November 14, 2017
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Mathematics Test20
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Question 1 
There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is...
A  36 
B  34 
C  32 
D  38 
Question 2 
Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90o or concave if the internal angle is 270o. If the number of convex corners in such a polygon is 25, the number of concave corners must be...
A  0 
B  22 
C  21 
D  20 
Question 2 Explanation:
In this kind of polygon, the number of convex angles will always be exactly 4 more than the number of concave angles (why?).
Note : The number of vertices should be even. Hence the number of concave and convex corners should add up to an even number. This is true only for the answer choice 21.
Question 3 
In the figure below, the rectangle at the corner measures 10 cm ? 20 cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?
A  10 cm 
B  50 cm 
C  None of these. 
D  40 cm 
Question 4 
If the product of n positive real numbers is unity, then their sum is necessarily
A  a positive integer

B  a multiple of n 
C  never less than n 
D  equal to n + n 1 
Question 4 Explanation:
The best way to do this is to take some value and verify. E.g. 2, 1/2 and 1. Thus, n = 3 and the sum of the three numbers = 3.5. Thus options 1, 2 and 4 get eliminated.
Question 5 
A  5 
B  3 
C  4 
D  2 
Question 5 Explanation:
Using log a – log b = log a/b, 2 / (y–5) = (y–5)/(y–3.5) where y = 2x Solving we get y = 4 or 8 i.e. x = 2 or 3. It cannot be 2 as log of negative number is not defined (see the second expression).
Question 6 
Given that 1? v? 1, 2? u? 0.5 and 2? z? 0.5 and w = vz/u, then which of the following is necessarily true?
A  2? w? 0.5 
B  4? w? 4 
C  4? w? 2 
D  –0.5? w? 2 
Question 6 Explanation:
u is always negative. Hence, for us to have a minimum value of vz/u, vz should be positive. Also for the least value, the numerator has to be the maximum positive value and the denominator has to be the smallest negative value. In other words, vz has to be 2 and u has to be –0.5. Hence the minimum value of vz/u = 2/–0.5 = –4. For us to get the maximum value, vz has to be the smallest negative value and u has to be the highest negative value. Thus, vz has to be –2 and u has to be –0.5. Hence the maximum value of vz/u = –2/–0.5 = 4.
Question 7 
How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and unit’s place respectively, exist such that x < y, z < y and x ? 0?
A  240 
B  285 
C  320 
D  245 
Question 7 Explanation:
If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values. Thus with y = 2, a total of 1 X 2 = 2 numbers can be formed. With y = 3, 2 X 3 = 6 numbers can be formed. Similarly checking for all values of y from 2 to 9 and adding up we get the answer as 240.
Question 8 
In the figure given below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ? ACD = y degrees and ? AOD = x degrees such that x = ky, then the value of k is ...
A  2 
B  3 
C  None of these 
D  1 
Question 8 Explanation:
If y = 10o,
? BOC= 10o (opposite equal sides)
? OBA= 20o (external angle of BOC ? )
? OAB= 20 (opposite equal sides)
?AOD = 30o (external angle of AOC ? )
Thus k = 3
Question 9 
In a triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, then AP : QC is equal to ...
A  3 : 2 
B  4 : 1 
C  3 : 8 
D  1 : 1 
Question 10 
A vertical tower OP stands at the center O of a square ABCD. Let h and b denote the length OP and AB respectively. Suppose ? APB = 60o then the relationship between h and b can be expressed as ...
A  2h2 = b2 
B  3b2 = 2h2 
C  2b2 = h2 
D  3h2 = 2b2 
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