{"id":33940,"date":"2019-08-23T18:26:18","date_gmt":"2019-08-23T12:56:18","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=33940"},"modified":"2019-08-23T18:26:18","modified_gmt":"2019-08-23T12:56:18","slug":"geometry-questions-for-rrb-ntpc-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/geometry-questions-for-rrb-ntpc-set-2-pdf\/","title":{"rendered":"Geometry Questions for RRB NTPC Set-2 PDF"},"content":{"rendered":"

Geometry Questions for RRB NTPC Set-2 PDF<\/strong><\/span><\/h2>\n

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.<\/p>\n

Download Geometry Questions for RRB NTPC Set-2 PDF<\/a><\/p>\n\n

Take a free mock test for RRB NTPC<\/a><\/p>\n

Download RRB NTPC Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>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 ?<\/p>\n

a)\u00a08 m<\/p>\n

b)\u00a015.6 m<\/p>\n

c)\u00a017 m<\/p>\n

d)\u00a018.5 m<\/p>\n

Question 2:\u00a0<\/b>5: 18 is the ratio of the length and perimeter of a rectangle. What would be the ratio of its length and breadth ?<\/p>\n

a)\u00a04: 3<\/p>\n

b)\u00a03: 5<\/p>\n

c)\u00a05: 4<\/p>\n

d)\u00a04: 7<\/p>\n

Question 3:\u00a0<\/b>Exterior angle of a regular polygon is 72. What would be the sum of its interior angles ?<\/p>\n

a)\u00a0360<\/p>\n

b)\u00a0480<\/p>\n

c)\u00a0520<\/p>\n

d)\u00a0540 –<\/p>\n

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Question 4:\u00a0<\/b>What is the sum of the opposite angles in a cyclic-quadrilateral?<\/p>\n

a)\u00a090 degrees<\/p>\n

b)\u00a0180 degrees<\/p>\n

c)\u00a0360 degrees<\/p>\n

d)\u00a0250 degrees<\/p>\n

Question 5:\u00a0<\/b>What will be the sum of the interior angles of a polygon with 10 sides?<\/p>\n

a)\u00a01440<\/p>\n

b)\u00a01340<\/p>\n

c)\u00a01500<\/p>\n

d)\u00a01600<\/p>\n

Question 6:\u00a0<\/b>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?<\/p>\n

a)\u00a040 sq. meter.<\/p>\n

b)\u00a0160 sq.meter.<\/p>\n

c)\u00a0110 sq. meter<\/p>\n

d)\u00a0100 sq. meter<\/p>\n

Question 7:\u00a0<\/b>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?<\/p>\n

a)\u00a09 cm<\/p>\n

b)\u00a010 cm<\/p>\n

c)\u00a011 cm<\/p>\n

d)\u00a012 cm<\/p>\n

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Question 8:\u00a0<\/b>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<\/p>\n

a)\u00a058 sq.m<\/p>\n

b)\u00a068 sq.m<\/p>\n

c)\u00a036 sq.m<\/p>\n

d)\u00a028 sq.m<\/p>\n

Question 9:\u00a0<\/b>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<\/p>\n

a)\u00a0Rs 8,420<\/p>\n

b)\u00a0Rs 16,128<\/p>\n

c)\u00a0Rs 16,218<\/p>\n

d)\u00a0Rs 15,128<\/p>\n

Question 10:\u00a0<\/b>There are 7 pentagons and hexagons in a chart If the total number of sides is 38 then the number of pentagons is<\/p>\n

a)\u00a03<\/p>\n

b)\u00a02<\/p>\n

c)\u00a05<\/p>\n

d)\u00a04<\/p>\n

Question 11:\u00a0<\/b>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<\/p>\n

a)\u00a0253<\/p>\n

b)\u00a046<\/p>\n

c)\u00a0223<\/p>\n

d)\u00a0211<\/p>\n

Download General Science Notes PDF<\/a><\/p>\n\n

Question 12:\u00a0<\/b>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?<\/p>\n

a)\u00a08 cm<\/p>\n

b)\u00a010 cm<\/p>\n

c)\u00a012 cm<\/p>\n

d)\u00a015 cm<\/p>\n

Question 13:\u00a0<\/b>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?<\/p>\n

a)\u00a016 : 5<\/p>\n

b)\u00a05 : 3<\/p>\n

c)\u00a03 : 1<\/p>\n

d)\u00a04 : 3<\/p>\n

Question 14:\u00a0<\/b>The exterior angle of a regular polygon is twice its interior angle. What is the number of sides of the polygon?<\/p>\n

a)\u00a06<\/p>\n

b)\u00a05<\/p>\n

c)\u00a04<\/p>\n

d)\u00a03<\/p>\n

Question 15:\u00a0<\/b>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?<\/p>\n

a)\u00a01 : 4<\/p>\n

b)\u00a02 : 1<\/p>\n

c)\u00a04 : 1<\/p>\n

d)\u00a03 : 1<\/p>\n

RRB NTPC Free Mock Test<\/a><\/p>\n

RRB NTPC Previous Papers (Download PDF)<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

