Algebra Review Test - 1 - Practice Mode

1

Directions for Questions 1 to 3: These are based on the functions defined below

Q(a, b) = Quotient when a is divided by b
R²(a, b) = Remainder when a is divided by b
R(a, b) = a²/b²
SQ(a, b) = √(a – 1)/√(b – 1)

Q. SQ (5, 10) – ? > 0

A. (8/3) R (5, 10) B. R² (5, 10) + Q (5, 10) C. R² (5, 10)/2 D. 1/2 {R(2, 3) + SQ(17, 26)}

Correct Answer: D. 1/2 {R(2, 3) + SQ(17, 26)}

Explanation:

Step 1: Calculate SQ(5,10)

Using the definition SQ(a,b)=b−1a−1:

SQ(5,10)=10−1

5−1

=9

4

=32

So the inequality simplifies to:

32−?>0⟹?<32

Step 2: Evaluate Standard Multiple-Choice Options

Option (a): 38⋅R(5,10)
R(5,10)=10252=10025=41
38×41=32
- Check: 32−32=0≯0 (This makes it equal to 0).
Option (b): R2(5,10)+Q(5,10)
R2(5,10)=5andQ(5,10)=0
5+0=5
- Check: 32−5=−4.33≯0.
Option (c): 2R2(5,10)
25=2.5
- Check: 32−2.5=−1.83≯0.
Option (d): Any value strictly less than 32 (e.g., 21, 31, 0, or R(5,10)=0.25).

32−0.25=0.4167>0✓

2

Directions for Questions 1 to 3: These are based on the functions defined below

Q(a, b) = Quotient when a is divided by b
R²(a, b) = Remainder when a is divided by b
R(a, b) = a²/b²
SQ(a, b) = √(a – 1)/√(b – 1)

Q. SQ (a, b) is same as

A. bQ(a, b) + R²(a) B. √(R(a, b) – 1) C. [R{(a – 1), (b – 1)}] D. √R(a,1) – 1/√R(b,1)

Correct Answer: C. [R{(a – 1), (b – 1)}]

Explanation:

Step-by-Step Derivation

1. Given definition:

$$\text{SQ}(a, b) = \frac{\sqrt{a - 1}}{\sqrt{b - 1}} = \sqrt{\frac{a - 1}{b - 1}}$$

2. Evaluate $R(a-1, b-1)$ using $R(x, y) = \frac{x^2}{y^2}$:

$$R(a-1, b-1) = \frac{(a - 1)^2}{(b - 1)^2}$$

3. Take the square root:

$$\sqrt{R(a-1, b-1)} = \sqrt{\frac{(a - 1)^2}{(b - 1)^2}} = \frac{a - 1}{b - 1}$$

$$\implies \mathbf{\text{SQ}(a, b) = \sqrt{R(a-1, b-1)}}$$

Standard Multiple-Choice Match

In standard test options for this question:

(A) $b \cdot Q(a, b) + R^2(a)$
(B) $\sqrt{R(a, b) - 1}$
(C) $\sqrt{R(a - 1, b - 1)}$ (Correct Answer)
(D) $\sqrt{R(a, 1) - 1} \Big/ \sqrt{R(b, 1) - 1}$

3

Directions for Questions 1 to 3: These are based on the functions defined below

Q(a, b) = Quotient when a is divided by b
R²(a, b) = Remainder when a is divided by b
R(a, b) = a²/b²
SQ(a, b) = √(a – 1)/√(b – 1)

Q. Which of the following relations cannot be false?

A. R(a, b) = R²(a, b) · Q(a, b) B. a² · Q(a, b) = b² · R²(a, b) C. a = R²(a, b) + y · Q(a, b) D. SQ(a, b) = R(a, b) · R²(a, b)

Correct Answer: C. a = R²(a, b) + y · Q(a, b)

Explanation:

By definition of integer division:

Dividend=(Divisor×Quotient)+Remainder

Using the given functional definitions:

Quotient=Q(a,b)
Remainder=R2(a,b)

Substituting these gives the fundamental arithmetic identity:

a=b⋅Q(a,b)+R2(a,b)

Evaluating the Standard Multiple-Choice Options

(A) a=b⋅Q(a,b)+R2(a,b) → ALWAYS TRUE (Cannot be false).
(B) R(a,b)=R2(a,b)⋅Q(a,b) → FALSE (b2a2=Remainder×Quotient).
(C) a2⋅Q(a,b)=b2⋅R2(a,b) → FALSE (Does not hold in general).
(D) a=R2(a,b)⋅Q(a,b) → FALSE (Ignores the divisor b).

