Net Percentage Change in Volume, Geometry or Percentage

Percentage change in area of Polygons – Geometry or Percentage?

PERIMETER:

The continuous line forming the boundary of a closed geometrical figure is called the perimeter. For example, the perimeter of a square with side ‘a’ is 4a and perimeter of a circle (i.e. circumference), with radius ‘r’, is 2πr. The unit of perimeter is same as the unit of the side, thus if there is a change of r% in side then percentage change in perimeter would be r%.

Example 1.

If the length and breadth of a rectangle are increased by 20%, then what is the percentage change in its perimeter?

Solution. Perimeter of rectangle = 2(L + B);

If length and breadth are increased by 20%, the new length and breadth will be, (1.2L) and (1.2B).

The new perimeter = 2 × (1.2L + 1.2B)⇒1.2 × 2(L + B) = 1.2 (Perimeter of the original rectangle)

Thus, the new perimeter is 1.2 times the original perimeter i.e. the perimeter has increased by 20%.

Example 2.

If the radius of a circle is decreased by 30%, then what is the percentage change in its circumference?

Solution. Circumference = 2 πr;

If the radius is decreased by 30%, the new radius will be 0.7r.

The new Circumference = 2π (0.7r)

⇒ = 0.7 × (2 πr) = 0.7 (Circumference of original circle)

Thus, the new circumference is 0.7 times the older circumference i.e. the circumference has decreased by 30%.

“If there is a change of r% in every side of a regular polygon or the radius of a circle, then the overall change in perimeter of the polygon or circumference of the circle would be r%” 

AREA:

The measure of the space enclosed within the boundary of a two dimensional geometrical figure is called its area. For example the area of square is (side)² and area of a circle, with radius ‘r’, is πr². The unit of area is square of the unit of the side in a given figure. 

Example 3.

If the radius of a Sphere is increased by 40%, then what is the change in the surface area?

Solution.  Surface area of sphere = 4 πr²;

If the radius is increased by 40%, the new radius will be 1.4r.

Therefore, surface area of the new sphere = 4π(1.4r)² 

⇒ 1.96 × (4πr²) = 1.96 (Surface area of the original sphere)

Thus, the new surface area is 1.96 times the original surface area i.e. it has increased by 96 %.

NOTE: As seen above, the term ‘4π’ is constant in the surface areas of both the spheres and thus, would cancel out when you compare the areas of two spheres. The change in surface area is due to the term r². i.e. (1.96)² = 1.96, which means an increase of 96%.

“If the side/radius of a figure is changed by r%, then the area will change by (r + r + r2/100)%”

Do remember that you need not take care of the whether there is increase or decrease as the sign will take care of the same. 

VOLUME:

The amount of space that a three-dimensional object occupies is called its volume. For example, the volume of a cube is (side)³ and volume of a sphere, with radius ‘r’, is 4/3 πr³. The unit of the volume is cube of the unit of the side in a given object.

Example 4.

If the radius of a hemisphere is increased by 10%, then what is the percentage change in the volume?

Solution. Volume of sphere = 2/3 πr³; If the radius is increased by 10% then the new radius = 1.1r;

New volume = 2/3 π(1.1r)³ 

⇒1.331 × (2/3 πr³)

The new volume is 1.331 times of the original volume i.e. it has increased by 33.1%.

NOTE: As seen above, the c term ‘2/3π’ is constant for the volumes of the original and the new hemi-spheres and would cancel out when we compare the volumes of two hemi-spheres. The change in volume is due to the term ‘r3’ i.e. (1.1)³ = 1.331, which means an increase of 33.1%.

Example 5.

Find out the percentage change in the area and volume of a sphere, when its radius is increased by 10%.

Solution. You can directly apply the above techniques as:

Change in area = (1.1)² = 1.21 i.e. 21 percent increase.

Change in the volume = (1.1)³ = 1.331 i.e. 33.1 percent increase

Learning:

When one of the side/perimeter/area/volume of a figure/object is changed by a certain percentage, and you have to find the percentage change in any of the remaining three, it is an advanced-level question of Percentages and not Geometry.

What is the Page Number?

CAT is known for asking tricky questions from a large variety of topics. One commonly asked question-type is where you are asked to determine the number of pages added twice/missed when the sum of pages is given.

Before we start the topic, let us first recall a basic relation related to natural numbers, i.e. the sum of first 'n' natural numbers is given by, 1+ 2 + 3 + 4 +……+ n 

So, what will be the sum of first 40 natural numbers?

From the above formula, we can calculate, 1 + 2 + 3+….…+ 40

Example 1:

A book has 60 pages numbered 1, 2, 3, 4………60. What is the sum of all the page numbers of this book?

Solution:

Here, we have to find the sum of the first 60 natural numbers. Using the formula given above,

Example 2:

The sum of the pages of a book is 630. Find the number of pages in this book.