(l – 5)(b+3) = lb – 9 …. (1)<\/p>\n

(l+ 3)(b+2) = lb + 67 …… (2)<\/p>\n

(1) => lb – 5b + 3l -15 = lb – 9<\/p>\n

or 3l – 5b = 6<\/p>\n

(2) = > lb + 3b + 2l + 6 = lb + 67<\/p>\n

or 2l + 3b = 61<\/p>\n

from these equations, l = 17 m<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

l: 2l + 2b = 5 : 18<\/p>\n

l : 2b = 5 : 8<\/p>\n

l : b = 5 : 4<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

The interior angle of the polygon = 180 – 72 = 108<\/p>\n

Let there be x sides in the polygon.<\/p>\n

108 x = (2x – 4)* 90<\/p>\n

So, x = 5 (pentagon)<\/p>\n

Sum of the interior angles = 5 x 108 = 540<\/p>\n

4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Summation of the opposite angles in a cyclic quadrilateral is equal to 180 degrees.<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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
\nHence, sum will be = 1440<\/p>\n

6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let\u2019s say length and breadth of the rectangle are 5x and 2x
\nSo perimeter will be 14x = 28 meter ; x = 2 meter
\nHence, length = 10 meter ; breadth = 4 meter
\nArea = 40 sq.meter<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The length of the hypotenuse will be =$\\sqrt{ 6^2 + 8^2} = 10$<\/p>\n

8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

The area of the bigger rectangle is 18 x 13<\/p>\n

The are of the smaller rectangle is 16 x 11<\/p>\n

The required value is 18 x 13 – 16 x 11 = 58 sq.m<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The sides of the rectangle are 48 and 14 (by Pythagoras\u00a0theorem)<\/p>\n

Total cost of grass cutting is 48 x 14 x 24 = \u00a016,128<\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let there be x pentagons and 7-x hexagons.<\/p>\n

Total sides = 5x + 6(7-x) = 38<\/p>\n

Solving we get x = 4<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

The number of ways of drawing straight lines from 23 points is $ C_2^{23}$<\/p>\n

solving, we get the number of ways = $\\frac{23!}{21!2!}$<\/p>\n

= 253<\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the length of the rectangle be L and the breadth of the rectangle be B.
\n(L-4)(B+2) = LB – 18 => 2L – 4B – 8 = -18 => 4B – 2L = 10
\n(L+2)(B-2) = LB – 14 => 2B – 2L – 4 = -14 => 2L – 2B = 10
\nSolving the two equations, we get, B = 10 cm and L = 15 cm
\nSo, the original breadth of the rectangle is 10 cm.<\/p>\n

13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the length be 3k. Perimeter = 16k
\nLet the length be L
\nSo, 2L + 2*3k = 16k => 2L + 6k = 16k => L = 5k
\nSo, ratio of length to breadth = 5k : 3k = 5 : 3<\/p>\n

14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Exterior angle + Interior angle = 180
\n=> 3*(Interior angle) = 180
\n=> Interior angle = 180\/3 = 60 degrees
\nInterior angle of a regular polygon = (2n – 4)*90\/n = 60
\n=> 180n – 360 = 60n
\n=> 120n = 360
\n=> n = 3<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Area of the original triangle = \u00bd * Base * Height
\nArea of the new triangle = \u00bd * (2*Base) * (2*Height) = 4 * (1\/2 * Base * Height) = 4 * Area of the original triangle<\/p>\n

So, the required ratio = 4 : 1<\/p>\n\n

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We hope this Geometry Questions set-2\u00a0 pdf\u00a0 for RRB NTPC Exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Take a free mock test for RRB NTPC Download RRB NTPC Previous Papers PDF Question 1:\u00a0If […]<\/p>\n","protected":false},"author":41,"featured_media":33949,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603,1601],"tags":[1637,2052,1647,1619],"class_list":{"0":"post-33940","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-ntpc","11":"category-rrb-ntpc-je","12":"tag-rrb-exam","13":"tag-rrb-maths","14":"tag-rrb-mocks","15":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":33949,"alt_text":"Geometry Questions for RRB NTPC Set-2 PDF","caption":"Geometry Questions for RRB NTPC Set-2 PDF","description":"Geometry Questions for RRB NTPC Set-2 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