Logical Reasoning Questions and Answers For RRB NTPC Exams

RRB NTPC - Co Ordinate Geometry - Problems and Solutions

  • 26. The co-ordinates of the end points of a diameter AB of a circle are A(-6, 8) and B(-10, 6). Find the  co-ordinates of its centre. 
   A.) -8, 7
   B.) -7, 8
   C.) 7, -8
   D.) -8, -7

Answer: Option 'A'

The center O is the mid point of AB. 
Co-ordinates of O are [(-6-10)/2, (8+6)/2] 
= -16/2, 14/2 
= (-8, 7)

  • 27. Find the co-ordinates of a point P which divides the join of A(5, -4) and B(10, 8) in the ratio 3 : 2. 
   A.) 9, 5
   B.) 7, 8
   C.) 8, 7
   D.) 9, -7

Answer: Option 'C'

Required point is P = (mx2 + nx1)/m+n, (my2 + ny1)/m+n 
P((3(10) + 2(5))/5, (3(15) + 2(-5))/5) 
=(30 + 10)/5, (45-10)/5 
= P(8, 7)

  • 28. A point C divides the join of A(2,5) and B(3,9) in the ratio 3 : 4. The co-ordinates of C are:
   A.) (15/7, 46/7)
   B.) (15/7, 47/7)
   C.) (15/7, 40/7)
   D.) None of these

Answer: Option 'B'

x1=2, x2=3 and y1=3, y2=7
= [(3 × 3 + 4 × 2)/7, (3 × 9 + 4 × 5)/7] 
= 9+6/7, 27+20/7 
= (15/7, 47/7)

RRB NTPC Free Mock Test 2019
  • 29. The end points of a line segment AB are A(-6, 4) and B(12,24). Its midpoint is:
   A.) (14, 4)
   B.) (-4, 14)
   C.) (3, 14)
   D.) (-3, 14)

Answer: Option 'D'

Midpoint is C((-6+12)/2, (4+24)/2) 
(-6/2, 28/2) = (-3, 14)

  • 30. The vertices of a ΔABC are A(-6, 18), B(12, 0) and C(9, −21). The centroid of ΔABC is: 
   A.) (5, 1)
   B.) (5, -1)
   C.) (4, -1)
   D.) None of these

Answer: Option 'B'

CENTROID OF A TRIANGLE 
The point of intersection of all the medians of a triangle is called its centroid. 
If A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ΔABC,
then the co-ordinates of its centroid are [1/3(x1 + x2 + x3), 1/3(y1 + y2 + y3)]
= [1/3(-6+12+9), 1/3(18+0-21)] 
= [15/3, -3/3] 
= (5, -1)

  • 31. The vertices of a quadrilateral ABCD are A(0, 0), B(3,3), C(3, 6) and D(0, 3). Then , ABCD is a 
   A.) square
   B.) parallelogram
   C.) rectangle
   D.) rhombus

Answer: Option 'B'

AB2 = (3-0)2 + (3-0)2 = 18 
BC2 = (3-3)2 + (6-3)2 = 9 
CD2 = (0-3)2 + (3-6)2 =18 
AD2 = (0-0)2 + (3-0)2 = 9 
AB = CD = √18 => 3√2, 
BC = AD = √9 
AC2 = (3-0)2 + (6-0)2 = 9 + 36 = 45 
BD2 = (0-3)2 + (3-3)2 = 9 + 0 = 9 
AC ≠ BD 
ABCD is a parallelogram.

  • 32. The points A(1, -3), B(13, 9), C(10, 12) and D(-2, 0) taken in order are the vertices of 
   A.) square
   B.) rhombus
   C.) parallelogram
   D.) rectangle

Answer: Option 'D'

AB2 = (13-1)2 + (9+3)2
= 122 + 122 = 288. 
BC2 = (10-13)2 + (12-9)2
= -32 + 32 = 9+9 = 18 
CD2 = (10+2)2 + (12-0)2
= 122 + 122 = 288 
AD2 = (-2-1)2 + (0+3)2 = (9+9) =18 
AB = CD and BC = AD 
AC2 = (10-1)2 + (12+3)2
= 92 + 152 = 81 + 225 = 306. 
BD2 = (-2-13)2 + (0-9)2
= 225 + 81 = 306 
AC = BD 
ABCD is a rectangle.

RRB NTPC Free Mock Test 2019
  • 33. If for a line m = tanϑ < 0, then 
   A.) ϑ is acute
   B.) ϑ is obtuse
   C.) ϑ = 90°
   D.) ϑ = 60°

Answer: Option 'B'

m = tanϑ < 0 => is obtuse

  • 34. If for a line m = tanϑ > 0, then 
   A.) ϑ is acute
   B.) ϑ is obtuse
   C.) ϑ = 90°
   D.) ϑ = 60°

Answer: Option 'A'

m = tanϑ < 0 => is acute

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