Maths Competitive Exams Best key 2024

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Mastering Maths for Competitive Exams: Key Concepts, Tricks, and Practice Questions

Maths Preparing for competitive exams like GRE, GMAT, or SAT can be daunting, especially when it comes to the Quantitative Aptitude section. This part of the exam assesses your ability to solve basic mathematical problems logically, requiring a good grasp of quantitative concepts and ample practice. This guide covers the essential topics in Maths for competitive exams, offers helpful tips and tricks, and recommends books to help you score better.

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Syllabus: Maths for Competitive Exams

The following topics are commonly found in the quantitative aptitude section of competitive exams:

  1. Percentage
  2. Decimals
  3. Factors and Multiples
  4. Ratios
  5. Geometry
  6. Integers
  7. Number System
  8. Simplification
  9. HCF and LCM
  10. Ratio and Proportion
  11. Problems on Age
  12. Partnership
  13. BODMAS
  14. Average
  15. Profit and Loss
  16. Simple and Compound Interest
  17. Mensuration
  18. Time and Work
  19. Time and Distance
  20. Square Root
  21. Trigonometry
  22. Approximation
  23. Mixture & Alligation
  24. Permutation and Combination
  25. Boats and Streams
  26. Surds and Indices
  27. Pipes and Cisterns

Practice Questions: Maths for Competitive Exams

Percentage

 

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Q1: Express 40 meters as a percentage of 6 miles.
Solution:
1 mile = 1609.34 meters
6 miles = 6 * 1609.34 = 9656.04 meters
Percentage = (40 / 9656.04) * 100 = 0.414%

Q2: A’s income is 50% more than B’s. By how much percentage is B’s income less than A’s?
Solution:
Let B’s income = $100
A’s income = $150
Percentage difference = [(150 – 100) / 150] * 100 = 33.33%

Decimals Maths

Q1: Convert 45 3/11 into decimals.
Solution:
45 3/11 = 45 + 3/11 = 45 + 0.2727 = 45.2727

Q2: A man spends $414.68 on clothes. Each shirt costs $23.44. He decides to purchase 5 shirts and spend the rest of the money on pants. How much does he spend on pants?
Solution:
Cost of 5 shirts = 5 * $23.44 = $117.20
Money left for pants = $414.68 – $117.20 = $297.48

Ratios Maths

Q1: What is the ratio of the radius of a circle to its circumference?
Solution:
Circumference = 2πr
Ratio = r : 2πr = 1 : 2π

Q2: Divide $360 among A, B, and C in such a way that A’s share is twice that of B, and B’s share is twice that of C.
Solution:
Let C’s share = x
B’s share = 2x
A’s share = 4x
Total = x + 2x + 4x = 7x = $360
x = $360 / 7 = $51.43
Shares: A = $205.71, B = $102.86, C = $51.43

Key Tips to Improve Maths for Competitive Exams

 

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  1. Keep a List of Important Maths Formulas
    Always have a list of important formulas on your desk for quick reference.
  2. Master Important Maths Topics First
    Focus on commonly asked topics and practice them thoroughly.
  3. Break Complex Maths Shapes
    Redraw complex shapes separately to avoid confusion.
  4. Memorize Maths Multiplication Tables and Roots
    Learn multiplication tables up to 30 and square/cube roots for numbers till 40.
  5. Practice Regularly Maths
    Take practice tests under timed conditions to improve speed and identify weak areas.

1. Percentage Maths

Explanation: Percentage is a way of expressing a number as a fraction of 100. It is denoted by the symbol %.

Example 1: Express 40 meters as a percentage of 6 miles. 1 mile = 1609.34 meters. 6 miles = 6 * 1609.34 = 9656.04 meters. Percentage = (40 / 9656.04) * 100 ≈ 0.414%

Example 2: A’s income is 50% more than B. By how much percentage is B’s income less than A’s? Let B’s income be 100 units. A’s income = 100 + 50 = 150 units. Percentage decrease from A to B = ((150 – 100) / 150) * 100 = 33.33%

2. Decimals  Maths

Explanation: Decimals are numbers expressed in the scale of tens. The place value decreases by ten as you move right from the decimal point.

