GRE Coaching Math: Lesson 1 -> Solving of Linear Equations ~ InSTaMeChEng

Let us now learn how to solve linear equations in three variables. The system of linear equations is generally in the form

a₁x+b₁ y+c₁z = d₁
a₂x+b₂ y+c₂z = d₂
a₃x+b₃ y+c₃z = d₃

To solve these equations, we select a variable, either x or y or z (usually z is chosen), and eliminate it from the system of equations.
We then obtain simultaneous linear equations in two variables x and y, which we solve as we did in section 2.3. After obtaining the solutions to x and y, we substitute these values in any one of the original equations to obtain the value of z.

Regarding questions on solving 3 Simultaneous, when presented with 3 equations (with 3 variables), what is the easiest way to combine them?

First, you don't often get 3 simultaneous equations on this test, so don't worry about it too much.
Second, take a look at what the problem wants before you decide how to approach it. You won't be asked to solve for all three variables, the way you would in math class. Often they'll want some combination of variables (for example, what is x+y?). You should keep in mind then that you are trying to solve for the combination x+y, NOT for x and y individually (that takes a lot longer and sometimes, depending on the problem, it's actually impossible to solve for x and y individually even when you can solve for the combination x+y).

Then look at your three equations and try to visualize how you could manipulate them to leave you with the specific combination that you want. Don't spend too much time at this - just dive into it if you don't see anything obvious. But do take 10-15 seconds to see if you can figure something out.
The problems in the in-action problem sets are not trying to mimic GMAT problems - they're actually more like pure math textbook problems, designed to see whether you have a clear understanding of the concept than to test whether you could solve it. So I wouldn't worry too much about something like that if it's really giving you trouble, because the "real" math could be lot more complicated procedurally (in this instance) than what the test would give you. For example, with simultaneous equations on the test, you often don't have to multiply the equations by anything before you add or subtract - whereas, in real math, you almost always have to do that.

Just focus in on what they asked and aim for that. In other words the focus should be on the question to clearly understand what they have asked.

Also remember that you may able to find a solution to most of the problems by the method of substitution like VYC or VIC namely variables of your choices (VYC) or variables in the answer choices (VIC) technique. If the answer choices are in terms of variables, you can just try substitute in real numbers and do arithmetic instead of algebra! As we know all the numbers are Real.

Let us look at these examples:

Example 1

Solve the linear equations.

x+2y+2z=11         ----------- (1)
2x+y+z=7             ------------ (2)
3x+4y+z=14         ------------ (3)

To eliminate z from (1), (2) and (3), we multiply equations 2 and 3 by 2.

2∗ (2x+y+z=7)
2∗ (3x+4y+z=14)
   4x+2y+2z=14        ------- (4)
   6z+8y+2z=28        ------- (5)

Subtracting 1 from 4 and 1 from 5, we get
4x+ 2y+ 2z=14
-x ± 2y ± 2z=-11
____________
         3x= 3 =>  x=3/3 => x=1
6x+ 8y + 2z =28
-x ± 2y ± 2z =-11
__________
   5x+6y = 17
Substitute x=1 in this equation.
  5∗1+6y = 17 => 6y =17-5 =>     6y =12 =>     y =12/6 =>   y =2
Substitute x=1, y=2 in equation 1.
   x+2y+2z = 11
1+2 ∗2+2z = 11
     1+4+2z = 11
             2z = 11-5 =>  2z = 6 =>  z = 6/2 =>   z = 3
Solution set = {(1,2,3)}.

