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Problems on Quant and Rate Change : CAT Quantitative Aptitude

Oct 9, 2018 12:22 IST

    The topic of ‘Rate of Change’ requires a basic understanding of initial calculus and it has been seen that CAT does throw a problem once in a while on this topic Let’s see the solved examples to understand the basic nature of problems falling under this topic.


    The radius of a circle increases at a rate of 0.4 cm/s. Then, the rate of increase of the area when the radius is 2 cm is:



    Let the area be A cm2 and the radius r cm.

    That is the area is increasing  times as fast as the radius.


    So the area is increasing   times as fast as the radius. But the radius is increasing at 0.4 cm/s.

    therefore The area is increasing at 


    Water is flowing of a conical funnel having a semi-vertical angle of 45º. When the depth of water is 4 cm, the water level is falling by 0.2 cm/s. The volume rate of flow of the water will equal:

    Let the volume of water be V cm3 when the depth of water is h cm and the radius r cm.

    (h always equals r because the semi-vertical angle is 45º).

    Thus the volume of water is decreasing  times faster than the radius.


    Thus V is decreasing 16∏times faster than the height when h = 4 cm. But the height is decreasing by 0.2 cm/s.

    therefore The flow rate of water

    Therefore, the correct choice is option [3].

    Two lines represented by the equations: 3x + 2y = 7  and 12y = 17 - 18 x  are drawn on a Cartesian plane. These lines, then, will:

    [1] have a point in common

    [2] be perpendicular to each other

    [3] be parallel to each other

    [4] be collinear


    Thus the lines are parallel because the gradient of each line is the same. Therefore, the correct choice is option [3].


    If a line is such that it passes through the points A(2, 5) and B(3, 10), then the slope and y-intercept of this line will, respectively, be:

    [1] -5 and 5

    [2]  5 and -1

    [3]  -5 and -10

    [4]  5 and -5


    The point (2, 5) lies on the line     y= mx=c

    therefore   5 = 2m + c          …(i)

    The point (3, 10) lies on the line     y= mx=c       

    therefore 10 = 3m+c                …(ii)

    Subtracting (i) from (ii), m = 5

    Substituting m = 5 in (i), 5 = 2x5x+c

    c= -5

    Therefore The equation of the line is y = 5x-5

    Therefore, the correct option is [4].

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