## 43rd SSC CGL level Solution Set, 5th on topic Mensuration

This is the 43rd solution set of 10 practice problem exercise for SSC CGL exam and 5th 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 43, Mensuration 5*, 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:

**Before start,**go through the tutorials**Basic concepts on Geometry 1**,**, Basic concepts on Geometry 2**or any other short but good material to refresh your concepts if you so require.**Basic concepts on Geometry 3****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 15 minutes.**When the time limit of 15 minutes is over,**mark up to which you have answered,**but go on to complete the set.****At the end,**refer to the answers to the questions to mark your score at 15 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.**Identify and analyze**the problems that**you couldn't do**to learn how to solve those problems.**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.**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.**Give a gap**before you take a 10 problem practice test again.

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

**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

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****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*

**If you like, **you may * subscribe* to get latest content from this place.

### 43rd solution set - 10 problems for SSC CGL exam: topic Mensuration - Answering time 15 mins

**Problem 1.**

An area of a square km of land had a rainfall of 2cm over some time. If 50% of the raindrops were collected in a pool with a 100m by 10m base, what would be increase in water level in the pool?

- 1 km
- 10 km
- 1 m
- 10 m

**Solution 1 - Problem analysis and execution**

The volume of the water due to 2cm rainfall over an area of 1 square km of land is,

$V=0.02\times{1000}\times{1000}=20000$ m$^3$.

50% of this volume is, 10000 m$^3$.

As base area of the pool collecting this volume of water is, $100\times{10}=1000$ m$^2$, the water in the pool will rise by,

$h=\displaystyle\frac{10000}{1000}=10$ m.

**Answer:** Option d: 10 m.

**Key concepts used:** **Cuboid volume concepts -- unit equalization.**

**Note** : One should take care to convert all the values to the same unit before starting any simplification process. For example, a km and a cm have been converted here to a metre for final simplication. This simple * precautionary step* must be taken for error free solution result.

**Problem 2.**

Water is flowing at the rate of 3 km per hour through a pipe of circular cross-section of internal diameter 20 cm into a circular cistern of diameter 10 m and depth 2 m. In how much time would the cistern be filled?

- 2 hours 40 minutes
- 1 hour
- 1 hour 40 minutes
- 1 hour 20 minutes

**Solution 2 - Problem analysis and execution**

Pipe cross-section area is,

$A_{pipe}=0.01{\pi}$ m$^2$.

Volume of water flowing through this cross-section area in 1 hour at the speed of 3 km per hour is,

$V_{hour}=3000\times{0.01\pi}=30\pi$ m$^3$.

Volume of cylindrical cistern is,

$V_{cistern}={\pi}\times{5^2}\times{2}=50\pi$ m$^3$.

So the time required for the cistern to be filled by the pipe is,

$T=\displaystyle\frac{50\pi}{30\pi}=\frac{5}{3}$ hr $=\text{1 hour 40 minutes}$.

**Answer:** Option c : 1 hour 40 minutes.

**Key concepts used:** Flow volume concept -- cylindrical volume concept -- unit equalization.

**Problem 3.**

If the perimeter of a circle, a square and an equilateral triangle are same and their areas are C, S and T respectively which of the following is true?

- $C=S=T$
- $S \lt C \lt T$
- $C \gt S \gt T$
- $C \lt S \lt T$

**Solution 3 - Problem analysis**

The radius $r$ of the circle, the side length $a$ of the square and the side length $d$ of the equilateral triangle are the elements deciding the perimeter and the areas. Equality of the perimeters gives rise to three equality relationships between these three primary elements, and these relationships need to be used for comparison of the three areas.

**Solution 3 - Problem execution**

As the perimeters are equal, we have,

$2{\pi}{r}=4a=3d$,

This gives,

$d=\displaystyle\frac{4}{3}a$,

$a=\displaystyle\frac{1}{2}{\pi}{r}$, and

$d=\displaystyle\frac{2}{3}{\pi}{r}$.

We will now use these relationships in comparing the areas. The areas of the circle, square and the equilateral triangle are,

$C={\pi}{r^2}$,

$S=a^2$, and

$T=\displaystyle\frac{\sqrt{3}}{4}.d^2$.

**Comparison of C and S**

$S=\displaystyle\frac{\pi}{4}{\pi}.r^2=x.C$, where $x=\displaystyle\frac{\pi}{4}=\frac{3.1472}{4} \lt 1$ as $\pi=3.1472$.

So,

$C \gt S$.

**Comparison of C and T**

$T=\displaystyle\frac{\sqrt{3}}{4}.d^2$

$=\displaystyle\frac{\sqrt{3}}{4}.\frac{4}{9}{\pi}{\pi}r^2$

$=\displaystyle\frac{\pi}{3\sqrt{3}}.C=y.C$, where $y=\displaystyle\frac{\pi}{3\sqrt{3}}=\frac{3.1472}{3\times{1.7}}=\frac{3.1472}{5.1} \lt 1$

So $C \gt T$.

