text{SP} – \text{CP}}{\text{CP}} \times 100$$

Formula (Loss %):$$\text{Loss \%} = \frac{\text{CP} – \text{SP}}{\text{CP}} \times 100$$

Variables:

• CP: cost price
• SP: selling price

Example MCQ:
Q13. A pen is bought for ₹400 and sold for ₹460. What is the profit percentage?

A) 10%
B) 12%
C) 15%
D) 20%

Answer: C) 15%

Explanation:
Profit = 460 − 400 = 60$$\text{Profit \%} = \frac{60}{400} \times 100 = 15\%$$

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Formula 14 – SP from CP using Profit % / Loss %

Formula (profit):$$SP = CP \times \left(1 + \frac{p}{100}\right)$$

Formula (loss):$$SP = CP \times \left(1 – \frac{p}{100}\right)$$

Variables:

• CP: cost price
• SP: selling price
• p: profit% or loss%

Example MCQ:
Q14. A book is bought for ₹560 and sold at a loss of 15%. What is the selling price?

A) ₹420
B) ₹448
C) ₹476
D) ₹520

Answer: C) ₹476

Explanation:$$SP = 560 \times \left(1 – \frac{15}{100}\right) = 560 \times 0.85 = 476$$

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Formula 15 – Discount Percentage

Formula:$$\text{Discount \%} = \frac{MP – SP}{MP} \times 100$$

Variables:

• MP: marked price
• SP: selling price after discount

Example MCQ:
Q15. A shirt with marked price ₹2000 is sold for ₹1600. What is the discount percentage?

A) 15%
B) 20%
C) 22%
D) 25%

Answer: B) 20%

Explanation:
Discount = 2000 − 1600 = 400$$\text{Discount \%} = \frac{400}{2000} \times 100 = 20\%$$

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Formula 16 – Successive Discounts

For two discounts $d_1\%$d1​% and $d_2\%$d2​% on the same MP:

Net discount %:$$d_{\text{net}} = d_1 + d_2 – \frac{d_1 d_2}{100}$$

Variables:

• d₁, d₂: discount percentages
• d_net: overall discount percentage

Example MCQ:
Q16. A shop gives two successive discounts of 10% and 20% on an article. What is the net discount?

A) 28%
B) 29%
C) 30%
D) 31%

Answer: A) 28%

Explanation:$$d_{\text{net}} = 10 + 20 – \frac{10 \times 20}{100} = 30 – 2 = 28\%$$

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6. Simple & Compound Interest – Important Formulas

Formula 17 – Simple Interest (SI)

Formula:$$SI = \frac{P \times R \times T}{100}$$

Variables:

• P: principal (initial amount)
• R: rate of interest (per annum in %)
• T: time in years

Example MCQ:
Q17. What is the simple interest on ₹5000 at 8% per annum for 3 years?

A) ₹1000
B) ₹1100
C) ₹1200
D) ₹1300

Answer: C) ₹1200

Explanation:$$SI = \frac{5000 \times 8 \times 3}{100} = \frac{5000 \times 24}{100} = 1200$$

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Formula 18 – Amount at Simple Interest

Formula:$$A = P + SI = P \left(1 + \frac{RT}{100}\right)$$

Variables:

• A: total amount after time T
• P, R, T: as above

Example MCQ:
Q18. What is the amount after 3 years if ₹5000 is invested at 8% p.a. simple interest?

A) ₹6000
B) ₹6200
C) ₹6300
D) ₹6500

Answer: B) ₹6200

Explanation:
We already found SI = ₹1200$$A = P + SI = 5000 + 1200 = 6200$$

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Formula 19 – Amount at Compound Interest (Annual)

Formula:$$A = P \left(1 + \frac{R}{100}\right)^T$$

Variables:

• A: amount after T years
• P: principal
• R: yearly rate%
• T: time in years

Example MCQ:
Q19. What will be the amount on ₹8000 at 10% p.a. compound interest after 2 years (annual compounding)?

A) ₹8800
B) ₹9600
C) ₹9680
D) ₹9800

Answer: C) ₹9680

Explanation:$$A = 8000 \left(1 + \frac{10}{100}\right)^2 = 8000 \times 1.1^2 = 8000 \times 1.21 = 9680$$

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Formula 20 – Present Value (Discounting)

Formula:$$P = \frac{F}{\left(1 + \frac{R}{100}\right)^T}$$

Variables:

• P: present value
• F: future amount
• R: rate%
• T: time in years

Example MCQ:
Q20. The amount ₹12100 is due after 2 years at 10% p.a. compound interest. What is the present value?

