Download CAT Quant Questions PDF<\/a><\/p>\n Take a free mock test for CAT<\/a><\/p>\n Join 100 Days CAT Course<\/a><\/p>\n Question 1:\u00a0<\/b>A wooden box \u00a0of thickness 0.5 cm, length 21 cm, width 11 cm and height 6 cm is painted on the inside. The expenses of painting are Rs. 70. What is the rate of painting per square centimetres?<\/p>\n a)\u00a0Rs 0.7<\/p>\n b)\u00a0Rs 0.5<\/p>\n c)\u00a0Rs 0.1<\/p>\n d)\u00a0Rs 0.2<\/p>\n Question 2:\u00a0<\/b>The figure shows a circle of diameter AB and radius 6.5 cm. If chord CA is 5 cm long, find the area of triangle ABC. a)\u00a060 sq. cm<\/p>\n b)\u00a030 sq. cm<\/p>\n c)\u00a040 sq. cm<\/p>\n d)\u00a052 sq. cm<\/p>\n Question 3:\u00a0<\/b>The sum of the areas of two circles, which touch each other externally, is $153\\pi$. If the sum of their radii is 15, find the ratio of the larger to the smaller radius.<\/p>\n a)\u00a04<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>In ABC, points P, Q and R are the mid-points of sides AB, BC and CA respectively. If area of ABC is 20 sq. units, find the area of PQR.<\/p>\n a)\u00a010 sq. units<\/p>\n b)\u00a05\u221a3 sq. units<\/p>\n c)\u00a05 sq. units<\/p>\n d)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>In the adjoining figure, points A, B, C and D lie on the circle. AD = 24 and BC = 12. What is the ratio of the area of CBE to that of ADE? a)\u00a01 : 4<\/p>\n b)\u00a01 : 2<\/p>\n c)\u00a01 : 3<\/p>\n d)\u00a0Data insufficient<\/p>\n Download CAT 2022 Syllabus PDF<\/a><\/p>\n Question 6:\u00a0<\/b>A rectangular plank $\\sqrt{10}$ metre wide, is placed symmetrically along the diagonal of a square of side 10 metres as shown in the figure. The area of the plank is:<\/p>\n <\/p>\n a)\u00a0$10(\\sqrt{20}-1)$ sq.mt<\/p>\n b)\u00a0$10(\\sqrt{5}-1)$ sq.mt<\/p>\n c)\u00a0$10\\sqrt{20}-1$ sq.mt<\/p>\n d)\u00a0None<\/p>\n Question 7:\u00a0<\/b>A right circular cylinder has a height of 15 and a radius of 7. A rectangular solid with a height of 12 and a square base, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. Liquid is then poured into the cylinder such that it reaches the rim. The volume of the liquid is<\/p>\n a)\u00a0147(5\u03c0-8)<\/p>\n b)\u00a0180(\u03c0-5)<\/p>\n c)\u00a049(5\u03c0-24)<\/p>\n d)\u00a049(15\u03c0-8)<\/p>\n Question 8:\u00a0<\/b>A cylinder, a Hemi-sphere and a cone stand on the same base and have the same heights. The ratio of the areas of their curved surface is:<\/p>\n a)\u00a0$2:2:1$<\/p>\n b)\u00a0$2:\\sqrt{2}:1$<\/p>\n c)\u00a0$\\sqrt{2}:3:1$<\/p>\n d)\u00a0None of the above<\/p>\n Question 9:\u00a0<\/b>A right circular cylinder has a radius of 6 and a height of 24. A rectangular solid with a square base and a height of 20, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. If water is then poured into the cylinder such that it reaches the rim, the volume of water is:<\/p>\n a)\u00a0288(\u03c0 – 5)<\/p>\n b)\u00a0288(2\u03c0 – 3)<\/p>\n c)\u00a0288(3\u03c0 – 5)<\/p>\n d)\u00a0None of the above<\/p>\n Question 10:\u00a0<\/b>Your friend\u2019s cap is in the shape of a right circular cone of base radius 14 cm and height 26.5 cm. The approximate area of the sheet required to make 7 such caps is<\/p>\n a)\u00a06750 sq cm<\/p>\n b)\u00a07280 sq cm<\/p>\n c)\u00a08860 sq cm<\/p>\n d)\u00a09240 sq cm<\/p>\n Take a free CAT online mock test<\/a><\/p>\n Question 11:\u00a0<\/b>Radius of a spherical balloon, of radii 30 cm, increases at the rate of 2 cm per second. Then its curved surface area increases by:<\/p>\n a)\u00a0120\u03c0<\/p>\n b)\u00a0480\u03c0<\/p>\n c)\u00a0600\u03c0<\/p>\n d)\u00a0None of the above<\/p>\n Question 12:\u00a0<\/b>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 