Algebra questions basically involve modeling word problems into equations and then solving them. Some of the very basic formula 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}) - (a
^{2}- b^{2}) = (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}) - (a
^{3}+ b^{3}+ c^{3}- 3 a b c) = (a + b + c) (a^{2}+ b^{2}+ c^{2}- a b - b c - c a) - If a + b + c = 0, then a
^{3}+ b^{3}+ c^{3}= 3 a b c - For a quadratic equation ax
^{2}+ bx + c = 0,

**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 (n^{2} + 1) = 41 n

=> 20 n^{2} – 41 n + 20 = 0

=> 20 n^{2} – 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 and its reverse is 88 and the difference of the digits is 4, with the tens place being larger than the units place. Find the number.

**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.

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