## CAT - Quantitative Aptitude

Nishant C
3 weeks ago
**Quantitative Aptitude** |
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### One side of a staircase is to be closed in by rectangular planks from the floor to each step. The width of each plank is 9 inches and their height are successively 6 inches, 12 inches, 18 inches and so on. There are 24 planks required in total. Find the area in square feet. (more)

One side of a staircase is to be closed in by rectangular planks from the floor to each step. The

width of each plank is 9 inches and their height are successively 6 inches, 12 inches, 18 inches

and so on. There are 24 planks required in total. Find the area in square feet.

sammy babar
7 months ago
**Quantitative Aptitude** |
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### i am not getting nxt question set after solving one set can anyone help me ???? (more)

next question paper is not getting

Sammy, If you are registered for CAT, that topic may only have one set. Please try with other topics. Can you let me know which topic did you attempt?

Nikhil Sharma
7 months ago
**Quantitative Aptitude** |
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### Questions on http://www.takerz.in/catstore.htm (more)

http://www.takerz.in/images/travel/Screenshot (67).png Click on the link and solve my doubts. Find more at http://www.takerz.in/catstore.htm

Hi Nikhil, Please do not use external links.. Add images directly.

Nikhil Sharma
8 months ago
**Quantitative Aptitude** |

### Quadratic function f(x) - see details (more)

I am having doubt in this question Please solve the question by clicking on the link

f(x) and g(x) are both continuous functions (polynomials)

f(x) for x = 2, will be = 0

-2^{2} + 2a + b = 0

2a + b = 4 ... (1)

also f(x) has only 1 root, hence discriminant = 0

a^{2} - 4*(-1)(b) = 0

a^{2} + 4b = 0

b =- a^{2} / 4

replacing value of b in (1)

2a - a^{2} / 4 = 0

8a - a^{2} = 0

a = 0 or a = 8

a = 0, b = 0

the function will be x^{2} = 0, a constant function (But question says that the function is quadratic)

hence a = 8

b = -16

f(x) = -x^{2} + 8x - 16 .... (2)

g(x) > 0 except for 2 <= x <= 3

thus g(x) has roots at x = 2 and x = 3

g(2) = 0

2^{2} + 2c + d = 0

2c + d = -4 ... (3)

g(3) = 0

3c + d = -9 ... (4)

from 3 and 4

c = -5

d = 6

g(x) = x^{2} - 5x + 6 ... (5)

since h(x) is max [g(x), h(x)]

for h(x) we need to find the intersection of f(x) and g(x)

from 2 and 5

x^{2} - 5x + 6 = -x^{2} + 8x - 16

2x^{2} - 13x + 22 = 0

The solution is imaginary

the functions never intersect

g(x) is always > f(x)

h(x) = g(x)

value of x at which h(x) is minimum

g'(x) = 0

2x - 5 = 0

x = 5/2