Problems on Numbers Questions and Solutions

  • 15. 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. 15
    2. 20
    3. 25
    4. 35
    Answer :

    Option B

    Explanation:

    Let the numbers be a, b and c.
    Then,
    \begin{aligned}
    a^2 + b^2 + c^2 = 138
    \end{aligned}
    and (ab + bc + ca) = 131
    \begin{aligned}
    (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
    \end{aligned}
    = 138 + 2 x 131 = 400

    \begin{aligned}
    => (a + b + c) = \sqrt{400} = 20.
    \end{aligned}

  • 16. The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is

    1. 20
    2. 23
    3. 27
    4. 150
    Answer :

    Option B

    Explanation:

    We know
    \begin{aligned}
    (x + y)^2 = x^2 + y^2 + 2xy
    \end{aligned}

    \begin{aligned}
    => (x + y)^2 = 289 + 2(120)
    \end{aligned}

    \begin{aligned}
    => (x + y) = \sqrt{529} = 23
    \end{aligned}

  • 17. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number

    1. 8
    2. 10
    3. 12
    4. 14
    Answer :

    Option A

    Explanation:

    => 3(2x+9) = 75
    => 2x+9 = 25
    => x = 8

  • 18. Difference between a two-digit number and the number obtained by interchanging the two digits is 36, what is the difference between two numbers

    1. 2
    2. 4
    3. 8
    4. 12
    Answer :

    Option B

    Explanation:

    Let the ten digit be x, unit digit is y.
    Then (10x + y) - (10y + x) = 36
    => 9x - 9y = 36
    => x - y = 4.

  • 19. if the sum of \begin{aligned} \frac{1}{2} \end{aligned} and \begin{aligned} \frac{1}{5} \end{aligned} of a number exceeds \begin{aligned} \frac{1}{3} \end{aligned} of the number by \begin{aligned} 7\frac {1}{3} \end{aligned}, then number is

    1. 15
    2. 20
    3. 25
    4. 30
    Answer :

    Option B

    Explanation:

    Seems a bit complicated, isnt'it, but trust me if we think on this question with a cool mind then it is quite simple...
    Let the number is x,
    then, \begin{aligned} (\frac{1}{2}x + \frac{1}{5}x) - \frac{1}{3}x = \frac{22}{3} \end{aligned}

    \begin{aligned}
    => \frac{11x}{30} = \frac{22}{3}
    \end{aligned}

    \begin{aligned}
    => x = 20
    \end{aligned}

Please Like Us