CAT Linear Equations

Question
A system of equations has 3 equations
3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.
If the system of equations has no solution, find k.

Explanation
This is a very interesting question. If we can generate one of the equations from the other two, we can then say that the system has infinite solutions. If we can generate one of the equations from the other two, but with the constant part alone being different, this would be akin to having parallel lines when we are dealing with 2 variables and that would result in the system having no solutions. So, lets look for that.

So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) =  2x + 3y + kz.
We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y ) =  2x + 3y . Then said a, b would give us k.

Or, we are effectively solving for
3a + 4b = 2
4a + 5b = 3

Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.

Now, k = 5a + 3b = 10 -3 = 7.

If k = 7, this system of equations would result in no solutions.

If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + 7z = 6.
First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.

Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has no solution.

Yeah, the more mechanical, rather deceptively straight-forward-looking determinant method is also there. But where is the joy in that.

CAT Linear Equations

Question

(|x| -3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?

1. 4
2. 10
3. 6
4. 8

Answer: Choice (B)

Explanatory Answer

Product of two integers is 12. Further |y| + 4 > 4 . So, we know that one of the numbers has to be greater than 4.

So, we can have

1 x 12 => x = +4, y = +8

2 x 6 => x = +5, y = +2

3 x 4 => x = +6, y = 0

4 + 4 + 2 = 10 possible solutions

CAT Linear Equations

Another question from the book Quantitative Aptitude for CAT . 

Question

1.       x + |y| = 8, |x| + y = 6.How many pairs of x, y satisfy these two equations?

A.      2

B.      4

C.      0

D.      1

Answer: Choice (D)

Explanatory Answer

We start with the knowledge that the modulus of a number can never be negative, though the number itself may be negative.

The first equation is a pair of lines defined by the equations

y = 8 – x        ------- (i) {when y is positive}

y = x – 8        ------- (ii) {when y is negative}

With the condition that x ≤ 8 (because if x becomes more than 8, |y| will be forced to be negative, which is not allowed)

The second equation is a pair of lines defined by the equations:

y = 6 – x        ------- (iii) {when x is positive}

y = 6 + x        ------- (iv) {when x is negative}

with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be negative.

On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and (iv) are parallel (same slope).

Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv) holds good only when x is negative

Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This satisfies both equations.

Only one solution is possible for this system of equations.

Interesting question from yahoogroups

This one was posted on our yahoogroups. The discussion was also interesting. This is one of my favourite questions. Probably the best question that has been posted to the groups this past month. Just thought I would capture the essence of the mail thread on this post.

Question

Let the cost of 3 apples and 4 oranges be Rs 21. If Anjli can buy at most 4 apples and 3 oranges with Rs 20 and no more fruit, then the maximum amount that could be left with Anjli will be about?

First set of replies

· Ans is 2.50 /- . correct me if i am wrong

· If each apple cost 33paisa and oranges 5 rupees then Anjali can buy 3 apples and 4 apples for 21then 4 apples and 3 oranges will cost 16.2 which means that she can have 3.8 rs. For such problems u need to assign maximum values to more quantities and then should proceed

· say let me follow with yo approach! even then ! i can assume value 5.1 for b? right ? and 4*(5.1) + 3*(0.2) = 21. so here b = 5.1 and a=0.2 and 4(0.2) + 3(5.1) will be 16.1 and the answer would be 3.9 right ?? Kindly any one reply to this !

· Orange-Rs. 4, Apple-Rs. 1.66. Balance- Rs. 1.33, Hope You got it

Question recap (this mail sent by yours truly, must confess that I fell for the trap)

Let the cost of 3 apples and 4 oranges be Rs 21. If Anjli can buy at most 4 apples and 3 oranges with Rs 20 and no more fruit, then the maximum amount that could be left with Anjli will be about?

I would go for pricing apples at 0, price per orange would be Rs. 5.25 Cost of 4 apples and 3 oranges = 3* 5.25 + 0 = Rs.15.75, which leaves us with Rs. 4.25

Of course, this assumes that the price of apple cannot be negative :-)

Correct Solution - This is a beautiful solution.

all are missing one point that with 20 she can not buy any more fruit ..means the money left is less than price (minimum) of any og the two friuts...otherwise she can buy more friuts, so max amount of money left can be just higher than cost of chaper fruit.

Now 4x+3y =21

also 3x+4y+left money =20

3x+4y+y=20(leftmoney=cost of cheaper fruit, y will be chaper as on exchange of qunatity total value decreases or 4x+3y>3x+4y..........x>y)

4x+3y =21

3x+5y =20

solving this y=17/11=1.545 which is the answer

This is why this is a beautiful question (final reply from question-provider)

@sony, thank you so much, thats exactly what we were missing

a)1 b)1.25 c)2 d)2.25 e)1.5

these are the options given,

from the options given i was ignoring the costs that include denominations other than normally what we see, but i'm getting the answer as .75(with apple=2 and orange=3.75)

from your solution the answer is 1.5, i guess thats the correct one from the options.

Posted by

Rajesh B Director, 2IIM
CAT correspondence and classroom coaching @ Chennai, Mumbai, Bangalore