CAT Previous Year Questions: Mensuration Notes | Study Quantitative Aptitude (Quant) - CAT
CAT: CAT Previous Year Questions: Mensuration Notes | Study Quantitative Aptitude (Quant) - CAT
The document CAT Previous Year Questions: Mensuration Notes | Study Quantitative Aptitude (Quant) - CAT is a part of the CAT Course Quantitative Aptitude (Quant).
Try yourself:If the rectangular faces of a brick have their diagonals in the ratio 3 : 2 √3 : √15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is
[2019]
Explanation
From the given data, we know l2 + b2 = 9k ...(1) b2 + h2 = 12k ...(2) l2 + h2 = 15k ...(3) Adding (1) , (2) and (3) we get 2(l2 + b2 + h2) = 36 k l2 + b2 + h2 = 18k ...(4) Subtracting (1), (2) , (3) from (4) we get h2 = 9k l2 = 6k b2 = 3k Thus, ratio of shortest edge to longest edge = √3 : 3 = 1 : √3.
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Try yourself:Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is
[2019]
Explanation
Construct an Equilateral triangle and cut of equal lengths from all three sides. So we obtain a hexagon and three equilateral triangles of side 'a' in return. We know that a hexagon comprises of 6 Equilateral Triangles. Since the side of the hexagon is also 'a', we obtain 6 equilateral triangles in return. Thus, we have a total of 6 + 3 = 9 equilateral triangles of side 'a' Ratio of area of Hexagon with Area of Triangle = 6: 9 = 2: 3
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Try yourself:The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is
[2019]
Explanation
It is given that the diagram is a square pyramid and all 4 sides of the triangle are of equal sides. Now, we need to plug in a right triangle and apply Pythagoras theorem to find the vertical height of the triangle. Draw a vertical from the apex of the pyramid and draw a line to the base along the equilateral triangle. Connect the points in the base to form a right triangle. The hypotenuse would be the slant height of the pyramid, which is the altitude of the equilateral triangle. Length of the base = 10 cms Hypotenuse = Altitude of the equilateral triangle = x 20 = 10cms Therefore, (Vertical height)2 = (10√3)2 - 102 = 200 Vertical height = √200 = 10√2 cms
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Try yourself:A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is
[2019]
Explanation
The question is "A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is" Hence, the answer is 1026(1 + π)
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*Answer can only contain numeric values
Try yourself:A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is:
[TITA 2018]
Correct Answer : 198 cubic ft
Explanation
From the diagram, △OAD is similar to △OBC Volume of the remaining part = Volume of OBC – Volume OAD Volume of the remaining part = (1/3 x 22/7 x 4 × 4 × 12 ) - (1/3 x 22/7 x 1 × 1 × 3) Volume of the remaining part = 22/7 x (64 − 1) Volume of the remaining part = 198 cubic ft
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Try yourself:In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is
[2018]
Explanation
Given Area (ABCD) = 72 sq cm, CD = 9cms, AD = 16cms Area of Parallelogram = Base × Height CD × 16 = 72 CD = (72/16) cms We extend CD till P such that ∠APD = 90° So, again by applying the area formula considering CD as base CD × AP = Area of Parallelogram ABCD 9 × AP = 72 AP = 8 cms We have AP = 8 cms, AD = 16 cms We observe that the ratio of the sides form a 1 : √3 : 2 triangle Therefore DP = 8√3 cms Area (△APD) = 1/2 × AD × DP = 1/2 × 8 × 8√3 sq cms Area (△APD) = 32 √3 sq cms
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Try yourself:From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
[2018]
Explanation
Given that ABCD be the rectangle of area of 768 sq cm. A semicircular part with diameter AB and area 72π sq cm is removed From the area given we can find the radius r as r2 = 144 r = 12 cm The diameter AB is 24 since r = 12 So we have a rectangle ABCD AB × BC = 768 24 × BC = 768 BC = 768/24 BC = 32 Now we have to find the perimeter of the left over portion (shaded area) Perimeter = AD + DC + BC + 1/2 perimeter of the circle Perimeter = 32 + 24 + 32 + 2πr/2 (r = 12 cm) Perimeter = 88 + 12π cm The perimeter of the leftover portion, in cm, is 88 + 12π
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Try yourself:The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio
[2018]
Explanation
Given that the area of a rectangle and the square of its perimeter are in the ratio 1: 25, we have to find the ratio of the lengths of the shorter and longer sides of the rectangle i.e. b : l The area of the rectangle = lb The square of its perimeter = [2(l + b)]2 ⇒ lb : [2(l + b)]2 = 1 : 25 ⇒ lb : 4[l2 + 2lb + b2] = 1 : 25 ⇒ 25 lb = 4[l2 + 2lb + b2] ⇒ 25 lb = 4l2 + 8lb + 4b2 ⇒ 4l2 - 17lb + 4b2 = 0 ⇒ 4l2 - 16lb - lb + 4b2 = 0 ⇒ 4l[l- 4b] – b [l- 4b] = 0 ⇒ [l - 4b][4l - b] = 0 By this we can find that the ratio of the shorter and longer sides of the rectangle b : l is in the ratio 1: 4 In other way ,if we want to deal with only one variable and do not want to factorize divide this 4l2 - 17lb + 4b2 = 0 by l2 throughout Since we have to find ratio of b : l, b/l can be substituted with x 4 - 17x + 4x2 = 0 Now x can be found by simplifying it
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Try yourself:A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?
[2018]
Explanation
Given that a parallelogram ABCD has an area 48 sq cm. The length of CD is 8 cm and that of AD is s cm Since the area given here is 48 sq.cm the height can be found as 6 sq.cm We have to find which of the following options is necessarily true Since this (shaded portion) is a right angled triangle with 6 as its height and s as its hypotenuse so we can assure that s ≥ 6 and s cannot be less than 6 so we can eliminate the other options. By this we can come to the conclusion that s ≥ 6
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Try yourself:A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to:
[2017]
Explanation
Given that a solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. We have to find the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube. Let us take the volume of the 5 cubes to be 1x3 , 1x3 , 8x3 , 27x3 , 27x3 Volume of the original cube = x3(1 + 1 + 8 + 27 + 27) Volume of the original cube = 64x3 Sides of the original cube = ∛64x3 = 4x Similarly, sides of the 5 smaller cubes = x , x , 2x , 3x , 3x . Surface Area of a cube = 6a2 Surface Area of the original cube = (4x) × (4x) = 16x2 Surface area of the smaller cubes = x2 , x2 , 4x2 , 9x2 , 9x2 Sum of the surface areas of the smaller cubes = x2(1 + 1 + 4 + 9 + 9) = 24x2 Change in surface area = 24x2 – 16x2 = 8x2 %change = Hence the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to 50%.
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*Answer can only contain numeric values
Try yourself:A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9 π cm3. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is:
(TITA 2017)
Correct Answer : 6
Explanation
Given that the height of the cylinder is 3 cm, while its volume is 9π cm3 Volume of the cylinder ⇒ πr2h = 9π cm3 r2 = 9 × 3 or r = √3 So the diameter is 2√3 cm A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. Since OPQ is the right angled triangle We can find the OP = 1 cm. We have to find the vertical distance of the topmost point of the ball from the base of the cylinder. Since OP = 1, to reach the topmost point still it has to go 2 cm from the point O. The vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is 2 + 1 + 3 = 6 cm
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The document CAT Previous Year Questions: Mensuration Notes | Study Quantitative Aptitude (Quant) - CAT is a part of the CAT Course Quantitative Aptitude (Quant).
x 20 = 10
cms