Example 2

Solve the equations

3x-4y =6z-16
4x -y -z = 5
x=3y+2(z-1)

Rearrange the terms to obtain the general form.
3x-4y-6z = -16 ------ (1)
4x-y-z = 5 ------ (2)
x-3y-2z = -2 ------ (3)

We will eliminate z from the system of equations by multiplying equations (2) and (3) by 6 and 3, respectively. We thus obtain

6 ∗ (4x-y-z = 5 )
3 ∗ ( x-3y-2z = -2 )

24x-6y-6z = 30 ---------(4)
3x-9y-6z = -6 ----------(5)

Subtracting (1) from (4) we get

24x-6y-6z = 30

6z =

21x-2y = 46 ----------(6)

Subtracting (1) from (5) we get
3x - 9y - 6z = -6

-3x

6z =

       -5y = 10
          y = 10/-5
          y = -2

Substituting y = -2 in (6) we get
21x - 2∗(-2) =46
21x + 4 =46
21x = 46-4
21x = 42=> x = 42/21=> x = 2

Substituting x = 2, y = -2 in equation (1) we get
=>3∗2-4∗(-2) -6z = -16
=>6 + 8 - 6z = - 16

=> 14 - 6z = - 16

=> - 6z = - 16-14

=> - 6z = -30

=> z = - 30/-6

=> z = 5

∴ x = 2, y = -2 z = 5
Solution set = {(2, -2, 5)}

Example 3

Solve the equations.
x-y/5 = 6         (1)
y-z/7 = 8         (2)
z-x/2 = 10       (3)

x-y/5 = 6         (1)
y-z/7 = 8         (2)
z-x/2 = 10       (3)

(1) can be written as

x-y/5 = 6
x = 6+y/5
x= (30+y)/5

Substituting for x in equation (3) we get
z -1/2(30+y)/5 = 10
z-(30+y/10) = 10
(10z-30-y)/10 = 10
10z - 30 - y =10x10
10z - y - 30 =100
10z - y = 100+30
10z-y = 130 ------ (4)

Equation (2) becomes

y -z/7 = 8
(7y-z)/7= 8
7y - z = 8 ∗7
7y - z = 56 ------------ (5)

-y + 10z = 130   ------------(4)
7y - z = 56        ------------ (5)

Multiplying 4 by 7
7(-y + 10z = 130)
Adding -7y +70z = 910
                7y - z = 56
         _______________
                  69z = 966
                     z = 966/69
                     z = 14

Substitute z = 14 into equation (5)

7y - 14 = 56
7y = 56 + 14
7y = 70
y = 70/7
∴       y = 10

Substitute y = 10 in equation (1)
x -10/5 = 6
x - 2 = 6
x = 6+2 =>     x = 8

Solution set ={( 8, 10, 14 )}

Example 4

Solve the equations.

(y+z)/4 = (z+x)/3 = (x+y)/2
x + y + z = 27

We need to reduce these equations to a recognizable form, such as

a₁x + b₁y + c₁ z = d₁
a₂x + b₂y + c₂ z = d₂
a₃x + b₃y + c₃ z = d₃

Consider =(y+z)/4 = (z+x)/3

Cross multiplying 3(y + z) = 4(z + x)
                          3y + 3z = 4z + 4x
                           4x + 4z - 3y - 3z = 0

                      ∴            4x - 3y + z = 0               (1)
Consider (z+x)/3 = (x+y)/2
Cross multiplying
2(z + x) = 3(x + y)
2z + 2x = 3x + 3y
3x + 3y - 2z - 2x = 0

         x + 3y - 2z = 0         (2)
We now have the following equations

4x - 3y + z = 0      (1)
x - 3y - 2z = 0      (2)
   x + y + z = 27    (3)

We eliminate y from these equations. Multiplying equation (3) with 3 we get
3∗ (x + y + z = 27)
3x + 3y + 3z = 81         (4)
Adding (1) and (4)

  4x - 3y + z = 0
3x + 3y + 3z = 81
______________
     7x +4z = 81           (5)
Adding (1) and (2) we get
4x - 3y + z = 0
  x + 3y -2z = 0
_____________
        5x- z = 0
           5x = z

Substituting z = 5x in (5) we get
7x + 4 ∗5x = 81
   7x + 20x = 81
           27x = 81
             x = 81/27
               x = 3
z = 5x
z = 5∗3
z = 15

Substitute x = 3, z = 15 in equation (1)
4x - 3y + z = 0
4∗3 - 3y +15 = 0
12 - 3y + 15 = 0
    - 3y + 27 = 0
             -3y = -27
                y = -27/-3
                y = 9

Solution set = {( 3, 9, 15 )}

EXERCISES:
Results will be posted on the comments.

