Math Problem Statement

8. Find the sum: 8, -4, 2, -........ ∞ (INFINITY) 9. In how many ways a committee, consisting of 5 men and 6 women, can be formed from 8 men and 10 women? 10. Find the rate of interest required for Rs. 7440 to earn 760 in 14 months.

Solution

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Q8. Find the sum:

8,4,2, 8, -4, 2, -\dots\ \infty

This is an infinite geometric series. The terms alternate between positive and negative, and the first term (aa) is 8. The common ratio (rr) is found by dividing the second term by the first:
r=48=12r = \frac{-4}{8} = -\frac{1}{2}

The formula for the sum of an infinite geometric series is:
S=a1r,if r<1S_\infty = \frac{a}{1 - r}, \quad \text{if } |r| < 1

Substituting the values:
S=81(12)=81+12=832=163S_\infty = \frac{8}{1 - \left(-\frac{1}{2}\right)} = \frac{8}{1 + \frac{1}{2}} = \frac{8}{\frac{3}{2}} = \frac{16}{3}

Answer for Q8: 163\frac{16}{3}


Q9. In how many ways a committee, consisting of 5 men and 6 women, can be formed from 8 men and 10 women?

This is a combination problem where we need to select 5 men from 8 and 6 women from 10. The number of ways to choose rr objects from nn objects is given by:
(nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

For men:
(85)=8!5!(85)!=8×7×63×2×1=56\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

For women:
(106)=10!6!(106)!=10×9×8×74×3×2×1=210\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

The total number of ways is the product of the two combinations:
(85)(106)=56210=11760\binom{8}{5} \cdot \binom{10}{6} = 56 \cdot 210 = 11760

Answer for Q9: 11,760 ways


Q10. Find the rate of interest required for Rs. 7440 to earn Rs. 760 in 14 months.

This is a problem of Simple Interest (SI). The formula for SI is:
SI=PRT100SI = \frac{P \cdot R \cdot T}{100}

Where:

  • P=7440P = 7440 (Principal)
  • SI=760SI = 760 (Simple Interest)
  • T=1412=76T = \frac{14}{12} = \frac{7}{6} years (time in years)
  • R=?R = ? (Rate of interest)

Rearranging the formula to solve for RR:
R=SI100PTR = \frac{SI \cdot 100}{P \cdot T}

Substituting the values:
R=760100744076=76000744076=76000674407R = \frac{760 \cdot 100}{7440 \cdot \frac{7}{6}} = \frac{76000}{7440 \cdot \frac{7}{6}} = \frac{76000 \cdot 6}{7440 \cdot 7}

Simplify step by step:
R=45600052080=8.75%R = \frac{456000}{52080} = 8.75\%

Answer for Q10: 8.75%


Let me know if you'd like further details or clarification.

Here are 5 related questions to expand your understanding:

  1. What is the general condition for the convergence of an infinite geometric series?
  2. How can combinations be used to solve problems involving groups with specific constraints?
  3. What is the difference between simple interest and compound interest?
  4. How would the sum of the series in Q8 change if the common ratio were positive?
  5. How can permutation formulas help in forming committees with specific roles?

Tip: For combination problems, always ensure the sum of selected individuals doesn't exceed the total available pool.

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Math Problem Analysis

Mathematical Concepts

Infinite Geometric Series
Combinatorics
Simple Interest

Formulas

Sum of infinite geometric series: S∞ = a / (1 - r), |r| < 1
Combination formula: C(n, r) = n! / [r!(n - r)!]
Simple Interest formula: SI = (P × R × T) / 100

Theorems

Convergence of infinite geometric series (|r| < 1)
Combinatorial selection principles
Simple Interest calculation

Suitable Grade Level

Grades 9-12