Directions : Solve the series to answer the following questions. Series I: 60…

2023

Directions : Solve the series to answer the following questions.
Series I: 60 120 24 48 9.6 19.2
Series II: 100 W X Y Z 32
Note: Both series follows the same patterns

If roots of quadratic equation are Y/W and (W - X) / 16, then find the quadratic equation whose variable is r?

  1. A.

    r2 - 90r/5 + 7/11 = 0

  2. B.

    r2 - 32r/5 + 8/25 = 0

  3. C.

    r2 - 87r/4 + 3 = 0

  4. D.

    r2 - 52r/5 + 4 = 0

  5. E.

    None of the above

Attempted by 2 students.

Show answer & explanation

Correct answer: D

Concept

For two number series that follow the same alternating multiply/divide rule, first extract the rule from the fully known series, then apply it term-by-term to the partly hidden series. Once the required values are known, build the quadratic from its roots using the identity: if a and b are the roots, the equation is r2 − (a + b)r + ab = 0.

Application

Step 1 — find the pattern from Series I.

  1. 60 × 2 = 120

  2. 120 ÷ 5 = 24

  3. 24 × 2 = 48

  4. 48 ÷ 5 = 9.6

  5. 9.6 × 2 = 19.2

So the rule alternates: ×2, then ÷5, then ×2, then ÷5, then ×2.

Step 2 — apply the same rule to Series II (start 100, end 32).

  1. W = 100 × 2 = 200

  2. X = 200 ÷ 5 = 40

  3. Y = 40 × 2 = 80

  4. Z = 80 ÷ 5 = 16

  5. check: 16 × 2 = 32 ✓ (matches the given last term)

Step 3 — compute the two roots.

  • First root = Y / W = 80 / 200 = 2/5 = 0.4

  • Second root = (W − X) / 16 = (200 − 40) / 16 = 160 / 16 = 10

Step 4 — form the quadratic.

  • Sum of roots = 2/5 + 10 = 52/5

  • Product of roots = (2/5) × 10 = 4

Equation: r2 − (52/5)r + 4 = 0, i.e. r2 − 52r/5 + 4 = 0.

Cross-check

Substitute r = 10: 100 − (52×10)/5 + 4 = 100 − 104 + 4 = 0 ✓. Substitute r = 0.4: 0.16 − (52×0.4)/5 + 4 = 0.16 − 4.16 + 4 = 0 ✓. Both roots satisfy it.

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