Let 's' be the number of sides of a regular polygon. What is the least value of 's' for which the number of points of intersection of the diagonals is more than the sum of the numbers of sides and number of diagonals of the polygon?
Answer: B This problem is worked out by physical counting. When s=5, the polygon is a pentagon, Diagonal =5 and sides = 5.
Points of intersection of diagonal =5+5 10.
Thus, 10 = 5(diagonal)+5(sides) When s=6, the polygon is a Hexagon, Diagonal =9 and sides = 6.
Points of intersection of diagonal = 13 +6 = 19.
Thus, 19 > 9(Diagonal)+6 (Sides) Thus, Choice 'B' is the answer.
Q. No. 2:
A cuboid of size 5.2m x 13m x 39m is completely cut into 'p' small cubes. If p has the minimum possible value, what is the total surface area of each cube?
Answer: D We can get small cubes from a cuboid by taking each side as a common factor of the 3 dimensions. The least number of identical cubes would correspond to the HCF of the 3 dimensions. Side of the cubes formed => HCF(5.2, 13,39) = 2.6m Number of cubes formed = [(5.2)(13)(39)]/[(2.6)(2.6)(2.6)] = 150. Surface area of each cube = 6*(2.6)(2.6) = 6(6.76) sq m Total surface area of all such cubes formed is (150)(6)(6.76) = 6084 sq m.
Q. No. 3:
A solid cube into three solids such that two of them are cuboids with two dimensions exactly half the bigger cube and the third dimension same as the original cube. The third solid has to dimensions same as the original cube. Find the ratio of total surface area of the smaller cubes and the other solid.
Answer: A let the cube of side x. sides of smaller cuboid = x/2, x/2, x. Other solid will be a cuboid with one side as x/2 and other two sides as x each. Then surface area of the cuboid with side x/2, x/2, x => 2[(x*x/2) + (x/2 * x/2) + (x * x/2)] = 5x2/2 Surface area of the cuboid with sides x, x, x/2 => 2[(x*x)+(x*x/2)+(x*x/2)] = 4x2 The required ratio = 5x2/2:4x2 = 5:8
Q. No. 4:
A child consumed an ice cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. If the height of the ice-cream cone was 8cm, what was the height of the remaining ice cream cone ?
Answer: D It is two dimensional figure of the cone ADF is the part which is left. AG = 8cm. Let, GC = r cm, EF =r1 cm, AE = h cm. AE/EF = AG/GC [Traingle AEF and AGC are similar] h/r1 = 8/r => r1 = hr/8 Volume of smaller cone = 1/3 * 22/7 * (hr/8)2 * h Volume of bigger cone = 1/3 * 22/7 * r2 * 8 Smaller cone is 12.5% of bigger cone. => h = 4.
Q. No. 5:
In a rocket shape firecrackers, explosive powder is to be filled up inside the metallic enclosure. The metallic enclosure is made up of cylindrical base and conical top with the base of radius 8cm. The ratio of height of cylinder and cone is 5:3. A cylindrical hole is drilled through the metal solid with height one-third the height of metal solid. What should be the radius of the hole, so that volume of the hole (in which gun powder is to be filled up) is half of the volume of metal solid after drilling ?
Answer: A As the ratio of height of cylindrical base to that of the conical top is 5:3. Let their hugs be 5k and 3k respectively. Let R be the radius of the cylindrical hole. Given that the height of the cylindrical hole = 8k/3 => (x/3 *R2 * 3k) + (x*R2* 5k) = 6xkR2 Given that the volume of the solid left after the hole is made = 2(Volume of the cylindrical base). => 6xkR2/ 3 = 2xkR2 = x(r)2 * 8k/3 => r = √3 /2* R But, it is given that, R = 8 => r = √3 /2* 8 = 4√3 cm.
Q. No. 6:
A right circular cone is enveloping a right circular cylinder such that the base of the cylinder rests on the base of the cone. If the radius and the height of the cone is 4cm and 10 cm respectively, then the largest possible curved surface area of the cylinder of radius 'r' is :
Answer: B Let this be 2-dimensional figure AE=h and EF= r ED = (10-h) [∴AD -AE] Triangle AEF and Triangle DAC are similar, => AF/EF = AD/CD => h/r = 10/4 => h = 5r/2 Height of cylinder (H) = (10 - 5r/2) Curved surface area = 2πr (10 - 5r/2) = 5πr(4-r)
