Enroll to MAH-CET Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>The edge of an ice cube is 14 cm. The volume of the largest cylindrical ice cube that can be fit into it?<\/p>\n a)\u00a02200 cu. cm<\/p>\n b)\u00a02000 cu. cm<\/p>\n c)\u00a02156 cu. cm<\/p>\n d)\u00a02400 cu. cm<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>The edge of an ice cube is 14 cm. The volume of the largest cylindrical ice cube that can be formed out of it is<\/p>\n a)\u00a02200 cu. cm<\/p>\n b)\u00a02000 cu. cm<\/p>\n c)\u00a02156 cu. cm<\/p>\n d)\u00a02400 cu. cm<\/p>\n e)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>A rope makes 125 rounds of a cylinder with base radius 15 cm. How many times can it go round a cylinder with base radius 25 cm?<\/p>\n a)\u00a0100<\/p>\n b)\u00a075<\/p>\n c)\u00a080<\/p>\n d)\u00a065<\/p>\n e)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>A right circular cylindrical tank has the storage capacity of 98808 ml. If the radius of the base of the cylinder is three-fourth of the height, what is the diameter of the base ?<\/p>\n a)\u00a028 cms<\/p>\n b)\u00a056 cms<\/p>\n c)\u00a021 cms<\/p>\n d)\u00a058 cms<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>The respective ratio of curved surface area and total surface area of a cylinder is 4:5. If the curved surface area of the cylinder is 1232 cm2, what is the height ? (in cm)<\/p>\n a)\u00a028 cm<\/p>\n b)\u00a034 cm<\/p>\n c)\u00a036 cm<\/p>\n d)\u00a030 cm<\/p>\n e)\u00a0None of these<\/p>\n Take Free MAH-CET mock tests here<\/a><\/p>\n Question 6:\u00a0<\/b>The sum of the radius and height of a cylinder is 18 metre. The total surface area of the cylinder is 792 sq. metre, what is the volume of the cylinder ? (in cubic metre)<\/p>\n a)\u00a01848<\/p>\n b)\u00a01440<\/p>\n c)\u00a01716<\/p>\n d)\u00a01724<\/p>\n e)\u00a01694<\/p>\n Question 7:\u00a0<\/b>If the volume and curved surface area of a cylinder are 616 $m^3$ and 352 $m^2$ respectively what is the total surface area of the cylinder (in $m^2$)<\/p>\n a)\u00a0429<\/p>\n b)\u00a0419<\/p>\n c)\u00a0435<\/p>\n d)\u00a0421<\/p>\n e)\u00a0417<\/p>\n Question 8:\u00a0<\/b>The sum of the radius and height of a cylinder is 42 cm. Its total surface area is 3696 cm 2. What is the volume of cylinder ?<\/p>\n a)\u00a017428 cubic cm<\/p>\n b)\u00a017248 cubic cm<\/p>\n c)\u00a017244 cubic cm<\/p>\n d)\u00a017444 cubic cm<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b><\/p>\n Read the following information and answer the questions based on it. Question 9:\u00a0<\/b>How many cuboids shall have only one colour ?<\/p>\n a)\u00a012<\/p>\n b)\u00a016<\/p>\n c)\u00a022<\/p>\n d)\u00a028<\/p>\n e)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>How many cuboids shall have only two colours red and green in their two sides?<\/p>\n a)\u00a08<\/p>\n b)\u00a012<\/p>\n c)\u00a016<\/p>\n d)\u00a020<\/p>\n e)\u00a0None of these<\/p>\n Question 11:\u00a0<\/b>How many cuboids shall not any colour?<\/p>\n a)\u00a0No any<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a06<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>How many cuboids shall have all the three colours?<\/p>\n a)\u00a08<\/p>\n b)\u00a010<\/p>\n c)\u00a012<\/p>\n d)\u00a014<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>The six faces of a wooden cube of side 6 cm are labelled A, B, C, D, E and F respectively. Three of these faces A, B, and C are each adjacent to the other two, and are painted red. The other three faces are not painted. Then, the wooden cube is neatly cut into 216 little cubes of equal size. How many of the little cubes have no sides painted?