Solve the following equations

1. x + 3y + 4z = 14
x + 2y + z = 7
2x + y + 2z = 2

2. x + 4y + 3z = 17

3x + 3y + z = 16
2x + 2y + z = 11

3. x - 2y + 3z = 2

2x - 3y + z = 1
3x - y + 2z = 9

4. 5x + 2y = 14

y - 6z = -15
x + 2y+z = 0

5. y-z / 3 = y-x / 2= 5z - 4x

y + z = 2x + 1

GMAT / GRE / CAT type problems:

Ben wanted to know how old his friend Mr. Stone was. Mr. Stone told him, "I started school when I was 4-1/2 year old, and I stayed at the same school for 1/6 of my life. I was in the army 1/5 of my life and when I left the army I spent 1/4 of my life as a clerk. Now I have spent 1/3 of my life in retirement." How old is Mr. Stone?

For such problems, use a variable for the age. Call it S .
Currently Mr. Stone is ‘S’ yrs old.
He was 4.5 yrs old when he started school.---- 4.5
For 1/6 of his life , he was in school ------ (1/6) x S
in the army 1/5 of my life -------- (1 / 5) x S
1/4 of my life as a clerk ---------- (1 /4) x S

spent 1/3 of my life in retirement ----- (1 / 3) x S

That was his life to date.

Whatever he has done in life thus far must = his age.

S = 4.5 + (S / 6) + (S / 5) + ( S / 4) + (S / 3)

Now this is a fraction simplification problem.

The LCD for 6, 5, 4 and 3 is 60

\ (S / 6) + (S / 5) + ( S / 4) + (S / 3) is now written as (10 S + 12 S + 15 S + 20 S) / 60

Plug this back in the first equation S = 4.5 + (57S / 60)

Subtract 57 S / 60 from both sides.

S – (57S / 60) = 4.5 => 3S / 60 = 4.5 => S / 20 = 4.5 => S = 20 x 4.5 gives S = 90 years.

A cashier mentally reversed the digits of one customer's correct amount of change and this gave the customer an incorrect amount of change. Id the cash register contained 45 cents more than it should have as a result of this error, which of the following could have been the correct amount of change in cents.
A 14 B 45 C 54 D 65 E 83

One can express any two-digit number as a sum of its tens digit plus its ones digit by remembering to multiply the tens by 10.

For example, 45 is 4*10 + 5.

So if we want to say that flip-flopping the digits is the same as adding 45, we'd say that:

10x + y is the initial number (x is the tens digit and y is the units)
10y + x is the new number (now y is the tens digit and x is the units)
And by reversing we're adding 45, so we'd have: 10x + y = 10y + x + 45

Combine like terms to get: 9x = 9y + 45

Divide by 9 to simplify: x = y + 5

so we know that there's a difference of 5 between the digits, so we could have:
16 & 61 ; 27 & 72 ; 38 & 83 ; 49 & 94
The only option that's here in the answer choices is 83, so that's correct. Choice E

The remainder when m + n is divided by 12 is 8, and the remainder when m – n is divided by 12 is 6. If m > n, then what is the remainder when mn divided by 6?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Tricky Approach: Pick numbers m and n accordingly. For example (m + n) = 8 and (m - n) = 6. Thus, m = 7 and n = 1. Therefore, mn = 7 which leaves a remainder of 1 when divided by 6.

Classic Approach:
(m + n) is of the form (12a + 8), where a is any non-negative integer.
(m - n) is of the form (12b + 6), where b is any non-negative integer.

Therefore, m = [(12a + 8) + (12b + 6)]/2 = (6a + 6b + 7)
Therefore, n = [(12a + 8) - (12b + 6)]/2 = (6a - 6b + 1)

Therefore, mn = (6a + 6b + 7)(6a - 6b + 1) = (36a² - 36ab + 6a +36ab - 36b² + 6b + 42a - 42b + 7) = (36a² - 36b² + 48a -36b + 6 + 1) = Multiple of 6 + 1

Hence mn leaves a remainder of 1 when divided by 6.

