SSC CGL level Solution Set 42, Mensuration 4

42nd SSC CGL level Solution Set, 4th on topic Mensuration

SSC CGL solution set 42 mensuration 4

This is the 42nd solution set of 10 practice problem exercise for SSC CGL exam and 4th on topic Mensuration. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set.

We will repeat here the method of taking a 10 problem test if you have not gone through it already. And if you have not taken the test yet, you can refer to SSC CGL level Question Set 42, Mensuration 4, and then after taking the test come back to this solution.

Method for taking this 10 problem test and get the best results from the test set:

  1. Before start, go through the tutorials Basic concepts on Geometry 1, Basic concepts on Geometry 2, Basic concepts on Geometry 3 or any other short but good material to refresh your concepts if you so require.
  2. Answer the questions in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.
  3. When the time limit of 12 minutes is over, mark up to which you have answered, but go on to complete the set.
  4. At the end, refer to the answers to the questions to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.
  5. Identify and analyze the problems that you couldn't do to learn how to solve those problems.
  6. Identify and analyze the problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledge improve it by referring to only that part of concept from the best source you get hold of. You might google it. If it is because of your method of answering, analyze and improve those aspects specifically.
  7. Identify and analyze the problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.
  8. Give a gap before you take a 10 problem practice test again.

Important: both practice tests and mock tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.

Resources that should be useful for you

You may refer to:

7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests or section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

Tutorials that you should refer to

Basic concepts on Geometry 1 lines points and triangles

Basic concepts on Geometry 2 quadrilaterals polygons squares

Basic and rich concepts on Geometry 3 Circles

Basic and rich Geometry concepts part 4 Arc angle subtending concept proof

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42nd solution set - 10 problems for SSC CGL exam: topic Mensuration - Answering time 12 mins

Problem 1.

A wooden box measures 20cm by 12cm by 10cm. Thickness of the wood is 1cm. Volume of the wood to make the box (in cm$^3$) is,

  1. 519
  2. 2400
  3. 960
  4. 1120

Solution 1 - Problem analysis and execution

The volume of the wood will be the outer volume of the cuboid box minus the inner volume of the holllow cuboid space which has dimensions 1cm less than the outer dimensions, that is, 19cm by 11cm by 9cm.

So the volume of the wood $=20\times{12}\times{10} - 19\times{11}\times{9}=2400-1881=519$ cm$^3$.

Answer: Option a: 519.

Key concepts used: Cuboid volume concepts.

Problem 2.

If a right circular cone is separated into solids of volumes $V_1$, $V_2$ and $V_3$ by two planes parallel to the base, which also trisect the altitude then $V_1 : V_2 : V_3$ is,

  1. 1 : 7 : 19
  2. 1 : 2 : 3
  3. 1 : 4 : 6
  4. 1 : 6 : 9

Solution 2 - Problem analysis and visualization

The figure below defines the problem.

ssc cgl solution set 42 mensuration 4-2

The two planes parallel to the base and passing through the lines FQG and DPE trisect the height AR and also the triangle sides AB and AC. Thus,

Height of top cone, $AP=\frac{1}{3}AR$, and

height of second cone AFG, $AQ=\frac{2}{3}AR$, where AR is the height of the main cone.

Similarly because of the similarity of the three triangles $\triangle ARC$, $\triangle AQG$ and $\triangle APE$,

Base radius of the top cone, $PE=\frac{1}{3}RC$, and

base radius of the middle cone, $QG=\frac{2}{3}RC$, where RC is the base radius of the main cone.

We are to find ratio of the volumes of three parts created by the two sections parallel to the base.

If we assume the volumes of the three parts starting from the top as $V_1$, $V_2$ and $V_3$ and three cones of heights AP, AQ and AR as $V_{1c}$, $V_{2c}$ and $V_{3c}$,

$V_1 = V_{1c}$,

$V_2=V_{2c} - V_{1c}$, and

$V_3=V_{3c} - V_{2c}$.

Solution 2 - Problem execution second stage

The volume of a cone is,

$V=\frac{1}{3}{\pi}{r^2}{h}$, where $r$ is the base radius and $h$ is the height.