<\/p>\n a)\u00a0125<\/p>\n b)\u00a0135<\/p>\n c)\u00a091<\/p>\n d)\u00a0108<\/p>\n e)\u00a0100<\/p>\n Question 14:\u00a0<\/b>A colourless cube is painted blue and then cut parallel to sides to form two rectangular solids of equal volume. What percentage of surface area of each of new solids is not painted blue?<\/p>\n a)\u00a025<\/p>\n b)\u00a016<\/p>\n c)\u00a020<\/p>\n d)\u00a018<\/p>\n Question 15:\u00a0<\/b>The intersection of two cubes cannot be<\/p>\n a)\u00a0cube<\/p>\n b)\u00a0triangle<\/p>\n c)\u00a0rectangle<\/p>\n d)\u00a0none of these<\/p>\n Question 16:\u00a0<\/b>The cost of the paint is 36.50 per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint outside of a cube having 8 feet each side?<\/p>\n a)\u00a0692<\/p>\n b)\u00a0768<\/p>\n c)\u00a0876<\/p>\n d)\u00a0972<\/p>\n Question 17:\u00a0<\/b>Consider a cuboidal underground tunnel of length 500 m whose cross-section is given in the figure. If $1 m^3$ of concrete costs 1000 rupees, find the amount of money needed to build the\u00a0Tunnel.<\/p>\n a)\u00a0$8(4-\\pi)10^6 $ rupees<\/p>\n b)\u00a0$64(4-\\pi)10^6 $ rupees<\/p>\n c)\u00a0$16(4-\\pi)10^6 $ rupees<\/p>\n d)\u00a0$32(4-\\pi)10^6 $ rupees<\/p>\n Question 18:\u00a0<\/b>The length, breadth and height of a rectangular cuboid are in the ratio 1: 2 : 3. If the length, breadth and height are increased by 100% each then what would be the increase in the volume of the cuboid ?<\/p>\n a)\u00a03 times<\/p>\n b)\u00a07 times<\/p>\n c)\u00a010 times<\/p>\n d)\u00a09 times<\/p>\n Question 19:\u00a0<\/b>Ram, a farmer, managed to grow shaped- watermelons inside glass cases of different shapes. The shapes he used were: a perfect cube, hemi-spherical, cuboid, cylindrical along with the normal spherical shaped watermelons. Thickness of the skin was same for all the shapes. Each of the glass cases was so designed that the total volume and the weight of the all the water- melons would be equal irrespective of the shape. a)\u00a0Cube<\/p>\n b)\u00a0Hemi-sphere<\/p>\n c)\u00a0Cuboid<\/p>\n d)\u00a0Cylinder<\/p>\n e)\u00a0Normal spherical<\/p>\n Question 20:\u00a0<\/b>A cube of side 12 cm is painted red on all the faces and then cut into smaller cubes, each of side 3 cm. What is the total number of smaller cubes having none of their faces painted?<\/p>\n a)\u00a016<\/p>\n b)\u00a08<\/p>\n c)\u00a012<\/p>\n d)\u00a024<\/p>\n Get 5 MAH-CET mocks at just Rs.299<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Radius of the cylinder = r =\u00a0$\\frac{14}{2}$ = 7 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Radius of the cylinder = r = $\\frac{14}{2}$ = 7<\/p>\n Height of the cylinder = h =14<\/p>\n Volume = Pi x r2 <\/sup>x h<\/p>\n = $\\frac{22}{7}$ x 7 x 7 x 14 = 2156<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the required number of rounds be x.<\/p>\n More radius, less rounds (Inverse Proportion)<\/p>\n Radius in cm == \u00a0Round<\/p>\n 15 == 125<\/p>\n 25 == x<\/p>\n \u27f9 x=(15 * 125)\/25 = 75 rounds<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Volume if cylinder = \u03c0$(R)^2$h<\/p>\n R is the radius and h is the height of cylinder<\/p>\n Volume = 98808 ml<\/p>\n It is given that R = $\\frac{3}{4}$h<\/p>\n $\\frac{4}{3}$ $(R)^3$ \u03c0 = 98808<\/span><\/p>\n R = 28.68 <\/span><\/span><\/p>\n Diameter = 28.68 \u00d7 2 ~58<\/span><\/span><\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let R and H be the radius and height of cylinder<\/p>\n Curved Surface area of cylinder = 2$\\barwedge$RH = 1232<\/p>\n ($\\barwedge$RH= 616 sqr cm)<\/p>\n Total surface area of cylinder = 2$\\barwedge$RH + 