4

Directions for Questions 4 to 7: Answer the questions based on the following information:

W(a, b) = least of a and b
M(a, b) = greatest of a and b
N(a) = absolute value of a

Q. Find the value of 1 + M [y + N {– W (x, y)}, N {y + W (M (x, y), N (y))}] given that x = 2 and y = – 3.

A. 0 B. 1 C. 2 D. 3

Correct Answer: C. 2

Explanation:

Definitions: $W(a, b) = \min(a, b)$, $M(a, b) = \max(a, b)$, $N(a) = \vert{}a\vert{}$.

Inner Term 1:
$W(x, y) = W(2, -3) = -3$
$N(-W(x, y)) = N(3) = 3$
$y + N(-W(x, y)) = -3 + 3 = 0$
Inner Term 2:
$N(y) = N(-3) = 3$
$M(x, y) = M(2, -3) = 2$
$W(M(x, y), N(y)) = W(2, 3) = 2$
$N(y + 2) = N(-3 + 2) = N(-1) = 1$
Outer Term:
$M(0, 1) = 1$
$1 + M(...) = 1 + 1 = 2$

Correct Answer: (c) 2

5

Directions for Questions 4 to 7: Answer the questions based on the following information:

W(a, b) = least of a and b
M(a, b) = greatest of a and b
N(a) = absolute value of a

Q. Given that a > b, then the relation M {N (x), W (x, y)} = W [x, N {M (x, y)}] does not hold if
(a) x > 0, y < 0, |x| > |y|

A. x > 0, y < 0, |x| > |y| B. x > 0, y < 0, |y| > |x| C. x > 0, y > 0 D. x < 0, y < 0

Correct Answer: D. x < 0, y < 0

Explanation:

Given: $x > 0, y < 0, \vert{}y\vert{} > \vert{}x\vert{}$ (Option b).

Let $x = 2, y = -5$:

LHS: $M\{N(x), W(x, y)\} = M\{2, -5\} = 2$
RHS: $M(x, y) = M(2, -5) = 2 \implies N(2) = 2$
$W[x, N\{M(x, y)\}] = W[2, 2] = 2$

Testing $x < 0, y < 0$ (Option d): Let $x = -2, y = -5$:

LHS: $M\{2, -5\} = 2$
RHS: $M(-2, -5) = -2 \implies N(-2) = 2 \implies W[-2, 2] = -2$
$\text{LHS} \neq \text{RHS}$
Correct Answer: (d) x < 0, y < 0

6

Directions for Questions 4 to 7: Answer the questions based on the following information:

W(a, b) = least of a and b
M(a, b) = greatest of a and b
N(a) = absolute value of a

Q. Which of the following must be correct for x, y < 0

A. N (W (x, y)) ≤ W (N (x), N (y)) B. N (M (x, y)) > W (N (x), N (y)) C. N (M (x, y)) = W (N (x), N (y)) D. N (M (x, y)) < M (N (x), N (y))

Correct Answer: C. N (M (x, y)) = W (N (x), N (y))

Explanation:

Given: $x < 0, y < 0$.

Let $x = -5, y = -2$:

$N(x) = 5, N(y) = 2$
$M(x, y) = -2 \implies N(M(x, y)) = N(-2) = 2$
$W(N(x), N(y)) = W(5, 2) = 2$

Thus, $N(M(x, y)) = W(N(x), N(y))$ always holds true for negative numbers.

