Mensuration.......[competitive. exam prep.... topic 27]
Dear Readers,
We are providing you Important Short Tricks on Mensuration Questions which are usually asked in Bank Exams. Use these below given short cuts to solve questions within minimum time. These shortcuts will be very helpful for your upcoming All Bank Exam 2016.To make the chapter easy for you all, we are providing you all some Important Short Tricks to Mensuration Questions which will surely make the chapter easy for you all.
About Mensuration :-
Mensuration questions are an important part of Quant paper in banking exam. Mensuration question asked in banking exam related to Perimeter and Area.
About Perimeter & Area:-
To put simply, Area measures the area of shape i.e. the space that shape takes up. Perimeter is the measurement of the boundary of the figure.
1. Square:- A square four sided polygon characterized by right angles and sides of equal length.
Area = side2,
Perimeter = 4side
2. Rectangle:- A four sided flat shape with straight sides where all interior angles are right angles (90°). Also opposite sides are parallel and of equal length.
Area = Length × Breadth,
Perimeter = 2(L+B)
3. Circle;- Circle is the locus of points equidistant from a given point, the center of the circle. The common distance from the center of the circle to its points is called radius
Area = π(radius)2,
Perimeter = 2πradius
Let’s look at some questions asked:-
Que 1. The length of rectangular plot is 20 m more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per meter is Rs. 5300, what is the length of the plot?
Sol. Cost = Rate × Perimeter
perimeter = 5300/26.50=200
2[L+B]=200
2[L+L-20]=200
L-10=50
L=60
Que 2. If the length is to be increased by 10% and breadth of a rectangle plot to be decreased by 12%, then find % change in area?
Sol:- In such questions, use formula: Increment in area = [L%+B%+{(L%+B%)/100)}]
Increment in area = 10 – 12(because breadth is decreased) {+ 10× (-12/100)} = -3.2%
Que 3. The length of rectangular plot is increased by 60%. By what percentage should the width be decreased to maintain the same area?
Sol:- In such questions, use formula:
Required % decrease in breadth = [%change in L{100/(100+%change in L)}]
Required % decrease in breadth = 60 (100/160) = 37.5%
Que 4. If radius of circle is increased by 5% find the percentage change in its area.
In such questions, use formula:
Sol:-
Change in area = (2x + x2/100) %
Change in area = 2×5 + 52/100 = 10 + ¼ = 10.25% increment.
Note: In such questions, negative sign implies decrements while positive sign shows increment.
Que 5. The circumference of a circle is 100 cm. Find the side of the square inscribed in the circle.
Sol. Always use this formula, Side of square inscribed in a circle of radius r = Root 2
circumefrence = 100
2πr=100
r = 50/π
Side of square = Root 2 (50/π)
Note: Similar formula:
1) Area of largest triangle inscribed in a semi-circle of radius r = r2
2) Area of largest circle that can be drawn in a square of side x = π(x/2)2
Que 6. The length and breadth of the floor of the room are 20 by 10 feet respect. Square tiles of 2 feet are to be laid. Black tiles are laid in first row on all sides, white tiles on 1/3rd of remaining sides and blue tiles on the rest. How many blue tiles are required?
Sol:- Side of a tile = 2 feet ⇒ Area of 1 tile = 22 = 4 sq ft. —– (1)
Length left after lying black tile on 4 sides = 20 – 4
⇒ Area left after black tiles = (20 – 4)×(10-4) = 96 sq. ft.
Area left after white tiles = 2/3×96 = 64 sq. ft.
⇒ Area for blue tiles = 64 sq. ft.
Number of blue tiles = 64/4 = 16 (using 1)
Que 7. A cow is tethered in the middle of the field with a 14 ft long rope. If the cow grazes 100 sq. ft. per day, then approximate time taken to graze the whole field?
Sol: Here, the rope of cow is like radius
Area = π (14)2
No. of days = (Area of field)/Rate of cow = π (14)2/100 = 6 days(approx)
Que 8. A circle and a rectangle have same perimeter. Sides of rectangle are 18 by 26 cm. What is the area of circle?
Sol:- 2πr = 2 (18 + 26) ⇒ r = 14 cm
Area = π r2 = 616 cm2
Que 9. What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle?
Sol:-
Area of triangle = ½ × L × B
Area of rectangle = L × B
Area of a rectangle: area of a triangle = L × B: ½ × L × B = 2: 1
Que 10. In a rectangular plot, a cow is tied down at a corner with a rope of 14m long. Find the area that cow can graze?
