Ð B = 2 ÐC ( \ Ð C = Ð B )
and Ð B = 2 Ð A
Hence Ð A = Ð C = x (say )
then ÐB = 2 x
and 2x + x + x = 180 gives x = 45^{0}
i.e. hence Ð A = Ð C = 45^{0} and ÐB = 2 x = 90^{0}
\ side opposite Ð B is the largest and sides opposite Ð A and ÐC are equal.
Hence AB = BC
\ AB^{2} = BC^{2}
\ AB^{2} < BC^{2} + AC^{2}
Hence B.
Two important facts namely
(a) 0 < x < 1 implies x to be a positive proper fraction and
(b) x  : =
Since x < 1. \ x^{2} < x. \ x2 < 1 ( \ x > 1 ) and hence > 1.
(Since the reciprocal of proper positive fraction is more than 1).
\ > x2
x ^{2} > x2
Hence A.
Given 0 < a < b < c
The sum a + b can be small. i.e. 1 + .1 or 1.1 or it can be very large,
i.e. 1 + 1000 i.e. 1001.
Similarly c  a can be very small. 1.2  1 i.e. .2 or can be very large,
i.e. 100001  1 i.e. 100000.
Hence D.
OR Mathematically
a + b ³ c  a or a + b < c  a
If a + a ³ c  b or a + a < c  b
If 2a ³ or a <
If c and b are very close then c  b or can be very small but if c is very
large compared to b, c  b or can be very large
a can be more than or less than or even equal to .
Hence D.
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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
