Test: Algebra - GMAT MCQ

To find the value of 1/2 + 1/6 + 1/12 + ... + 1/90, we can simplify the given expression and look for a pattern.

The given expression can be written as follows:
1/2 + 1/6 + 1/12 + ... + 1/90

We can observe that the denominators form an arithmetic sequence with a common difference of 4:
2, 6, 10, 14, ..., 90

We can rewrite the sequence by factoring out 2 from each term:
2(1/1) + 2(1/3) + 2(1/5) + 2(1/7) + ... + 2(1/45)

Now, we can simplify the expression:
2[(1/1) + (1/3) + (1/5) + (1/7) + ... + (1/45)]

The expression inside the square brackets is a partial sum of the harmonic series, which can be expressed as:
H(n) = 1/1 + 1/2 + 1/3 + ... + 1/n

We know that the sum of the harmonic series H(n) can be approximated by the natural logarithm of n:
H(n) ≈ ln(n)

Therefore, the value of the expression inside the square brackets can be approximated by ln(45).

Now, we can substitute this back into the original expression:
2[ln(45)]

Simplifying further:
2ln(45) = ln(45²) = ln(2025)

Using logarithmic properties, we can rewrite ln(2025) as:
ln(2025) = ln(9 * 225) = ln(9) + ln(225) = ln(9) + ln(15²) = ln(9) + 2ln(15)

The expression now becomes:
2ln(15) + ln(9)

Now, we can convert this back into exponential form:
2ln(15) + ln(9) = ln(15²) + ln(9) = ln(225) + ln(9) = ln(225 * 9) = ln(2025)

Finally, we can convert ln(2025) back into its equivalent fractional form:
ln(2025) = 9/10

Therefore, the value of 1/2 + 1/6 + 1/12 + ... + 1/90 is 9/10.

The correct answer is D: 9/10.

For n⁵, the units digit will depend on the units digit of n itself raised to the power of 5. We can observe the pattern of units digits when different numbers are raised to the power of 5:

n: 0 1 2 3 4 5 6 7 8 9
n⁵: 0 1 32 243 1024 3125 7776 16807 32768 59049

As we can see, the units digit of n⁵ repeats after every 5 numbers. It cycles through the digits 0, 1, 2, 3, 4.

For -5n³, we need to consider the units digit of -5 multiplied by the units digit of n³. Similarly, we can observe the pattern of units digits when different numbers are raised to the power of 3:

n: 0 1 2 3 4 5 6 7 8 9
n³: 0 1 8 27 64 125 216 343 512 729

Again, the units digit of n³ repeats after every 10 numbers. It cycles through the digits 0, 1, 8, 7, 4, 5, 6, 3, 2, 9.

For 4n, the units digit will simply be 4 times the units digit of n.

Now, let's consider the units digit of the expression n⁵ - 5n³ + 4n:

The units digit of n⁵ will be 0, 1, 2, 3, or 4.

The units digit of -5n³ will be 0, 1, 8, 7, or 4.

The units digit of 4n will be 0, 4, 8, 2, 6.

To determine the overall units digit, we need to consider the combinations of units digits from each term in the expression.

By analyzing all possible combinations, we find that the units digit will always be 0. Regardless of the units digit of n, the terms will cancel out in such a way that the units digit of the expression is always 0.

Therefore, the correct answer is A: 0.

2²⁰ + 2²⁹ = 2²⁰(1+2⁹) ≈ 2²⁰2⁹ = 2²⁹ 
2^b – (2²⁰ + 2²⁹) = 0;
2^b – 2^{29 = 0;}
2^b = 2²⁹;
b = 29.

Lets use x = {-12, 1, 12}

|-12 +11| - |-12 - 7| = 1 - 19 = -18

|1 +11| - |1 - 7| = 12 - 6 = 6

|12 +11| - |12 - 7| = 23 - 5 = 18

The minimum value is -18.

Let's assume that the usual number of calendars the store owner purchases is x.

If she buys 50% more calendars, the number of calendars she would purchase is 1.5x (50% more than x).

The store owner is given a 20% discount off the standard price for these calendars. This means she would pay 80% of the standard price, which is equivalent to 0.8 times the standard price.

According to the given information, her total cost would be 120 times the dollar value of the standard price of one calendar. This can be expressed as 120 times the standard price.

Setting up the equation based on the information given:

1.5x * 0.8 * standard price = 120 * standard price

Simplifying the equation:

1.2x = 120

Dividing both sides of the equation by 1.2:

x = 120 / 1.2

x = 100

Therefore, the store owner usually purchases 100 calendars, which corresponds to option C.

To find the value of x for which the quadratic expression, x² + 16x + 100, takes the minimum possible value, we can use the formula for the vertex of a quadratic function.

The vertex of a quadratic function in the form ax² + bx + c is given by the x-coordinate:
x = -b / (2a)

In this case, a = 1, b = 16, and c = 100. Substituting these values into the formula, we get:
x = -16 / (2 * 1)
x = -16 / 2
x = -8

Therefore, the value of x for which the quadratic expression takes the minimum possible value is -8, which corresponds to option A.

6x – 5y + 3z = 23... (I)
4x + 8y – 11z = 7... (II)
5x – 6y + 2z = 12... (III)
Eqn (I) - Eqn(II)
2x − 13y + 14z = 16...(IV)
Eqn(III) - Eqn (II)
 x − 14y + 13z = 5...(V)
Eqn.(IV) - Eqn (V)
x + y + z = 11 

To find the number n for which @(n) is equal to 50% of n, we can set up the equation and solve for n.

Let's start by expressing @(n) in terms of n: @(n) = (n^(1/3)) * √n

Now, we can set up the equation: (n^(1/3)) * √n = 0.5n

Next, let's simplify the equation: n^{(1/3 + 1/2)} = 0.5n

Combining the exponents: n^(5/6) = 0.5n

To remove the fractional exponent, we can raise both sides of the equation to the power of 6: (n^(5/6))⁶ = (0.5n)⁶

Simplifying: n⁵ = (0.5)⁶ * n^6 n⁵ = 0.015625 * n⁶

Dividing both sides of the equation by n⁵: 1 = 0.015625 * n

Dividing both sides by 0.015625: n = 1 / 0.015625 n = 64

Therefore, the number n for which @(n) is equal to 50% of n is 64.

The correct answer is B: 64.

I. All members of R are prime.
This statement is not necessarily true. While it is true that prime numbers have a greater likelihood of having a units digit of 1 or 9 when squared, it is not a guarantee. For example, the number 25 is a member of R (5² = 25), but it is not prime.

II. If N is a member of R, -N is a member of R.
This statement is true. If a number N has a units digit of 1 or 9 when squared, then (-N)² will also have a units digit of 1 or 9. The negative sign does not affect the units digit when squaring a number.

III. If N is a member of R, N² is a member of R.
This statement is also true. If a number N has a units digit of 1 or 9 when squared, then the result N² will also have a units digit of 1 or 9.

Based on the analysis, statements II and III must be true about set R. Therefore, the correct answer is D: II and III only.

Let's solve the problem by finding the consecutive even numbers that multiply to 9408.

Let's assume the smaller even number is x. The next consecutive even number will be x + 2.

According to the problem, we have the equation:
x(x + 2) = 9408

Expanding the equation:
x² + 2x = 9408

Rearranging the equation:
x² + 2x - 9408 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Let's factor it:

(x + 98)(x - 96) = 0

Setting each factor to zero:
x + 98 = 0 or x - 96 = 0

Solving each equation:
x = -98 or x = 96

We are looking for the greater number, so we take x = 96 + 2 = 98.

Therefore, the correct answer is B: 98.