25 SSC CGL Mocks – Just Rs. 149<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Question 1:\u00a0<\/b>If a metallic cone of radius 30 cm and height 40 cm is melted and recast into metallic spheres of radius 10 cm, find the number of spheres formed.<\/p>\n a)\u00a05<\/p>\n b)\u00a09<\/p>\n c)\u00a06<\/p>\n d)\u00a012<\/p>\n Question 2:\u00a0<\/b>find the\u00a0volume\u00a0of a cube if the\u00a0perimeter of one face of the cube is 30 cm.<\/p>\n a)\u00a0400.175<\/p>\n b)\u00a0472.175<\/p>\n c)\u00a0362.725<\/p>\n d)\u00a0421.875<\/p>\n Question 3:\u00a0<\/b>A solid spherical copper ball, whose diameter is 42 cm, is melted and converted into a wire having diameter equal to 42 cm. The length of the wire is<\/p>\n a)\u00a028<\/p>\n b)\u00a014<\/p>\n c)\u00a0$\\large\\frac{56}{3}$<\/p>\n d)\u00a0$\\large\\frac{28}{3}$<\/p>\n Question 4:\u00a0<\/b>The radii of two cylinders are in the ratio 2:3. Their curved surface areas are in the ratio 4:7. Then find the ratio between their volumes.<\/p>\n a)\u00a08:23<\/p>\n b)\u00a07:24<\/p>\n c)\u00a08:21<\/p>\n d)\u00a023:8<\/p>\n Question 5:\u00a0<\/b>The radii of two cylinders are in the ratio 3:5 and their Curved Surface Areas are in the ratio 5:8. Find the ratio between their volumes.<\/p>\n a)\u00a09:24<\/p>\n b)\u00a08:23<\/p>\n c)\u00a025:24<\/p>\n d)\u00a024:8<\/p>\n SSC CGL Previous Papers Download PDF<\/a><\/p>\n SSC CHSL PREVIOUS PAPERS DOWNLOAD<\/a><\/p>\n Question 6:\u00a0<\/b>The base of a right prism, whose height is 2 cm, is a square. If the total surface area of the prism is 10 cm2, then its volume is:<\/p>\n a)\u00a0$3\\ cm^{3}$<\/p>\n b)\u00a0$1\\ cm^{3}$<\/p>\n c)\u00a0$2\\ cm^{3}$<\/p>\n d)\u00a0$4\\ cm^{3}$<\/p>\n Question 7:\u00a0<\/b>The radius of a wire is decreased to one third. If volume remains the same, length will increase by:<\/p>\n a)\u00a06 times<\/p>\n b)\u00a01 time<\/p>\n c)\u00a03 times<\/p>\n d)\u00a09 times<\/p>\n Question 8:\u00a0<\/b>What is the ratio of volume of cylinder to the volume of cone if their heights are in the ratio of 3:2 (i.e height of cylinder to height of cone) and the radius of the cylinder is 1.428 times the radius of the cone.<\/p>\n a)\u00a025:49<\/p>\n b)\u00a049:50<\/p>\n c)\u00a050:49<\/p>\n d)\u00a049:25<\/p>\n Question 9:\u00a0<\/b>A solid spherical copper ball, whose diameter is 14 cm, is melted and converted into a wire having diameter equal to 14 cm. The length of the wire is<\/p>\n a)\u00a027 cm<\/p>\n b)\u00a0$\\frac{16}{3}$<\/p>\n c)\u00a015 cm<\/p>\n d)\u00a0$\\frac{28}{3}$<\/p>\n Question 10:\u00a0<\/b>A cylindrical well of height 20 metres and radius 14 metres is dug in a field 72 metres long and 44 metres wide. The earth taken out is spread evenly on the field. What is the increase (in metre) in the level of the field?