**Comparison of S and T**

As $S=\displaystyle\frac{3.1472}{4}C$, and

$T=\displaystyle\frac{3.1572}{5.1}C$,

$S \gt T$, as the denominator of $T$ is greater than that of $S$.

Finally then the required relationship is,

$C \gt S \gt T$.

**Answer:** c: $C \gt S \gt T$.

**Key concepts used:** Area and perimeter concept of circle, square equilateral triangle -- substitution technique -- squares of small two digit numbers, $17^2=289$ helped us to get the value of $\sqrt{3}$ -- variable value comparison technique -- coefficient comparison technique -- fraction comparison technique -- basic algebraic concepts -- efficient simplification.

**Note:** We didn't need to compute the values of $S$ and $T$ again for comparison as these values in terms of $C$ already determined their relative values. Though the problem is conceptually simple it might take longer than desired time if systematic solution path is not followed. Efficient simplification concepts help.

**Problem 4.**

The diameter of the front wheel of an engine is $2x$ cm and that of rear wheel is $2y$ cm. To cover the same distance, find the number of times the rear wheel will revolve when the front wheel revolves $n$ times.

- $\displaystyle\frac{nx}{y}$ times
- $\displaystyle\frac{yn}{x}$ times
- $\displaystyle\frac{n}{xy}$ times
- $\displaystyle\frac{xy}{n}$ times

#### Solution 4 - Problem analysis and execution

The distance traversed by each wheel would be number of revolutions times the perimeter of the wheel. As the distance traversed is same, we have,

$n\times{2\pi}\times{x} = m\times{2\pi}\times{y}$, where $m$ is the number of revolutions of the rear wheel. So,

$m = \displaystyle\frac{nx}{y}$ times.

**Answer:** Option a: $\displaystyle\frac{nx}{y}$.

**Key concepts used:** Visualization -- circular to linear distance transformation -- ratio and proportion.

**Problem 5.**

What part of a ditch 48 m long, 16.5 m broad and 4 m deep can be filled by digging a cylindrical tunnel of diameter 4 m and length 56 m (use $\pi = \displaystyle\frac{22}{7}$)?

- $\displaystyle\frac{1}{9}$
- $\displaystyle\frac{2}{9}$
- $\displaystyle\frac{8}{9}$
- $\displaystyle\frac{7}{9}$

**Solution 5 - Problem analysis and execution**

The volume of the cuboid ditch is,

$V_{ditch}=48\times{16.5}\times{4}=48\times{66}$ m$^3$.

The volume of the earth dug from the tunnel is,

$V_{tunnel}={\pi}\times{2^2}\times{56}=88\times{8}$ m$^3$.

The part of ditch filled by this amount of earth is the ratio of the two volumes,

$\displaystyle\frac{V_{tunnel}}{V_{ditch}}=\frac{88\times{8}}{48\times{66}}=\frac{2}{9}$.

**Answer:** Option b: $\displaystyle\frac{2}{9}$.

**Key concept used:** Cuboid volume concept -- cylinder volume concept -- efficient simplification.

**Problem 6.**

The difference in areas of two squares drawn on two line segments of different lengths is 32 sq cm. Find the length of the greater line segment if one is longer than the other by 2 cm.

- 7 cm
- 16 cm
- 11 cm
- 9 cm

**Solution 6 - Problem analysis and execution**

If the lengths of the two line segments are $a$ cm and $b$ cm, the difference in areas of the two squares is,

$a^2 - b^2 = (a +b)(a-b)=32$, where $a$ is the longer line segment.

As $a-b=2$, we have,

$2(a+b)=32$,

Or, $a + b=16$.

Adding this result with the equation of $a-b=2$,

$2a=18$,

Or, $a=9$ cm.

**Answer:** Option d : 9 cm.

**Key concepts used:** Area of a square -- Basic algebraic concepts -- linear equations in two variables.

**Problem 7.**

The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in cm$^2$) of the triangle is,

- 48
- 144
- 72
- 24

** Solution 7 - Problem visualization and solution**

We need to use here the relationship between the area of a triangle and the area of the triangle formed by the medians of the original triangle. The relationship is,

$A_t=\displaystyle\frac{4}{3}A_m$ where $A_t$ is Area of a triangle and $A_m$ is Area of the triangle formed by the medians.

As the values of three medians are given we will find the area of the triangle formed by the medians using Heron's formula,

$A =\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semi-perimeter of the triangle and $a$, $b$ and $c$ are the side lengths.

In case of the triangle of medians the semi-perimeter is,

$s_m=\displaystyle\frac{9+12+15}{2}=18$cm

So the area of the triangle of medians is,

$A_m=\sqrt{18(18-9)(18-12)(18-15)}$

$=54$ cm$^2$.

So the area of the original triangle is,

$A_t=\displaystyle\frac{4}{3}A_m$

$=72$ cm$^2$.

