Quantitative Aptitude - SPLessons

Number Problems

Chapter 8

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Number Problems

shape Introduction

A set of numbers are put in the form of a puzzle. We need to analyze the given conditions and by assuming the unknown numbers equations are written to them. By those equations solve the unknown values. These usually involve around framing the complex statements about a single number. There is basic rule i.e. “start with ‘X'”.


shape Methods

A. Types of Numbers:

1. Natural Numbers: Counting numbers 1, 2, 3, 4, 5,….. are called natural numbers.

2. Whole Numbers: All counting numbers together with zero form the set of whole numbers. Thus,

    (i) 0 is the only whole number which is not a natural number.

    (ii) Every natural number is a whole number.


3. Integers: All natural numbers, 0 and negatives of counting numbers i.e., {…, – 3 , – 2 , – 1 , 0, 1, 2, 3,…..} together form the set of integers.

    (i) Positive Integers: {1, 2, 3, 4, …..} is the set of all positive integers.

    (ii) Negative Integers: {- 1, – 2, – 3,…..} is the set of all negative integers.

    (iii) Non-Positive and Non-Negative Integers: 0 is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while {0, – 1 , – 2 , – 3 , …..} represents the set of non-positive integers.


4. Even Numbers: A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.

5. Odd Numbers: A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.

6. Prime Numbers: A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself. Prime numbers up to 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Prime numbers Greater than 100: Let be a given number greater than 100. To find out whether it is prime or not, we use the following method:

Find a whole number nearly greater than the square root of p. Let k > square root of p. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime. e.g,, We have to find whether 191 is a prime number or not. Now, 14 > square root of 191. Prime numbers less than 14 are 2, 3, 5, 7, 11, 13. 191 is not divisible by any of them. So, 191 is a prime number.

7. Composite Numbers: Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.


Co-primes: Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes.


Example 1:
Factor: \(4x^{2}\) + \(12x\) + \(9\)

Solution:

    Since \(4x^{2}\) = \((2x)^{2}\) and 9 = \(3^{2}\), we guess \((2x + 3)^{2}\)

    Test: using the square formula, \((2x + 3)^{2}\) = \(4x^{2}\) + \(12x\) + \(9\)

    Therefore, \((2x + 3)^{2}\) is the answer.


Example 2:
Factor: \(9x^{2}\) + \(24xy\) + \(16y^{2}\)

Solution:

    Since \(9x^{2}\) = \((3x)^{2}\) and \(16y^{2}\) = \((4y)^{2}\), we guess \((3x – 4y)^{2}\)

    Test: using the square formula, \((3x – 4y)^{2}\) = \(9x^{2}\) + \(24xy\) + \(16y^{2}\)

    Therefore, \((3x – 4y)^{2}\) is the answer.



Example 1:
Factor: \(x^{2} – 9\)

Solution:

    Write \(x^{2} – 9\) as \(x^{2} – 3^{2}\)

    By using the formula, we get \((x + 3)(x – 3)\)


Example 2:
Factor: \(16x^{4} – 81y^{4}\).

Solution:

    Write \(16x^{4} – 81y^{4}\) as \((4x^{2})^{2} – (9y^{2})^{2}\)

    By the formula, we get \((4x^{2} + 9y^{2})\) \((4x^{2} – 9y^{2})\)

    Now, \((4x^{2} + 9y^{2})\) is a sum of squares (not factorable),

    but we can factor \((4x^{2} – 9y^{2})\) further as a diffrence of squares again!

    Thus, \((4x^{2} + 9y^{2})\) \((4x^{2} – 9y^{2})\) = \((4x^{2} + 9y^{2})\) \(((2x)^{2} – (3y)^{2})\)

    By the diffrence of squares, we get.

    \((4x^{2} + 9y^{2})\)\((2x + 3y)(2x – 3y)\)



Example 1:
Factor: \(x^{3} + 27\).

Solution:

    \(x^{3} + 27\) as \(x^{3} + 3^{3}\)

    By the formula, we get

    \((x + 3)(x^{2} + 3x + 9)\)


Example 2:
Factor: \(8x^{3} – 125\).