Sol. The area that cow can graze can be illustrated as shaded area:
Here, the shaded area is quarter of a circle with radius 14m,
Area of grazed field = ¼ × π (14)2 = 308 m2
How to Use this Trick
Check whether π=22/7 has been used in the formula for finding out the Particular Area, Curved Surface Area, Total Area, Volume, etc. If it is so, then
mensuration chart
Here are examples to explain the chart given above.
Example 1:
Find the surface area of a sphere whose volume is 4851 cubic meters.
a) 1380 m2
b) 1360 m2
c) 1368 m2
d) 1386 m2
Using the Trick:
We know that surface area of a sphere = 4πr2
It means ‘π’ has been used in finding out the surface area of the sphere.
We can easily see that only ‘1386’ from the given options is divisible by ‘11’
Hence, surface area of the sphere = 1386 m2
Example 2:
The radius and height of a right circular cylinder are 14 cm & 21 cm respectively. Find its volume.
a) 12836 cm3
b) 12736 cm3
c) 12936 cm3
d) 12837 cm3
Using the Trick
The know that volume of Cylinder = πr2h
We must check divisibility by ‘11’. Here, both ‘12936’ and ‘12837’ are divisible by 11. But you also notice that radius (14 cm) & height (21 cm) are both multiples of 7. So the option divisible by ‘7’ is your answer.
Hence, volume of a right circular cylinder = 12936 cm3 (since this is the only option divisible by 7)
We can test this as follows:
Volume of the given cylinder
= (22/7) × 14 × 14 × 14 × 21 cm3
= 22 × 14 × 14 × 3 cm3
Volume must be divisible by ‘7’.
Example 3:
The radius and height of a right circular cone are 7 cm & 18 cm respectively. Find its volume.
a) 814 cm3
b) 624 cm3
c) 825 cm3
d) 924 cm3
Using the Trick:
The option should be divisible by ‘11’ because ‘π’ has been used in finding its volume. One of the parameters is a multiple of 7 without being a higher power. So we must go through fundamentals.
Now, volume of a right circular cone = (1/3)πr2h
= (1/3) × (22/7) × 7 × 7 × 18 cm3
= 22 × 7 × 6
Clearly, we need an answer that is a multiple of 11, 7 as well as 3.
Among the given options, 814, 825 and 924 are all multiples of 11. However, we see that only one option is divisible by 7. So this is the correct answer.
Hence, volume of the given cone = 924 cm3
Example 4:
Find the circumference of a circle whose radius is 49 cm.
a) 208 cm
b) 288 cm
c) 308 cm
d) 407 cm
Using the Trick:
The option should be divisible by ‘11’ because ‘π’ has been used in finding its circumference. One of the parameters is a higher power of 7. Thus, we need to find the only option that is a multiple of 7. If, however, we find more than one option that is a multiple of 7, we need to go through fundamentals.
Among the options, 308 and 407 are both multiples of 11. However, only 308 is a multiple of 7. So circumference of the circle = 308 cm.
We can test this as follows:
Circumference of circle = 2πr cm
= 2 × (22/7) × 7 × 49 cm
= 2 × 22 × 7 cm
Remember: If there is only one parameter equal to ‘7’ or multiple of ‘7’ and this parameter is not in a higher power in the formula, the answer will not be divisible by ‘7’
Example 5:
Find the curved surface area of a right circular cylinder whose radius & height are 14 cm & 50 cm respectively.
a) 3300 cm2
b) 3420 cm2
c) 4440 cm2
d) 4400 cm2
Solution:
Curved surface area of a right circular cylinder = 2πrh
Here only one parameter (r = 14 cm) is a multiple of 7 (without being a higher power of 7). This parameter is used only as ‘r’ and not in its higher powers. So we see that our answer will not be a multiple of 7. However, the presence of π means that it will still be a multiple of 11.
Among the given options, 3300 and 4400 are both multiples of 11. We also see that both are not multiples of 7 either. However, we can see that none of our parameters are multiples of 3, so curved surface area cannot be a multiple of 3 either. So our answer cannot be 3300.
Therefore curved surface area of right circular cylinder must be 4400 cm2.
We can test this as follows:
Curved surface area of a right circular cylinder = 2πrh
=2 × (22/7) × 14 × 50
=2 × 22 × 2 × 50
= 4400 cm3
NOTE:
The whole trick is based on the multiplication & divisibility by 11 & 7. So you are advised to use this trick very carefully using your quick mental mathematics. Please learn divisibility rules for both 11 and 7 for this purpose – they are both pretty simple! Please do not waste time solving these types of questions.
