This quick trick to solve Mensuration problems is totally based on your thinking. It involves next to no calculations. This trick will save you precious time in your IBPS Clerk or SBI PO exams.

The trick will help you get answers in just a few seconds. Solve the questions in detail first to understand this method more clearly. Once you get acquainted with the mechanisms of this trick, you will find it extremely helpful.

Below are given types of questions where this trick can be applied.

**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

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 m^{2
}b) 1360 m^{2
}c) 1368 m^{2
}d) 1386 m^{2}

**Using the Trick:**

We know that surface area of a sphere = 4πr^{2}

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 m^{2}

** **

**Example 2:**

The radius and height of a right circular cylinder are 14 cm & 21 cm respectively. Find its volume.

a) 12836 cm^{3
}b) 12736 cm^{3
}c) 12936 cm^{3
}d) 12837 cm^{3}

**Using the Trick:**

The know that volume of Cylinder = πr^{2}h

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 cm^{3} (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 cm^{3}

= 22 × 14 × 14 × 3 cm^{3}

⇒ 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 cm^{3
}b) 624 cm^{3
}c) 825 cm^{3
}d) 924 cm^{3}

**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)πr^{2}h

= (1/3) × (22/7) × 7 × 7 × 18 cm^{3}

= 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 cm^{3}

** **

**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 cm^{2
}b) 3420 cm^{2
}c) 4440 cm^{2
}d) 4400 cm^{2}

**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 cm^{2}.

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 cm^{3}

**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.**