2$\\barwedge(R)^2$ = 2$\\barwedge$R(H+R)<\/span><\/p>\n it is given that CurvedSurfaceArea\/TotalSurfaceArea = 4\/5<\/span><\/p>\n 1232\/(1232\u00a0+ 2$\\barwedge(R)^2$)<\/span> = 4\/5<\/span><\/p>\n R = 7 cm <\/span><\/p>\n 2$\\barwedge$RH = 1232 2$\\barwedge$7H = 1232<\/p>\n 44H =1232<\/span> \u00a0<\/span><\/p>\n \u00a0<\/span><\/p>\n 6)\u00a0Answer\u00a0(E)<\/strong><\/p>\n let the height and radius of cylinder be H mtr and R mtr<\/p>\n R + H = 18<\/p>\n total surface area of cylinder = 2$\\barwedge$RH + 2$\\barwedge(R)^2$ = 792<\/p>\n R(H + R) = $\\frac{792×7}{22×2}$<\/p>\n R = 7 mtr<\/p>\n H = 18-7 = 11 mtr<\/p>\n volume = $\\frac{22}{7}(R)^2(H)$<\/p>\n Volume = 1694 cubic mtr<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Volume of a cylinder=$\\pi \\times r^{2} \\times h$ <\/p>\n <\/p>\n <\/p>\n <\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total surface area of cylinder<\/p>\n => $2 \\pi r h + 2 \\pi r^2 = 3696$<\/p>\n => $2 \\pi r (r + h) = 3696$<\/p>\n $\\because (r + h) = 42$ \u00a0 [Given]<\/p>\n => $2 \\times \\frac{22}{7} \\times r \\times 42 = 3696$<\/p>\n => $44 \\times 6 \\times r = 3696$<\/p>\n => $r = \\frac{3696}{44 \\times 6} = 14$ cm<\/p>\n => $h = 42 – 14 = 28$ cm<\/p>\n $\\therefore$ Volume of cylinder = $\\pi r^2 h$<\/p>\n = $\\frac{22}{7} \\times 14 \\times 14 \\times 28$<\/p>\n = $17248 cm^3$<\/p>\n 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Number of cuboids which have only 1 colour are the middle cuboids in all the faces. Also, there are 2 types of each faces.<\/p>\n 2*(B-2)*(H-2)+2*(H-2)*(L-2).<\/p>\n =2*(4-2)*(3-2)+2*(3-2)*(5-2)+2*(5-2)*(4-2).<\/p>\n = 2*2*1 + 2*1*3 + 2*3*2.= 4 + 6 + 12.<\/p>\n =22.<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Number of cuboids which\u00a0have only two colours red and green in their two sides are the middle cuboids at the corner edges. There are 4 such edges which have combination of red and green colour.<\/p>\n Number of required cuboids = $(5-2) \\times 4$<\/p>\n = $3 \\times 4=12$<\/p>\n => Ans – (B)<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of cuboids which do not have any colour = $(5-2) \\times (4-2) \\times (3-2)$<\/p>\n = $3 \\times 2 \\times 1=6$<\/p>\n => Ans – (D)<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The number of cuboid which will have all the three colours are the corner pieces.<\/p>\n Thus, 8 cuboids will have all the three colours.<\/p>\n => Ans – (A)<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Since A, B and C are adjacent faces. If we remove them, the resultant solid will also be a cube with side 5.<\/p>\n Hence total number of cubes unpainted = $5^3$ = 125<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Assuming the sides of the cube = a, then surface painted blue = 6a$^2$<\/p>\n Now after cutting into half the blue surface = 3a$^2$ for each solid and a new colourless surface with area a$^2$ is generated.<\/p>\n Hence the total surface area\u00a0 = 3a$^2$ + a$^2$ = 4a$^2$<\/p>\n % surface area\u00a0not painted blue =\u00a0$\\ \\dfrac{\\ a^2}{4a^2}\\times\\ 100$ = 25<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Cubes are a 3-D figure, and so, the intersection of two 3-D figure must be a 2-D figure.<\/p>\n So, Cube cannot be the intersection of two cubes.