If there is a ‘None of these’ among the given options, then don’t use this trick.
Check whether π=22/7 has been used in the formula for finding out the Particular Area, Curved Surface Area, Total Area, Volume, etc. If it is so, then
mensuration chart
Here are examples to explain the chart given above.
Example 1:
Find the surface area of a sphere whose volume is 4851 cubic meters.
a) 1380 m2
b) 1360 m2
c) 1368 m2
d) 1386 m2
Using the Trick:
We know that surface area of a sphere = 4πr2
It means ‘π’ has been used in finding out the surface area of the sphere.
We can easily see that only ‘1386’ from the given options is divisible by ‘11’
Hence, surface area of the sphere = 1386 m2
Example 2:
The radius and height of a right circular cylinder are 14 cm & 21 cm respectively. Find its volume.
a) 12836 cm3
b) 12736 cm3
c) 12936 cm3
d) 12837 cm3
Using the Trick
The know that volume of Cylinder = πr2h
We must check divisibility by ‘11’. Here, both ‘12936’ and ‘12837’ are divisible by 11. But you also notice that radius (14 cm) & height (21 cm) are both multiples of 7. So the option divisible by ‘7’ is your answer.
Hence, volume of a right circular cylinder = 12936 cm3 (since this is the only option divisible by 7)
We can test this as follows:
Volume of the given cylinder
= (22/7) × 14 × 14 × 14 × 21 cm3
= 22 × 14 × 14 × 3 cm3
Volume must be divisible by ‘7’.
Example 3:
The radius and height of a right circular cone are 7 cm & 18 cm respectively. Find its volume.
a) 814 cm3
b) 624 cm3
c) 825 cm3
d) 924 cm3
Using the Trick:
The option should be divisible by ‘11’ because ‘π’ has been used in finding its volume. One of the parameters is a multiple of 7 without being a higher power. So we must go through fundamentals.
Now, volume of a right circular cone = (1/3)πr2h
= (1/3) × (22/7) × 7 × 7 × 18 cm3
= 22 × 7 × 6
Clearly, we need an answer that is a multiple of 11, 7 as well as 3.
Among the given options, 814, 825 and 924 are all multiples of 11. However, we see that only one option is divisible by 7. So this is the correct answer.
Hence, volume of the given cone = 924 cm3
Example 4:
Find the circumference of a circle whose radius is 49 cm.
a) 208 cm
b) 288 cm
c) 308 cm
d) 407 cm
Using the Trick:
The option should be divisible by ‘11’ because ‘π’ has been used in finding its circumference. One of the parameters is a higher power of 7. Thus, we need to find the only option that is a multiple of 7. If, however, we find more than one option that is a multiple of 7, we need to go through fundamentals.
Among the options, 308 and 407 are both multiples of 11. However, only 308 is a multiple of 7. So circumference of the circle = 308 cm.
We can test this as follows:
Circumference of circle = 2πr cm
= 2 × (22/7) × 7 × 49 cm
= 2 × 22 × 7 cm
Remember: If there is only one parameter equal to ‘7’ or multiple of ‘7’ and this parameter is not in a higher power in the formula, the answer will not be divisible by ‘7’
Example 5:
Find the curved surface area of a right circular cylinder whose radius & height are 14 cm & 50 cm respectively.
a) 3300 cm2
b) 3420 cm2
c) 4440 cm2
d) 4400 cm2
Solution:
Curved surface area of a right circular cylinder = 2πrh
Here only one parameter (r = 14 cm) is a multiple of 7 (without being a higher power of 7). This parameter is used only as ‘r’ and not in its higher powers. So we see that our answer will not be a multiple of 7. However, the presence of π means that it will still be a multiple of 11.
Among the given options, 3300 and 4400 are both multiples of 11. We also see that both are not multiples of 7 either. However, we can see that none of our parameters are multiples of 3, so curved surface area cannot be a multiple of 3 either. So our answer cannot be 3300.
Therefore curved surface area of right circular cylinder must be 4400 cm2.
We can test this as follows:
Curved surface area of a right circular cylinder = 2πrh
=2 × (22/7) × 14 × 50
=2 × 22 × 2 × 50
= 4400 cm3
NOTE:
The whole trick is based on the multiplication & divisibility by 11 & 7. So you are advised to use this trick very carefully using your quick mental mathematics. Please learn divisibility rules for both 11 and 7 for this purpose – they are both pretty simple! Please do not waste time solving these types of questions.
If there is a ‘None of these’ among the given options, then don’t use this trick.
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