The correct answer is A.

A tourist purchased a total of $1,500 worth of traveller’s checks in $10 and $50 denominations. During the trip the tourist cashed 7 checks and then lost all of the rest of the number of $10 checks cashed was one more or one less than the number of $50 checks cashed, what is the minimum possible value of the checks that were lost?

(A) $1,430 (B) $1,310 (C) $1,290 (D) $1,270 (E) $1,150

Given that:

a = number of checks worth 10 bucks and

b = number of checks worth 50 bucks

You also have 10a + 50b = 1500. Divide by 10 on each side and we get: a + 5b = 150.

Now, we are told that a number of 7 checks were cashed during the trip and the rest were lost. we are then expected to "minimize" the damage. Since the number of $10 checks cashed was one more or one less than the number of $ 50 checks cashed, you will pick the situation when there were more $50 checks cashed than $10 checks. This is because you want to get the most value out of the checks cashed, so you pick the more valuable one.

So, we get that 4 checks of $50 were cashed and 3 checks of $10 were cashed, or 4*50 + 3*10 = 200 + 30 = $230. The minimum possible value will be 1500 - 230 = 1270.

Even if we already have the answer, you can double-check to see that it is indeed correct: if 4 $10 checks and 3 $50 checks were cashed, then the guy cashed 40 + 150 = 190. This means that the poor tourist lost 1500 - 190 = 1310, which is greater than 1270.Choice B

At a certain restaurant, the ratio of the number of cooks to the number of waiters is 3 to 13. When 12 more waiters are hired, the ratio of the number of cooks to the number of waiters changes to 3 to 16. How many cooks does the restaurant have?

A. 4 B. 6 C. 9 D. 12 E. 15

The key here is setting up the equation. Since we don’t know the initial scale of the number of cooks(C) and waiters(W) , we can express this scale by “x”.

C / W = 3x / 13x.

Notice that whatever x is, the ratio will hold true. (x must be an integer, since you can’t have a portion of a cook, unless of course he chops his finger off by accident!)

“When 12 more waiters are hired” is the insertion of an absolute. Adding the 12 waiters, the new ratio becomes: C / W = 3x / (13x + 12)

“The ratio of the number of cooks to the number of waiters changes to 3 to 16” defines this new ratio: C / W = 3x / (13x + 12) = 3 / 16

Before we cross multiply and solve for x, we want to cancel out the 3’s in both the numerator. (More on this below.) After cross-multiplying, we get:

16x = 13x + 12 => 3x = 12 => x = 4

Answer A, right? Well, recall that x represents the scaling factor. The stimulus asks for the number of cooks, which we originally represented by 3x. So, 3*4 = 12 cooks. That’s 120 fingers. Choice D.

x^.5 + x^.25 = 20 What is/(are) the solution(s)?

Let x ^0.25 = y => x ^0.5 = Y²

Now the equation is Y² + Y -20 = 0 => y = 4 or -5 => x = 256 or 625

A real value for k is chosen so that the quadratic equation above has no real roots.Which of the following describes method to change the chosen value of k in order to guarantee that the equation has two different real roots?
a) Decrease k by 10
b) Decrease k by 15
c) Increase k by 7
d) Increase k by 10
e) Increase k by 13

If the discriminant is negative, then the roots are complex.
We need to make sure the discriminant is positive.

d = k^2 - 43.1 (4ac)

So, k^2 MUST BE greater than 43.1, hence k must be greater than sqrt(43.1) (approx. 6.56)
OR smaller than -sqrt(43.1) (approx. -6.56)

If the equation above has no real roots, we know that -6.56 < k < 6.56.

To make sure it gets out of the interval, we either need to add something greater than 13.12 or subtract something greater than 13.12.

In this case, only b) -15 satisfies our condition, hence being the correct answer.

Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month. The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did not save. If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?

(A)1/2 (B)1/3 (C)1/4 (D)1/5 (E)1/6

Let’s take a look at the algebra. Our first task: assigning a bunch of variables.

monthly take-home pay = t

monthly saved pay = s

What else do we know from those two variables?

yearly take-home pay = 12t

yearly saved pay = 12s

monthly pay not saved = t – s

Let’s start translating. Sentence 2 becomes:

12s = 3(t – s)

The second half of the question becomes:

s/t=?

