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# Algebra | Set -1

• Difficulty Level : Easy
• Last Updated : 02 Jun, 2021

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