Problems on Numbers

1.If one-third of one-fourth of a number is 15, then three-tenth of that number is:

1. 35
2. 36
3. 45
4. 54

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the number be x \\ Then, \frac{1}{3} of \frac{1}{4} of x = 15 \Leftrightarrow x = 15 \times 15 = 180 \\ So, required number \\ = (\frac{3}{10} \times 180) = 54 \end{aligned}

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

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2.Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

1. 9
2. 11
3. 13
4. 15

Answer:Option 1

Explanation:

Solution:

Let the three integers be x, x + 2 and x + 4.
Then, 3x = 2(x + 4) + 3 ⇔ x = 11.
∴ Third integer = x + 4 = 15.

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3.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?

1. 3
2. 4
3. 9
4. Cannot be determined

Answer:Option 1

Explanation:

Solution:

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

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4.A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

1. 18
2. 24
3. 42
4. 81

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the ten's and unit digit be x and \frac{8}{x} respectively \\ Then, (10 x + \frac{8}{x}) + 18 = 10 \times \frac{8}{x} + x \\ \Rightarrow 10 x^{2} + 8 + 18 x = 80 + x^{2} \\ \Rightarrow 9 x^{2} + 18 x - 72 = 0 \\ \Rightarrow x^{2} + 2 x - 8 = 0 \\ \Rightarrow (x + 4) (x - 2) = 0 \\ \Rightarrow x = 2 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{ten's}}\,{\text{and}}\,{\text{unit}}\,{\text{digit}}\,{\text{be}}\,x\,{\text{and}}\,\frac{8}{x}\,{\text{respectively}} \cr & {\text{Then}},\left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Rightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Rightarrow 9{x^2} + 18x - 72 = 0 \cr & \Rightarrow {x^2} + 2x - 8 = 0 \cr & \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr}$

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5.The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

1. 69
2. 78
3. 96
4. Cannot be determined

Answer:Option 1

Explanation:

Solution:

Let the ten's digit be x and unit's digit be y.
Then, x + y = 15 and x - y = 3 or y - x = 3.
Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6.
Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9.
So, the number is either 96 or 69.
Hence, the number cannot be determined.

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6.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:

1. 20
2. 30
3. 40
4. None of these

Answer:Option 1

Explanation:

Solution:

Let the numbers be a, b and c.
Then, a² + b² + c² = 138 and (ab + bc + ca) = 131.
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 138 + 2 x 131 = 400.
⇒ (a + b + c) = √400 = 20.

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7.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:

1. 3
2. 5
3. 9
4. 11

Answer:Option 1

Explanation:

Solution:

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

1. 3
2. 10
3. 17
4. 20

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the number be x \\ Then, x + 17 = \frac{60}{x} \\ \Rightarrow x^{2} + 17 x - 60 = 0 \\ \Rightarrow (x + 20) (x - 3) = 0 \\ \Rightarrow x = 3 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then}},\,x + 17 = \frac{{60}}{x} \cr & \Rightarrow {x^2} + 17x - 60 = 0 \cr & \Rightarrow (x + 20)(x - 3) = 0 \cr & \Rightarrow x = 3 \cr}$

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9.The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:

1. 380
2. 395
3. 400
4. 425

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the number be x and y \\ Then, x y = 9375 and \frac{x}{y} = 15 \\ \frac{x y}{(x / y)} = \frac{9375}{15} \\ \Rightarrow y^{2} = 625 \\ \Rightarrow y = 25 \\ \Rightarrow x = 15 y = (15 \times 25) = 375 \\ ∴ Sum of the number \\ = x + y = 375 + 25 = 400 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x\,{\text{and}}\,y \cr & {\text{Then}},\,xy = 9375\,{\text{and}}\,\frac{x}{y} = 15 \cr & \frac{{xy}}{{\left( {x/y} \right)}} = \frac{{9375}}{{15}} \cr & \Rightarrow {y^2} = 625 \cr & \Rightarrow y = 25 \cr & \Rightarrow x = 15y = \left( {15 \times 25} \right) = 375 \cr & \therefore {\text{Sum}}\,{\text{of}}\,{\text{the}}\,{\text{number}} \cr & = x + y = 375 + 25 = 400 \cr}$

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10.The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is: