Geometry Questions for RRB NTPC Set-2 PDF
Download RRB NTPC Geometry Questions and Answers set – 2 PDF. Top 15 RRB NTPC Geometry questions based on asked questions in previous exam papers very important for the Railway NTPC exam.
Download Geometry Questions for RRB NTPC Set-2 PDF
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Question 1: If the length of a rectangle is decreased by 5 metres and breadth increased by 3 metres, its area decreases by 9 $ m^{2} $. If its length is increased by 3 m and breadth by 2 m,its area increases by 67 $ m^{2} $. What is the length of a rectangle ?
a) 8 m
b) 15.6 m
c) 17 m
d) 18.5 m
Question 2: 5: 18 is the ratio of the length and perimeter of a rectangle. What would be the ratio of its length and breadth ?
a) 4: 3
b) 3: 5
c) 5: 4
d) 4: 7
Question 3: Exterior angle of a regular polygon is 72. What would be the sum of its interior angles ?
a) 360
b) 480
c) 520
d) 540 –
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Question 4: What is the sum of the opposite angles in a cyclic-quadrilateral?
a) 90 degrees
b) 180 degrees
c) 360 degrees
d) 250 degrees
Question 5: What will be the sum of the interior angles of a polygon with 10 sides?
a) 1440
b) 1340
c) 1500
d) 1600
Question 6: Length and breadth of a rectangle are in the ratio of 5:2, where the perimeter of the rectangle is 28 meter. What will be area of the rectangle?
a) 40 sq. meter.
b) 160 sq.meter.
c) 110 sq. meter
d) 100 sq. meter
Question 7: If the base and perpendicular of a right-angled triangle are 6 cm and 8 cm then what will be the length of the hypotenuse?
a) 9 cm
b) 10 cm
c) 11 cm
d) 12 cm
Question 8: There is a path 1 m wide outside a rectangular field of 16 m length and 11m breadth then the total area of the path is
a) 58 sq.m
b) 68 sq.m
c) 36 sq.m
d) 28 sq.m
Question 9: The diagonal of a rectangular field is 50 m and one of the sides is 48 m If the cost of cutting the grass of the field is Rs 24 per square metre then the total cost of cutting all grass of the rectangular field is
a) Rs 8,420
b) Rs 16,128
c) Rs 16,218
d) Rs 15,128
Question 10: There are 7 pentagons and hexagons in a chart If the total number of sides is 38 then the number of pentagons is
a) 3
b) 2
c) 5
d) 4
Question 11: The number of straight lines that can be drawn in a plane with 23 given points assuming that no three of them are collinear is
a) 253
b) 46
c) 223
d) 211
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Question 12: If the length of a rectangle is decreased by 4 cm and the breadth of the rectangle is increased by 2 cm, the area of the rectangle decreases by 18 $cm^2$. If the length of the rectangle is increased by 2 m and the breadth of the rectangle is decreased by 2 m, the area of the rectangle decreases by 14 $cm^2$. What is the original breadth of the rectangle?
a) 8 cm
b) 10 cm
c) 12 cm
d) 15 cm
Question 13: The ratio of the perimeter to the breadth of a rectangle is 16 : 3. What is the ratio of the length to breadth of the rectangle?
a) 16 : 5
b) 5 : 3
c) 3 : 1
d) 4 : 3
Question 14: The exterior angle of a regular polygon is twice its interior angle. What is the number of sides of the polygon?
a) 6
b) 5
c) 4
d) 3
Question 15: Each side of a right-angled triangle is doubled so as to form another right-angled triangle. What is the ratio of the area of the new triangle to the area of the original triangle?
a) 1 : 4
b) 2 : 1
c) 4 : 1
d) 3 : 1
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Answers & Solutions:
1) Answer (C)
(l – 5)(b+3) = lb – 9 …. (1)
(l+ 3)(b+2) = lb + 67 …… (2)
(1) => lb – 5b + 3l -15 = lb – 9
or 3l – 5b = 6
(2) = > lb + 3b + 2l + 6 = lb + 67
or 2l + 3b = 61
from these equations, l = 17 m
2) Answer (C)
l: 2l + 2b = 5 : 18
l : 2b = 5 : 8
l : b = 5 : 4
3) Answer (D)
The interior angle of the polygon = 180 – 72 = 108
Let there be x sides in the polygon.
108 x = (2x – 4)* 90
So, x = 5 (pentagon)
Sum of the interior angles = 5 x 108 = 540
4) Answer (B)
Summation of the opposite angles in a cyclic quadrilateral is equal to 180 degrees.
5) Answer (A)
As we know that summation of the interior angles of a triangle = (n-2)*180 where n is the number of sides of the polygon
Hence, sum will be = 1440
6) Answer (A)
Let’s say length and breadth of the rectangle are 5x and 2x
So perimeter will be 14x = 28 meter ; x = 2 meter
Hence, length = 10 meter ; breadth = 4 meter
Area = 40 sq.meter
7) Answer (B)
The length of the hypotenuse will be =$\sqrt{ 6^2 + 8^2} = 10$
8) Answer (A)
The area of the bigger rectangle is 18 x 13
The are of the smaller rectangle is 16 x 11
The required value is 18 x 13 – 16 x 11 = 58 sq.m
9) Answer (B)
The sides of the rectangle are 48 and 14 (by Pythagoras theorem)
Total cost of grass cutting is 48 x 14 x 24 = 16,128
10) Answer (D)
Let there be x pentagons and 7-x hexagons.
Total sides = 5x + 6(7-x) = 38
Solving we get x = 4
11) Answer (A)
The number of ways of drawing straight lines from 23 points is $ C_2^{23}$
solving, we get the number of ways = $\frac{23!}{21!2!}$
= 253
12) Answer (B)
Let the length of the rectangle be L and the breadth of the rectangle be B.
(L-4)(B+2) = LB – 18 => 2L – 4B – 8 = -18 => 4B – 2L = 10
(L+2)(B-2) = LB – 14 => 2B – 2L – 4 = -14 => 2L – 2B = 10
Solving the two equations, we get, B = 10 cm and L = 15 cm
So, the original breadth of the rectangle is 10 cm.
13) Answer (B)
Let the length be 3k. Perimeter = 16k
Let the length be L
So, 2L + 2*3k = 16k => 2L + 6k = 16k => L = 5k
So, ratio of length to breadth = 5k : 3k = 5 : 3
14) Answer (D)
Exterior angle + Interior angle = 180
=> 3*(Interior angle) = 180
=> Interior angle = 180/3 = 60 degrees
Interior angle of a regular polygon = (2n – 4)*90/n = 60
=> 180n – 360 = 60n
=> 120n = 360
=> n = 3
15) Answer (C)
Area of the original triangle = ½ * Base * Height
Area of the new triangle = ½ * (2*Base) * (2*Height) = 4 * (1/2 * Base * Height) = 4 * Area of the original triangle
So, the required ratio = 4 : 1
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