Questions from Previous CATs and other METs
II. The sphree completely fits into the cube
DO DB
DO . AOB is similar to COD
DC AB OB
DO
3
2 .
Hence, (2).
63. If the side of a cube is twice the radius of a sphere, which of the following is true?
I. The cube completely fits into the sphere.
II. The sphree completely fits into the cube.
(1) I only (2) II only (3) Both I and II (4) Neither I nor II
Sol. If a = 2r, then the sphere completely fits into the cube. Hence, (2).
64. A cylinder of base radius 5 cm and height 58 cm is inclined in such a manner that the smallest vertical distance from the ground to its top surface is 29 cms. Find the angle of inclinations.
(1) 15° (2) 30° (3) 45° (4) 60°
Sol. Midpoint theorem
2 1 58
sinx 29 x = 30°. Hence, (2).
65. A point on the largest side of a triangle is equidistant from the vertices of that side. A line passing through this point, and parallel to the smallest side intersects the longest and second longest sides in two points which are 17 cms apart. Find the length of the smallest side.
(1) 8.5 cm (2) 34 cm (3) 51 cm (4) Cant’ get
Sol. Midpoint theorem
ADE ABC
2
1
BC
DE AB
AD BC = 17 × 2 = 34 cms. Hence, (2).
66. A sphere is melted and half the liquid is used to form 11 identical cubes. Whereas the remaining half is used to form 7 equal smaller sphere. The ratio of the side of the cube to the radius of the new small spheres is:
67. The difference in the circumferences of 2 circles is a multiple of 6. Which of the following statement is/are true?
i. If the larger circle has a radius that is a multiple of 3, then the samller circle cannot have a radius that is a multiple of 3.
ii. If the smaller circle has a radius that is a multiple of 3, then the larger circle cannot have a raidus that is a multipel of 3.
iii. The smaller circle cannot have a radius that is half the radius of the larger circle.
(1) i and ii only (2) ii and iii only (3) iii only (4) None of these
68. A palm tree swings with the breeze in such a manner that the angle covered by its trunk is 14°. If the topmost point of the tree covers a distance of 33 meters, find the length of the
made by its trunk 33
360
69. A spherical balloon is blown until its diameter reaches 10 ft. Then it is continued to be blown until the diameter reaches 12 ft. By what proportion has the 2nd volume increaesd as compared to the first volume?
70. In the diagram below, what is BOD?
Given ODC = 30°, AOC = 50°. ‘O’ is the centre of the circle.
(1) 10° (2) 45° (3) 85° (4) 112.5°
Sol. DOC = 180 - 50 = 130°. DCA = 180 - (30 + 130) = 20°. OBC = 20° BOC = 180 - 40
= 140°BOD = 140 - 130 = 10°. Hence, (1).
71. In the figure alongside are two concentric circles, with a tangent to the inner circle forming a chord to the outer one. The length of the chord is 8 cm. If the radii of both circles are integers, what is the radius of the inner circle?
(1) 3 cm. (2) 4 cm. (3) 5 cm. (4) 6 cm.
Sol. The radius of the inner circle makes 90° angle with the tangent and cuts it in half. The only right-angled triangle with 4 as one side has 3 and 5 as the other sides. So the smaller circle’s radius is 3 cm.
(As all the sides are integres). Hence, (1).
72. The figure below is an equilateral triangle with incircle and circumcircle.
Find the (area of A) : (area of B) : (area of C). Sol. a - altitude of the triangle.
3
;2 3
a
a radius of incircle and circumcircle respectively..
Side of an equilateral triangle a 3
2 . Area of equilateral triangle 2 3
3a
. Area of the circumcircle
2
2
73. What is the arae of the largest triangle that can be fitted into a rectangle of length l and breadth b?
(1) lb/3 (2) lb/2 (3) 2lb/3 (4) 3lb/4
Sol. The maximum area of the triangle = lb/2 . Hence, (2).
74. What is the ratio of the area of an equilateral triangle to that of a circle circumscribing it?
(1) 1 : 3 3 (2) 3 3 : 4
(3) 3 3 : 4 (4) Depends on the side of the triangle Sol. Let side of triangle = a; so area of triangle 2
4
75. A circular park has sum its area and perimeter equal to 8. The diagonal of the park equals:
(1) 4 (2) 8 (3) 2 (4)
Sol. A circular park has sum of its Area + Perimeter = 8 r2 + 2 r = 8 where r = radius r2 + 2r - 8 = 0 r = 2 or r = -4 but radius cannot be negative r = 2 d = 4 Hence, (1).
76. Two circles x and y with centre A and B intersect at C and D. Area of circle X is 4 times area of circle Y. Then AB = ?
(1) 5r (2) 5 r (3) 3r (4) r
2 5
Sol. ACB = 90°. Angle at the point of intersection to the centre of the circles. BC = r. AC = 2r (as area of X = 4 area of Y) AB r24r2 5r . Hence, (2).