So the ratio of the volumes of the three cones is,

$V_{1c} : V_{2c} : V_{3c} = {PE^2}.{AP} : {QG^2}.{AQ} : {RC^2}.{AR}$

$=\left(\displaystyle\frac{1}{3}\right)^3 : \left(\displaystyle\frac{2}{3}\right)^3 : 1$

$=1 : 8 : 27$, a total of 36 portions.

Finally then the required ratio of the three parts,

$V_{1} : V_{2} : V_{3}=(V_{1c}) : (V_{2c}-V_{1c}) : (V_{3c}-V_{2c})$

$=1 : (8-1) : (27-8)$

$=1 : 7 : 19$.

Answer: Option a : 1 : 7 : 19.

Key concepts used: Volume concepts of cone -- similar triangle properties -- ratio concepts.

Problem 3.

The largest sphere that can be carved out of a cube of side 7cm has a volume (in cm$^3$),

  1. 481.34
  2. 179.67
  3. 543.72
  4. 718.66

Solution 3 - Problem analysis and execution

The core problem concept is shown in the figure below in which the largest circle is drawn inside a square of side length 7cm. This depicts the state in a plane through the cube, perpendicular to the top face of the cube and through the centre. The side length of the square and the radius of the circle will have the same relationship as the side length of the cube and the radius of the largest sphere inscribed in it.

ssc cgl solution set 42 mensuration 4-3

The largest sphere that can be carved out of a cube will have then its radius as half the side length of the cube. In this case then the radius of the sphere is $\frac{7}{2}$cm.

Volume of the sphere is,

$V_{sphere}=\displaystyle\frac{4}{3}{\pi}{\left(\displaystyle\frac{7}{2}\right)}^3$

$=\displaystyle\frac{4}{3}\times{\displaystyle\frac{22}{7}}\times{\displaystyle\frac{7^3}{8}}$

$=\displaystyle\frac{11\times{49}}{3}$

$=\displaystyle\frac{539}{3}$

$=179.67$.

Answer: b: 179.67.

Key concepts used: Visualization -- volume of sphere -- 3D inscribing concept -- efficient simplification -- delayed calculation.

Note: We follow the concept of efficient simplification by delaying multiplication of numbers as late as possible and at the last stage canceling out the factors between the denominator and the numerator to reduce the calculation to its minimum. In this case we needed to multiply 49 by 11 following the rule of multiplication by 11 (adding unit's digit and ten's digit and pushing the unit's digit of the result between the two digits of the original number and adding the ten's digit of the result, if any, to the ten's digit of the original number). Finally division of 539 by 3 was a direct mental division approximating 0.66666.... to 0.67. Practice of efficient simplification saves valuable seconds and also ensures accuracy of the result.

Problem 4.

The height of a right prism with a square base is 15cm. If the total surface area of the prism is 608 sq cm its volume (in cm$^3$) is,

  1. 910
  2. 920
  3. 980
  4. 960

Solution 4 - Problem analysis and visualization

The following figure describes the problem.

ssc cgl solution set 42 mensuration 4-4

A right prism is a three dimentional solid shape with a polygon base that remains same at all horizontal cross-sections along the height perpendicular to the base. In our case, as the base is a square, the top face is also a square and all horizontal cross-sections parallel to the base are squares. If the base were a triangle, the prism would have been a right triangular prism.

Thus volume of a right square prism is,

$V_{prism}=\text{Base}\times{\text{Height}}=15d^2$ where $d$ is the side length of the square.

Also the total area of the right prism in this case is,

$A_{prism}=2{d}^2+4\times{15}{d}=608$,

Or, $d^2 +30d -304=0$,

Or, $(d-8)(d+38)=0$.

As $d$ cannot be negative, $d=8$.

So the volume of the right prism is,

$V_{prism}=8^2\times{15}=960$ cm$^3$.

Answer: Option d: 960.

Key concepts used: Visualization -- right square prism volume and area concepts -- factorization of a quadratic equation in a single variable -- basic algebraic concepts.

Problem 5.