<\/p>\n a)\u00a06.67<\/p>\n b)\u00a03.56<\/p>\n c)\u00a05.61<\/p>\n d)\u00a04.83<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n Question 11:\u00a0<\/b>If the diameter of a circle is the side of a square, then the ratio of the areas of the circle and the square is<\/p>\n a)\u00a0$\\pi : 1$<\/p>\n b)\u00a0$\\pi : 4$<\/p>\n c)\u00a0$\\pi : 2$<\/p>\n d)\u00a0$\\pi : 8$<\/p>\n Question 12:\u00a0<\/b>The area of the iron sheet required to prepare a cone 40 cm high with base radius 9 cm is (Take $\\pi = \\frac{22}{7}$)<\/p>\n a)\u00a0${\\large\\frac{9900}{7}}$<\/p>\n b)\u00a0${\\large\\frac{9800}{7}}$<\/p>\n c)\u00a0${\\large\\frac{18000}{7}}$<\/p>\n d)\u00a0${\\large\\frac{11000}{7}}$<\/p>\n Question 13:\u00a0<\/b>find the area of an equilateral triangle if the height of the triangle is 24 cm.<\/p>\n a)\u00a0$192\\sqrt{3} cm^2$<\/p>\n b)\u00a0$175\\sqrt{3} cm^2$<\/p>\n c)\u00a0$178\\sqrt{3} cm^2$<\/p>\n d)\u00a0$164\\sqrt{3} cm^2$<\/p>\n Question 14:\u00a0<\/b>Three sides of a triangular meadow are of length 28 m, 45 m and 53 m long respectively. Find the cost of sowing seeds(in rupees per sq.m) in the meadow at the rate of 12 rupees per sq.m.<\/p>\n a)\u00a07560<\/p>\n b)\u00a06860<\/p>\n c)\u00a07960<\/p>\n d)\u00a07860<\/p>\n Question 15:\u00a0<\/b>In an equilateral triangle,if h-R=15 cm where h=height of the triangle and R=circumradius then what is the area of the triangle ?<\/p>\n a)\u00a0$275\\sqrt{3}$<\/p>\n b)\u00a0$225\\sqrt{3}$<\/p>\n c)\u00a0$500\\sqrt{3}$<\/p>\n d)\u00a0$675\\sqrt{3}$<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n SSC CHSL Free Mock Test<\/a><\/p>\n Question 16:\u00a0<\/b>In an equilateral triangle,if h-R=5 cm where h=height of the triangle and R=circumradius then what is the area of the triangle ?<\/p>\n a)\u00a0$50\\sqrt{3}$<\/p>\n b)\u00a0$100\\sqrt{3}$<\/p>\n c)\u00a0$75\\sqrt{3}$<\/p>\n d)\u00a0$25\\sqrt{3}$<\/p>\n Question 17:\u00a0<\/b>Find the area(in sq.cm) of the sector subtending 72\u00b0 at the centre of a circle with a radius of 35cm ?<\/p>\n a)\u00a0690<\/p>\n b)\u00a0734<\/p>\n c)\u00a0770<\/p>\n d)\u00a0792<\/p>\n Question 18:\u00a0<\/b>Find the percentage increase in its area, if the length and breadth of a rectangle increases by 30% and 25% each.<\/p>\n a)\u00a054.25<\/p>\n b)\u00a048.75<\/p>\n c)\u00a058.25<\/p>\n d)\u00a062.5<\/p>\n Question 19:\u00a0<\/b>A wire of length 704 cm is used to form a circle. What will be the area$(in\u00a0 m^2)$ of the circle formed?<\/p>\n a)\u00a039424<\/p>\n b)\u00a03.9424<\/p>\n c)\u00a039246<\/p>\n d)\u00a036424<\/p>\n Question 20:\u00a0<\/b>Find the maximum possible area$(in\u00a0 cm^2)$ of a triangle which can be formed using a thread of length 36 cm.?<\/p>\n a)\u00a0$45$<\/p>\n b)\u00a0$39\\sqrt{3}$<\/p>\n c)\u00a0$36\\sqrt{3}$<\/p>\n d)\u00a0$36$<\/p>\n 1500+ Free SSC Questions & Answers<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, a metallic cone of radius 30 cm and height 40 cm = $\\dfrac{1}{3} \\pi * 30^2 * 40 = 12000 \\pi cm^3$<\/p>\n metallic spheres of radius 10 cm<\/p>\n Volume of sphere = $\\dfrac{4}{3} \\pi R^3$<\/p>\n = $\\dfrac{4}{3} \\pi * 10^3 = \\dfrac{4000 \\pi}{3} cm^3$<\/p>\n => Required no. of spheres = $\\dfrac{12000 \\pi}{\\dfrac{4000 \\pi}{3}}$<\/p>\n = 9<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that one face of a cube is a square with perimeter 30 cm<\/p>\n Let each side of the cube be $a$<\/p>\n => 4 $a$ = 30<\/p>\n => $a$ = 7.5<\/p>\n Volume of cube = $7.5^3$<\/p>\n = 421.875 $cm^3$<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, spherical copper ball, whose diameter is 42 cm<\/p>\n Radius of spherical copper ball(r) = 21 cm<\/p>\n Volume of sphere = $\\frac{4}{3}\\pi r^3$<\/p>\n Radius of Cylindrical wire(R) = 21cm<\/p>\n Let length of wire be ‘L’<\/p>\n Volume of cylinder = $\\pi R^2L$<\/p>\n Here Volume of sphere = Volume of sphere<\/p>\n $\\large\\frac{4}{3}$ $\\pi r^3$ = $\\pi R^2L$<\/p>\n $\\large\\frac{4}{3}$ $\\times21\\times21\\times21$ = 21$\\times$21$\\times$L<\/p>\n $\\large\\therefore$ Length of the wire ‘L’ = $28$ cm<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given $\\dfrac{r_1}{r_2} = \\dfrac{2}{3}$ \u21d2 $\\dfrac{2}{3} \\times \\dfrac{h_1}{h_2} = \\dfrac{4}{7}$<\/p>\n \u21d2 $\\dfrac{h_1}{h_2} = \\dfrac{6}{7}$<\/p>\n Volume $= \\pi r^2 h$<\/p>\n \u21d2 $\\dfrac{\\pi r_1^2 h_1}{\\pi r_2^2 h_2} = ((\\dfrac{r_1}{r_2})^2) \\times (\\dfrac{h_1}{h_2})$<\/p>\n = $\\dfrac{4}{9} \\times \\dfrac{6}{7} = \\dfrac{8}{21}$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given $\\dfrac{r_1}{r_2} = \\dfrac{3}{5}$ \u21d2 $\\dfrac{3}{5} \\times \\dfrac{h_1}{h_2} = \\dfrac{5}{8}$<\/p>\n \u21d2 $\\dfrac{h_1}{h_2} = \\dfrac{25}{24}$<\/p>\n Volume $= \\pi r^2 h$<\/p>\n \u21d2 $\\dfrac{\\pi r_1^2 h_1}{\\pi r_2^2 h_2} = ((\\dfrac{r_1}{r_2})^2) \\times (\\dfrac{h_1}{h_2})$<\/p>\n = $\\dfrac{9}{25} \\times \\dfrac{25}{24} = \\dfrac{9}{24}$<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let side of base = $a$ cm and height = $h=2$ cm<\/p>\n Total surface area of prism = Curved surface area + (base+top) area<\/p>\n => $10$ = Perimeter of base $\\times$ height + $2\\times$ area of base<\/p>\n => $(4\\times a\\times2)+(2\\times a^2)=10$<\/p>\n => $a^2+4a-5=0$<\/p>\n => $a^2+5a-a-5=0$<\/p>\n => $a(a+5)-1(a+5)=0$<\/p>\n => $(a+5)(a-1)=0$<\/p>\n => $a=1$\u00a0 \u00a0 \u00a0 $[\\because a$ cannot be negative.$]$<\/p>\n $\\therefore$ Volume = Base area $\\times$ height<\/p>\n = $(1)^2\\times2=2$ $cm^3$<\/p>\n => Ans – (C)<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let radius of wire is $r=3$ cm and length = $h=1$ cm<\/p>\n => Volume of cylinderical wire = $\\pi r^2h$<\/p>\n = $\\pi\\times(3)^2\\times1=9\\pi$ $cm^2$<\/p>\n New radius = $r’=\\frac{1}{3}\\times3=1$ cm<\/p>\n Let new length = $h’$ cm<\/p>\n If volume remains the same, => $\\pi (r’)^2\\times(h’)=9\\pi$<\/p>\n => $(1)^2\\times(h’)=9$<\/p>\n => $h’=9$<\/p>\n $\\therefore$ Length was increased by = $\\frac{h’}{h}=9$<\/p>\n => Ans – (D)<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Radius of cylinder=$\\frac{10}{7}$(radius of cone)<\/p>\n $\\frac{r_{cyli}}{r_{cone}}$=$\\frac{10}{7}$<\/p>\n Ratio of heights $\\frac{h_{cyli}}{h_{cone}}$=$\\frac{1}{2}$ 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Radius of spherical copper ball(r) = 7cm<\/p>\n Volume of sphere = $\\frac{4}{3}\\prod r^3$<\/p>\n Radius of Cylindrical wire(R) = 7cm<\/p>\n Let length of wire be ‘L’<\/p>\n Volume of cylinder = $\\prod R^2L$<\/p>\n Here Volume of sphere = Volume of sphere<\/p>\n $\\frac{4}{3}\\prod r^3$ = $\\prod R^2L$<\/p>\n $\\frac{4}{3}\\times7\\times7\\times7$ = 7$\\times$7$\\times$L<\/p>\n $\\therefore$ Length of the wire ‘L’ = $\\frac{28}{3}$ cm<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Increase in the level of the field is the height of field (cuboidal shape) when volume of well (cylinderical) is equal to the volume of field (cuboidal).