Math Problem Statement

The marks scored in Mathematics by A, B, C and D form a geometric progression in that order. If the marks scored by C were 30 5/9% less than the sum of the marks scored by A and B, then the marks scored by D were what percent more than the marks scored by B?

Solution

Let's carefully analyze the problem and solve it step by step:

We are told that the marks scored by A, B, C, and D are in geometric progression (GP). Let's denote their marks as:

  • A=aA = a
  • B=arB = ar
  • C=ar2C = ar^2
  • D=ar3D = ar^3

where aa is the first term and rr is the common ratio of the GP.

Step 1: Relationship between marks of C and sum of A and B

It is given that the marks scored by C are 3059\frac{5}{9}% less than the sum of the marks scored by A and B. This can be mathematically written as:

C=(A+B)×(13059100)C = (A + B) \times \left(1 - \frac{30 \frac{5}{9}}{100}\right)

First, convert 305930 \frac{5}{9}% into a decimal: 3059%=2759%=275900=0.305530 \frac{5}{9}\% = \frac{275}{9} \% = \frac{275}{900} = 0.3055

Thus, the relation becomes: ar2=(a+ar)×(10.3055)ar^2 = (a + ar) \times (1 - 0.3055) ar2=(a+ar)×0.6945ar^2 = (a + ar) \times 0.6945

Simplifying: ar2=0.6945a(1+r)ar^2 = 0.6945a(1 + r) r2=0.6945(1+r)r^2 = 0.6945(1 + r)

Step 2: Solving the quadratic equation

Expand and rearrange: r20.6945r0.6945=0r^2 - 0.6945r - 0.6945 = 0

We can solve this quadratic equation using the quadratic formula: r=b±b24ac2ar = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=0.6945b = -0.6945, and c=0.6945c = -0.6945.

Substitute the values: r=(0.6945)±(0.6945)24(1)(0.6945)2(1)r = \frac{-(-0.6945) \pm \sqrt{(-0.6945)^2 - 4(1)(-0.6945)}}{2(1)} r=0.6945±0.4823+2.7782r = \frac{0.6945 \pm \sqrt{0.4823 + 2.778}}{2} r=0.6945±3.26032r = \frac{0.6945 \pm \sqrt{3.2603}}{2} r=0.6945±1.80562r = \frac{0.6945 \pm 1.8056}{2}

Thus, the two possible values for rr are: r1=0.6945+1.80562=1.25r_1 = \frac{0.6945 + 1.8056}{2} = 1.25 r2=0.69451.80562=0.5555r_2 = \frac{0.6945 - 1.8056}{2} = -0.5555

Since the common ratio in a geometric progression should be positive, we take r=1.25r = 1.25.

Step 3: Percentage comparison of marks between D and B

Now, we want to find what percentage the marks scored by D are more than the marks scored by B. The marks scored by D are ar3ar^3 and the marks scored by B are arar. The percentage increase is given by:

Percentage Increase=DBB×100=ar3arar×100=r3rr×100=(r21)×100\text{Percentage Increase} = \frac{D - B}{B} \times 100 = \frac{ar^3 - ar}{ar} \times 100 = \frac{r^3 - r}{r} \times 100 = (r^2 - 1) \times 100

Substitute r=1.25r = 1.25: Percentage Increase=(1.2521)×100=(1.56251)×100=0.5625×100=56.25%\text{Percentage Increase} = (1.25^2 - 1) \times 100 = (1.5625 - 1) \times 100 = 0.5625 \times 100 = 56.25\%

Thus, the marks scored by D are 56.25% more than the marks scored by B.

Conclusion:

The answer is 56.25%\boxed{56.25\%}.

Would you like more details on any part of this explanation?

Here are some related questions:

  1. How is a geometric progression defined and used in solving problems?
  2. Can you explain more about the quadratic formula and its application here?
  3. How does the concept of percentage increase work in real-life problems?
  4. What are the properties of geometric progressions that help in calculating ratios between terms?
  5. How do you handle decimal percentages like 3059%30 \frac{5}{9}\% in problems?

Tip: In geometric progression, the ratio between any two consecutive terms remains constant, which makes it easy to model growth or decay situations.

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Math Problem Analysis

Mathematical Concepts

Geometric Progression
Percentage Calculations
Quadratic Equations

Formulas

C = (A + B) * (1 - 30 5/9%)
r^2 - 0.6945r - 0.6945 = 0
Percentage Increase = (r^3 - r) / r * 100

Theorems

Quadratic formula

Suitable Grade Level

High School