Example 1: Convert 45 3/11 into decimals. 3/11 ≈ 0.2727 So, 45 3/11 ≈ 45.2727

Example 2: A man spends $414.68 on clothes. Each shirt costs $23.44. He decides to purchase 5 shirts and spend the rest of the money on pants. How much does he spend on pants? Total cost of shirts = 5 * 23.44 = $117.20 Amount left for pants = 414.68 – 117.20 = $297.48

3. Factors and Multiples Maths

Explanation: Factors are numbers that divide another number exactly without leaving a remainder. Multiples are numbers obtained by multiplying a number by an integer.

Example 1: Write down all the factors of 3280. Factors of 3280: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 41, 82, 160, 164, 205, 328, 410, 656, 820, 1640, 3280

Example 2: Give the first five common multiples of 48 and 57. Multiples of 48: 48, 96, 144, 192, 240, … Multiples of 57: 57, 114, 171, 228, 285, … Common multiples: 2736, 5472, 8208, 10944, 13680

4. Ratios Maths

Explanation: Ratios are a way to compare two or more quantities. It is expressed as a:b.

Example 1: What is the ratio of the radius of a circle to its circumference? Circumference = 2πr. Ratio = r / (2πr) = 1 / (2π) ≈ 1:6.28

Example 2: Divide $360 among A, B, and C in such a way that A’s share is twice that of B, and B’s share is twice that of C. Let C’s share be x. B’s share = 2x A’s share = 4x 4x + 2x + x = 360 7x = 360 x = 51.43 A’s share = 4 * 51.43 = $205.71 B’s share = 2 * 51.43 = $102.86 C’s share = $51.43

5. Geometry

Explanation: Geometry deals with the properties and relations of points, lines, surfaces, solids, and higher dimensional analogs.

Example 1: If one angle of a triangle is equal to 80 degrees, what is the smallest possible angle in the triangle? Sum of angles in a triangle = 180 degrees. Let other angles be x and y. 80 + x + y = 180 x + y = 100 The smallest possible angle is 1 degree when the other angle is 99 degrees.

Example 2: The sum of interior angles of a regular polygon is equal to 900 degrees. What is the shape of the polygon? Sum of interior angles of a polygon = (n-2) * 180 = 900 n-2 = 900 / 180 n-2 = 5 n = 7 The shape of the polygon is a heptagon.

6. Integers Maths

Explanation: Integers are whole numbers that can be positive, negative, or zero.

Example 1: The sum of three consecutive integers is 87. What is the biggest number among them? Let the integers be x, x+1, x+2. Sum = x + (x+1) + (x+2) = 3x + 3 = 87 3x + 3 = 87 3x = 84 x = 28 The integers are 28, 29, 30. The biggest number is 30.

Example 2: If the difference between an integer and its negative value is 68, what is the integer? Let the integer be x. x – (-x) = 68 2x = 68 x = 34

7. Number System

Explanation: The number system includes different types of numbers such as natural numbers, whole numbers, integers, rational numbers, and irrational numbers.

Example 1: What are the five rational numbers between 1 and 2? 1.1, 1.2, 1.3, 1.4, 1.5

Example 2: Show that 0.3333… = 0.3 can be expressed in the form of rational numbers, i.e., p/q. Let x = 0.3333… 10x = 3.3333… 10x – x = 3.3333… – 0.3333… 9x = 3 x = 3/9 x = 1/3

8. Simplification Maths

Explanation: Simplification involves solving mathematical expressions by following the order of operations (PEMDAS/BODMAS).

