  In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.

#### 1. A number is as much greater than 36 as is less than 86. Find the number?

```
Let the number be x.
Then, x - 36 = 86 - x
=> 2x = 86 + 36 = 122
=> x = 61.
Hence, the required number is 61.
```

#### 2. Find a Number Such that when 15 is subtracted from 7 times the number, the Result is 10 more than twice the number?

```Let the number be x. Then,
7x - 15 = 2x + 10
=> 5x = 25 =>x = 5.
Hence, the required number is 5.
```

#### 3. The sum of a rational number and its reciprocal is 13/6. Find the number?

```Let the number be x.
Then, x + (1/x) = 13/6
=> (x2 + 1)/x = 13/6
=> 6x2 – 13x + 6 = 0
=> 6x2 – 9x – 4x + 6 = 0
=> (3x – 2) (2x – 3) = 0
=>  x = 2/3 or x = 3/2
Hence the required number is 2/3 or 3/2.
```

#### 4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh of the other by 8, find the smaller number?

```
Let the numbers be x and (184 - x).
Then, (X/3) - ((184 – x)/7) = 8
=> 7x – 3(184 – x) = 168
=> 10x = 720 => x = 72.
So, the numbers are 72 and 112.
Hence, smaller number = 72.
```

#### 5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers?

```Let the number be x and y.
Then,
x – y = 11      ----(i)
and
1/5 (x + y) = 9
=> x + y = 45   ----(ii)
Adding (i) and (ii), we get: 2x = 56 or x = 28.
Putting x = 28 in (i), we get: y = 17.
Hence, the numbers are 28 and 17.
```

#### 6. If the sum of two numbers is 42 and their product is 437, then find the absolute difference between the numbers?

```Let the numbers be x and y. Then,
x + y = 42 and xy = 437
x - y = sqrt[(x + y)2- 4xy]
= sqrt[(42)2 - 4 x 437 ]
= sqrt[1764 – 1748]
= sqrt
= 4.
Required difference = 4.
```

#### 7. The sum of two numbers is 16 and the sum of their squares is 113. Find the numbers?

```Let the numbers be x and (15 - x).
Then, x2 + (15 - x)2= 113
=>  x2 + 225 + X2 - 30x = 113
=>  2x2 - 30x + 112 = 0
=>  x2 - 15x + 56 = 0
=>  x - 7) (x - 8) = 0
=>  x = 7  or  x = 8.

So, the numbers are 7 and 8.
```

#### 8. The average of four consecutive even numbers is 27. Find the largest of these numbers?

```Let the four consecutive even numbers be
x, x + 2, x + 4 and x + 6.
Then, sum of these numbers = (27 x 4) = 108.
So, x + (x + 2) + (x + 4) + (x + 6) = 108
=> 4x = 96
=> x = 24.
Largest number = (x + 6) = 30.
```

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

```Let the numbers be x, x + 2 and x + 4.
Then, X2+ (x + 2)2 + (x + 4)2 = 2531 => 3x2+ 12x - 2511 = 0
=>  X2 + 4x - 837 = 0
=> (x - 27) (x + 31) = 0
=>  x = 27.
Hence, the required numbers are 27, 29 and 31.
```

#### 10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers?

```Let the numbers be x and y,
such that x > y Then, 3x - 4y = 5 ...(i)
and  x + y) - 6 (x - y) = 6 =>  -5x + 7y = 6     …(ii)
Solving (i) and (ii), we get: x = 59 and  y = 43.
Hence, the required numbers are 59 and 43.
```

#### 11. The ratio between a two-digit number and the sum of the digits of that number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?

```Let the ten's digit be x.
Then, unit's digit = (x + 3).
Sum of the digits = x + (x + 3) = 2x + 3.
Number = l0x + (x + 3) = llx + 3.
Then,  11x+3 / 2x + 3  = 4 / 1
=> 1lx + 3 = 4 (2x + 3)
=>  3x = 9
=>  x = 3.
Hence, required number = 11x + 3 = 36.
```

#### 12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted from the number, its digits are interchanged. Find the number?

``` Let the ten's digit be x.
Then, unit's digit = (9 - x).
Number = l0x + (9 - x) = 9x + 9.
Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x.
therefore,
(9x + 9) - 63 = 90 - 9x
=> 18x = 144
=> x = 8.
So, ten's digit = 8 and unit's digit = 1.
Hence, the required number is 81.
```

#### 13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator. And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.?

```Let the required fraction be x/y. Then,
x+1 / y+1 = 2 / 3  =>  3x – 2y = - 1   …(i)
and
x – 1 / y – 1 = 1 / 2 =>  2x – y = 1 …(ii)
Solving (i) and (ii), we get : x = 3 , y = 5
therefore, Required fraction = 3 / 5.
```

#### 14. 50 is divided into two parts such that the sum of their reciprocals is 1/ 12.Find the two parts?

```Let the two parts be x and  (50 - x).
Then, 1 / x + 1 / (50 – x) = 1 / 12
=> (50 – x + x) / x ( 50 – x) = 1 / 12
=> x2 – 50x + 600 = 0
=> (x – 30) ( x – 20) = 0
=> x = 30 or x = 20.
So, the parts are 30 and 20.
```

#### 15. If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the numbers?

```Let the numbers be x, y and z. Then,
x+ y = 10      ...(i)
y + z = 19     ...(ii)
x + z = 21     ...(iii)
Adding (i) ,(ii) and (iii),
we get:  2 (x + y + z ) = 50
=>  (x + y + z) = 25.
Thus, x= (25 - 19) = 6;
y = (25 - 21) = 4;
z = (25 - 10) = 15.

Hence, the required numbers are 6, 4 and 15.
```