From the given diagram it is easy to see that a = 1; b = 1 and c = 3
(Note: a, b, c are integers)
\ b  a = 1 (1) = 1 + 1 = 2 and b  c = 1  3 = 2 and 2 > 2.
Hence A.
OR
One can even conclude without calculation that b > a
(since b is to the right of a), i.e. b  c < 0 i.e. negative.
\ b  1 > b  c
Hence A .
( x + y)^{2} = x^{2} + 2xy + y^{2}
Hence x^{2} + 2xy + y^{2} > x^{2} + y^{2}
Since the extra term 2xy on the left hand side of the inequality in positive.
( \ x > 0 and y > 0).
Hence A.
Even if the student does not know much about polygons he can consider two very common regular polygons namely equilateral triangle whose exterior angle is 120^{0} and a square whose exterior angle is 90^{0} and arrive at the conclusion that as the number of sides increases (from 3 to 4) the exterior angle decreases from 120^{0} to 90^{0}). Hence the exterior angle of a regular pentagon (five sides) will be more than an exterior angle of a regular hexagon ( 6 sides).
Hence A.
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Index
Test 1
Section 1 : Verbal Section
Section 2 : Quantitative Section
Section 3 : Analytical Section
Section 4 : Quantitative Section
Section 5 : Verbal Section
Section 6 : Analytical Section
Section 7 : Verbal Section
Answer Key To Test 1
Answer Explanation To Test 1
Section 1 : Verbal Section
Section 2 : Quantitative Section
Section 3 : Analytical Section
Section 4 : Quantitative Section
Section 5 : Verbal Section
Section 6 : Analytical Section
Section 7 : Verbal Section
Test
2