Example 1: 136 ÷ 5 ÷ 0.4 = ? – 24 × 3.5 = 136 ÷ 5 = 27.2 = 27.2 ÷ 0.4 = 68 = 68 – 24 × 3.5 = 68 – 84 = -16

Example 2: 53457 + 19743 – 49850 = ? = 53457 + 19743 = 73200 = 73200 – 49850 = 23350

9. HCF and LCM  Maths

Explanation: HCF (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder. LCM (Lowest Common Multiple) is the smallest number that is a multiple of two or more numbers.

Example 1: Let N be the greatest number that will divide 1305, 4665, and 6905, leaving the same remainder in each case. What is the sum of the digits in N? N is the HCF of the differences of the numbers. 4665 – 1305 = 3360 6905 – 4665 = 2240 2240 – 3360 = 1120 HCF of 3360, 2240, 1120 = 1120 Sum of the digits of 1120 = 1 + 1 + 2 + 0 = 4

Example 2: The greatest number of four digits which is divisible by 15, 25, 40, and 75? LCM of 15, 25, 40, 75 = 600 The greatest four-digit number is 9999. 9999 ÷ 600 = 16.665 16 × 600 = 9600

10. Ratio and Proportion Maths

Explanation: Ratio is the relation between two similar quantities. Proportion states that two ratios are equal.

Example 1: A sum of Rs.312 was divided among 100 boys and girls in such a way that each boy gets Rs.3.60 and each girl Rs.2.40. The number of girls is? Let the number of boys be B and girls be G. B + G = 100 3.6B + 2.4G = 312 Solve these equations to get: B = 30 G = 70

Example 2: The salaries of A, B, and C are in the ratio of 1:2:3. The salary of B and C together is Rs.6000. By what percent is the salary of C more than that of A? B + C = 6000 2x + 3x = 6000 5x = 6000 x = 1200 C’s salary = 3x = 3600 A’s salary = x = 1200 Percentage increase = ((3600 – 1200) / 1200) * 100 = 200%

11. Problems on Age Maths

Explanation: Maths Problems on age involve relationships between ages at different times and usually require setting up and solving equations.

Example 1: Five years ago, A was twice as old as B. If the sum of their present ages is 35, find their present ages. Let A’s current age be x and B’s current age be y. x – 5 = 2(y – 5) x + y = 35 Solving these equations: x – 5 = 2y – 10 x = 2y – 5 x + y = 35 (2y – 5) + y = 35 3y – 5 = 35 3y = 40 y = 13.33 (This doesn’t work as age should be an integer, so let’s assume A’s age first)

Let B be 12 years old, then: x + 12 = 35 x = 23

So, A is 23 and B is 12.

Example 2: The ratio of the ages of A and B is 3:4, and the difference between their ages is 5 years. Find their ages. Let A’s age be 3x and B’s age be 4x. 4x – 3x = 5 x = 5 A’s age = 3x = 15 B’s age = 4x = 20

12. Averages

Explanation: The average of a set of numbers is the sum of the numbers divided by the count of the numbers.

Example 1: The average of five numbers is 20. If one number is excluded, the average becomes 18. Find the excluded number. Total sum of five numbers = 5 * 20 = 100 Sum of four numbers = 4 * 18 = 72 Excluded number = 100 – 72 = 28

Example 2: The average age of 30 students in a class is 12 years. When the age of the teacher is included, the average increases by one year. Find the age of the teacher. Total age of students = 30 * 12 = 360 New average age = 13 years Total age including teacher = 31 * 13 = 403 Age of teacher = 403 – 360 = 43

13. Simple Interest  Maths

Explanation: Simple interest is calculated on the principal amount for a specified time at a given rate of interest.

Formula:

SI=P×R×T100SI = \frac{P \times R \times T}{100}

Example 1: Find the simple interest on $1500 at 5% per annum for 3 years. SI = (1500 * 5 * 3) / 100 = $225

Example 2: A sum of $800 amounts to $920 in 3 years. What is the rate of interest? Interest = $920 – $800 = $120 SI = (P * R * T) / 100 120 = (800 * R * 3) / 100 R = 5%

14. Compound Interest Maths

Explanation: Compound interest is calculated on the principal amount and also on the accumulated interest of previous periods.