A) ₹9000
B) ₹9500
C) ₹10000
D) ₹11000

Answer: C) ₹10000

Explanation:$$P = \frac{12100}{(1.1)^2} = \frac{12100}{1.21} = 10000$$

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7. Time, Speed & Distance – Important Formulas

Formula 21 – Basic Relation

Formula:$$D = S \times T$$

Variables:

• D: distance
• S: speed
• T: time

Example MCQ:
Q21. How long will it take to travel 150 km at 50 km/h?

A) 2 hours
B) 2.5 hours
C) 3 hours
D) 3.5 hours

Answer: C) 3 hours

Explanation:$$T = \frac{D}{S} = \frac{150}{50} = 3 \text{ hours}$$

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Formula 22 – Unit Conversion (km/h ↔ m/s)

Formulas:$$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$$$$1 \text{ m/s} = \frac{18}{5} \text{ km/h}$$

Variables:

• Units: km/h and m/s

Example MCQ:
Q22. Convert 54 km/h into m/s.

A) 12 m/s
B) 14 m/s
C) 15 m/s
D) 18 m/s

Answer: C) 15 m/s

Explanation:$$54 \times \frac{5}{18} = 54 \div 18 \times 5 = 3 \times 5 = 15 \text{ m/s}$$

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Formula 23 – Relative Speed (Trains / Opposite & Same Direction)

Formulas:

• Opposite direction: $$S_{\text{rel}} = S_1 + S_2$$
• Same direction: $$S_{\text{rel}} = |S_1 – S_2|$$

Variables:

• S₁, S₂: speeds of two moving objects
• S_rel: relative speed

Example MCQ:
Q23. Two trains move in opposite directions at 40 km/h and 50 km/h. What is their relative speed?

A) 10 km/h
B) 40 km/h
C) 50 km/h
D) 90 km/h

Answer: D) 90 km/h

Explanation:$$S_{\text{rel}} = 40 + 50 = 90 \text{ km/h}$$

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Formula 24 – Train Crossing a Pole

Formula:$$T = \frac{L}{S}$$​

Variables:

• T: time to cross
• L: length of train (in meters)
• S: speed in m/s

Example MCQ:
Q24. A train 150 m long runs at 54 km/h. How much time does it take to cross a pole?

A) 8 s
B) 10 s
C) 12 s
D) 15 s

Answer: B) 10 s

Explanation:
First convert speed:$$54 \text{ km/h} = 15 \text{ m/s}$$$$T = \frac{L}{S} = \frac{150}{15} = 10 \text{ seconds}$$

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Formula 25 – Boats & Streams (Upstream / Downstream)

Formulas:$$S_{\text{up}} = u – v$$$$S_{\text{down}} = u + v$$

Variables:

• u: speed of boat in still water
• v: speed of stream
• S_up: upstream speed
• S_down: downstream speed

Example MCQ:
Q25. A boat’s speed in still water is 12 km/h and the stream speed is 3 km/h. How long will it take to go 27 km upstream?

A) 2 hours
B) 2.5 hours
C) 3 hours
D) 3.5 hours

Answer: C) 3 hours

Explanation:$$S_{\text{up}} = 12 – 3 = 9 \text{ km/h}$$$$T = \frac{D}{S_{\text{up}}} = \frac{27}{9} = 3 \text{ hours}$$

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8. Time & Work – Extra but Important

Formula 26 – One Person’s Work Rate

Formula:
If a person finishes work in x days:$$\text{1-day work} = \frac{1}{x}$$

Variables:

• x: days to complete the work
• 1-day work: fraction of work done in one day

Example MCQ:
Q26. If A alone can finish a job in 12 days, what fraction of the work does A do in one day?

A) 1/10
B) 1/12
C) 1/15
D) 1/20

Answer: B) 1/12

Explanation:
One day’s work = 1 / total days = 1/12.

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Formula 27 – Two Persons Working Together

Formula:
If A finishes in x days and B in y days:

• Combined 1-day work: $$\frac{1}{x} + \frac{1}{y}$$
• Total time together: $$T = \frac{1}{\frac{1}{x} + \frac{1}{y}}$$

Variables:

• x, y: individual times
• T: time together

Example MCQ:
Q27. A can finish a work in 12 days and B in 18 days. In how many days will they finish working together?

A) 6 days
B) 7.2 days
C) 8 days
D) 9 days

Answer: B) 7.2 days

Explanation:$$\text{1-day work} = \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36}$$$$T = \frac{1}{5/36} = \frac{36}{5} = 7.2 \text{ days}$$

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Formula 28 – Finding Third Person’s Time (from Together)

Concept:
Total rate = sum of individual rates.

