Bank Clerk level Solved question Set 3, surface and volume measurement Mensuration 1

3rd Bank clerk level Solved Question Set, 1st on surface and volume measurement Mensuration

This is the 3rd solved question set of 10 practice problem exercise for Bank Clerk exams and the 1st on topic surface and volume measurement Mensuration. It contains,

1. 1st question set on Mensuration for Bank Clerk level exams to be answered in 10 minutes (10 chosen questions)
2. Answers to the questions, and
3. Detailed solutions explaining concepts and showing how to solve the problems quickly in mind with minimum writing.

For maximum gains, the test should be taken first, and then the solutions are to be referred to. But more importantly, to absorb the concepts, techniques and reasoning explained in the solutions, one must solve many problems in a systematic manner using the conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

3rd Question set - 10 problems for Bank Clerk exams: 1st on topic Mensuration - answering time 10 mins

Q1. Area of a circle is equal to the area of a rectangle. The breadth and perimeter of the rectangle are equal to 8.5m and 42m. Area of the circle is then (in sq m),

1. 146.25
2. 104.25
3. 128.25
4. 116.25
5. None of the above

Q2. A cistern 6m long and 4m wide contains water up to the height of 1m 25cm. What is the total area of the wet surface (in sq m)?

1. 42
2. 52
3. 49
4. 64
5. None of the above

Q3. The ratio between the length and breadth of a rectangular land is 5 : 3 respectively. If its perimeter is 48m, what will be its area in sq. m?

1. 115
2. 116
3. 120
4. Cannot be determined
5. None of the above

Q4. The height and base of a triangle are equal to the length and breadth of a rectangle respectively. If the perimeter of the rectangle is 86m and the difference between the length and breadth is 5m, what is the area of the triangle (in sq m}?

1. 216
2. 256
3. 242
4. 228
5. 224

Q5. What would be the cost of building a 7m wide garden around a circular field with diameter equal to 280m if the cost per sq m for building the garden is Rs. 21? (in Rs.)?

1. Rs. 1,11,624
2. Rs. 1,56,242
3. Rs. 2,06,118
4. Rs. 2,48,521
5. None of the above

Q6. Sum of perimeters of two squares is 80cm. If a side of the second square is three times a side of the first square, what is the area of the second square (in cm$^2$)?

1. 144
2. 36
3. 81
4. 324
5. 225

Q7. A circle with center at A has radius 7m more than a second circle with center at B. If the area of the circle with center at A is 1386 sq m, what is the difference between the circumference of the two circles (in m)?

1. 44
2. 56
3. 82
4. 70
5. 96

Q8. The parallel sides of a trapezium are in the ratio of 2 : 3 and their shortest distance is 12 cm. If the area of the trapezium is 480 sq cm, the longer of the two parallel sides is of length,

1. 48cm
2. 36cm
3. 56cm
4. 42cm

Q9. Diameter of a roller is 2.4m and its length is 1.68m. If it takes 1000 complete revolutions once over to level a field, the area of the field is (in sq m),

1. 12672
2. 12762
3. 11768
4. 12671

Q10. A slice from a circular pizza of diameter 14 inches is cut in such a way that the slice forms a sector angle of $45^0$. What is the area of the upper flat surface of the cut slice of the pizza (in sq inches)?

1. 17.75
2. 17.5
3. 19.25
4. 18.25
5. 19.75

Answers to the questions

Q1. Answer: Option e: None of the above.

Q2. Answer: Option c: 49.

Q3. Answer: Option e: None of the above.

Q4. Answer: Option d: 228.

Q5. Answer: Option e: None of the above.

Q6. Answer: Option e : 225.

Q7. Answer: Option a: 44.

Q8. Answer: Option a: 48cm.

Q9. Answer: Option a: 12672.

Q10. Answer: Option c: 19.25 sq inches.

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3rd solution set - 10 problems for Bank clerk exams: 1st on topic Mensuration - answering time 10 mins

Q1. Area of a circle is equal to the area of a rectangle. The breadth and perimeter of the rectangle are equal to 8.5m and 42m. Area of the circle is then (in sq m),

1. 146.25
2. 104.25
3. 128.25
4. 116.25
5. None of the above

Solution 1: Problem solving by area and perimeter relations

The perimeter of the rectangle is,

$\text{Perimeter }=2(\text{Breadth }+\text{ Length})$,

Or, $42=17+2\times{\text{Length}}$, two times breadth is 17m

Or, $\text{Length}=12.5$m, half of 25m is 12.5m.

So, Area of the rectangle which is equal to the area of the circle is,

$\text{Length}\times{\text{Breadth}}=12.5\times{8.5}=106.25$ sq m.

Answer: Option e: None of the above.

Key concepts used: Perimeter of a rectangle -- Area of a rectangle -- Solving in mind.

Note: You need not use the formula for area of a circle.

Q2. A cistern 6m long and 4m wide contains water up to the height of 1m 25cm. What is the total area of the wet surface (in sq m)?

