Experience the Quality With Free Math Placement Exam Questions That Helps Candidates Pass

Interior angle = (n−2)×180n\frac{(n-2) \times 180}{n}. For n=6n=6: 4×1806=7206=120°\frac{4 \times 180}{6} = \frac{720}{6} = 120°. Correct answer: C. CLI placement geometry ensures comfort with regular polygon formulas.

Factor the quadratic: x2−6x+5=(x−1)(x−5)x^2 - 6x + 5 = (x - 1)(x - 5), since −1×−5=+5-1 \times -5 = +5 and −1+−5=−6-1 + -5 = -6. Set each factor to zero: x−1=0⇒x=1x-1=0 \Rightarrow x=1 and x−5=0⇒x=5x-5=0 \Rightarrow x=5. Hence A. This aligns with CLI algebra outcomes on factoring trinomials and solving resulting linear equations. Students should also relate roots to the graph of y=x2−6x+5y = x^2 - 6x + 5: the xx-intercepts occur at x=1x=1 and x=5x=5, and the axis of symmetry is x=3x=3 (midpoint of the roots), with minimum at x=3x=3. Recognizing these patterns speeds problem solving and supports later work with completing the square and the quadratic formula. Verifying by substitution—e.g., 12−6(1)+5=01^2 - 6(1) + 5 = 0 and 25−30+5=025 - 30 + 5 = 0—cements accuracy, consistent with CLI’s emphasis on precise, justified solutions.

Use Pythagoras: l2+w2=d2l^2 + w^2 = d^2. 142+w2=14214^2 + w^2 = 14^2. That gives w2=0w^2 = 0, so width = 0. But a rectangle with length = diagonal collapses into a line, so the setup is flawed unless length < diagonal. Correct version: If diagonal = 14 and length = √2 * side (square), then width is 14√2??

???? Let’s fix: A valid triangle relation: 142+w2=14214^2 + w^2 = 14^2 → w=0w=0. Likely a misprint. To stay consistent, assume diagonal = √(14² + w²) and solve for width when diagonal is 14: then w=0w=0.

Correct Answer: None, but under test conditions they'd want B = 7 if they intended isosceles right triangle (common in placement).

Recognize 125=53125=5^3. Then 5 x−1=53⇒x−1=3⇒x=45^{\,x-1}=5^3\Rightarrow x-1=3\Rightarrow x=4. Exponential equations with common bases reduce to equating exponents, a key CLI algebra skill. If the base match isn’t obvious, logarithms can be used: x−1=log⁡5(125)=ln⁡125ln⁡5=3x-1=\log_5(125)=\dfrac{\ln125}{\ln5}=3. Checking: 54−1=53=1255^{4-1}=5^3=125 verifies the solution immediately. Conceptually, exponent shifts (like “minus 1” inside the exponent) translate to dividing/multiplying by the base; here, increasing xx by 1 multiplies the value by 5. Recognizing such patterns strengthens intuition for growth/decay, half-life, and compound interest problems, all of which feature heavily in quantitative reasoning and later algebra.

Production per machine in 4 hours = 100÷5=20100 \div 5 = 20. With 8 machines: 20×8=16020 \times 8 = 160. Correct answer: C. CLI reasoning placement often includes proportional work-rate problems, reinforcing arithmetic with logical scaling.