## Monday, December 12, 2011

### Speed Distance Time : TANCET Quant

Speed Distance Time is one of the topics from which you should expect a question either in the quantitative reasoning section or in the data sufficiency section of the TANCET MBA.

Here is a relatively easy quantitative reasoning question from this topic

Question
It takes Kiran 50 minutes lesser to cover a distance if she travels three times as fast as she does now. How long will she take to cover the distance if she travels twice as fast as she does now?
1. 50 minutes
2. 75 minutes
3. 150 minutes
4. 100 minutes
5. 37.5 minutes

Correct Answer : Choice 5. 37.5 minutes

Let Kiran's present speed be S km/min. And let her take t minutes to reach her destination.

If Kiran travels three times as fast as she does now i.e., at 3S, she will cover the same distance in 1/3 of the time i.e., in t/3 minutes.

So, she is saving (t - t/3) minutes = 2t/3 minutes.
The question states that she is saving 50 minutes.
Therefore, 2t/3 = 50
or t = 75 minutes.

If she travels at S km/min, she will reach her destination in 75 minutes.

So, if she travels at 2S km/min, she will take half that time = 75/2 minutes = 37.5 minutes.

## Tuesday, November 8, 2011

### TANCET Math - Ratio Proportion

Hi

Ratio proportion is an often tested area in the quant section of the TANCET MBA exam. I have posted a medium level difficulty question on Ratios.

Question

The sum of the ages of 3 friends is 90 years. If 5, 3 and 2 years are subtracted respectively from the ages of the oldest, middle and the youngest the ratio of their ages becomes 8 : 5 : 3. What is the age of the oldest of the 3 friends?
1. 40
2. 48
3. 28
4. 17
5. 45

Correct Answer : 45 years. Choice 5.

Let the age of the 3 friends be a, b and c - a the oldest, b the middle one and c the youngest.

Therefore, a + b + c = 90

The question states that a - 5 : b - 3 : c - 2 :: 8 : 5 : 3
So, a - 5 + b - 3 + c - 2 = 8k + 5k + 3k
Or a + b + c - 10 = 16k ... eqn (1)
Substituting a + b + c = 90 in eqn (1), we get 90 - 10 = 16k
or 16k = 80
Hence, k = 5.

a - 5 = 8k
or a - 5 = 40
Therefore, a = 40 + 5 = 45 years.

## Friday, September 16, 2011

### Angles in a triangle : TANCET Geometry

Geometry, specifically triangles is an oft tested area in the TANCET quant section.

Here is a relatively easy question on triangles.

If the largest angle in a triangle is 70o, what is least possible value of the smallest angle of the triangle?

1) 69
2) 1
3) 40
4) 39
5) 41

The sum of the interior angles of a triangle is 180.

The question states that the largest of the angles is 70.

We are interested in finding the least possible value of the smallest of the three angles.

The least value for the smallest angle will be when the other two angles are as large as possible.

The largest value that the other two angles can take is 70 each.

Therefore, the least value of the smallest of the angles = 180 - 70 - 70 = 40

You can access more questions in Geometry by visiting this link

## Monday, August 22, 2011

### TANCET Quant : Linear Equations, Word Problems

Here is a word problem that requires you to frame a pair of linear equations, solve the equations to arrive at an answer.

Rajesh is 10 years younger to Baskar. 10 years back, Rajesh's age was two-thirds that of Baskar's. How old is Baskar now?
1. 30
2. 40
3. 20
4. 16
5. 28

Correct Answer is Choice (2). 40 years.

Let the present age of Baskar be 'b' and that of Rajesh be 'r'.
So, r = b - 10 .... eqn (1).

10 years back Rajesh was (r - 10) years old. 10 years back Baskar was (b - 10) years old.
The question states that 10 years back Rajesh was two thirds as old as Baskar was.

i.e., (r - 10) = (2/3)*(b - 10) .... eqn (2).
Cross multiplying, we get 3(r - 10) = 2(b - 10)
or 3r - 30 = 2b - 20 .... eqn (2)

From eqn (1) we can substitute r as (b - 10) in eqn (2)
So, 3(b - 10) - 30 = 2b - 20
or 3b - 30 - 30 = 2b - 20
or b = 40.

The present age of Baskar is 40 years.

## Wednesday, May 18, 2011

### Word problems : Linear Equations

You should anticipate a question that expects you to convert information presented in words into an equation and solve it.

They are commonly referred to as "word problems". The key to getting those questions right is to try and decode the information in the sentence and frame equations accordingly.

Here is a question that will give you an idea of how to frame a linear equation from the information given in the question statement.

If the result of seven more than four times a number is tripled, the resultant number is three more than fifteen times the number. What is the number?
1. 6
2. 7
3. 4
4. 15
5. 12

## Thursday, January 20, 2011

### TANCET Math : Averages

Averages is an oft repeated topic in TANCET quant section. You may also get a variant of questions from the arithmetic mean and weighted average concept as Data Sufficiency Questions in TANCET.

Here is a sample averages question. A rather easy one.

Average weight of 25 boys in a class is 48 kgs. The average weight of the class of 40students is 45 kgs. What is the average weight of the 15 girls in the class?
1. 44 kgs
2. 42 kgs
3. 40 kgs
4. 39 kgs
5. 42.5 kgs

## Saturday, January 15, 2011

### Speed Distance Time Question TANCET

Speed, Distance and Time is an important topic from a TANCET quant perspective. You could expect to get one to two questions from this topic.

The core concept in this topic is the relation between these three parameters
Distance = speed * time.

The second point to keep in mind while solving questions from this topic is to check if the units of the three parameters match.
i.e., if distance is measured in kilometres and time in hours, speed should be in km/hr. Alternatively, if speed is in kilometres and time in minute, speed should be in km/min.

Here is an easy to moderate level difficulty question in this topic.

A bus travels at 40 kmph for the first 100 kms, 60 kmph for the next 100 kms and 48 kmph for the final 200 kms, what is the average speed over the total distance?
1. 48 kmph
2. 50 kmph
3. 49 kmph
4. 52 kmph
5. 50.4kmph

The concept tested in this question is that of average speeds.

The answer to this question is 48 kmph.

Average speed for the first 200kms = 2ab/(a + b) = 2 * 40 * 60/(40 + 60) = 48 kmph.

Bus travels equal distances at two different speeds, average speed = 2ab/(a + b)

Now, if we break the overall journey into two stretches of 200 kms each, the bus travels the first 200kms at 48kmph, and the second at 48 kmph. Overall, average speed = 48 kmph.

Alternatively, you could find out the total time taken to cover the entire distance of 400 kms

Time taken for the first 100 kms = 100 / 40 = 10/4 hours

Time taken for the second 100 kms = 100/60 = 10/6 hours

Time taken for the last 200 kms = 200/48 = 50/12.

Adding up all three, we get 10/4 + 10/6 + 50/12 = 100/12 hours.

Average speed = total distance / total time

= 400 /(100/12) = 4800/100 = 48 kmph