Solution:

    \(8x^{3} – 125\) as \((2x)^{3} – (5)^{3}\)

    By the formula, we get

    \((2x – 5)(4x^{2} + 10x + 25)\)

shape Formulae

  • \((a+b)^2 = a^2 + b^2 +2ab\)
  • \((a-b)^2 = a^2 + b^2 -2ab\)
  • \(a^2 – b^2 = (a+b)(a-b)\)
  • \((a+b)^2\) – \((a-b)^2\) = \(4ab\)
  • \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
  • \((a^3 + b^3) = (a + b)(a^2 – ab + b^2)\)
  • \((a^3 – b^3) = (a-b)(a^2 + ab + b^2)\)
  • \((a^3 + b^3 + c^3 – 3abc) = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)\)
  • \(When \ a + b + c = 0, \ then \ (a^3 + b^3 + c^3) = 3abc\)

shape Samples

1. Find a number such that 20 is subtracted from 9 times the number, the result is 10 more than fourth the number?

Solution:

    Let the number to be found be \(x\)

    When 20 is subtracted from 9 times a number = 9\(x\) – 20

    The result is 10 more than fourth of the number = 4\(x\) + 10

    Therefore, 9\(x\) – 20 = 4\(x\) + 10

    ⇒9\(x\) – 4\(x\) = 20 + 10

    ⇒5\(x\) = 30

    ⇒\(x\) = \(\frac{30}{5}\)

    ⇒\(x\) = 6

    Therefore, the unknown number is 6


2. If the sum of two numbers is 12 and their product is 11. Find the difference between the numbers?

Solution:

    Given that sum of two numbers is 12, and product is 11,

    Let a + b = 12 and ab = 11

    As, \((a+b)^2\) – \((a-b)^2\) = \(4ab\)

    By substituting the values,

    ⇒\((12)^2\) – \((a-b)^2\) = 4 x 11

    ⇒144 – \((a-b)^2\) = 44

    ⇒144 – 44 = \((a-b)^2\)

    ⇒\((a-b)^2\) = 100

    ⇒(a – b) = 10

    Now consider,

    a + b = 12 ———-(i)

    a – b = 10 ———-(ii)

    By solving equations (i) and (ii)

    the two numbers are a = 11 and b = 1

    Therefore, difference between the numbers = 11 – 1 = 10


3. If the numbers are added in pairs, the sum equals 12, 17, and 25. Find the numbers?

Solution:

    Let the numbers be a, b, and c. Then

    Given that

    a + b = 12 ———(i)

    b + c = 17 ———(ii)

    c + a = 25 ———(iii)

    By adding equations (i), (ii), and (iii)

    a + b + b + c + c + a = 12 + 17 + 25

    ⇒2(a + b + c) = 54

    ⇒(a + b + c) = \(\frac{54}{2}\)

    ⇒(a + b + c) = 27 ———-(iv)

    Now substitute (i), (ii), and (iii) in (iv)

    c = 27 – 12 = 15

    a = 27 – 17 = 10

    b = 27 – 25 = 2

    Therefore, the unknown values are 10, 2 and 15 respectively


4. The average of four consecutive even numbers is 36. Find the smallest and largest number?

Solution:

    Let the four consecutive four numbers be

    \(x\), \(x\) + 2, \(x\) + 4, \(x\) + 6

    Given that \(\frac{(x + x + 2 + x + 4 + x + 6)}{4}\) = 36

    ⇒4\(x\) + 12 = 36 x 4

    ⇒4\(x\) + 12 = 144

    ⇒4\(x\) = 144 – 12

    ⇒4\(x\) = 132

    ⇒\(x\) = \(\frac{132}{4}\)

    ⇒\(x\) = 33

    Therefore smallest number is \(x\) = 33

    Largest number is \(x\) + 6 = 33 + 6 = 39


5. The sum of squares of three consecutive odd numbers is 2531. Find the numbers?

Solution:

    Let the three consecutive three odd numbers be

    \(x\), \(x\) + 2, \(x\) + 4

    Given that \(x^2 + (x + 2)^2 + (x + 4)^2\) = 2531

    \(x^2\) + \(x^2\) + 4 \(x\) + 4 + \(x^2\) + 16 + 8\(x\) = 2531

    ⇒3\(x^2\) + 20 + 12\(x\) = 2531

    ⇒3\(x^2\) + 12\(x\) – 2531 + 20 = 0

    ⇒3\(x^2\) + 12\(x\) – 2511 = 0

    Divide the whole equation by 3

    ⇒\(x^2\) + 4\(x\) – 837 = 0

    By factorising the equation,

    ⇒\(x^2\) + 31\(x\) + 27\(x\) – 837 = 0

    ⇒\(x(x + 31)\) – \(27(x + 31)\) = 0

    ⇒\((x – 27)\)\((x + 31)\)

    ⇒\(x\) = 27 or \(x\) =-31

    Therefore \(x\) = 27 (as negative signs cannot be considered, 31 is eliminated)

    Hence, the unknown numbers are 27, 29, and 31.