Correct Answer: (c) N (M (x, y)) = W (N (x), N (y))

7

Directions for Questions 4 to 7: Answer the questions based on the following information:

W(a, b) = least of a and b
M(a, b) = greatest of a and b
N(a) = absolute value of a

Q. For what value of x is W (x² + 2x, x + 2) < 0?

A. –2 < x < 2 B. –2 < x < 0 C. x < –2 D. Both (2) and (3)

Correct Answer: D. Both (2) and (3)

Explanation:

Given: $W(x^2 + 2x, x + 2) < 0$.

$W(a, b) < 0$ implies at least one of the terms must be strictly negative ($a < 0$ or $b < 0$).

Condition 1: $x + 2 < 0 \implies x < -2$
Condition 2: $x^2 + 2x < 0 \implies x(x + 2) < 0 \implies -2 < x < 0$

Combining both valid regions yields $x < -2$ or $-2 < x < 0$.

Correct Answer: (d) Both (2) and (3)

8

It is given that $ (a^{n-3}+a^{n-5}b^{2}+\cdots+b^{n-3})pq=0 $, where $ p $ and $ q\neq0 $ and $ n $ is odd, then

$ \dfrac{a^n-b^n}{a^n+b^n}\cdot\dfrac{a+b}{a-b}= \ ? $

A. 1 B. – 1 C. 3/2 D. 0

Correct Answer: D. 0

Explanation:

Given: $(a^{n-3} + a^{n-5}b^2 + \dots + b^{n-3})pq = 0$ with $p, q \neq 0$ and $n$ is odd. Since $p, q \neq 0$, the series inside must be $0$:

$$a^{n-3} + a^{n-5}b^2 + \dots + b^{n-3} = 0$$

For odd $n$, this represents the polynomial expansion of $\frac{a^{n-1} - b^{n-1}}{a^2 - b^2}$. Since this equals $0$, we must have $a^{n-1} = b^{n-1}$, which implies $a = -b$ (or $a + b = 0$).

Substituting $a + b = 0$ into the target expression:

$$\frac{a^n - b^n}{a^n + b^n} \cdot \frac{a + b}{a - b} = \frac{a^n - b^n}{a^n + b^n} \cdot 0 = 0$$

Correct Answer: (d) 0

9

If $ f=\dfrac{1}{\log_{2}\pi}+\dfrac{1}{\log_{4.5}\pi} $,

which of the following is true?

A. f > 4 B. 2 < f < 4 C. 1 < f < 2 D. 0 < f < 1

Correct Answer: C. 1 < f < 2

Explanation:

Using the change of base property $\frac{1}{\log_b a} = \log_a b$:

$$f = \log_\pi 2 + \log_\pi 4.5 = \log_\pi (2 \times 4.5) = \log_\pi 9$$

Since $\pi \approx 3.1416$:

$\pi^1 = \pi \approx 3.14$
$\pi^2 \approx 9.87$

Because $3.14 < 9 < 9.87$, it follows that $1 < \log_\pi 9 < 2$.

Correct Answer: (c) 1 < f < 2

10

If px + qy > rx + sy, and y, x, p, q, r, s > 0 and if x < y, then which of the following must be true?

A. p/q > q/r B. p – q > r – s C. p + q > r + s D. p + q < r + s

Correct Answer: C. p + q > r + s

Explanation:

Given: $px + qy > rx + sy$ with $x, y, p, q, r, s > 0$ and $x < y$.

Rearranging the inequality:

$$px - rx > sy - qy \implies x(p - r) > y(s - q)$$

Since $x < y$ and both are positive, dividing by positive terms yields:

$$p - r > \frac{y}{x}(s - q)$$

Because $\frac{y}{x} > 1$, this relationship enforces that:

$$p - q > r - s$$

Correct Answer: (b) p – q > r – s

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Algebra Review Test 1

Step 1: Calculate SQ(5,10)

Step 2: Evaluate Standard Multiple-Choice Options

Step-by-Step Derivation

Standard Multiple-Choice Match

Evaluating the Standard Multiple-Choice Options

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Subtotal:

Algebra Review Test 1

Step 1: Calculate SQ(5,10)

Step 2: Evaluate Standard Multiple-Choice Options

Step-by-Step Derivation

Standard Multiple-Choice Match

Evaluating the Standard Multiple-Choice Options