<\/p>\n Radius of well = $R=14$ m and height = $H=20$ m<\/p>\n Length of field = $l=72$ m and width = $b=44$ m<\/p>\n Let height = $h$ m<\/p>\n => Volume of cuboid = Volume of cylinder<\/p>\n Now, volume of cuboid\u00a0 = (Area of rectangle – Area of circle) $\\times$ height<\/p>\n => $(lb-\\pi R^2)\\times h=\\pi R^2H$<\/p>\n => $[(72\\times44)-(\\frac{22}{7}\\times14^2)]\\times(h)=\\frac{22}{7}\\times(14)^2\\times20$<\/p>\n => $(3168-616)h=44\\times280$<\/p>\n => $h=\\frac{44\\times280}{2552}\\approx4.83$ m<\/p>\n => Ans – (D)<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the radius be $x$. Since the diameter is equal to the side of square<\/p>\n => Side of square = $2x$<\/p>\n Ratio of area of circle to that of square<\/p>\n = $\\large\\frac{\\pi (x)^2}{(2x)^2}$<\/p>\n = $\\large\\frac{\\pi}{4}$<\/p>\n => Required ratio = $\\pi : 4$<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We know that the<\/p>\n Surface area of cone= $\\large\\frac{22}{7}$ $\\times r\\times(r+L)$<\/p>\n where L=slant height L= $\\sqrt{(r)^{2} + (h)^{2}}$<\/p>\n Here when $r=9$ and $h=40$,<\/p>\n L= $\\sqrt{(9^{2} + (40)^{2})}$ = $\\sqrt{1681}$ = 41 cm<\/p>\n Area, A= ${\\large\\frac{22}{7}}$ $\\times9\\times(9+41)$<\/p>\n A=\u00a0 ${\\large\\frac{9900}{7}}(cm)^{2}$<\/p>\n 15000 Questions – Free SSC Study Material<\/a><\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, In an equilateral triangle all the angles are equal to 60\u00b0<\/p>\n In $\\triangle$ADC<\/p>\n => $tan \\angle ACD = \\frac{AD}{DC}$<\/p>\n => $tan 60\u00b0 = \\frac{24}{DC}$<\/p>\n => DC = $\\frac{24}{\\sqrt{3}} = 8\\sqrt{3}$<\/p>\n => BC = 2*DC = $16\\sqrt{3}$<\/p>\n Area of $\\triangle$ ABC = $\\frac{\\sqrt{3}}{4} * side^2$<\/p>\n = $\\frac{\\sqrt{3}}{4} * (16\\sqrt{3})^2$<\/p>\n = $192\\sqrt{3} cm^2$<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given the sides of triangle are 28 m, 45 m and 53 m<\/p>\n Since, $28^2 + 45^2 = 784+2025 = 2809 = 53^2$<\/p>\n => The given sides are of right angled triangle as,<\/p>\n $c^2 + b^2 = a^2$where a, b, c are the sides of the triangle.<\/p>\n => Area of triangular field = $\\large\\frac{1}{2}$ $ \\times 28 \\times 45= 630 m^2$<\/p>\n => Cost of sowing seeds = $12 \\times 630 = Rs. 7560 rupees\/m^2$<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In an equilateral triangle,all the points such as orthocentre,centroid,circumcenter coincide. s=$30\\sqrt{3}$ 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In an equilateral triangle,all the points such as orthocentre,centroid,circumcenter coincide. $\\sqrt{3}*s\/2$=15<\/p>\n s=$10\\sqrt{3}$<\/p>\n Area of an equilateral triangle=$\\sqrt{3}(s^{2})\/4$<\/p>\n =$75\\sqrt{3}$<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We know that,<\/p>\n Area of sector of a circle = $\\frac{x}{360}*\\pi *r^{2}$<\/p>\n Where $x$ is the angle subtended at the centre and $r$ is the radius of the circle<\/p>\n Area =$\\frac{72}{360}\\times\\pi \\times35^{2}$ = 770 sq.cm<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let $l$ and $b$ be the initial values of length and breadth respectively.