RRB Quant Practice Questions – Geometry (17 – 04 – 2018)
1. In ΔABC, ∠B = 5∠C and ∠A = 3∠C, then the measure of ∠C is
a) 45°
b) 30°
c) 20°
d) 5°
Explanation: We know
that in a triangle,
Sum
of angles = 180°
⇒ ∠A + ∠B + ∠C = 180°
⇒ 3∠C + 5∠C + ∠C = 180°
⇒ 9∠C = 180°
⇒ ∠C = 180/9 =
20°
∴ ∠C = 20°
2. The sum of two
angles of a triangle is 116° and their difference is 24°. The measure of
smallest angle of the triangle is:
a) 38°
b) 28°
c) 46°
d) 64°
Answer: C)
Explanation: Let the
angles are x° and y° .
The
sum of two angles of a triangle is 116o
x°
+ y° = 116° ....(i)
The
difference of two angles of a triangle is 24°.
x°
- y° = 24° .....(ii)
Adding
eq.(i) and eq.(ii), we will get,
2x°
= 116° + 24°
⇒ x° = 140°/2 = 70°
Then,
y° = 116° - 70° = 46°.
Another
third angle of the triangle = 180° - 116° = 64°
∴ The
smallest angle of the triangle is 46°.
3. In the given
figure, x = ?
a) α + β –
γ
b) α - β +
γ
c) α + β +
γ
d) α + γ –
β
Answer: C)
Explanation: Let ∠BOC = t,
∴∠AOC = β – t
External
angle of triangle = Sum of internal opposite angle
∴ t + α = x1 and β –t + γ = x2
Where,
x1 + x2 = x
Adding
the two equations we get,
α
+ β + γ = x
4. If the angles of a
right angled triangle are in Arithmetic progression, then the angles are in the
ratio
a) 3 : 5 :
7
b) 0.5 : 1
: 2
c) 1 : 2 :
3
d) 2 : 3 :
4
Answer: C)
Explanation: Let the
angles of the triangle be (a - d), (a) and (a + d).
Now
sum of all the interior angles of a triangle = 180°
⇒ (a - d) +
(a) + (a + d) = 180°
⇒ 3a = 180
⇒ a = 60°
We
know in a right angle triangle the largest angle is 90°
⇒ a + d = 90
⇒ d = 30°
So,
the other angle is a – d = 60 – 30 = 30°
∴ the ratio
of the angles is 1 : 2 : 3
5. In the figure, ΔODC
~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Value of ∠DOC, ∠DCO and ∠OAB respectively are: Given DC || AB
a) 65°, 65°
and 65°
b) 45°, 55°
and 55°
c) 55°, 45°
and 65°
d) 55°, 55°
and 55°
Answer: D)
Explanation: DOB is a
straight line.
Therefore,
∠DOC + ∠COB = 180°
⇒ ∠DOC = 180° - 125°
=
55°
In
ΔDOC,
∠DCO + ∠CDO + ∠DOC = 180°
(Sum
of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70° + 55° = 180°
⇒ ∠DCO = 55°
It
is given that ΔODC ~ ΔOBA.
∴ ∠OAB = ∠OCD
[Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
∴ ∠OAB = ∠OCD
[Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
a) 3X + 7y
= 29
b) 3X – 7y
= 1
c) 3X + 7y
= 1
d) 3X – 7y
= 29
Answer: C)
Explanation: We know
that equation of line passed through point (x, y)
⇒ y = mx + c
Given
: m = – 3/7, (x, y) = (5, -2)
⇒ – 2 = - (3/7) × 5 + c
⇒ – 2 = -15/7 + c
⇒ c = 15/7 – 2 = 1/7
⇒ y = -3x/7
+ 1/7
⇒ 3X + 7Y =
1
Hence,
3X + 7y = 1 will be the equation of the line whose slope is – 3/7 and it passes
through the point (5, – 2)
7. Equation of the
straight line parallel to x– axis and also 6 units below x–axis is:
a) x = -6
b) x = 6
c) y = -6
d) y = 6
Answer:
C)
Explanation:
We know that the equation of a straight line parallel to x-axis at a distance b
from it is y = b.
Therefore,
the equation of a straight line parallel to x-axis at a distance 6 units below
the x-axis is y = -6
8. The graphs of 5x +
3y = 12 and 2x + 4y = 0 intersect at the point
a) (12/7, –24/7)
b) (–12/7,
–24/7)
c) (–12/7,
24/7)
d) None of
these
Answer: D)
Explanation: Given,
5x
+ 3y = 12 and 2x + 4y = 0
To
find their point of intersection we need to solve both the equations.
2x
+ 4y = 0
⇒ x = – 2y
Substitute
the above value in 5x + 3y = 12
⇒ 5(– 2y) + 3y = 12
⇒ – 10y + 3y = 12
⇒ – 7y = 12
⇒ y = – 12/7
⇒ x = (– 2)(– 12/7) = 24/7
∴ The point
of intersection of the lines 5x + 3y = 12 and 2x + 4y = 0 is (24/7, – 12/7)
9. The coordinates of
one end point of a diameter of a circle are (2, 4) and the coordinates of its
centre are (1, 2). Find the co-ordinates of other end of the diameter.
a) (2, 1)
b) (1, 0)
c) (3, 1)
d) (0, 0)
Answer: D)
Explanation: We know
that, if line segment formed by joining points A (x1, y1)
and B (x2, y2) is divided by point P (x3, y3)
to a ratio of m: n then:
x3
= (mx2 + nx1)/(m + n) and y3 = (my2
+ ny1)/(m + n)
The
center divides the diameter into two equal parts.
∴ m : n = 1
: 1
Given,
co-ordinates of center is (1,2) and co-ordinates of one end of diameter is (2,4).
Let
the coordinates of other end of the diameter is (a, b)
=>
1 = (1*a + 1*2)/(1 + 1) and 2 = (1*b + 1*4)/(1 + 1)
=>
a = 0 and b = 0
∴
co-ordinates of the other end = (0, 0)
10. Find the area of
the triangle formed by the line x + 4y = 3 and the coordinate axes.
a) 1/3 sq.
unit
b) 4/9 sq.
unit
c) 4/5 sq.
unit
d) 9/8 sq.
unit
Answer: D)
Explanation: By
rearranging the equation and plotting the graph we get, x = 3 – 4y
Thus
the triangle formed by the points are (0,0), (3,0) and (0, ¾)
⟹ Area of
triangle = ½ × base × height
⟹ Area of
triangle = ½ × 3 × ¾
⟹ Area of
triangle = (9/8) sq. Unit
