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New GRE Numeric Entry: Practice Question of the Week #11

This is one of the new question types for the quantitative section, where you type in your answer into a box on the screen. No choices to plug in or guess from makes this problem type a little tougher than multiple choice. We’ll post the answer and explanation for this question tomorrow, along with a link to try the question on the Magoosh website so you can familiarize yourself with the new format. Good luck!

The figure shows the graph of the equation y=k-x^2, where k is a constant. If the area of triangle ABC is 1/8, what is the value of k?


Update: Here’s the answer/explanation post!

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5 Responses to New GRE Numeric Entry: Practice Question of the Week #11

  1. Huzefa August 10, 2011 at 2:20 am #

    I got 1/4, by taking the area of triangle 1/2xbasexheight=1/8. basexheight will be 1/4. Hence, k = 1/4

  2. Ahmed August 9, 2011 at 8:26 pm #

    Co-ordinates of A, B and C are (0,k), (k^0.5,0) and (-k^0.5,0) respectively. So BC= 2√k and OA=K. It yields area of triangle ABC = 0.5* 2√k * k = k^(3/2)

    k^(3/2)= 1/8

    =>(2√k)^3 = 1

    => 2√k=1


    • R B October 2, 2013 at 11:32 pm #


      You are 100% perfect with your answer.

  3. Ashok Kumar August 9, 2011 at 7:43 pm #

    I too get 1/4 and used the same logic as above. But I have one doubt. Should I enter the answer as “1/4” in the text box or can I also enter “0.25” in the text box?

  4. Mohamed August 9, 2011 at 3:37 pm #

    I got 1/4

    y-intercept when x =0, y-intercept = k

    x-intercepts = √k and -√k

    area of tri = 1/2 * 2√k * k = 1/8

    k√k = 1/8 (squaring both sides)

    k^3 = 1/64

    k = 1/4

    to check: 1/4 – (√1/4)^2 = 0

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