<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total surface area of cube =\u00a0$6\\times\\ 8^2\\ =384$<\/p>\n Amount of paint required =\u00a0$\\ \\frac{\\ 384}{16}=24\\ kgs$<\/p>\n Cost of 24kgs paint =\u00a0$24\\times\\ 36.5\\ =876$<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Area of the tunnel to be filled with concrete = 16*8-$\\pi\\ \\left(32\\right)$<\/p>\n Length of the length along which the tunnel has to be filled with concrete = 500m<\/p>\n Volume =\u00a016*8-$\\pi\\ \\left(32\\right)$*500<\/p>\n = $32(4-\\pi)$*500<\/p>\n Cost of concrete per\u00a0$1 m^3$ = 1000<\/p>\n Cost of\u00a0$32(4-\\pi)$*500 =\u00a0$32(4-\\pi)$*500*1000<\/p>\n =$16(4-\\pi)*10^{6}$<\/p>\n C is the correct answer.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given the length, breadth and height of a rectangular cuboid are in the ratio 1: 2 : 3.<\/p>\n Let the length, breadth and height be k, 2k, and 3k units where ‘k’ is a constant.<\/p>\n Volume of the cuboid(V’) = $l\\times b\\times h$ units = $k\\times 2k\\times 3k$\u00a0 = $6k^3$ units<\/p>\n Given\u00a0the length, breadth and height are increased by 100% each\u00a0.<\/p>\n Then the new length, breadth and height will be doubled.<\/p>\n So new length, breadth and height of the rectangular cuboid are 2k, 4k, and 6k units.<\/p>\n New Volume of the cuboid (V’) = $2k\\times 4k\\times 6k = 48k^3$ units.<\/p>\n Increase in the volume of the cuboid = V’ – V = $48k^3 – 6k^3 = 42k^3 = 7 [6k^3] = 7V$<\/p>\n Therefore the volume has increased 7 times the initial volume.<\/p>\n 19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let the volume of watermelon = $V$<\/p>\n Total surface area = $S$<\/p>\n Thickness of the skin = $t$<\/p>\n => Volume usable for juice = $V – S t$<\/p>\n Hence, if total surface area is minimum then usable volume of the watermelon will be max.<\/p>\n For equal volume, sphere has the least surface area.<\/p>\n Ans – (E)<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total number of cubes will be = $\\frac{12 \\times 12 \\times 12}{3 \\times 3 \\times 3}$ = 64
\nThe length ,breadth and height of a rectangular piece of wood in the 4cm,3cm, 5cm respectively
\nOpposite side of 5cm x 4 cm pieces are coloured in red colour
\nOppsite sides 4cm x 3 cm ,are cloured in blue
\nRest 5 cm x 3 cm are coloured in green in both sides
\nNow the piece is cut in such way that a cuboid of 1cm x 1cm x 1cm will be made<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Q4_K7aT4uw.png\")
<\/figure>\n
\nA customer wants to but water-melons for making juice, for which the skin of the water- melon has to be peeled off, and therefore is a waste. Which shape should the customer buy?<\/p>\n
\nHeight of the cylinder = h =14
\nVolume = Pi x r2\u00a0<\/sup>x h
\n=\u00a0$\\frac{22}{7}$ x 7 x 7 x 14 = 2156<\/p>\n
\n<\/span><\/p>\n
\n<\/span>H = 28 cm<\/span><\/p>\n
\nwhere $r$ and $h$ are the radius and height of the cylinder.
\n$\\pi \\times r^{2} \\times h$ = $616 m^{3}$
\nCurved Surface Area of Cylinder=$2\\times \\pi \\times r \\times h$=$352 m^{2}$
\n$\\pi \\times r \\times h$=$176$
\nReplacing $\\pi \\times r \\times h$ in Volume formula we get,
\n$ r \\times 176$=$616$
\n$r=3.5 m$
\nTotal Surface Area = Curved Surface Area + 2$\\times$ Area of base
\n=$352 + 2\\times pi \\times r^{2}$
\n=$352 + 2\\times pi \\times 3.5^{2}$
\n=$352+77$
\n=$429 m^{2}.$
\nHence Option A is the correct answer.<\/span><\/span><\/span><\/span><\/span><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/9984_sITUqT2.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/9984_UcxL21R.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/9984_TQAAUqx.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/9984.PNG\")
<\/figure>\n
\nCubes with 3 sides painted = cubes at corner sides = 8
\nCubes with 2 sides painted = cubes on lines = 24
\nCubes with 1 sides painted = cubes on middle of surface = 24
\nSo cubes with no side painted = 64 – (8+24+24) = 8<\/p>\n