The base of a right pyramid is an equilateral triangle of side $10\sqrt{3}$ cm. If the total surface area of the pyramid is $270\sqrt{3}$ sq cm, its height is,

  1. $10\sqrt{3}$ cm
  2. $12$ cm
  3. $10$ cm
  4. $12\sqrt{3}$ cm

Solution 5 - Problem analysis and visualization

The following figure represents the problem visual.

ssc cgl solution set 42 mensuration 4-5

By definition a right pyramid is a solid shape with $(n+1)$ faces where $n$ is the number of sides of the polygon base. Each vertex of the polygon base is connected to a single apex point with perpendicular to the base from the apex point passing through the centre of gravity of the polygon base.

In our right triangular pyramid, base $\triangle BCD$ is an equilateral triangle with side length as $10\sqrt{3}$ cm. All three slanting faces are isosceles triangles of same height $AQ=h_1$, say, and height of the pyramid $h$, say, is perpendicular to the base passing through the centroid G of the equilateral base. Each median such as DQ bisects opposite side BC and is perpendicular to it (as $\triangle BCD$ is an equilateral triangle).

The area of the equilateral base is,

$A_{base}=\displaystyle\frac{\sqrt{3}}{4}{d^2}$, where $d$ is the side length.

In this case,

$A_{base}=300\times{\displaystyle\frac{\sqrt{3}}{4}}=75\sqrt{3}.$

Total area of the three slanting faces is,

$A_{slanting}=3\times{\frac{1}{2}}\times{10\sqrt{3}}\times{h_1}$

$=15\sqrt{3}{h_1}$.

Total surface area of the pyramid is thus,

$A_{surface}=75\sqrt{3} + 15\sqrt{3}{h_1}=270\sqrt{3}$,

Or, $15h_1=195$,

Or, $h_1=13$.

Solution 5 - Problem execution second stage

Median langth of the base is,

$DQ=\sqrt{(10\sqrt{3})^2 - (5\sqrt{3})^2}=\sqrt{300-75}=15$

As centroid G divides the median in a 2 : 1 ratio,

$GQ=\frac{1}{3}\times{15}=5$.

Finally in right $\triangle AGQ$, height of pyramid,

$h=\sqrt{{h_1}^2 - GQ^2}=\sqrt{13^2 - 5^2}=12$ cm.

Answer: Option b: $12$ cm.

Key concept used: Right pyramid concepts -- Area of equilateral triangle concept -- median section ratio at centroid concept -- Pythagoras theorem.

Problem 6.

If a metallic cone of base radius 30cm and height 45cm is melted and recast into metallic spheres of radius 5cm find the maximum number of spheres that can be made,

  1. 81
  2. 41
  3. 40
  4. 80

Solution 6 - Problem analysis and execution

To get the maximum number of spheres of radius 5cm that can be made out of metal melted from the cone of base radius 30cm and height 45cm we need to divide the volume of the cone by the volume of a single sphere.

Volume of the cone,

$V_{cone}=\frac{1}{3}{\pi}\times{30^2}\times{45}$

$=900\times{15{\pi}}$

Volume of a single sphere is,

$V_{sphere}=\frac{4}{3}{\pi}\times{5^3}=\displaystyle\frac{500}{3}{\pi}$

So the required number of spheres is,

$N=\displaystyle\frac{V_{cone}}{V_{sphere}}$

$=\displaystyle\frac{900\times{15}\times{3}}{500}$

$=81$.

Answer: Option a : 81.

Key concepts used: Visualization -- volume of sphere and cone concepts -- efficient simplification.

Problem 7.

The diameter of a circular wheel is 7m. How many revolutions will it make in travelling 22km?

  1. 1000
  2. 100
  3. 400
  4. 500

Solution 7 - Problem visualization and solution

With each complete revolution of the wheel, the distance traversed will be equal to the perimeter of the wheel which is,

$\text{Perimeter}=2{\pi}\times{\displaystyle\frac{7}{2}}=22$ m.

So to cover a distance of 22 km or 22000 m, 1000 revolutions will be required.

Answer: a: 1000.

Key concepts used: Circle perimeter -- cirular to linear distance mapping.

Problem 8.

