Test: Simplifying Expression - GMAT MCQ

= [(1296)-(1156)]/35

= 140/35

= 4

9986*9986=squre of (10000-14)
=100000000-280000+196
=99720196

 where  x =1/4

= [ (1/4)² + 13(1/4) + 36 ] / (1/4) + 4

= [ 1/16 + 13/4 + 36 ] / 17/4

= [ ( 1 + 52 + 576) /16 ] x 4/16

= 629/16 x 4/17

= 37/4 

Step 1: Nature of Roots of Quadratic Equation Theory

D is called the Discriminant of a quadratic equation.
D = b² - 4ac for quadratic equations of the form ax² + bx + c = 0.
If D > 0, roots of such quadratic equations are Real and Distinct
If D = 0, roots of such quadratic equations are Real and Equal
If D < 0, roots of such quadratic equations are Imaginary. i.e., such quadratic equations will NOT have real roots.
In this GMAT sample question in algebra, we have to determine the largest integer value that ‘k’ can take such that the discriminant of the quadratic equation is positive.

Step 2: Compute maximum value of k that will make the discriminant of the quadratic equation positive

The given quadratic equation is x² - 6x + k = 0.
In this equation, a = 1, b = -6 and c = k
The value of the discriminant, D = 6² - 4 * 1 * k
We have compute values of 'k' for which D > 0.
i.e., 36 – 4k > 0
Or 36 > 4k or k < 9.

The question is "What is the highest integer value that k can take?"
If k < 9, the highest integer value that k can take is 8.

Solution : (a²+b²)(a+b)(a-b) + 2b⁴

= (a²+b²)(a² - b²) + 2b⁴

= (a⁴ - b⁴) + 2b⁴

since a> b > 0

set is a =2 b = 1

value = 15+2 = 17

Step 1: Question statement and Inferences

I:

(a + b)² = x

a² + b² + 2ab = x

Similarly,

II:

(a – b)² = y

a² + b² – 2ab = y

Step 2:Finding required values

If we subtract y from x, we have

x – y = (a² + b² + 2ab) – (a² + b² -2ab) = 4ab

We have eliminated a² and b². We are very close to the answer.

Step 3: Calculating the final answer

The question requires us to find the expression for 2ab. We have already obtained the expression for 4ab.

Lets divide both sides by 2,

Step 1: Question statement and Inferences

This question is based on the concept that n² – 1 = (n – 1)(n + 1)

Step 2 & 3: Finding required values and calculating the final answer

Apply the n² – 1 concept to the numerator:

Apply the n² – 1 concept again to the 16² – 1: = (16-1)(16+1)=15*17

Cancel the 15 × 17 from top and bottom:

Convert 16 to 2⁴ :

Convert (2⁴)² to 2⁸ :

We need to find the value of 2b.

Given equation is:

(We can safely cancel b from both sides since it is given that b≠0)

Given equation is  

x+xy² = z³

This can be rewritten as:

x(1+y²) = z³  …………….. (I)

We need to find the option statement whose value is not equal to 1.

Now let us look at each option.

Option A:

This is same as

 …………….. Expression A

From (I), we know that the numerator and denominator of Expression A are equal.

Therefore the value of Expression A is 1.

Option B:

This can be rewritten as:

We already saw in the previous option that this is equal to 1.

Option C:

This can be rewritten as: 

From I, we can rewrite this as:

 ………….. Expression C

It is given that y is a non-zero number. Therefore, the value of  (1+y²) is always greater than 1.

So, Expression C cannot be equal to 1.

Option D:

This can be rewritten as:

From options A and B, we already know that this expression is equal to 1.

Option E:

We already know that this expression is equal to 1.

Correct Answer: C

Step 1: Question statement and Inferences

Set up the fraction without the brackets so you can simplify it.

Step 2 & 3: Simplifying the fraction and calculating the final answer

Turn the bottom negative exponent into a fraction:

Multiply the numerator and denominator with 3⁷:

(3⁻⁴+3⁻⁵+3⁻⁵+3⁻⁶+3⁻⁶)×3⁷

Distribute the 3⁷ among the other exponents:

(3³+3²+3²+3¹+3¹)

Simplify the remaining numbers:

(27+9+9+3+3)

51

51 = 17 * 3

Answer: Option (C)

Steps 1 & 2: Understand Question and Draw Inferences

We need to find the sum of the squares of x² and y².

Square of x² = (x²)² = x⁴

Square of y² = (y²)² = y⁴

Therefore we need to find the value of the expression x⁴ + y⁴

Step 3: Analyze Statement 1

Statement (1) says:

x⁸ – y⁸ = 6

We know that:  x⁸ – y⁸ = (x⁴ + y⁴)(x⁴ – y⁴ )

[Formula used:   (a² - b²) = (a + b)(a - b)

This means that,(x⁴ + y⁴)(x⁴ – y⁴ ) = 6 ……………… (I)

However, we don’t know the value of the expression (x⁴ – y⁴ )

Therefore statement 1 alone is not sufficient to arrive at a unique answer

Step 4: Analyze Statement 2

Statement (2) says:

(x⁴ – y⁴ ) = 2 ……………………(II)

However, this alone doesn’t help us in finding the value of (x⁴ + y⁴)

Therefore statement 2 alone is not sufficient to arrive at a unique answer.

Step 5: Analyze Both Statements Together (if needed)

Now let us look at both the statements together.

 Using (II) in (I), we get:

(x⁴ + y⁴) * 2 = 6

(x⁴ + y⁴) = 3

Statement (1) and statement (2) together are sufficient to arrive at a unique answer.

Correct Answer: C

Step 1: Question statement and Inferences

The fraction is equal to zero when the numerator is equal to zero.

Step 2 & 3: Simplifying the fraction and calculating the final answer

For the fraction to equal zero, (x² + x⁴ -x⁶) has to equal 1 to subtract with the 1 on its left:

The only way that (x² + x⁴ -x⁶) can equal 1 is if x = 1 or –1. There are two possible values of x:  1 and –1.

Note that x cannot equal 2 or –2, making the denominator equal 0. This would result in a non-real number, which is not equal to 0.

Answer: Option (C)

Step 1: Question statement and Inferences

We have to replace x everywhere in the given expression for P with −1/x

Step 2: Finding required values

Replacing x with −1/x

 , we get Q.

Multiplying the numerator and the denominator with -1, we have

Multiplying the numerator and the denominator with x³, we have

Step 3: Calculating the final answer

Notice the similarity between P and Q.

Are P and 1/Q  equal? No.

How about  - 1/Q ?

In other words, PQ = -1

Answer: Option (B)

Given equation is:

(Let’s call the above equation as I)

We need to find the value of 2x + 3

On the RHS of I, notice that:

1. The numerator: 70y+70=70(y+1)
2. The denominator: y+y2=y(1+y)

Therefore I can written as:

Observe that you can cancel the common term   in the numerator and denominator.

Similarly, you can cancel the common term   in the denominator of LHS and the denominator of RHS.

(Note that it is given that y>2 and therefore neither y nor y+1 can be zero. So we can safely cancel these terms out.)

(x−2)²+6=70

•  
• (x−2)²=64
•  
• (x−2)=±8

Case 1:

(x - 2) = 8

• x = 10
• 2x + 3 = 23

Case 2:

(x - 2) = -8

• x = -6
• 2x + 3 = -9

In the given options, -9 is given.

Correct Answer: A