Improve Your Calculation Speed - Percentage Change

We need to calculate 21% of 578. A good approximation of a percentage change can be done using the 10-1 approach.

In the 10-1 approach, one starts by calculating the rounded off values representing 10% and 1% of the number. Now 10% of 578 is 58 (rounded off) and 1% is 6 (rounded off).

Now we can calculate any percentage of this number by using 10% and 1%

To calculate 21% of this number, we start by getting 20%, which is 10%*2 = (58 × 2) = 116. We now add 1% i.e. 6 in this and get the answer as 122.

To calculate 19%, we subtract 1% from 20% and get the answer as 116 – 6 = 110.

 Now let’s say a problem requires us to calculate 52% of 281. Its 10% is 28 and 1% is 3, both being rounded up values. First we find 50% of this number i.e. 140 (half of the number) and then 2% of this will be added in this i.e. 6 to get the answer as 146.

Knowledge of percentage change is also important for cracking data interpretation questions.

The formula that we use for percentage change from P to Q is: 100 * (Q – P)/P. 

In the pressure of an exam, especially when we are working with larger numbers, we take more time when we write.

Let’s try to do this mentally.

 We have to find the percentage change from 271 to 353. Here the difference between the two numbers are 82 and 10% of the base value i.e. 271 is 27. How many 27s can fit into 82?

Three times of 27 is 81. Hence it is approximately 30%.

Let us take another example, say find percentage change from 911 to 938.

The difference between the two is 27. In this case 10% is 91 and 1% is 9. Multiplying 9 by 3 we get 27. Hence percentage change is 3%.

To conclude, the 10-1 approach can be used to be able to calculate percentage change mentally, thereby saving precious time in an exam!

Geometry Bring in Focus: PYTHAGORAS

Pythagorean Theorem:

The Pythagorean theorem, also known as 'Pythagoras' theorem, states that, “the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides”. The theorem can also be written as an equation relating the lengths of the sides a, b and c, often called the "Pythagorean equation”. Although Pythagoras theorem is applicable in case of a right-angled triangle only, yet it has a lot of direct and indirect applications.

The basic formula to calculate hypotenuse is a² + b² = c², where c represents the length of the hypotenuse and a and b the lengths of the triangle's other two sides.

If the length of both a and b are known, then c can be calculated as-

Classification of triangles on Pythagoras theorem:

• If a² + b² < c², then the triangle would be an obtuse angled triangle.
• If a² + b² > c², then the triangle would be an acute angled triangle.

Pythagorean Triplets:

In a set of 3 positive integers (a, b, c) which satisfies Pythagoras theorem, the greatest number would always represent the hypotenuse and any of the two smaller numbers could either be the base or hypotenuse.

The following is a list of common Pythagorean triplets, which are asked in the exams directly or indirectly:

(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53)

A primitive Pythagorean triplet is one in which a, b and c are co-prime.

Note: If all the numbers in the triplet is multiplied by any constant number, then the resulting numbers would also make a Pythagorean Triplet. As you know that, 3, 4, 5 is a triplet, so when all the three terms are multiplied with any constant number, that will also make a triplet.

Thus, (6,8,10), (9,12,15), (12,16,20), …(30,40,50), all these will be triplets.

Similarly, (10,24,26), (15,36,39), (20,48,52), (25,60,65) will also be Pythagoras triplets as (5,12,13) is a triplet; and all these are multiples of the same. You can learn the application of the Pythagoras theorem in the following examples.

Example 1:

Find the area of triangle with sides 12cm, 35cm, 37cm.

Solution:

You know that it is a Pythagorastriplet, so the hypotenuse must be 37 i.e. longest side.

The rest 2 sides are perpendicular and base.

Area of triangle= ½ × Base × Height
Area of triangle = ½ × 12 × 35 = 210

Example 2:

Find the area of a right-angled triangle of hypotenuse 91 cm and height 35 cm.

Solution:

The given triangle is right angled, so its sides must form a Pythagoras triplet. You know (5, 12, 13) is a triplet, and by observing the lengths of the given side, there’s a common factor i.e. 7. When it is given that hypotenuse is 91 (i.e. 13 × 7) and height is 35 (i.e. 5 × 7), you can say that the base must be 12 × 7 = 84.

Area of triangle = ½ × Base × Height
⇒½ × 91 × 84 = 3822 cm²

Example 3:

Find the area of rectangle ABCD.

Solution:

As you can see from the equations, this question involves complex calculation. When you know (28, 45, 53) is a triplet, then twice of that i.e. (56, 90, and 106) will also be a triplet. As 90 and 106 are already given to us, the third side will be 56 (i.e. 28 × 2) because ∠ADC=90°, as ABCD is a rectangle. So, AD = 56 cm.

Area of rectangle = (56 x 90) = 5040 cm².

Note: Here that having knowledge of the triplets can really make the calculationsin triangle-based questions very easy. Do remember these triplets, those will prove to be handy in papers.

Quick Review: Time, Speed & Distance

The most important relationship between these three quantities, and possibly the only one which needs to be known is,

Distance = Speed × Time

Speed = Distance traveled / Time

Time = Distance traveled / Speed

Average speed, if equal distances are covered at a km/hr and b km/hr is 2ab/a+b

If two bodies are moving in the same direction at a speed of a and b respectively, then their relative speed is the difference of the two speeds.

If two bodies are moving in the opposite directions at a speed of a and b respectively, then their relative speed is a + b.

Two objects A and B moving along a circular path in the same direction, having started simultaneously and from the same point traveling at speeds of a and b, will meet again when the faster object has gained one full circle over the slower object, i.e. when the relative speed |a-b|completes one full round. The two objects will again meet at the starting point at a time, which is the LCM of the time taken for each of the objects individually to complete one round.

If the length of a train is L meters and the speed of the train is S m/s, then the time taken by the train to pass a stationary man/pole is L/S sec.

If the length of the train is L₁ and its speed is S m/s and the length of a platform (stationary object of comparable length) is L₂, then the time taken by the train to cross the platform is (L₁ + L₂)/S sec.

If the lengths of 2 trains are L₁ (faster) and L₂ (slower) m, and their speeds are S₁ and S₂ m/s resp., then the time taken by the faster train to overtake the slower train is L₁+L₂/ S₁+S₂ sec, and the time taken for the trains to cross each other is L₁+L₂/ S₁+S₂ sec.

If the average speed of a train, without stoppages, is S₁ km/hr and the speed with stoppages is S₂, then Stoppage time (in min/hr) = S₁-S₂/S₁×60.

Problems about boats in stream: 

if the speed of the boat in still water is say B kmph and if the speed at which the stream is flowing is W kmph,

1. When the boat is traveling with the stream the speed of the boat = (B + W) kmph
2. When the boat is traveling against the stream the speed of the boat = (B – W) kmph.
3. If the upstream is denoted as U and downstream is denoted as D then
4. B = D+U/2,  W= D-U/2

Important Distance and Time Conversions:

• 1 km = 1000 meter
• 1 meter = 100 cm
• 1 hour = 60 min
• 1 min = 60 sec
• 1 hour = 3600 sec
• 1 km/hr = (1×1000/1×3600) = 5/18 m/sec.

If A beats B by x metres or s seconds, then the speed of B is x/s metres/sec.

If the length of a circular track is L m, and if A and B take x and y sec. respectively, to complete one round, then both of them will meet at the starting point after LCM (x, y) sec.

If the length of a circular track is L m, and if the speeds of A and B are x m/sec. and y m/sec respectively, then the time after which both of them will meet at a point other than the starting point is L/x-y sec, if they are running in the same direction and L/x+y, if running in the opposite direction.