4 cm and 10 cm respectively, then the largest possible curved surface area of the cylinder of radius r is:<\/p>\n a)\u00a0$20\u03c0r^{2}$<\/p>\n b)\u00a0$5\u03c0r(4 – r)$<\/p>\n c)\u00a0$5\u03c0r(r – 4)$<\/p>\n d)\u00a0$5\u03c0r(2 – r)$<\/p>\n Question 13:\u00a0<\/b>A spherical metal of radius 10 cm is molten and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by:<\/p>\n a)\u00a01000 times<\/p>\n b)\u00a0100 times<\/p>\n c)\u00a010 times<\/p>\n d)\u00a0No change<\/p>\n e)\u00a0None of the above<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Inside dimensions of cube are as follows: 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n According to given dimensions, triangle will be a right angled triangle. CAT Online Most Trusted Courses<\/a><\/p>\n CAT Previous Year Papers PDF<\/a><\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given: 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n As we know, the triangle joining midpoints of sides will divide it in 4 similar traingles of equal area.<\/p>\n So area will be = $\\frac{20}{4} = 5$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n As we know angles of same sectors are equal 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n <\/p>\n In the given diagram AB=$\\sqrt{10}$ m<\/p>\n Given that PQRS is a square and the plank is placed symmetrically $\\triangle$BPA and $\\triangle$AQC will be isosceles right triangles.<\/p>\n So PA=PB=$\\frac{\\sqrt{10}}{\\sqrt{2}}$=$\\sqrt{5}$ m<\/p>\n PQ= PA+AQ<\/p>\n AQ= PQ-PA=10-$\\sqrt{5}$ m<\/p>\n We know that AQ=QC ($\\triangle$AQC is isosceles right triangle)<\/p>\n So AC=$\\sqrt{2}$AQ=$\\sqrt{2}$*(10-$\\sqrt{5}$) m<\/p>\n Now we can calculate area of plank<\/p>\n Area of ABCD= AB*AC=\u00a0$\\sqrt{10}$*$\\sqrt{2}$(10-$\\sqrt{5}$)=10($\\sqrt{20}$-1) sq. mt<\/p>\n <\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Volume of liquid = Volume of cylinder – Volume of rectangular solid<\/p>\n <\/p>\n Volume of cylinder = $\\pi*r^{2}*h$<\/p>\n =$\\pi*7^{2}*15$ = $735\\pi$<\/p>\n Volume of rectangular solid = Area of square base * height<\/p>\n In square ABCD,\u00a0 AC = $\\sqrt{2}$*AB\u00a0 \u00a0 \u00a0 \u00a0 so AB = $\\frac{14 }{\\sqrt{2}}$ = $7\\sqrt{2}$<\/p>\n Volume of rectangular solid = Area of square base * height\u00a0 = $(AB)^{2}$*height = $(7\\sqrt{2})^{2}*12$<\/p>\n =\u00a0 $98*12$ = $1176$<\/p>\n So Volume of liquid = Volume of cylinder – Volume of rectangular solid<\/p>\n = $735\\pi$ – $1176$ = $147(5\u03c0-8)$<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The cylinder, hemisphere and cone stand on the same base and have the same height. Let the radius of the three solids be $r$ and the height be $h$.<\/p>\n Height of the hemisphere, $h$ = $r$ (Radius)<\/p>\n Curved surface area of the cylinder = $2*\\pi*r*r$ = $2*\\pi*r^2$ 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that the radius of cylinder = 6 cm.\u00a0The rectangular solid with a square base is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall.<\/p>\n Therefore, the diagonal of square base =\u00a0the diameter of circular base<\/p>\n Hence, a$\\sqrt{2}$ = 2*6 = 12 => a = $6\\sqrt{2}$ cm.<\/p>\n The volume of water = Volume of the cylinder – Volume of the\u00a0rectangular solid<\/p>\n $\\Rightarrow$ $\\pi*6^2*24$ – $(6\\sqrt{2})^2*20$<\/p>\n $\\Rightarrow$ $864*\\pi – 1440$<\/p>\n $\\Rightarrow$ $288(3\\pi – 5)$<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n A cone is formed when the 2 edges of the sector of a circle are joined.<\/p>\n We have to find the central angle subtended to find out the area of the sheet required to make the cone. The slant height of the cone will be equal to the radius of the sector of the circle.