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# CAT Quantitative Aptitude Problems: Equations and Inequations

Nov 20, 2018 15:26 IST

Equations and Inequations is an important topic for the MBA entrance exam preparation, especially for the CAT Exam. The CAT quantitative aptitude section carries one to two questions from this concept every year to test the mental aptitude of the candidates. To enhance the understanding of this topic, application problems are provided in the exercise below. Taking a look at few examples will enable the aspirants grasp the core of the concept easily to prepare for the upcoming CAT exam.

The Problem 1 is a simple example which will throw some light upon the basics of equations and to help you understand the basics of equation formation and solution. The other two problems (2 and 3) are based on the same set of data and are from CAT 2007 exam.

Read on to find out the problem questions from this topic and prepare well for the CAT quantitative aptitude section:

Problem 1:
Two candles of equal height but of different thickness are lit simultaneously. It is then observed that the first candle burns-off completely in 6 hours and the second one burns-off completely in 8 hours. Find when the first candle will be exactly half the height of the second candle. Solution:
Let the initial height of the candles be equal to 1 unit. If the candles have burned for x hours for the first candle to be half the height of the second candle, we can write

Height of the first candle left = Height of the second candle = Now, we know that first candle is half the height of the second candle, therefore

That is, the first candle will be exactly half of the second in hours.

Option 4 is the correct option.

Information for 2 and 3:

Let a1=p and b1=q, where pa and q are positive quantities.

Define an=pbn-1,   bn= abn-1 when n is even and n>1 ......(i)

and     an= pan-1,            bn= qan-1 when n is odd and n>1.........(ii)

Problem 2: Considering the information given above, which of the following best describes an+bn for even n?

Solution 2:

an=pbn-1,             bn= abn-1 when n is even and n>1 ......(i)

an= pan-1,            bn= qan-1 when n is odd and n>1.........(ii)

Using the above equations, we find (even and odd n) related values of

a2,a4,a6,a8,a10,a12

a3,a5,a7,a9,a11,a13 and similarly all the values of bx

It is very important to clearly understand where and when to use this concept of equations and inequations   Now, let us find the corresponding values of (an+bn) when n is even and also when n is odd.

Therefore, for (a2+b2), since n should be even, then a2=pb1=qb1=q(q)

a2+a2=p(q)+q(q)

Now, putting n=2, across the given choices, we come to realize that all the choices given out the same result, hence we cannot elimate any of the given choices to get a unique answer.

Now, we put n= 4 and find (a4+b4), then a4=pb3=p(qa2)=p2q2

And b4=qb3=q(qa2)=pq3, a4+b4=p2q2+pq2(p+q).........(xx)

Now, putting n=4 in the given choices, we get as following:

Choice  : , which is not same as equation (xx)

Choice  : , same as equation (xx) hence could be the answer

Choice  : , which is not same as equation (xx)

Choice  : , which is not same as equation (xx)

Therefore, the answer is choice 

Problem 3:

Taking into information given for problem 2 and 3 , and for , find the smallest odd n such that an+bn<0.01?

 7

 9

 11

 13

Solutions:

To find the smallest odd value of n such that an+bn<0.01 or an+bn<1/100, we check for the values of 'n' form the choices till we get the matching result.

Now, substituting the first option in the above inequation, we get Similarly, a9+b9=p(a8+b8)

And a11+b11= P(a10+b12)

a13+b13 = p(a12+b12)

Again, (a6+b6) or (a8+b8) Or (a10+b10) etc. can be obtained because for an even value of n, from choice  (please refer to solution of Problem 2 above) as it gives the value of (an + bn)

Now,  Therefore, choice  is again the correct answer.

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