Solution:

Here, let us consider that there are 'n' pages in the book whose sum is 630

⇒ n(n + 1) = 1260 ⇒ n2 + n –1260 = 0 which is a quadratic equation and its roots are 35 and -36. Since, the number of pages of the book cannot be negative, so the answer is 35. We can also solve this question by hit and trial; We have n(n+1) = 1260

As ≈ n2 = 1260 (To simplify the calculations we have taken n and n + 1 as equal)

⇒ n = 35 as the square of 35 is 1225 and we have reduced n + 1 to n.

The above illustrations were to familiarize with the concept. Now, let us see some advanced level questions from this topic that is more relevant for CAT and other competitive examinations.

Example 3:

The page numbers of a book were added and the sum obtained was 1053, but one page was added twice. Find the page number which was added twice?

Solution:

Let the total pages were 'n'

Hence 1 + 2 + 3 + 4+ ………….+ n = 1053

⇒ n ( n +1) = 2106.As ≈ n2 = 2106 (To simplify calculations, we have taken n and n + 1 as equal) Now, perfect square closest to 2106 is 2116, which is the square of 46. Here, one page was added twice. It means that 1053 is greater than the actual sum; therefore, 2106 is also greater than the actual value. So, we will not take n = 46, as it will further increase the sum. So, it has to be a number, which appears immediately before 46 i.e. 45. Hence, we can conclude that 45 pages were added. Now, the sum of 45 pages will be1 + 2 + 3 +……….+ 45.

Since the given sum is 1053, so the page number added twice was 1053 - 1035 = 18.

Example 4:

A teacher asked his students to find the sum of all pages of their book. One student gave the answer as 640.The teacher told him that he has not added one page. Find the page number which the student forgot to add?

Solution:

Let the total pages are 'n'. We have

1 + 2 + 3 + ……………+ n = 640

⇒ n(n + 1) = 1280 ⇒ ≈ n2 = 1280.

Now, the perfect square closest 1280 is 1296, which is the square of 36. Since the student did not add one page, so 640 is lesser than the actual sum and 1280 is also less than the actual value. Hence, we will take n = 36.

Hence, initially, 36 pages were added. So, their sum will be- 

As the given sum is 640, so the student forgot to add page number 666 - 640 = 26.

I double my speed, why will I take half the time?

The most important relationship in this concept is:

·         Distance = Speed × Time

·         Speed = Distance ÷ Time

·         Time = Distance ÷ Speed

Note: The units used in the three quantities should either be in SI or MKS system.

Now let us discuss a problem:

If I double my speed, why will I take half the time?

The reason is that, speed and time has an inverse relationship, provided the distance remains constant.

See the following example to understand this concept:

·         If A travels 1500 m at 5 kmph, then the time taken
= 1500m ÷ 5kmph = 1.5 ÷ 5 = 0.3 hr.
= 0.3 × 60 min = 18 min.

·         If A doubles his speed, i.e. 10 kmph, then time taken
= 1.5 ÷ 10 hr = 0.15 hr.
= 0.15 × 60 min = 9 min, which is half of 18 min.

Note: If the distance is constant, then the ratio of speeds of the moving objects is reverse of the ratio of the time taken by those and vice versa.

Example 1:

The ratio of speeds of two men, A and B is 1:2. Find the ratio of time taken by them to travel 10 km each.

Solution:

We know that, for equal distance traveled, the ratio of their time taken is reverse of the ratio of their speeds.

So answer will be 2:1.

Example 2:

A runs a race of 1 km at a speed of 6 kmph and B at a speed of 12 kmph. How many meters start can B give A?

Solution:

A covers 1 km in 1/6 hrs = 10 min, and B in 1/12 hrs = 5 min.

Since, A takes 5 min longer than B, the start that B can give A is the distance traveled by A in 5 min which is one-half of a km i.e. 500 meters.

Or, we can simply use the fact that whenever speeds are in the ratio 1:2, the start that the faster one can give to the slower one in a d km race is always half of d.

Example 3:

A man travels a total distance of 5 km. After every one km he doubles his speed. If he covers 5 km in 15hrs and 30 min, find his speed during the first 1km.

Solution:

The speeds at which the man traveled are in the ratio, 1 : 2 : 4 : 8 : 16.
Thus, time taken to travel every km is in the ratio 16 : 8 : 4 : 2 : 1.
The total time taken to travel 5km = 15 hrs 30 min = 15.5 hrs.
Therefore, the time taken to travel the first 1 km = 16/31 × 15.5 hrs = 8 hrs. 

So, speed during the first one km = 1/8 kmph

Example 4:

A man travels for 1hr at a speed of 20 kmph and for the next one hour he doubles his speed. Find the average speed at which he traveled for two hours.

Solution:

Average speed = total distance/total time.
Total distance traveled in two hours = 20 + 40 = 60 km.

Therefore, average speed = 60/ = 30 kmph

Mystery of TSD with the help of Ratio

The concept of ratio is a very important tool while solving questions in any aptitude-based paper. A lot of calculations can be avoided by applying basic concepts of ratio. In this article we will learn to apply ratio to solve Time, Speed and Distance questions.

Let us learn the concept with the help of some examples:

Example 1:

If I travelled at 3/4^th of my average speed and reached 25 minutes late, what is the time that I usually take to reach my destination?