77. From a circular paper a man makes two conical caps. The surface area of the two are in the ratio 2 : 1. He then covers the face of the caps with other circular pieces of paper. The ratio of the area of these pieces is:
78. OA = R. What is the ratio of area of circle X and Y?
. For squares AHGE and ABCD, diagonals are in the ratio 3 : 8 sides are in ratio 3 : 8. Let sides be 3x and 8x Area of ABCD = 8x × 8x = 64x2 ...(i)
81. What is the angle in degrees made by a sector the ratio of whose area with the area of the semicircle is equal to 1 : 10?
(1) 36 (2) 18 (3) 24 (4) 9
Sol. We know that area of the sector/area of the circle = angle of the sector/360. Thus we get 1/10 = angle of the sector / 180 (as the second curve is a semi-circle). Thus the angle of the sector = 18 degrees. Hence, (2).
82. The length of the spiral string below can be
(1) 12 a (2) 14 a (3) 18 a (4) 20 a
Sol. Consider arc ABC. ABC > 2 a and ABC > 2 (2a). Greater than circle of radius ‘a’ less than circle of radius ‘2a’. 2 a < ABC < 2 a. Similarly, 2 (2a) < CDE < 2 (3a)
2 (3a) < EFG < 2 (4a). Adding, 2 (6a) < ABC + CDE + EFG < 2 (9a) Only one choice 14 a lies between 12 a and 18 a. Hence, (2).
83. The sum of the perpendicular sides in a right angled triangle is 18. Find the length of the hypotenuse when the area of the triangle is maximum.
(1) 10 2 (2) 9 2 (3) 164 (4) None of these
Sol. Maximum area at a = b = 9. Hypotenuse = 92. Hence, (2).
84. Find 1 + 2, if x = 60°.
(1) 60° (2) 120° (3) 150° (4) 90°
Sol. For a cyclic quadrilateral opposite angles add up to 180.
180 - (1 +2) + 180 - 60 = 180 1 +2 = 180 - 60 = 120° Hence, (2)
89. The heights of two right triangles are in the ratio 1 : 2 the bases are in the ratio 2 : 1 and the ratio of smaller to the longer side of the triangles other than the hypotenuses are in the ratio 2 : 1 then:
(1) the hypotenuse are equal (2) the areas are equal (3) the triangles are congruent (4) all of these
Sol. Given,
Solving (I) and (II), we get 3 a4x, and
3
b5x . Hence, the required ratio = 4/5. Hence, (3).
91. A circle passes through the vertex of an equilateral triangle and is tangent to the opposite side at the midpoint. What is the ratio in which the circle cuts the other side?
(1) 1
Sol. AC + AB = 5AD. AC - AD = 8 AC = 8 + AD 8 + AD + AB = 5AD.
4AD = 8 + AB AB = 4AD - 8. Now, AC2 = AB2 + BC2 = AB2 + AD2. (8 + AD2) = (4AD - 8)2 + AD2. 64 + AD2 + 16AD = 16AD2 + 64 - 64AD + AD2 0 = 16AD2 - 80AD 0 = AD - 5 AD = 5 AB = 12 AC = 13
Area = AB × AD = 12 × 5 = 60. Hence, (1).
93. In the figure, the area of parallelogram ABCD is 24. What is the area of BFE, if CE = 2BC and AB = BF?
(1) 24 (2) 72 (3) 48 (4) 36
Sol. AB = BF. CE = 2BC. Area of parallelogram ABCD = base × height. AB × h1 = 24
1
24 AB h
Area of BFE h2BF h2AB 2
1 2
1
1 2 1
2 24 12 2
1
h h h h
.BHC and BGE are similar triangles. 3
3 1
3 1
2 2
1 2
1
h
h h
h h h x x
Area of BDE = 12 × 3 = 36. Hence, (2).
94. PQ = PR and QS = TR. Which of the following is not necessarily true?
(1) SPT + PST = SPT + PTS (2) QPS = RPT
(3) PRS PQT (4) PST is an equilateral triangle.
Sol. Clearly (1) and (2) are truePRS and PQT by SAS test of congruence. Hence, (3) is true. PST is an isosceles triangle but not necessarily an equilateral triangle. Hence, (4).
95. The medians of a triangle ABC intersect the sides AB and AC at D and E. CE and BD
(Median divides triangle into equal areas).
Also, inBEC, area of BEM = 1/3 area of BEC (Same altitude and EM EC 3
96. Given the triangle ABC and its side a, b, c then the length of the median mb which is a median on side b is
Take a point D such that AD = CD, thus ABCD is parallelogram since the diagonals bisect each other. And we know in parallelogram sum of squares of diagonal is equl to sum of square of sides.
Thus (2mb)2 + b2 = 2(a2 + c2), thus (2 2 2 2 2)1/2 2
1 a c b
mb . Hence, (2).