1. 42
2. 52
3. 49
4. 64
5. None of the above

Solution 2: Problem solving using wall surface area of rectangular enclosed space

Perimeter of the horizontal cross-section of the cistern is,

$2(6+4)=20$m

So the surface area of the wet portion of the cistern wall is,

$20\times{1.25}=25$ sq m.

Surface area of the bottom of the cistern is,

$6\times{4}=24$ sq m.

So the total surface area of the wet portion of the cistern is,

$25+24=49$ sq m.

Answer: Option c: 49.

Key concepts used: Surface area of the walls of rectangular enclosed space -- Surface area of wet portion of a cistern -- Solving in mind.

The problem is simple enough to be solved in mind quickly in a few tens of seconds.

Note: even in such a simple problem, you need to be aware of the context, that is, the wet bottom of the cistern.

Q3. The ratio between the length and breadth of a rectangular land is 5 : 3 respectively. If its perimeter is 48m, what will be its area in sq. m?

1. 115
2. 116
3. 120
4. Cannot be determined
5. None of the above

Solution 3: Problem solution by ratio concepts and area and perimeter of a rectangle 

By HCF reintroduction technique on ratios, we have the actual values of length and breadth as, $5x$ m and $3x$ m respectively where $x$ is the cancelled out HCF in the ratio.

Perimeter of the rectangle,

$2(5x+3x)=16x=48$m, as given,

Or, $x=3$.

So length is 15m and breadth is 9m, and the area, 135 sq m.

Answer: Option e: None of the above.

Key concepts used: Ratio concepts -- HCF reintroduction technique -- Area and perimeter of a rectangle -- Solving in mind.

The problem could easily be solved in mind.

Q4. The height and base of a triangle are equal to the length and breadth of a rectangle respectively. If the perimeter of the rectangle is 86m and the difference between the length and breadth is 5m, what is the area of the triangle (in sq m}?

1. 216
2. 256
3. 242
4. 228
5. 224

Solution 4: Problem solving using area and perimeter of a rectangle and area of a triangle

Let height of the triangle and length of the rectangle be $L$ m and base of the triangle and breadth of the rectangle be $B$ m.

So perimeter of rectangle is,

$2(L+B)=86$m,

Or, $L+B=43$m.

Given,

$L-B=5$m.

So $L=24$m and $B=19$m.

Thus area of the triangle is,

$\displaystyle\frac{1}{2}(LB)=12\times{19}=228$ sq m.

Answer: Option d: 228.

Key concepts used: Area and perimeter of a rectangle -- Linear equations in two variables -- Area of a triangle -- Solving in mind.

Following the simple to understand process, solution could be reached wholly in mind.

Q5. What would be the cost of building a 7m wide garden around a circular field with diameter equal to 280m if the cost per sq m for building the garden is Rs. 21? (in Rs.)?

1. Rs. 1,11,624
2. Rs. 1,56,242
3. Rs. 2,06,118
4. Rs. 2,48,521
5. None of the above

Solution 5: Problem solving using area of a circular ring

Area of the garden is the difference between areas of two concentric circles, the larger with radius, $140+7=147$m and the smaller circular field of radius 140 m.

It is then,

$\text{Garden area}=\pi(147^2-140^2)$, area of a circle is, $\pi R^2$ where $R$ is the radius,

$=\displaystyle\frac{22}{7}(147+140)(147-140)$, by the often used algebraic relation, $a^2-b^2=(a+b)(a-b)$

$=22\times{287}$

$=6314$ sq m.

As cost per sq m is Rs. 21, the total cost will be,

$C=126280+6314=\text{Rs. }132594$.

Answer: Option e: None of the above.

Key concepts used: Area of a circular ring -- Area of circles -- Basic algebraic concepts.

To ensure accuracy we needed to evaluate products, $22\times{287}$ and $6314\times{21}$ by hand calculation. Still the time taken was comfortably within a minute.

Q6. Sum of perimeters of two squares is 80cm. If a side of the second square is three times a side of the first square, what is the area of the second square (in cm$^2$)?

1. 144
2. 36
3. 81
4. 324
5. 225

Solution 6: Problem analysis and solution by perimeter and area of squares

Assuming length of a side of first square as $x$ cm, length of a side of the second square becomes $3x$ cm and the sum of their perimeters,

$4x+12x=16x=80$,

Or, $x=5$ cm

So the length of a side of second square is 15cm and its area, $15^2=225\text{ cm}^2$.

Answer: Option e : 225.

Key concepts used: Perimeter of a square -- Area of a square -- Solving in mind.

The problem was simple enough to solve mentally and quickly.

Q7. A circle with center at A has radius 7m more than a second circle with center at B. If the area of the circle with center at A is 1386 sq m, what is the difference between the circumference of the two circles (in m)?

