QUANTITATIVE APTITUDE

22. (D)
23. (C)
24. (D) x – 3y ≥ 0 x + y ≥ – 2 3x – y ≤ – 2
4x ≥ – 4 (from equations 2 and 3)
x ≥ – 2
x ≤ 2
25. (A) Draw DX. As can be seen easily,
2
AX = DX (Isosceles∆ ). 45°
∴ AX = 2
45°
AB = 2 + 3 = 5 cm
26. (D) From Lagrange’s mean value theorem,
there is c in (a, b), such that:
fb ( )– fa ( )
= f’(c)
b– a
Here, f(a) = f(3) = 6
f(b) = f(9) = 2
f’(c) or f’(a) = −
− =
2 6 2
–
9 3 3
27. (D) v = 1 2
πr h for a cone
3
or h = 3 ]
v
2 ....... equation 1
πr
given
Amount of canvas = curved area = S
OBJECTIVE-TYPE QUESTIONS
= π rl = π r 2 2
( r + h )
S2 = 2 2
π r (r2 + h2 ) = 2 2 2 9v
π r ( r + )
2 4
π r
Let S2 = z
dz
= π2 ( – )
dr 4
2
3 18v
r
2 3
π r
and dz
2
2
2 54v
12r
+
2 4
dr
π r
Put dz
= 0
dr
4r3 2
18v – 2 3 = 0
π r
2r6 9v 2
= 2
π
9v 2 = 2 2
π r 6
∴ dz
dr
2 = π2 ( )
A X 3
B
2
=π
2 2 (12r2 2 6
12π r
+ )
2 4 =π
π r
2 (24r2 )
= positive quantity
z (i.e. S2 2 6
) has minimum value if 9v = 2π r
D
2 2
2
1
2
3
C
πr h
i.e. 9( )
2
2
= 2
3
2
π r6 i.e. h2 = 2r2 i.e. h
= 2
r
* Such answers must be tabulated and learnt for ready
reference.
28. (C) G = n, H = n n n
Sn 29. (C) x = 2
a± a – b = 2
a+ a – b and 2
a−a – b
and y = 2
c± c – d = 2
c+ c – d and 2
c−c – d
2
a+ a −b
=
a– a
2
– b
c c d +
2
−
2
c– c – d
By reversing componendo and dividendo,
a c
=
2 2
a − b c −d
Squaring, a
a b
2
2 =
– c
2
2 , i.e. a
c – d
2d = bc2 30. (D)
31. (D) Given B< A≤ C,
E< B≤ D < G≤F B
B< F≤A B
H < B
From this information,
H< B< F ≤ A≤ C ..... (1)
( < QZ , ) ( A< P)
and E < B ≤ D < G ≤ F ..... (2)
A is false from (2), B from (1), C from (1 and 2)
D “may be” true
32. (D) From 1 and 2
33. (D) From 1 and 2
34. (C) B is true, A and D are false, C “may be” true
35. (D) In any case, since x, y, z >1,
1 1 1
, , < 1 (i.e. negative)
x y z
(–) × (–) × (–) = – (negative quantity)
36. (B) The ends of diameter AC are:
A(0, 0) and C(2, 2)
Equation of the circle with
ends of diameter as (x , y ) 1 1
and (x , y ) is:
2 2
(x – x 1 ) (x – x 2 ) + (y – y 1 ) (y – y 2 ) = 0
(x – 0) (x – 2) + (y – 0) (y – 2) = 0
x 2 – 2x + y 2 – 2y = 0
x 2 + y 2 = 2(x + y)
D(0,2) C(2,2)
A(0,0) B(2,0)