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.

Explanatory Answer

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.

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

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

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.

Explanatory Answer

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.
Correct answer (1)

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

For additional questions on Speed, Distance and Time, click here

TANCET Quant : Percents

Here is a relatively easy question from the topic Percents.

Question

If A is 40% more than B and C is 68% more than B, what percent is C more than A?
1. 28%
2. 17%

3. 20%
4. 10.8%
5. 54%

TANCET Quant : Progressions

AP, GP (Arithmetic Progressions and Geometric Progressions) are two topics from which you could expect to get at least one question.

An interesting topic and the kind of questions that you can expect in the TANCET quant section from these two topics will be rather elementary.

The AP topic has two formulae to remember.

The first one is to find the nth or the last term 'l' of an AP : l = a + (n - 1)d, where 'a' is the first term, n is the number of terms and 'd' is the common difference.

The second formula is to find the sum of first 'n' terms of an AP. Sn = (n/2)(2a + (n - 1)d)

Here is a question that tests your understanding of these concepts.

If the first term of an AP is 12 and the 7th term is 36, what is the sum of these 7 terms?

1. 240

2. 168

3. 336

4. 84

5. 121

Explanatory Answer

The first term a = 12

The 7th term l = 36.

We know that l = a + (n - 1)d

or 36 = 12 + (7 - 1)d

or 24 = 6d

or d = 4.

Sum of these first 7 terms = (n/2)(a + (n - 1)d), where n = 7, a = 12, d = 4

Substituting we get, (7/2)(2*12 + (7 - 1)4)

= (7/2)(24 + 24)

= 7 * 24

= 168

Choice (2) is the correct answer.

Speed Time Distance : TANCET Math

Speed, Time and Distance is an interesting topic from which you could expect to get one question in the math section and probably a question in the data sufficiency section of the TANCET MBA test.

The concept is extremely simple : Distance = Speed * Time
The place where you need to be careful is to make sure that the units that you use for the different quantities have integrity i.e., if the unit for speed used is km/hr, then you should use hour as the unit for time.

Here is a simple question in speed time distance.
Question
Traveling at 4/5ths of her usual speed, Sonal reached office 20 minutes late. How long did she take today?
1. 60 minutes
2. 80 minutes
3. 100 minutes
4. 64 minutes
5. 96 minutes

Explanatory Answer
She travels at 4/5th of her usual speed.
Let her usual speed be 's' km/min. So, today she travels at 4/5s.
Let the usual time she takes be 't' minutes. So, today she will take (t + 20) minutes.
As the distance traveled in both the cases is the same, we can equate them
i.e., st = 4/5s (t + 20)
dividing both sides by 's' we get t = 4/5(t + 20)
or 5t = 4t + 80
or t = 80 minutes.
t is the usual time take.
The time taken today is t + 20 = 100 minutes.

Note: 1. We used km/min as the unit for speed as the unit of time given in the question is in minutes.
2. we did not stop at finding the value of 't'. The question was about finding the time taken today = t + 20.

You can get additional questions on speed time and distance at this link.