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1 km = 1000 m
= 5/8 miles
Geometry
The sections on calculation and mensuration have
their own portion of inquiries in the CAT and other
MBA selection tests
...
Coming up next is a thorough assortment of formulae
in view of two-layered and three-layered figures:
1 mile = 1760 yd
= 5280 ft
1 inch = 2
...
100 kg = 1 quintal
D
...
2 pounds
(approx
...
Coming up next is
an intense point
...
In the Online CAT, the
Quantitative Aptitude segment has comprised of a
normal of 15-20% inquiries from these parts
...
Subsequently, the understudy is encouraged to
guarantee that he/she concentrates on this part
totally and completely
...
The most effective method to Prepare for
Quantitative Aptitude for CAT
Right point: A point whose action is 90 degrees
...
The important expertise expected for doing great in
this part is the capacity to apply the formulae and
hypotheses
...
The understudy is
encouraged to recall the formulae in this section with
the goal that he can tackle every one of the inquiries
in light of this part
...
1 m = 100 cm = 1000
mm
B
...
37
inches
Heartless point: A point whose action is greater than
90 degrees but under 180 degrees
...
Coming up
next is a heartless point
...
strengthening is assuming they are amounting to 180
degrees
...
Coming up next
is a reflex point
...
The following
(∠1 and ∠2) are upward points
...
In the figure beneath, ∠1 and ∠2 are
contiguous points
...
Investigate the accompanying figure:
1
2
5
Integral points: Two points whose actions add to 90
degrees ∠1 and ∠2 are corresponding points on the
grounds that together they structure a right point
...
Up to two points amount to
90 degrees, they would be called correlative
(regardless of whether they are not adjoining one
another)
...
The accompanying points ∠1
and ∠2 are advantageous points
...
The main condition for two points to be called
2
8
4
3
7
6
Points 3,4,5,8 are inside points
...
substitute inside points: Pairs of inside points on
inverse sides of the cross-over
...
Point 4 and point 8 additionally
substitute inside points
...
Consequently, in the figure we have: Angle 3 = Angle
5; Also Angle 4 = Angle 8
...
Points 2 and point 7 are substitutes for outside
points
...
Both the points in a couple of substitutes
outside points are equivalent
...
coinside points: When two lines are cut by a third line
(cross-over) co-inside points are between the sets of
lines on a similar side of the cross-over
...
In the given figure, points 3
and 8 are co-inside points
...
Additionally, ∠4 + ∠5 = 180
...
angle Bisector:
Calculation and Mensuration
relating points: These are sets of points that are in
comparable positions when two equal lines are
crossed by a cross-over
...
Likewise,
the sets of points, 1 and 4; 5 and 7; 6 and 8 are
comparing points
...
Hence, in the figure-∠1 = ∠4; ∠5 = ∠7; ∠2 =
∠3 and ∠6 = ∠8
...
One side should be normal (for example OY) and
these two points should be supplementary
...
One side should be normal (for example OY) and
these two points should be strengthened
...
e
...
(Point bisector is equidistant from the different sides
of the point
...
What is the value of x in the given figure?
Y
R
2
1
X
7x
O
Z
P
Angles on the side of a line: ∠1 + ∠2 + ∠3 = 180°
3x
O
Q
(a) 18o (b) 20o
(c) 28o
(d) None of these
2
1
3
2
...
If 2a + 3, 3a + 2 are complementary, then a = ?
(a) 17o (b) 20o
1
...
1
...
(a) 30o (b) 40o
(c) 50o
(d) 60o
O
C
D
(a) 40o (b) 50o
(c) 60o
1
...
Then, find the value of
x
...
Find the value of q
...
(a) 60o (b) 80o
(c) 90o
1
...
P
L1
x
46°
1
...
o
x°
(d) None of these
o
S
In the given diagram if BC||ED and ∠BAC = 70°,
then find the value of d and c
...
A
(c) 70o
(d) 80o
If AB||CD and AF||BE then the value of x is:
F
70
D
128
B
d
D
B
e
x
E
108°
A
c
b
T
Q
E
C
C
(a) 52o, 58o
(c) 44o, 36o
1
...
(b) 72o
(c) 88o
(d) 82o
In the figure if PQ||SR and ST||QR then x = ?
P
S
T
(a) 45o (b) 75o
x
40°
1
...
B
(c) 90o
(d) 100o
In the given figure, if AB||CD then the value of x
=?
A
x
R
30°
C
Q
y
C
B
D
F
(a) 100o
120°
B
x°
D
1
...
A
70°
B
E
E
(a) 135o
2y °
(b) 145o
(d) None of these
(c) 155o
F
y°
140°
x
C
1
...
T
D
(c) 110o
(d) 120o
If in the given figure, AB||CD and BC||DE, then
x =?
42°
A
P
Q
y
U
B
75°
C
X
68°
R
(a) 188
E
(b) 202
(d) 212o
o
o
(c) 208o
1
...
Then ∠ACB = ?
Q
A
(a) 95o
(c) 115o
C
B
(b) 105o
(d) 125o
Answer Key:
P
1 (a)
S
D
R
2 (b)
5 (a) 6 (b)
9 (b) 10 (a)
13 (b) 14 (d)
3 (a)
7 (b)
11 (c)
15 (c)
4 (b)
8(a)
12(a)
16 (c)
1
...
8
...
∠c = 180o – (70o + 52o) = 58o
= 180o or x = 18o
...
2
...
Draw line MON||AB||CD
b = 54o
A
B
60°
M
3
...
2a + 3 + 3a + 2 =
90o
5a + 5 = 90o
O
60°
N
50°
D
C
∠ABO = ∠BON [Alternate angles]
∠BON = 60o
5x +
17o + x + 13o =
180o
6x +
30o = 180o
4
...
10
...
Option (a) =
30o,
complementar
y angle of 30o
is 60o and 30o
is half of 60o
...
x+10°
x+10°
4x = 160o
x = 40o
=10
2!3!
Option (c) is correct
...
q + r = 180o
– (45o) = 135o
r+
o
125 = 180o ⇒
r = 55o q + 55o
= 135o q =
80o
7
...
5C2 =
12
...
If AB||CD then ∠CEB + ∠ABE = 180o
∠CEB = 108o
o
x = 180 – 82 = 98
o
o
108o + ∠ABE = 180o
o
∠ABE = 72o
19
...
∠QPR = ∠SRP = 40o
[Alternative angles]
x = ∠SRQ = ∠SRP + ∠PRQ [Alternative
angles]
x = 60o + 40o = 100o 15
...
Extend AB to E
C
B
o
x°
75°
C
x
E
D
BC DE , so ∠ABC = ∠AFD = 75o
CD BE, hence x + 75o =
35°
∠ABX = ∠CXE = 120o
exterior angle of ∆DXE)
180o
x = 120o + 35o [x is
x = 155o
16
...
∠ACB = 180o – (∠CAB + ∠CBA)
∠PAB + ∠RBA = 180o
PAB RBA
+
2
= 90
2
∠CAB + ∠CBA = 90o
∠ACB = 180o – (90o) = 90o
18
...
y = 10o
x + y = 50o + 10o = 60o
x = 105o