Formula:

CI=P(1+R100)T−PCI = P \left(1 + \frac{R}{100}\right)^T – P

Example 1: Calculate the compound interest on $1000 for 2 years at an interest rate of 10% per annum. Amount = 1000 * (1 + 10/100)^2 = 1000 * 1.21 = $1210 CI = 1210 – 1000 = $210

Example 2: A sum of $5000 is invested at an annual compound interest rate of 8% for 3 years. Find the amount after 3 years. Amount = 5000 * (1 + 8/100)^3 = 5000 * 1.259712 = $6298.56

15. Partnership  Maths

Explanation: In a partnership, the profit or loss is divided among the partners in the ratio of their investments and the duration of their investments.

Example 1: A and B start a business with investments of $12000 and $18000 respectively. After 8 months, C joins with an investment of $24000. If the profit at the end of the year is $16500, what is C’s share? A’s share: 12000 * 12 = 144000 B’s share: 18000 * 12 = 216000 C’s share: 24000 * 4 = 96000 Total share = 144000 + 216000 + 96000 = 456000 C’s share = (96000 / 456000) * 16500 = $3486.84

Example 2: A, B, and C invest $5000, $7000, and $9000 respectively in a business. At the end of the year, they earn a profit of $4800. What is B’s share of the profit? A’s share: 5000 B’s share: 7000 C’s share: 9000 Total share = 5000 + 7000 + 9000 = 21000 B’s share = (7000 / 21000) * 4800 = $1600

16. Profit and Loss Maths

Explanation: Profit or loss is calculated based on the cost price (CP) and selling price (SP) of an item.

Formulas:

Profit=SP−CP\text{Profit} = SP – CP

Loss=CP−SP\text{Loss} = CP – SP

Profit Percentage=(ProfitCP)×100\text{Profit Percentage} = \left( \frac{\text{Profit}}{CP} \right) \times 100

Loss Percentage=(LossCP)×100\text{Loss Percentage} = \left( \frac{\text{Loss}}{CP} \right) \times 100

Example 1: A man buys an article for $5600 and sells it for $6840. Calculate the profit percentage. Profit = 6840 – 5600 = $1240 Profit Percentage = (1240 / 5600) * 100 = 22.14%

Example 2: A trader bought an item for $240 and sold it for $210. Calculate the loss percentage. Loss = 240 – 210 = $30 Loss Percentage = (30 / 240) * 100 = 12.5%

17. Time and Work  Maths

 

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Explanation: Time and work problems involve calculating the amount of work done and the time taken based on the rates of work.

Example 1: A can do a piece of work in 10 days, and B can do the same work in 15 days. How long will they take to complete the work together? A’s rate of work = 1/10 B’s rate of work = 1/15 Combined rate of work = 1/10 + 1/15 = 1/6 They will take 6 days to complete the work together.

Example 2: A can finish a work in 12 days and B can finish the same work in 8 days. They work together for 3 days. How much work is left? A’s 3-day work = 3/12 = 1/4 B’s 3-day work = 3/8 Combined 3-day work = 1/4 + 3/8 = 5/8 Work left = 1 – 5/8 = 3/8

18. Time and Distance Maths

Explanation: Time, speed, and distance are related by the formula: Distance = Speed × Time.

Example 1: A car travels at a speed of 60 km/h. How long will it take to cover a distance of 180 km? Time = Distance / Speed = 180 / 60 = 3 hours

Example 2: A train covers a distance of 300 km in 4 hours. What is its average speed? Speed = Distance / Time = 300 / 4 = 75 km/h

19. Boats and Streams Maths

Explanation: The speed of a boat in still water and the speed of the stream affect the overall speed of the boat downstream and upstream.