Solution:

Although this question can be solved by making a simple equation, still the use of ratio can eliminate the calculation-work. Since, the current speed is 3/4^th the average speed, we can take the current speed as 3km/hr and the average speed as 4km/hr so that the ratio of the speeds is 3 : 4.

As the distance covered is same in both the cases, so the time taken will be in the ratio 4 : 3 (in the reverse ratio of speeds). It means that if today, I took 4 minutes then on an average I take 3 minutes i.e. I am one minute late.

Now apply simple chain rule that, if I am late for 1 minute, the usual time taken is 3 minutes and if I am 25 minutes late, the normal time taken will be 25 × 3 = 75 minutes.

Example 2:

A thief flees city A in a car towards city B on a stretch of straight road, 400 kilometers long, at a speed of 60 km/hr. In 30 minutes a police party leaves city A to chase the thief at 80 km/hr.

Find the distance travelled by the police when it catches thief.

Solution:

In this question the thief travelled for the first 30 minutes. The speed of thief is 60 km/hr, so he has covered 30 km in 30 minutes. The ratio of speeds of thief and police is 60 : 80 or 3: 4. Excluding the first 30 minutes, the thief and the police have run for the same time. So the distance covered is in the ratio of their speeds i.e. 3 : 4.

Since, the initial distance between the thief and the police if 30 km, so in total, thief and police have covered 90 km and 120 km respectively (to make the ratio 3 : 4).

Example 3:

Amit & Bimal are at a distance of 800 m. They start towards each other @ 20 & 24 kmph. As they start, a bird sitting on the cap of Amit, starts flying towards Bimal, touches Bimal & then returns towards Amit & so on, till they meet.

What is the distance traveled by the bird, if its speed is 176 kmph?

Solution:

In this question, Amit and Bimal are moving towards each other. So their relative speed is 20 + 24 = 44 km/hr. The speed of the bird is 176 km/hr. Now, the logic is simple. The bird flies for the same time as both Amit and Bimal take to meet.

Since the time taken by Amit and Bimal together and the bird is same, so the distance covered will be in the ratio of their speeds. The ratio of the speeds is 44 : 176 or 1 : 4.

Hence, if Amit and Bimal cover 800 m, the bird will cover 800 × 4 = 3200 m

Honest Trader but Dishonest Weights

This is a typical type of profit and loss question in which a trader claims that he sells a commodity at cost price or sells at a low profit percentage, but uses a false weight, thereby gaining/losing money.

The basic formula applied in that case is:

Profit % = (100 x error)/(true weight – error)

Error = True weight – False weight

You will understand the concept better with the help of following examples:

Example 1:

A dishonest tradesman claims to sell his goods at cost price, but uses a weight of 800 gm for 1 kg. Find his gain percentage.

1.15%     2. 20%     3.  25%     4.22%

Solution:

Say, price of 1 gm = Rs. 1, so the 1kg = Rs. 1000.

For 800 gm amount, he charges amount for whole 1kg,

This means that, for selling good worth Rs 800, he charges Rs. 1000.

Therefore, CP = Rs. 800, and SP = Rs. 1000

⇒Profit = Rs. 200

Profit percentage = (200/800) x 1000 = 25%

Using the formula:

Percentage Profit earned = (100 x error)/(true weight – error)

= (100 x 200)/(1000 – 200) = 25%

Example 2:

A dealer sells his goods at cost price but unfortunately his machine is faulty and he incurs a loss of 20%. How much does the shopkeeper sell if the machine reads 1kg?

1. 1250 grams    2. 1300 grams  3.1225 grams    4.  1200 grams

Solution:

Let price of 1gm = Rs.1, so price of 1kg = Rs. 1000

As he’s selling 1kg amount (falsely) = Rs. 1000,

So SP = Rs. 1000.

Let CP = X ; SP = CP – Loss ;  0.80 X = 1000

Solving, we get, X = Rs.1250.

This means that he gives 1250 gm for 1 kg.

Example 3:

A shopkeeper sells his good at a profit of only 4%. But he gives 800 gms for 1 kg. What is his real profit percentage?

1. 25%     2. 30%     3. 40%     4.  35%

Solution:

Say, 1000 g costs Rs. 1000. As per the shopkeeper, he sells at a profit of 4%.

Thus, the selling price of 1000 gms. is, 1000 × 1.04 = Rs. 1040.

But, the quantity he has actually given is 800 gm, which cost him Rs. 800.

Therefore, he makes a profit of Rs. 1040 – 800 = Rs. 240.

Hence, profit percentage = 100 × 240/800 = 240/8 = 30%.

Example 4:

A shopkeeper sells his good at a loss of 2%. But he gives 700 gms for 1 kg. What is his real loss percentage?

1. 35%     2.  40%    3. 30%    4.  45%

Solution:

Say 1000 g costs Rs. 1000. As per the seller, he sells at a loss of 2%.

Thus, the selling price of 1000 gms is 1000 × 0.98 = Rs. 980.

But, the quantity given is 700 gm, which cost him Rs. 700.

Therefore, profit = 980 – 700 = Rs. 280.

⇒ Profit % = 100 × 280/700 = 280/7 = 40%