<\/p>\n Therefore, the final values of length and breadth are $1.3l and 1.25b$ respectively.<\/p>\n Initial area of the rectangle = $lb$<\/p>\n Final area of the rectangle = $1.3\\times1.25 lb = 1.625 lb$<\/p>\n Therefore, required percentage = $\\large\\frac{(1.625-1)lb}{lb}\\times$100 = 62.5%.<\/p>\n So, the correct option to choose is D.<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the radius of the circle formed be \u2018r\u2019<\/p>\n The length of the wire will be the circumference of the circle.<\/p>\n So we get,<\/p>\n $2\\pi r=704$<\/p>\n $=>2\\times\\frac{22}{7}\\times r=704$<\/p>\n $r=112$ cm<\/p>\n Thus the area will be = $\\pi r^2=256\\times154$.<\/p>\n =$39424 cm^2$<\/p>\n =$3.9424 m^2$. [$1m^2 = 10000cm^2$]<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The perimeter of the triangle is given as 36 cm.<\/p>\n The triangle of a fixed perimeter will have the maximum possible area when all the sides are of equal length.<\/p>\n So each side should be of length 12 cm for the triangle to have maximum possible area(i.e. Equilateral triangle).<\/p>\n In this case the area will be,<\/p>\n $\\frac{\\sqrt{3}}{4}\\times a^2$<\/p>\n $\\frac{\\sqrt{3}}{4}\\times12\\times12$<\/p>\n $=36\\sqrt{3}$ $cm^2$<\/p>\n
\nVolume of metallic cone = $\\dfrac{1}{3} \\pi r^2 h$<\/p>\n
\nCurved Surface Area $= 2\\pi r h$
\n\u21d2 $\\dfrac{2 \\pi r_1 h_1}{2 \\pi r_2 h_2} = \\dfrac{4}{7}$<\/p>\n
\nCurved Surface Area $= 2\\pi r h$
\n\u21d2 $\\dfrac{2 \\pi r_1 h_1}{2 \\pi r_2 h_2} = \\dfrac{5}{8}$<\/p>\n
\nVolume of cone= $\\frac{1}{3}$ $\\times $ $\\pi $ $\\times $ $r_{cone} $ $\\times $ $r_{cone} $ $\\times h $
\nVolume of cylinder=$\\pi $ $\\times $ $ r_{cyli} $ $\\times $ $r_{cyli} $ $\\times h $
\nRequired ratio=50:49<\/p>\n
\nAD = 24 cm and ABC is an equilateral triangle<\/p>\n![](\"https:\/\/cracku.in\/blogs:\/\/lh3.googleusercontent.com\/ZG4QbGmP31YEuDC3I9odKk-5ZqA7tw3q5eRFT-RCI9p0VzmW_fvOk18VdAYZaKJS5Z-WVgPIibltPPBNSWmtVHoKRrRhk6IQ0KWiGB5WIX9orFH-S7KNOBCop-Z8CBvMalgtSWnN\")
![](\"https:\/\/cracku.in\/media\/uploads\/209428.png\")
<\/figure>\n
\nLet the triangle be PQR and the circumcentre be O. let median intersect QR at A
\nCentroid divides median in the ratio 2:1.PA=h,OA=R
\nOA=15 cm
\nTherefore (PO:OA)=2:1
\nPO:15=2:1
\nPO=30
\nPA=PO+OP
\nPA=30+15
\nPA=45
\nPA is also the altitude using it side of the triangle can be calculated.
\n$\\sqrt{3}*s\/2$=45<\/p>\n
\nArea of an equilateral triangle=$\\sqrt{3}(s^{2})\/4$
\n=$675\\sqrt{3}$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209425.png\")
<\/figure>\n
\nLet the triangle be PQR and the circumcentre be O. let median intersect QR at A
\nCentroid divides median in the ratio 2:1
\nOA=5 cm
\nTherefore (PO:OA)=2:1
\nPO:5=2:1
\nPO=10
\nPA=PO+OP
\nPA=10+5
\nPA=15
\nPA is also the altitude using it side of the triangle can be calculated.<\/p>\n