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%

Geometry

Wastage is not always a Waste

At times, in geometry questions, we have symmetrical figures inscribed in other symmetrical figures e.g. a circle inscribed in a square, a circle inscribed in an equilateral triangle etc. and we have to find the common/wastage areas of such figures. In such cases, the concept of wastage of area may be helpful. Just to explain what this ‘wastage area’ is all about.

Let’s take an example.

Example:

A square paper has an area of 484 sq. cm. The largest possible circle is cut from this paper, what % area of the paper is wasted?

Solution:

In this case the area of square is 484 sq. cm. Hence the side of the square will be 22 cm.

So the diameter of the circle will be 22 cm. Area of the inscribed circle = π r² = 121π cm².

The area wasted at the four corners = (484 – 121 π) cm².

%age area wasted = (484-121π)/484 × 100 = 21.5%

Let me also make it clear to you that as the question was asking about the percentage change, the actual sides or area does not matter i.e. even if you take the side of the square to be 2 or 1, the answer will remain the same. This has universal application, whenever the percentage wastage is asked; any figure that makes your calculations easier can be taken.

Now here the area of the big square is 484 cm². If we take any square with side of any length and a largest circle is cut inside this square, the area wasted would always be 21.5% and the area used/ inside the circle will be 78.5%.

Similarly, if the largest possible square is cut/drawn inside the circle, then the % area wasted will be equal to 36.3%. On the same lines, if the largest possible cube is cut from a sphere, then the percentage volume wasted is 63% or the volume of cube is 37% of the volume of the sphere.

Some other important universal results are as follows: 

Outer figure	Inner figure	Wastage %
Square	Largest circle	21.5
Circle	Largest square	36.36
Equilateral triangle	Largest circle	39
Circle	Largest equilateral triangle	59
Square	Largest equilateral triangle	67
Equilateral triangle	Largest Square	51
Semicircle	Largest Square	49

In case your brain cells are buzzing now with this new tip that you have learnt, and you wish to discover how some of the above percentages were discovered, you have your home task: try to figure out how the above values were derived; this should be fun exercise for those who love mathematical wonders.

*Due to symmetry, all the corners will have equal percentage wastage.

Reasoning

Who is the Liar?

The problems based on truth and lie or Sacha and Jhuta are very common in CAT and similar exams. You will be able to understand the related concepts better with the help of the following example.

Example 1:

In an isolated island there live two tribes:

• SACHAS
• JHUTAS

The SACHAS always speak the truth, whereas the JHUTAS never do.

One day SUNIL, a stranger in the island, met four tribesmen. He asked one of them whether the second was a SACHA or a JHUTA. The reply was JHUTA. Similarly, the second tribesman said that the third was a JHUTA and the third said the same about the fourth.

When he asked the fourth as to what would the first tribesman have said about the third, the reply once again was JHUTA. Who’s who?

Solution:

Before solving this question, here are some important deductions that you can make.

1. If you asked the question to any tribesman ‘Are you a “SACHA” or a “JHUTA”?’ from that island irrespective of the tribe he belongs to, his reply would always be ‘SACHA’, because

• if he were a ‘SACHA’, he will always speak the truth about himself, that he belongs to ‘SACHA’ tribe and
• if he were a ‘JHUTA’ he will lie about his actual tribe and, say that he belongs to ‘SACHA’ tribe.

2. If two tribesmen, A & B, belong to the same tribes, either both ‘SACHAS’ or both ‘JHUTAS’, and you asked the question from A ‘Is B a “SACHA” or a “JHUTA” ?’, his reply will always be ‘SACHA’, because

• If they were belongs to ‘SACHA’ tribe then they always speak the truth, that other one belongs to ‘SACHA’ tribe and
• if both belongs to ‘JHUTA’ tribe then they always lie about themselves as well as about others, and would say that other belongs to ‘SACHA’ tribe.

3. If two tribesmen, A & B belong to different tribes and you asked the question from A ‘Is B a “SACHA” or a “JHUTA”?’, his reply will always be ‘JHUTA’, because

• if A is ‘SACHA’ then he is speaking the truth about B, and answering that B is ‘JHUTA’ or
• if A is ‘JHUTA’ then he is telling alie about B, and answering that B is ‘JHUTA’

Note: In conclusion, we can say that,if two persons belong to same tribe they will always say ‘SACHA’ about other and if two belongs to different tribe always say ‘JHUTA’ about the other.

1. Now, coming back to the problem:
2. As the first tribesman told Sunil, that the second was a ‘JHUTA’, it means that first & second tribesmen belonged to different tribes.Similarly, the second and third tribesmen were also from different tribes.
3. Moreover, the third and fourth tribesmen were also from different tribes, and
4. But as there are only two tribes, it implies that the first and third belonged to the same tribe eitherboth ‘SACHAS’ or both‘JHUTAS’, and the second and fourth also belonged to the same tribe either both ‘JHUTAS’ or both ‘SACHAS’.
5. When Sunil asked from the fourth tribesman‘what would the first tribesman have said about the third?’, we know from the second point mentioned above, that the answer given by third would always be Sacha, as they belong to the same tribe. But as you asked this question from the fourth person, he will also manipulate the answer as per HIS OWN TRIBE. The fourth said ‘Jhoota’.
6. It means the fourth tribesman, himself must be a ‘JHUTA’ because if he belongs to tribe ‘SACHA’  then his reply would be‘SACHA’, as he know that fist and third tribes belongs to same tribe and same tribe always say ‘SACHA’ about other.
7. Hence, the first and third tribesmen were both ‘SACHAS’ and the second and fourth were both ‘JHUTAS’.