**Answer: **c**: **72.

**Key concepts used:** Rich concept of relationship between areas of a triangle and the triangle formed by its medians -- area of triangle from medians -- area of triangle from sides.

**Problem 8.**

The area of a triangle of side lengths 9 cm, 10 cm and 11 cm (in cm$^2$) is,

- $30$
- $30\sqrt{2}$
- $60\sqrt{2}$
- $60$

** Solution 8 - Problem analysis and execution**

As the area of the triangle based on the length of its sides is required, we need to use the well-known Heron's formula for solution,

$Area=\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter and $a$, $b$ and $c$ are the side lengths.

In this problem case semi-perimeter is,

$s=\displaystyle\frac{9+10+11}{2}=15$ and so,

$Area=\sqrt{15(15-9)(15-10)(15-11)}$

$=30\sqrt{2}$ cm$^2$.

**Answer:** Option b: $30\sqrt{2}$

**Key concepts used:** Triangle area from side lengths -- Heron's formula.

**Problem 9.**

The area of a circle is increased by 22 cm$^2$ by increasing its radius by 1 cm. The original radius of the circle is,

- 3 cm
- 5 cm
- 9 cm
- 7 cm

**Solution 9 - Problem analysis and execution**

If $r$ be the original radius, by the problem definition increase in area is,

$22={\pi}\left((r+1)^2 - r^2\right)$,

Or, $2r + 1=7$,

Or, $r=3$ cm.

**Answer:** Option a: 3 cm

**Key concepts used:** Area of circle -- change analysis -- basic algebraic concepts.

** Problem 10.**

The base of a right prism is a quadrilateral ABCD. Given that $AB=9$ cm, $BC=14$ cm, $CD=13$ cm, $DA=12$ cm and $\angle DAB=90^0$. If the volume of the prism be 2070 cm$^3$ then the area of the lateral surface of the prism is,

- 720 cm$^2$
- 810 cm$^2$
- 2070 cm$^2$
- 1260 cm$^2$

**Solution 10 - Problem analysis and visualization**

The following is the figure that depicts the problem situation.

By definition a prism has the same horizontal cross-section as its base along its height which is its axis. In our problem the base and the top face are then each a quadrilateral of side lengths $AB=9$ cm, $BC=14$ cm, $CD=13$ cm and $DA=12$ cm.

The key to the problem is the given information of $\angle DAB=90^0$.

This enables us to calculate the area of $\triangle DAB$ as,

$\text{Area of } \triangle DAB$

$=\displaystyle\frac{1}{2}\times{9}\times{12}$

$=54$ cm$^2$.

In addition, this key information gives us the length of diagonal DB that acts also as the third side of $\triangle DBC$. By Pythagoras theorem the length of DB is,

$DB = \sqrt{12^2 + 9^2}$

$=\sqrt{144+81}$

$=\sqrt{225}$

$=15$ cm.

With length of three sides of $\triangle DBC$ available we will use Heron's formula to deduce the area of the triangle.

#### Problem solving execution second stage

Semi-perimeter of $\triangle DBC$ is,

$s=\displaystyle\frac{13+14+15}{2}=21$ cm.

So the area of $\triangle DBC$ is,

$A_{\triangle DBC}=\sqrt{21(21-13)(21-14)(21-15)}$

$=84$ cm$^2$.

Thus the area of the base quadrilateral is,

$A_{base}=54+84=138$ cm$^2$.

And the volume of the prism is,

$V_{prism}=\text{Base}\times{\text{Height}}=138H=2070$,

Or, $H=\displaystyle\frac{2070}{138}=15$ cm.

Finally then the lateral surface area of the prism is,

$A_{lateral}=\text{Base perimeter}\times{\text{Height}}$

$=(9+14+13+12)\times{15}$

$=48\times{15}$

$=720$ cm$^2$

**Answer:** a: 720 cm$^2$.

**Key concepts used:** Deductive reasoning -- key information identification -- Pythagoras theorem -- prism volume and surface area concepts -- Heron's formula -- triangle area by sides.

### Other related question set and solution set on SSC CGL mensuration

**SSC CGL level Solution Set 88 on Mensuration 8**

**SSC CGL level Question Set 88 on Mensuration 8**

**SSC CGL level Solution Set 87 on Mensuration 7**

**SSC CGL level Question Set 87 on Mensuration 7**

**SSC CGL level Solution Set 86 on Mensuration 6**

**SSC CGL level Question Set 86 on Mensuration 6**

**SSC CGL level Question Set 43 on Mensuration 5**

**SSC CGL level Solution Set 42 on Mensuration 4**

**SSC CGL level Question Set 42 on Mensuration 4**

**SSC CGL level Solution Set 41 on Mensuration 3**

**SSC CGL level Question Set 41 on Mensuration 3**

**SSC CGL level Solution Set 27 on Mensuration 2**

**SSC CGL level Question Set 27 on Mensuration 2**

**SSC CGL level Solution Set 26 on Mensuration 1**

**SSC CGL level Question Set 26 on Mensuration 1**