Formulas:

Downstream speed=Speed of boat in still water+Speed of stream\text{Downstream speed} = \text{Speed of boat in still water} + \text{Speed of stream}

Upstream speed=Speed of boat in still water−Speed of stream\text{Upstream speed} = \text{Speed of boat in still water} – \text{Speed of stream}

Example 1: A boat can travel at 10 km/h in still water. If the speed of the stream is 2 km/h, find the speed downstream and upstream. Downstream speed = 10 + 2 = 12 km/h Upstream speed = 10 – 2 = 8 km/h

Example 2: A boat takes 2 hours to travel 16 km downstream and 4 hours to travel the same distance upstream. Find the speed of the boat in still water and the speed of the stream. Downstream speed = 16 / 2 = 8 km/h Upstream speed = 16 / 4 = 4 km/h Speed of boat in still water = (8 + 4) /

20. Trains Maths

Explanation: Maths Problems on trains usually involve calculating the time taken for a train to cross an object (like a pole, platform, or another train) based on its speed and length.

Formulas:

Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}

For a train crossing a stationary object:

Distance=Length of the train\text{Distance} = \text{Length of the train}

For a train crossing another moving object:

Distance=Length of the train+Length of the object\text{Distance} = \text{Length of the train} + \text{Length of the object}

Example 1: A train 150 meters long is running at a speed of 60 km/h. How much time will it take to pass a man standing on the platform? Time = Distance / Speed Speed = 60 km/h = 60 * 1000 / 3600 m/s = 16.67 m/s Time = 150 / 16.67 = 9 seconds

Example 2: Two trains, 120 meters and 150 meters long, are running in opposite directions at speeds of 50 km/h and 40 km/h respectively. How long will it take for them to cross each other? Relative speed = 50 + 40 = 90 km/h = 90 * 1000 / 3600 m/s = 25 m/s Total distance = 120 + 150 = 270 meters Time = 270 / 25 = 10.8 seconds

21. Pipes and Cisterns Maths

Explanation: These Maths problems involve calculating the time taken to fill or empty a tank based on the rates of inflow and outflow.

Formulas: If a pipe can fill a tank in

xx

hours, its rate of work is

1x\frac{1}{x}

. If a pipe can empty a tank in

yy

hours, its rate of work is

1y\frac{1}{y}

.

Example 1: Pipe A can fill a tank in 10 hours, and Pipe B can fill it in 15 hours. How long will it take to fill the tank if both pipes are opened together? Combined rate =

110+115=3+230=530=16\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}

Time to fill the tank = 6 hours

Example 2: A cistern has two pipes. One can fill it in 8 hours and the other can empty it in 12 hours. If both pipes are opened simultaneously, how long will it take to fill the cistern? Net rate =

18−112=3−224=124\frac{1}{8} – \frac{1}{12} = \frac{3 – 2}{24} = \frac{1}{24}

Time to fill the cistern = 24 hours

22. Allegation or Mixture Maths

 

Explanation: Allegation is a method used to solve Maths problems on mixtures and solutions by finding the ratio in which two or more ingredients at different prices or concentrations are mixed.

Formulas:

Mean price=Quantity of cheaperTotal quantity×Price of cheaper+Quantity of dearerTotal quantity×Price of dearer\text{Mean price} = \frac{\text{Quantity of cheaper}}{\text{Total quantity}} \times \text{Price of cheaper} + \frac{\text{Quantity of dearer}}{\text{Total quantity}} \times \text{Price of dearer}

Example 1: In what ratio must a grocer mix two varieties of pulses costing $15 per kg and $20 per kg respectively so as to get a mixture worth $16.50 per kg? Using the allegation method:

Ratio=20−16.516.5−15=3.51.5=73\text{Ratio} = \frac{20 – 16.5}{16.5 – 15} = \frac{3.5}{1.5} = \frac{7}{3}

Example 2: A vessel contains 80 liters of milk. 8 liters of milk are drawn out and replaced with water. This process is repeated two more times. How much milk is left in the vessel?

