**Problem 1 : **

The sum of 5 times a number and 8 is 48. Find the number.

**Solution : **

Let the number be ‘x’

Then,

5x + 8 = 48

Subtract 8 from both sides.

(5x + 8) - 8 = 48 - 8

5x = 40

Divide both sides by 5.

5x/5 = 40/5

x = 8

So, the number is 8.

**Problem 2 :**

A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297. Find the number.

**Solution :**

Let "x0y" be the required three digit number. (As per the given information, middle digit is zero)

"The sum of the other digits is 9" ----> x + y = 9 -----(1)

"Interchanging the first and third digits" --------> y0x

From the information given in the question, we can have

y0x - x0y = 297

(100y + x) - (100x + y) = 297

100y + x - 100x -y = 297

-99x + 99y = 297

-x + y = 3 --------(2)

Solving (1) & (2), we get x = 3 and y = 6

So,

x0y = 306

So, the required number is 306.

**Problem 3 :**

Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers.

**Solution :**

Let "x" and "y" be the required two numbers such that

x > y.

From the point "1/5th of a the greater equal to 1/3rd of the smaller", we have

(1/5)x = (1/3)y

3x = 5y

3x - 5 y = 0 --------(1)

From the point "their sum is 16", we have

x + y = 16 ----------(1)

Solving (1) and (2), we get x = 10 and y = 6.

So, the two numbers are 10 and 6.

**Problem 4 :**

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number.

**Solution :**

Let "xy" be the required number between 10 and 100. (Two digit number)

"A number between 10 and 100 is five times the sum of its digits"

From the information above, we have

xy = 5(x+y)

10x + y = 5x + 5y

5x - 4y = 0 -------(1)

"If 9 be added to it the digits are reversed"

From the information above, we have

xy + 9 = yx

10x + y + 9 = 10y + x

9x - 9y = -9

x - y = -1 ---------(2)

Solving (1) and (2), we get x = 4 and y = 5.

So, the required number is 45.

**Problem 5 :**

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

**Solution :**

Let "x" and "y" be the two numbers such that x > y

Given : One number is greater than thrice the other number by 2

So, we have x = 3y + 2 ---------(1)

Given : 4 times the smaller number exceeds the greater by 5

So, we have 4y = x + 5 ---------(2)

Plugging (1) in (2), we get

4y = 3y + 2 + 5

4y = 3y + 7

4y = 3y + 7

y = 7

Plugging y = 7 in (1), we get x = 3(7) + 2

Therefore x = 23

So, the two numbers are 23 and 7.

**Problem 6 :**

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

**Solution :**

Let "xy" be the required two digit number.

Given : Two digit number is 7 times the sum of its digits.

So, we have

xy = 7(x + y)

10x + y = 7x + 7y

10x + y = 7x + 7y

3x - 6y = 0

x - 2y = 0 --------(1)

Given : The number formed by reversing the digits is 18 less than the given number

So, we have

xy - yx = 18

(10x + y) - (10y + x) = 18

10x + y - 10y - x = 18

9x - 9y = 18

x - y = 2 --------(2)

Solving (1) and (2), we get x = 4 and y = 2

xy = 42

So, the required number is 42.

**Problem 7 : **

If a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.

**Solution : **

Let the number be ‘x’

Then x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’

In the above sentence we have 296 is multiplied by the constant "k", 75 is added to that. In this form , we consider the number 75 as remainder when the number x is divided by 296.

We want to find the remainder when we divide the number "x" by 37. To do this, we need to have 37 at the place where we have 296 in the above equation.

So we can write 296 as 37 times 8 and 75 as 37 times 2 plus 1. It has shown below.

x = 37 × 8k + 37 × 2 + 1

x = 37(8k + 2) + 1

So, the remainder is ‘1’ when the number ‘x’ is divided by 37.

**Problem 8 : **

Find the least number which when divided by 35, leaves a remainder 25, when divided by 45, leaves a remainder 35 and when divided by 55, leaves a remainder 45.

**Solution : **

For each divisor and corresponding remainder, we have to find the difference.

35 - 25 = 10

45 - 35 = 10

55 - 45 = 10

we get the difference 10 (for all divisors and corresponding remainders)

Now we have to find the L.C.M of (35,45,55) and subtract the difference from the L.C.M.

L.C.M of (35, 45, 55) = 3465

So, the required least number is

= 3465 - 10

= 3455

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