Rearrange the first equation to get s/t by itself on one side:

12s = 3(t – s)

12s = 3t – 3s

15s = 3t

s/t = 3 /15 = 1 / 5

Choice D

One can express any two-digit number as a sum of its tens digit plus its ones digit by remembering to multiply the tens by 10. For example, 45 is 4*10 + 5.
So if we want to say that flip-flopping the digits is the same as adding 45, we'd say that:
10x + y is the initial number (x is the tens digit and y is the units)
10y + x is the new number (now y is the tens digit and x is the units)
And by reversing we're adding 45, so we'd have:    10x + y = 10y + x + 45
Combine like terms to get: 9x = 9y + 45
Divide by 9 to simplify:           x = y + 5
so we know that there's a difference of 5 between the digits, so we could have:
16 & 61 ;               27 & 72 ;               38 & 83 ;               49 & 94
The only option that's here in the answer choices is 83, so that's correct. Choice E

In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 of the boys and exactly 1/4 of the girls walk to school. What is the greatest possible number of students in this class who walk to school?
A. 9 B. 10 C. 11 D. 12 E. 13

Since 1/3 boys > 1/4 girls you want to maximize no of boys
The no of girls has to be a multiple of 4 since we cannot have half girls and no of boys multiple of 3
Working backwards.
36-4 girls = 32 boys not divisible by three
36-8 girls = 28 boys not divisible by three
36 - 12 girls = 24 boys which is divisible by three
so 24 boys and 12 girls and a total of 24*1/3 + 12*1/4 walking to school = 11

Let # of boys be B, then # of girls will be 36 - B.

We need to maximize (B/3) + (36-B)/4 => B /3 + 9 – (B / 4) = 9 + B /12, so we should maximize B, but also we should make sure that (B/12) = 9 remains an integer (as it represent # of people it cannot be in decimals).

Hence the Max value of B

for which b/12 is an integer is for b = 24

(B cannot be 36 as wee are told that there are some # of girls among 36) => (B /12) + 9 = 2 + 9 = 11.

Answer: C.

At 8:00 total tips received by A & B were $57. A gave off $4 to the kitchen staff and B received an additional $3. If A's total collection was 12 more than B then what was B's final collection?

Let at 8:00 tips received by A = A and By B = B
So A + B = 57
NOw A's tip = A-4
and B's = B+3
A-4 = B+3 + 12=>A= B+19
putting in 1st
B+19 + B =57
2B = 57-19=38=>B = 19
So B's final collection = 19+3=22

If 18 identical machines required 40 days to complete a job, how many fewer days would have been required to do a job if 6 additional machines of the same type had been used from the beginning

ANS :10
6/x + 18/40 = 1
6/x = 22/40
x = 120/11 approx 10 days

5 is placed to the right of two – digit number, forming a new three – digit number. The new number is 392 more than the original two-digit number. What was the original two-digit number?

If the original number has x as the tens digit and y as the ones digit (x and y are integers less than 10) then we can set up the equation:
100x + 10y + 5 = 10x + y + 392
90x + 9y = 387
9(10x+y) = 387
10x + y = 43 ==> x = 4, y = 3
the original number is 43, the new number is 435

A tourist purchased a total of 30 traveller’s cheques in $ 50 and $100 denominations. The total worth of the traveller’s check is $1800. How many checks of $ 50 denominations can he spend so that average amount (arithmetic mean) of the remaining traveller’s checks is $ 80?

A) 4 B) 12 C) 15 D) 20 E) 24

Let 100$ notes be x

Then (30 – x)50 + 100x = 1800

1500 – 50x + 100x = 300 => x = 6

{6 * 100 + (24 – x) * 50} / (30 –x) = 80

Ie., 600 + 1200 – 50 x = 2400 – 80 x => 30x = 600 => x = 20

Answer is D

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InSTaMeChEng

Thursday, 22 August 2013