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# Problems on Numbers

1.

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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2.

A number exceeds by 25 from its $$\dfrac{3}{8}$$ part. Then the number is:

$$x - \dfrac{3}{8}x = 25\\ \Rightarrow x = 40$$

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3.

If $$\dfrac{1}{3}$$ of $$\dfrac{1}{4}$$ of a number is 15, then $$\dfrac{3}{10}$$ of that number is:

Let the number be $$x$$
Then, $$\dfrac{1}{3}$$ of $$\dfrac{1}{4}$$ of $$x = 15 \Leftrightarrow x =15 \times 12 = 180$$
So, required number
$$\left( \dfrac{3}{10} \times 180 \right) =54$$

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4.

A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

Let the middle digit be $$x.$$
Then, $$2x = 10$$ or $$x = 5$$. So, the number is either $$253 \text{ or } 352.$$
Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.
Hence, required number $$= 253$$

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5.

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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6.

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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7.

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

Let the ten's digit be $$x$$ and unit's digit be $$y$$
Then,
$$(10x + y) - (10y + x) = 36\\ \Rightarrow 9(x - y) = 36\\ \Rightarrow x - y = 4$$

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8.

Sum of a rational number and its reciprocal is $$\dfrac {13}{6}$$. Find the number.

Let,
$$x + \dfrac{1}{x} = \dfrac{13}{6}\\ \Rightarrow \dfrac{x^2+1}{x} = \dfrac{13}{6}\\ \Rightarrow 6x^2-13x+6 = 0\\ \Rightarrow 6x^2-9x-4x+6 = 0\\ \Rightarrow (3x-2)(2x-3)\\ \Rightarrow x = \dfrac{2}{3} \text{ or } \dfrac{3}{2}$$

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9.

The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

Let the numbers be $$a$$, $$b$$ and $$c$$.
Then,
$$a^2 + b^2 + c^2 = 138\\ \text{and}\\ ab+bc+ca = 138\\ (a + b + c)^2\\ = a^2 + b^2 + c^2 + 2(ab + bc + ca)\\ =138 + 2 \times 131\\ = 400\\ \therefore (a + b + c) = \sqrt{400} = 20$$

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10.

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

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

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