To see from questions 1 to 3, please visit the page "Solving Word Problems Involving Quadratic Equations"

**Question 4 :**

A girl is twice as old as her sister. Five years hence, the product of their ages (in years) will be 375. Find their present ages.

**Solution :**

Let "x" be the age of sister

"2x" be the age of girl

Five years hence,

age of sister = x + 5

Age of girl = 2x + 5

Product of their ages = 375

(x + 5)(2x + 5 ) = 375

2x^{2} + 5x + 10x + 25 - 375 = 0

2x^{2} + 15x - 350 = 0

2x^{2} - 20x + 35x - 350 = 0

2x(x - 10) + 35(x -10) = 0

(2x + 35)(x - 10) = 0

x = 10

Hence the age of sister is 10 years, age of girl = 2(10) = 20 years.

**Question 5 :**

A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametrically opposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from the two gates should the pole be erected?

**Solution :**

x^{2} + (x + 4)^{2} = 20^{2}

x^{2} + x^{2} + 2x(4) + 4^{2} = 400

2x^{2} + 8x + 16 - 400 = 0

2x^{2} + 8x - 384 = 0

x^{2} + 4x - 192 = 0

(x + 16)(x- 12) = 0

x + 16 = 0 and x - 12 = 0

x = -16 and x = 12.

Hence required distance is 12 m and 16 m.

**Question 6 :**

From a group of 2x^{2} black bees , square root of half of the group went to a tree. Again eight-ninth of the bees went to the same tree. The remaining two got caught up in a fragrant lotus. How many bees were there in total?

**Solution :**

Total number of bees = 2x^{2}

Number of bees went to tree = (1/2)√x^{2 } = x

Number of bees went to the same tree = (8/9)2x^{2 = }16x^{2}/9

Remaining number of bees = 2

x + (16x^{2}/9) + 2 = 2x^{2 }

9x + 16x^{2} + 18 = 18x^{2}

18x^{2}- 16x^{2 }- 9x - 18 = 0

2x^{2 }- 9x - 18 = 0

2x^{2 }- 12x + 3x - 18 = 0

2x(x - 6) + 3(x - 6) = 0

(2x + 3)(x - 6) = 0

x - 6 = 0, 2x + 3 = 0

x = 6, x = -3/2 (not possible)

Total number of bees = 2(6)^{2} = 2(36) = 72

Hence total number of bees = 72.

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