Algebra | Set -1

Algebra questions basically involve modeling word problems into equations and then solving them. Some of the very basic formulae that come handy while solving algebra questions are :

• (a + b) ² = a ² + b ² + 2 a b
• (a – b) ² = a ² + b ² – 2 a b
• (a + b) ² – (a – b) ² = 4 a b
• (a + b) ² + (a – b) ² = 2 (a ² + b ² )
• (a² – b² ) = (a + b) (a – b)
• (a + b + c) ² = a ² + b ² + c ² + 2 (a b + b c + c a)
• (a ³ + b ³ ) = (a + b) (a ² – a b + b ² )
• (a ³ – b ³ ) = (a – b) (a ² + a b + b ² )
• (a³ + b³ + c³ – 3 a b c) = (a + b + c) (a² + b² + c² – a b – b c – c a)
• If a + b + c = 0, then a³ + b³ + c³ = 3 a b c
• For a quadratic equation ax² + bx + c = 0,

Sample Problems

Question 1 : A number is as much greater than 46 as is less than 78. Find the number. Solution : In these type of questions, we simply add the two given numbers and divide it by 2 so as to obtain the required number. So, required number = (46 + 78) / 2 = 124 / 2 = 62   Long Method : Let the required number be ‘n’. => n – 46 = 78 – n => 2 n = 46 + 78 => 2 n = 124 => n = 62 Thus, the required number is 62.   Question 2 : Find a number such that when 55 is subtracted from 4 times the number, the result is 5 more than twice the number. Solution : Let the required number be ‘n’. => 4 n – 55 = 2 n + 5 => 2 n = 60 => n = 30 Thus, 30 is the required number.   Question 3 : The sum of a number and its reciprocal is 41 / 20. Find the number. Solution : Let the number be ‘n’. => n + (1/n) = 41 / 20 => 20 (n² + 1) = 41 n => 20 n² – 41 n + 20 = 0 => 20 n² – 16 n – 25 n + 20 = 0 => (5 n – 4) (4 n – 5) = 0 => n = 4/5 or 5/4 Thus, the required number is 4/5 or 5/4   Question 4 : The sum of two numbers is 132. If one-third of the smaller exceeds one-sixth of the larger by 8, find the numbers. Solution : Let the two numbers be ‘x’ an ‘y’ such that x > y. => x + y = 132 and (y/3) = (x/6) + 8 => x + y = 132 and 2 y – x = 48 => x = 72 and y = 60   Question 5 : The sum of two numbers is 24 and their product is 128. Find the absolute difference of numbers. Solution : Let the numbers be ‘x’ and ‘y’. => x + y = 24 and x y = 128 Here, we need to apply the formula (x + y) ² – (x – y) ² = 4xy => (24)² – (x – y) ² = 4 x (128) => (x – y) ² = (24)² – 4 x (128) => (x – y) ² = 576 – 512 => (x – y) ² = 64 => |x – y| = 8 Therefore, absolute difference of the two numbers = 8   Question 6 : The sum of a two digit number ‘n’ and the number obtained by interchanging digits of n is 88. The difference of the digits of ‘n’ is 4, with the tens place being larger than the units place. Find the number ‘n’. Solution : Let the number be ‘xy’, where x and y are single digits. => The number is 10x + y => Reciprocal of the number = yx = 10y + x => Sum = 11 x + 11 y = 11 (x + y) = 88 (given) => x + y = 8 Also, we are given that the difference of the digits is 4 and x > y. => x – y = 4 Therefore, x = 6 and y = 2 Thus, the number is 62.

Problems on Algebra | Set-2

Program on Algebra

• Find number of solutions of a linear equation of n variables
• Program to find number of solutions in Quadratic Equation
• Program to find the Roots of Quadratic equation
• Find the missing value from the given equation a + b = c
• Maximum and Minimum Values of an Algebraic Expression

This article has been contributed by Nishant Arora   Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.