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# Problems on Numbers

1.

The sum of the numbers is 264. If the first number be twice the second and third number be one-third of the first, then the second number is:

Let the second number be $$x$$. Then, first number $$= 2x$$ and third number $$= \dfrac{2x}{3}$$
$$2x + x + \dfrac{2x}{3} = 264\\ \Rightarrow \dfrac{11x}{3} = 264\\ \Rightarrow x = 72$$

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2.

The difference between a two-digit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1:2?

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let the ten's and unit's digits be $$2x$$ and $$x$$ respectively.
Then, $$(10 \times 2x + x) - (10x + 2x) = 36$$
$$9x = 36\\ x = 4$$
Required difference
$$= (2x + x) - (2x - x)\\ = 2x\\ = 8$$

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3.

A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:

Let the ten's digit be $$x$$ and unit's digit be $$y$$
Then, number $$10x + y$$
Number obtained by interchanging the digits $$= 10y + x$$
$$(10x + y) + (10y + x) = 11(x + y)$$
which is divisible by 11

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4.

Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?

$$x + y = 80\\ x -4y = 5\\ x = 65\\ y = 15$$

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5.

Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

If the number is $$x$$
$$x + 17 = \dfrac{60}{x}\\ x^2 + 17x - 60 = 0\\ (x + 20)(x - 3) = 0\\ x = 3, -20\\ \text{so } x = 3 (\because 3 \text{ is positive})$$