In an equilateral $\triangle ABC$ of side 10cm, the side BC is trisected at D. The length of AD then (in cm) is,

  1. $3\sqrt{7}$
  2. $\displaystyle\frac{7\sqrt{10}}{3}$
  3. $\displaystyle\frac{10\sqrt{7}}{3}$
  4. $7\sqrt{3}$

Solution 8 - Problem analysis and visualization

The following figure represents the problem.

ssc cgl solution set 42 mensuration 4-8

As D trisects BC of length 2 cm, $BD=DE=EC=\displaystyle\frac{10}{3}$ cm where E also is the second trisecting point on BC.

Also as median AP is the perpendicular bisector of side BC, it bisects section DE so that,

$DP=\displaystyle\frac{10}{6}=\frac{5}{3}$.

Median length of the equilateral $\triangle ABC$ is,

$AP=\sqrt{100-25}=5\sqrt{3}$.

Finally then in right $\triangle APD$,

$AD=\sqrt{DP^2 + AP^2}$

$=\sqrt{\displaystyle\frac{25}{9} + 75}$

$=\sqrt{\displaystyle\frac{700}{9}}$

$=\displaystyle\frac{10\sqrt{7}}{3}$.

Answer: Option c: $\displaystyle\frac{10\sqrt{7}}{3}$.

Key concepts used: Visualization -- equilateral triangle concepts -- Pythagoras theorem.

Problem 9.

The length of perpendiculars drawn from any point in the interior of an equilateral triangle to the respective sides are $p_1$, $p_2$ and $p_3$. The length of each side of the triangle is then,

  1. $\displaystyle\frac{2}{\sqrt{3}}(p_1 +p_2 +p_3)$
  2. $\displaystyle\frac{4}{\sqrt{3}}(p_1 +p_2 +p_3)$
  3. $\displaystyle\frac{1}{3}(p_1 +p_2 +p_3)$
  4. $\displaystyle\frac{1}{\sqrt{3}}(p_1 +p_2 +p_3)$

Solution 9 - Problem analysis and visualization

The following figure corresponds to visualization of the problem.

ssc cgl solution set 42 mensuration 4-9

The perpendicular lengths $p_1$, $p_2$ and $p_3$ are from the internal poind D to sides AB, BC and CA of the equilateral $\triangle ABC$ of side length, say, $d$.

So sum of areas of the three triangles $\triangle DAB$, $\triangle DBC$ and $\triangle DCA$ is,

$A=\displaystyle\frac{1}{2}d(p_1 + p_2 + p_3)$.

Again, the area of the equilateral $\triangle ABC$ is,

$A=\displaystyle\frac{\sqrt{3}}{4}d^2=\displaystyle\frac{1}{2}d(p_1 + p_2 + p_3)$.

Or, $d=\displaystyle\frac{2}{\sqrt{3}}(p_1+p_2+p_3)$.

Answer: Option a: $\displaystyle\frac{2}{\sqrt{3}}(p_1+p_2+p_3)$.

Key concepts used: Visualization -- area of equilateral triangle -- area of triangle -- technique of area decomposition into components.

Problem 10.

If arcs of same length in two circles subtend angles $60^0$ and $75^0$ at their centre, the ratio of their radii is,

  1. 3 : 4
  2. 3 : 5
  3. 4 : 5
  4. 5 : 4

Solution 10 - Problem analysis

By definition, in a circle of radius $r$ the length of an arc AB subtending an angle $\theta$ at the centre is,

$Arc_{length}\text{ of AB}=r\theta$, where $\theta$ is in radians.

The following is the figure that depicts the relation.

ssc cgl solution set 42 mensuration 4-10

This relation makes sense as we know, when the angle subtended by the arc is $2\pi$ the length of the arc is the perimeter of the circle which is, $2\pi{r}$.

Problem solving execution

Thus for same arc length in two circles of radii $r_1$ and $r_2$,

$r_1{\theta}_1=r_2{\theta}_2$,

Or, $r_1 : r_2 = \theta_2 : \theta_1 = 75^0 : 60^0 = 5 : 4$.

As the angles are in a ratio, the ratio of angles in radians will be same as ratio of angles in degrees.

Answer: d: 5 : 4.

Key concepts used: Arc length concept -- ratio concept.


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