Formula:$$\frac{1}{T_{\text{all}}} = \frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C}$$

Variables:

• T_all: time when A, B, C work together
• T_A, T_B, T_C: individual times

Example MCQ:
Q28. A can finish a work in 10 days, B in 20 days, and A, B, C together finish in 4 days. In how many days can C alone finish the work?

A) 8 days
B) 10 days
C) 12 days
D) 15 days

Answer: B) 10 days

Explanation:
Rates:$$\frac{1}{T_{\text{all}}} = \frac{1}{4}$$$$\frac{1}{T_A} = \frac{1}{10}, \quad \frac{1}{T_B} = \frac{1}{20}$$

So:$$\frac{1}{T_C} = \frac{1}{4} – \frac{1}{10} – \frac{1}{20}$$

LCM 20:$$\frac{1}{4} = \frac{5}{20}, \quad \frac{1}{10} = \frac{2}{20}, \quad \frac{1}{20} = \frac{1}{20}$$$$\frac{1}{T_C} = \frac{5 – 2 – 1}{20} = \frac{2}{20} = \frac{1}{10} \Rightarrow T_C = 10 \text{ days}$$

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9. Ages – Extra but Useful

Formula 29 – Linear Age Relation

Concept & Formula:
If present age = past age + years passed.

General form:$$\text{Future age} = \text{Present age} + \text{years}$$

Variables:

• Present, past, future ages

Example MCQ:
Q29. The present age of a father is 3 times the age of his son. After 10 years, the father will be twice the age of his son. What is the present age of the son?

A) 10 years
B) 15 years
C) 20 years
D) 25 years

Answer: A) 10 years

Explanation:
Let son’s present age = x, father = 3x.

After 10 years:
Son = x + 10, Father = 3x + 10

Given: father = 2 × son$$3x + 10 = 2(x + 10) \Rightarrow 3x + 10 = 2x + 20 \Rightarrow x = 10$$

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Formula 30 – Age Ratio Forming Equation

Concept:
If after k years, ages are in ratio $a:b$a:b:

Formula (equation form):$$\frac{x + k}{y + k} = \frac{a}{b}$$

Variables:

• x, y: present ages
• k: years after (or before, then use −k)
• a:b: given ratio

(We already tested age type above; skip second MCQ here to save space.)

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10. Mixture & Alligation – Extra but Useful

Formula 31 – Mean Price of a Mixture

Formula:$$\text{Mean price} = \frac{\sum (\text{quantity} \times \text{price})}{\sum \text{quantity}}$$

Variables:

• quantity: amount of each ingredient
• price: price per unit of each ingredient

Example MCQ:
Q30. 2 kg of rice at ₹50/kg is mixed with 3 kg of rice at ₹70/kg. What is the mean price per kg?

A) ₹58
B) ₹60
C) ₹62
D) ₹64

Answer: C) ₹62

Explanation:$$\text{Mean} = \frac{2 \times 50 + 3 \times 70}{2 + 3} = \frac{100 + 210}{5} = \frac{310}{5} = 62$$

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Formula 32 – Alligation Rule for Ratio

For two ingredients at prices $P_1, P_2$P1​,P2​ giving mean price M:

Formula:$$Q_1 : Q_2 = (P_2 – M) : (M – P_1)$$

Variables:

• P₁, P₂: prices of ingredients (P₂ > M > P₁)
• M: mean price
• Q₁, Q₂: quantities

Example MCQ:
Q31. In what ratio should teas costing ₹50/kg and ₹80/kg be mixed to get a mixture worth ₹70/kg?

A) 1 : 2
B) 2 : 1
C) 3 : 2
D) 2 : 3

Answer: A) 1 : 2

Explanation:$$Q_1 : Q_2 = (P_2 – M) : (M – P_1) = (80 – 70) : (70 – 50) = 10 : 20 = 1 : 2$$

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Formula 33 – Replacement Mixture (One Operation)

If from a vessel of volume V, we remove R and refill with water once:

Fraction of original liquid left after one replacement:$$\text{Fraction left} = \frac{V – R}{V}$$

Variables:

• V: total volume
• R: volume removed and replaced

Example MCQ:
Q32. A 60 L vessel is full of milk. 12 L is removed and replaced with water once. What percentage of milk is left?

A) 70%
B) 75%
C) 80%
D) 85%

Answer: C) 80%

Explanation:$$\text{Fraction left} = \frac{60 – 12}{60} = \frac{48}{60} = \frac{4}{5} = 0.8$$

So milk left = 80%.

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