RRB Quant Practice Questions – Geometry (17 – 04 – 2018)

RRB Quant Practice Questions – Geometry (17 – 04 – 2018)
1. In ΔABC, B = 5C and A = 3C, 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°
3C + 5C + C = 180°
9C = 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°
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
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
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°
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°

6. The equation of the line if its slope is – 3/7 and it passes through the point (5, – 2) is
a) 3X + 7y = 29
b) 3X – 7y = 1
c) 3X + 7y = 1
d) 3X – 7y = 29
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
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
d) None of these
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)
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