1. 44
2. 56
3. 82
4. 70
5. 96

Solution 7: Problem solving using area and perimeter of circles

Area of the circle with center at A has area,

$\pi a^2=\displaystyle\frac{22}{7}a^2=1386$, assuming $a$ as the radius of circle with center at A,

Or, $a^2=7\times{63}=7^2\times{3^2}$.

So, $a=21$m, and radius of circle with center at B is,

$b=21-7=14$m.

Difference of perimeter between the two circles is,

$\pi(2a-2b)=\displaystyle\frac{22}{7}(42-28)=44$m, perimeter of a circle of radius $r$ is $2\pi r$,

Answer: Option a: 44.

Key concepts used:  Area of a circle -- Perimeter of a circle -- Solving in mind.

The problem could easily be solved in mind in a few tens of seconds.

Q8. The parallel sides of a trapezium are in the ratio of 2 : 3 and their shortest distance is 12 cm. If the area of the trapezium is 480 sq cm, the longer of the two parallel sides is of length,

1. 48cm
2. 36cm
3. 56cm
4. 42cm

Solution 8: Problem solving using area of a trapezium and ratio concept

By reintroducing the cancelled out HCF in the ratio as $x$, we get the actual values of the two parallel sides of the trapezium as, $2x$ and $3x$ both in cm.

Area of a trapezium may be considered as the sum of the area of the rectangle formed by the smaller of the two parallel sides and the shortest distance between these two parallel sides, and the triangle formed with base as difference between two parallel sides, height as shortest distance between parallel sides and other two sides as the rest of the two sides of the trapezium. The following figure makes it clear.

The original trapezium is ABCD on left. On the right, the side DC is moved parallely to be positioned as AF, effectively moving the $\triangle DHC$ by cutting it out from the trapezium and placing as $\triangle AEF$. This completes the other half of the triangle ABF. Area of the trapezium on the left is, area of rectangle AEHD and the areas of two part triangles, $\triangle ABE$ and $\triangle DHC$. By merging the two parts now, the area of the trapezium can be equated to the area of the rectangle with length AD, the smaller of the two parallel sides, and height AE plus area of the $\triangle ABF$ with base as difference between the two parallel sides and same height AE.

Coming back to our problem,

Smaller side length is, $2x$, difference between the two parallel sides $x$, the base of the triangle, and height is 12cm.

Area of rectangle AEHD,

$12\times{2x}=24x$, and area of the $\triangle ABF$ is,

$\displaystyle\frac{1}{2}x\times{12}=6x$.

So total area of the trapezium is,

$30x=480$,

Or, $x=16$.

So the longer side is, $3x=48$cm.

Answer: Option a: 48cm.

Key concepts used: Area of trapezium -- Ratio proportion -- HCF reintroduction technique -- Solving in mind.

Though explanation took time and length, if you know it, solution is quick.

Q9. Diameter of a roller is 2.4m and its length is 1.68m. If it takes 1000 complete revolutions once over to level a field, the area of the field is (in sq m),

1. 12672
2. 12762
3. 11768
4. 12671

Solution 9: Problem solution by roller perimeter times revolutions giving length and roller length as breadth of the field

Cylindrical Roller perimeter with diameter $d=2.4$m is,

$\pi d=\displaystyle\frac{22}{7}\times{2.4}$m

This perimeter times number of revolutions of 1000 gives the length of the field, while the breadth of the field is given by the roller length. 

So area of the field is,

$\displaystyle\frac{22}{7}\times{2.4}\times{1000}\times{1.68}$

$=22\times{24}\times{24}$

$=22\times{576}$

$=12672$ sq m.

Answer: Option a: 12672.

Key concepts used: Rolled area concept -- Perimeter of a cylinder -- Revolutions of a circular ring gives distance the ring covers -- Visualization.

We needed to write the calculation intensive step and did the final product calculation by hand for accuracy.

Important to note: Mensuration problems technically are easy as only a few formulas are to be used, but unless you can visualize what really is happening with the shapes, you may not be able to solve some of the more involved problems.

Q10. A slice from a circular pizza of diameter 14 inches is cut in such a way that the slice forms a sector angle of $45^0$. What is the area of the upper flat surface of the cut slice of the pizza (in sq inches)?

1. 17.75
2. 17.5
3. 19.25
4. 18.25
5. 19.75

Solution 10: Problem solution by angle to area proportionality of a circle

$360^0$ of a circle covers its whole area, $\pi r^2$,

So by unitary method, $45^0$ will cover an area, 

$\displaystyle\frac{45}{360}(\pi r^2)$

$=\displaystyle\frac{1}{8}\times{\displaystyle\frac{22}{7}}\times{7^2}$

$=\displaystyle\frac{77}{4}$

$=19.25$ sq inches.

Answer: Option c: 19.25 sq inches.

Key concepts used: Area of a circle -- Area held by an arc of a circle -- Arc angle to area proportionality of a circle -- Solving in mind.

The problem was simple enough to be comfortably and quickly solved in mind.

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