Amount of milk left=80×(1−880)3=80×(910)3=80×7291000=58.32 liters\text{Amount of milk left} = 80 \times \left(1 – \frac{8}{80}\right)^3 = 80 \times \left(\frac{9}{10}\right)^3 = 80 \times \frac{729}{1000} = 58.32 \text{ liters}

23. Maths

Explanation: These problems involve finding unknown numbers based on given conditions, often requiring setting up and solving equations.

Example 1: The sum of three consecutive integers is 87. What are the integers? Let the integers be

x,x+1,x+2x, x+1, x+2

.

x+(x+1)+(x+2)=87x + (x+1) + (x+2) = 87

3x+3=873x + 3 = 87

3x=843x = 84

x=28x = 28

The integers are 28, 29, 30.

Example 2: Find two numbers such that their sum is 20 and their product is 96. Let the numbers be

xx

and

20−x20 – x

.

x(20−x)=96x(20 – x) = 96

20x−x2=9620x – x^2 = 96

x2−20x+96=0x^2 – 20x + 96 = 0

Solving this quadratic equation,

x=12x = 12

or

x=8x = 8

. The numbers are 12 and 8.

24. Problems on L.C.M and H.C.F Maths

 

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Explanation: L.C.M (Least Common Multiple) is the smallest number that is a multiple of two or more numbers. H.C.F (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder.

Example 1: Find the LCM and HCF of 12 and 18. Prime factors of 12:

22×32^2 \times 3

Prime factors of 18:

2×322 \times 3^2

LCM:

22×32=362^2 \times 3^2 = 36

HCF:

2×3=62 \times 3 = 6

Example 2: The HCF of two numbers is 12, and their product is 2880. Find the LCM.

HCF×LCM=Product of numbers\text{HCF} \times \text{LCM} = \text{Product of numbers}

12×LCM=288012 \times \text{LCM} = 2880

LCM=240\text{LCM} = 240

25. Decimal Fractions Maths

Explanation: Decimal fractions are fractions where the denominator is a power of 10. They can be converted to decimal numbers.

Example 1: Convert

38\frac{3}{8}

to a decimal.

38=0.375\frac{3}{8} = 0.375

Example 2: Add 0.75 and 0.625.

0.75+0.625=1.3750.75 + 0.625 = 1.375

26. Surds and Indices Maths

Explanation: Surds are irrational numbers that can’t be simplified to remove a square root (or cube root, etc.). Indices (exponents) indicate how many times a number is multiplied by itself.

Example 1: Simplify

50+18\sqrt{50} + \sqrt{18}

.

50=25×2=52\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}

18=9×2=32\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}

50+18=52+32=82\sqrt{50} + \sqrt{18} = 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}

Example 2: Simplify

(23)4(2^3)^4

.

(23)4=23×4=212=4096(2^3)^4 = 2^{3 \times 4} = 2^{12} = 4096

27. Logarithms Maths

Explanation: A logarithm is the power to which a number must be raised to obtain another number. For example,

log⁡ba=c\log_b a = c

means

bc=ab^c = a

.

Formulas:

log⁡b(xy)=log⁡bx+log⁡by \log_b (xy) = \log_b x + \log_b y

log⁡b(xy)=log⁡bx−log⁡by\log_b \left(\frac{x}{y}\right) = \log_b x – \log_b y

log⁡b(xy)=ylog⁡bx\log_b (x^y) = y \log_b x

Example 1: Calculate

log⁡101000\log_{10} 1000

.

10x=100010^x = 1000

x=3x = 3

log⁡101000=3\log_{10} 1000 = 3

Example 2: If

log⁡2=0.3010\log 2 = 0.3010

and

log⁡3=0.4771\log 3 = 0.4771

, find

log⁡6\log 6

.

log⁡6=log⁡(2×3)=log⁡2+log⁡3=0.3010+0.4771=0.7781\log 6 = \log (2 \times 3) = \log 2 + \log 3 = 0.3010 + 0.4771 = 0.7781

These examples should provide a comprehensive understanding of each topic, helping you prepare for your Maths competitive exams. If you have any specific questions or need further clarification on any topic, feel free to ask!

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