<\/p>\n Slant height = $\\sqrt{26.5^2+14^2}$ = $\\sqrt{898.25}$ Let us use $\\sqrt{900}$ for the ease of calculation. Area of sheet required to make 7 cap = 7*1320 = 9240 sq cm<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n It is given that radius, R = 30 cm.<\/p>\n Curved surface area, S= $4\\pi*R^2$<\/p>\n $\\dfrac{dS}{dt}$ =\u00a0$4\\pi*(2R)*\\dfrac{dR}{dt}$<\/p>\n It is given that $\\dfrac{dR}{dt}$ = 2.<\/p>\n Hence,\u00a0$\\dfrac{dS}{dt}$ =\u00a0$4\\pi*(2*30)*2$ = $480\\pi$. Hence, option B is the correct answer.<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Top face will look like the figure shown below.<\/p>\n Curved surface area of the cylinder = $2*\\pi*r*h$<\/p>\n To calculate height of the cylinder in terms of ‘r’, we can see that $\\triangle$ADC is similar to $\\triangle$EFC.<\/p>\n Therefore,<\/p>\n $\\dfrac{AD}{DC}=\\dfrac{EF}{FC}$<\/p>\n $\\Rightarrow$\u00a0$\\dfrac{10}{4} = \\dfrac{h}{4-r}$<\/p>\n $\\Rightarrow$\u00a0$h = \\dfrac{5}{2}(4-r)$<\/p>\n Therefore, the curved surface area of the cylinder =\u00a0$2*\\pi*r*\\dfrac{5}{2}(4-r)$ =\u00a0$5\u03c0r(4 – r)$.<\/p>\n Hence, option B is the correct answer.<\/p>\n 13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Radius of larger sphere = $R = 10$ cm<\/p>\n Let radius of each of the smaller spheres = $r$ cm<\/p>\n => $\\frac{4}{3} \\pi R^3 = 1000 \\times \\frac{4}{3} \\pi r^3$<\/p>\n => $10^3 = 1000 r^3$<\/p>\n => $r = \\sqrt[3]{1} = 1$ cm<\/p>\n Initial surface area of sphere = $4 \\pi R^2 = 4 \\pi \\times 100 = 400 \\pi$<\/p>\n Final surface area of 1000 spheres = $1000 \\times 4 \\pi r^2 = 1000 \\times 4 \\pi = 4000 \\pi$<\/p>\n $\\therefore$ Increase in surface area = $4000 \\pi – 400 \\pi = 3600 \\pi$<\/p>\n => $\\frac{3600 \\pi}{400 \\pi} = 9$ times<\/p>\n
\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/08\/11\/untitled_5.png\")
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\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/08\/18\/untitled_ZSZgbAr.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/ss.png\")
<\/figure>\n
\nLength = $21 – 2*(0.5) = 20$
\nWidth = $11 – 2*(0.5) = 10$
\nHeight = $6 – 0.5 = 5.5$ (As top surface is open)
\nSo total area painted = $2(20 \\times 5.5) + 2(10 \\times 5.5) + (10 \\times 20)$
\n$= 220 + 110 + 200$
\n$= 530$ sq. cm.
\nCost of painting 530 sq. cm. is 70
\nSo cost per sq. cm. = $\\frac{70}{530} = 0.1$ (nearly)<\/p>\n
\nSo BC = 12
\nAnd area = $\\frac{1}{2} \\times 12 \\times 5$
\n= 30<\/p>\n
\n$\\pi((r_1)^2 + (r_2)^2) = 153\\pi$
\nSo
\n$(r_1)^2 + (r_2)^2 = 153$
\nOr $((r_1) + (r_2))^2 – 2(r_1)(r_2) = 153$
\nOr\u00a0\u00a0$(r_1)(r_2) = 36$ and $(r_1) + (r_2) = 15$
\n$r_1 = 12$
\n$r_2 = 3$
\nRatio = 4<\/p>\n
\nHence angle B and angle D will be equal. Angle BCE and angle EAD will be equal.
\nSo triangles BCE and EAD will be similar triangles with sides ratio as 12:24 or 1:2.
\nArea will be in ratio of 1:4.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/qa2.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/55_f7rzxw6.PNG\")
<\/figure>\n
\nCurved surface area of the hemisphere = $2*\\pi*r^2$
\nCurved surface area of the cone = $\\pi*r*\\sqrt{r^2+r^2}$ = $\\pi*r*\\sqrt{r^2+r^2}$ = $\\pi*r^2*\\sqrt{2}$
\nRatio = $2:2:\\sqrt{2}$ = $\\sqrt{2}:\\sqrt{2}:1$
\nAs the answer is not among the given options, option D is the right answer.<\/p>\n
\nIt has been given that the height of the cone is 26.5 cm and the base radius is 14 cm.<\/p>\n
\n$\\sqrt{898.25}$ is very close in value to $\\sqrt{900}$<\/p>\n
\n$\\sqrt{900}$ = $30$ cm.
\nArea of sheet required to make 1 cap = Curve surface area of the cone = $(\\pi)rl$ = (22\/7)*14*30 =1320<\/p>\n<\/figure>\n