Algenbra

Let X Remain UNKNOWN only

In this article, we explain the non-conventional way to solve Algebra questions without formulating equations. These alternative methods will enable you to save a lot of time as compared to the traditional and conventional approaches. Further, using these methods you will avoid making silly errors that creep in when the calculation work is lengthy.

You will understand how to solve Algebra questions without ‘x’ with the help of following examples:

Example 1:

The sum to n terms of the infinite series 1. 3² + 2. 5² + 3. 7² +....is:

1.  n/6 (n+1)(6n² + 14n + 7)
2.  n/6 (n+1)(2n + 1)(3n + 1)
3.  4n³ + 4n² + n
4. None of these

Solution:

The series in the question is obtained by multiplying 1, 2, 3, 4, … and 3², 5², 7², 9², … i.e. all natural number starting with 1, and the squares of all odd natural numbers, except one. Here, you need to find the general formula that will hold true for all the values of n.

Now, for the formula to satisfy all values of n, it should be applicable for n = 1 and n = 2 also. You can see that the first term in the given series is 9, and the second term is 50. This means that the sum of first two terms will be 9 + 50 = 59. So, that option in which you get 9, for n = 1 and 59, for n = 2 will be the answer. Here, the 1^st option gives the required values, and hence is the answer.

Note: n = 1 will be used for rejection or elimination of options only (as two different series can start with same number), and n = 2 will confirm the answer.

Example 2:

The sum to n terms of the series 1/2 + 3/4 + 7/8 + 15/16 +........ is equal to

1. 2ⁿ - n - 1                    2.  1 - 2^-n
3. 2^-n + n - 1                  4.  2ⁿ - 1

Solution:

Here, the first term is 1/2 and sum of the first two terms is 1/2  + 3/4  = 5/4 . So, by taking n = 1, you should get ½ as the answer, and by taking n = 2, you should get 5/4 as the answer. Here, the third option satisfies both of these and hence it is the answer.

However, you should be careful as n = 1 may satisfy one more of the options. This condition should be used only for rejection or elimination of options and the answer should always be confirmed by taking n = 2.

Example 3:

The nth term of the series 1+ 4/5 + 7/5² + 10/5³+.... will be:

1. 3n+1/5^n-1                  2. 3n - 2/5^n-1
3. 3n - 1/5ⁿ                    4. 3n+2/5^n-1

Solution:

By taking n = 1, we should get the 1^st term i.e. 1, and by taking n = 2, we should get the second term i.e. 4/5. Here, only the second option satisfies both of those and hence it is the answer.

DI

The Answer is a NO but is the Data Sufficient?

Data sufficiency questions test your knowledge of basic math facts and skills coupled with reasoning, analytical and problem-solving abilities. Each data sufficiency item presents you with a question where you need to decide whether or not the information presented along with the question would be sufficient to answer the question.

The challenge in DS questions, as they are popularly called, is not question solving but rather establishing whether the question has a solution or not. A special array of five answer choices is provided, each of which categorizes the relationship between the question and the information provided in a different way. You must select the answer choice that describes this relationship accurately.

Let’s have a cursory look at these answer options which generally feature in this question type:

Give answer (A) If the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
Give answer (B) If the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
Give answer (C) If the data either in Statement I or in Statement II alone are sufficient to answer the question.
Give answer (D) If the data even in both Statements I and II together are not sufficient to answer the question.
Give answer (E) If the data in both Statements I and II together are necessary to answer the question.

One small tip for this question type: Students often confuse the different answer options, and end up marking the incorrect choice. Always double check whether you are marking the correct option, and do not assume that the examiner would always present the options in a default order.

Go through the answer options to check whether the order of statements is as expected.

Example 1: Is the product of two numbers greater than 100?

A. The sum of the two numbers is greater than 50.
B. Each of the numbers is greater than 10.

Solution: Statement A alone is not sufficient to answer the question and this can be proved by examples. If the two numbers are 30 and 31, their sum is greater than 50 and their product is greater than 100; but if the two numbers are 50 and 1, though their sum is greater than 50, their product is only 50, and less than 100. Statement B is sufficient. If both of the numbers are greater than 10, then their product must be greater than 10 x 10, or greater than 100.
Hence only second statement is sufficient to solve the question.

Example 2: Is x a prime number?

A. 91 <x< 97
B. x is a factor of 121

Solution: Here the first statement is sufficient to answer the question as we see that there is no prime number between 91 <x< 97. Hence ‘x’ is not a prime number. What do we learn from this question? Remember, even if a question has an answer as ‘no’, even then it is a valid answer.

In second statement, the factors of 121 are 1, 11 and 121. Here 1 and 121 are not prime numbers whereas 11 is a prime number. Hence in this case ‘x’ may or may not be a prime number.
Hence, only the first statement is sufficient to solve the question.

Example 3: Is x = - 5?

A. x² = 36
B. x is a natural number.

Solution: Here the question directly asks whether x is equal to – 5 or not. From statement A, we have x = 6 or – 6. In both the cases x is not equal to – 5. Hence first statement is sufficient to get the answer. Statement B says that x is a natural number. Since x is a natural number, it cannot be negative. Hence it is not equal to – 5. So, the second statement is also sufficient to solve the question.
Hence, both statements are independently sufficient to answer the question.

To conclude, it is very important to read the question carefully in the case of data sufficiency questions. One major mistake committed by a number of students is that when the answer has to be yes/no and normally whenever you get the answer as no, you mark the answer as insufficient.

Remember: ‘NO’ is also an answer for Data Sufficiency questions.