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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.

If the difference between a number and its reciprocal is \(\frac{24}{5}\), find the number.

Solution:

Let a number be x.

Its reciprocal is \(\frac{1}{x}\)

\(x-\frac{1}{x}=\frac{24}{5}\)

\(\frac{x^{2}-1}{x}=\frac{24}{5}\)

5x^{2} – 5 -24x = 0 ⇒ 5x^{2} – 24x – 5 = 0

5x^{2} – 25x + x – 5 = 0

5x(x – 5) + 1 (x – 5) = 0

(5x + 1)(x – 5) = 0

x = \(\frac{-1}{5}\), 5

∴ The number is \(\frac{-1}{5}\) or 5.

Question 2.

A garden measuring 12m by 16m is to have a pedestrian pathway that is ‘w’ meters wide installed all the way around so that it increases the total area to 285 m^{2}. What is the width of the pathway?

Solution:

Area of ABCD = 16 × 12 ^{2}

= 192 m^{2}

Area of A’B’C’D’ (12 + 2w)(16 + 2w)

192 + 32 w + 24 w + 4 w^{2} = 285

4w^{2} + 56w – 93 = 0

4w^{2} + 62w – 6w – 93 = 0

2w(2w + 31) – 3(2w + 31) = 0

(2w – 3)(2w + 31) = 0

w = 1.5 or \(\frac{-31}{2}\) = 15.5

w = – 15.5 cannot possible 3

∴ w = \(\frac{3}{2}\) = 1.5 m

(w cannot be (-ve))

The width of the pathway = 1.5 m.

Question 3.

A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the bus.

Solution:

Let x km/hr be the constant speed of the bus.

The time taken to cover 90 km = \(\frac{90}{x}\) hrs.

When the speed is increased bus 15 km/hr.

= \(\frac{90}{x+15}\)

It is given that the time to cover 90 km is reduced by \(\frac{1}{2}\) hrs.

The speed of the bus cannot be -ve value.

∴ The original speed of the bus is 45 km/hr.

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 the age of the girl be = 2y years

Her sister’s age is = y years

(2y + 5)(y + 5) = 375

2y^{2} + 5y+ 10y + 25 – 375 = 0

2y^{2} + 15y – 350 = 0

y = 10, y cannot be (-ve).

∴ Girls age is 2y = 20 years.

Her sister’s age = y = 10 years.

Question 5.

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

Solution:

PQ = 20 m

PX – XQ = 4 m …………… (1)

Squaring both sides,

PX^{2} + XQ^{2} – 2PX . QX = 16 (∵ ∠Q × p = 90°)

PQ^{2} – 2P × QX = 16

400 – 16 = 2PX × QX

384 = 2PX – QX

PX . QX = 192

∴ (PX + QX)^{2} = PX^{2} + QX^{2} + 2PX . QX

= 400 + 2 × 192

= 784 = 28^{2}

∴ PX + QX = 28

From (1) & (2) 2PX = 32 ⇒ PX = 16 m QX = 12 m

∴Yes, the distance from the two gates to the pole PX and QX is 12 m, 16m.

Question 6.

From a group of black bees 2x^{2}, 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 no. of bees = 2x^{2}

18x^{2} – 9x – 16x^{2} = 2 × 9

2x^{2} – 9x – 18 = 0

(x – 6)(2x + 3) = 0

x = 6, x = \(\frac{-3}{2}\) (it is not possible)

No. of bees in total = 2x^{2}

= 2 × 6^{2} = 72

Question 7.

Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice? (Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances).

Solution:

Let the person stand at a distance ‘d’ from 2nd gallery having 9 singers.

Given that ratio of sound intensity is equal to the square of the ratio of their corresponding distance.

∴ \(\frac{9}{4}=\frac{d^{2}}{(70-d)^{2}}\)

4d^{2} = 9(70 – d)^{2}

4d^{2} = 9(70^{2} – 140d + d^{2})

4d^{2} = 9 × 70^{2} – 9 × 140d + 9d^{2}

∴ 5d^{2} – 9 × 140d + 9 × 70^{2} = 0

5d^{2} = 1260d + 44100 = 0

d^{2} – 252d + 8820 = 0

= \(\frac{420}{2} \text { or } \frac{84}{2}\)

= 120 or 42

∴ The person stand at a distance 28m from the first and 42 m from second gallery.

Question 8.

There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at ₹ 3 and ₹ 4 per square metre respectively is ₹364. Find the width of the gravel path.

Solution:

Area of the flower bed = a^{2}

Area of the gravel path = 100 – a^{2}

Area of total garden =100

given cost of flower bed + gravelling = ₹ 364

3a^{2} + 4 (100 – a^{2}) = ₹ 364

3a^{2} + 400 – 4a^{2} = 364

∴ a^{2} = 400 – 364

= 36 ⇒ a = 6

width of gravel path =\(\frac{10-6}{2}=\frac{4}{2}\) = 2 cm

Question 9.

Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If I had your eggs, I would have earned ₹ 15”, to which the second replied: “If I had your eggs, I would have earned ₹ 6 \(\frac { 2 }{ 3 } \)”. How many eggs did each had in the beginning? Answer:

Number of eggs for the first women be ‘x’

Let the selling price of each women be ‘y’

Selling price of one egg for the first women = \(\frac { y }{ 100-x } \)

By the given condition

(100 – x) \(\frac { y }{ x } \) = 15 (for first women)

y = \(\frac { 15 }{ 100-x } \) ……(1)

x × \(\frac { y }{ (100-x) } \) = \(\frac { 20 }{ 3 } \) [For second women]

y = \(\frac { 20(100-x) }{ 3x } \) ……..(2)

From (1) and (2) We get

\(\frac { 15 }{ 100-x } \) = \(\frac { 20(100-x) }{ 3x } \)

45x^{2} = 20(100 – x)^{2}

(100 – x)^{2} = \(\frac{45 x^{2}}{20}\) = \(\frac { 9 }{ 4 } \) x^{2}

∴ 100 – x = \(\sqrt{\frac{9}{4} x^{2}}\)

100 – x = \(\frac { 3x }{ 2 } \)

3x = 2(100 – x)

3x = 200 – 2x

3x + 2x = 200 ⇒ 5x = 200

x = \(\frac { 200 }{ 5 } \) ⇒ x = 40

Number of eggs with the first women = 40

Number of eggs with the second women = (100 – 40) = 60

Question 10.

The hypotenuse of a right-angled triangle is 25 cm and its perimeter 56 cm. Find the length of the smallest side.

Solution:

AB + BC + CA = 56 cm

AC = 25 cm

AB + BC = 56 – 25 = 31

AB^{2} + BC^{2} = AC^{2}

(AB + BC)^{2} – 2AB . BC = AC2 [∵ a^{2} + b^{2} = (a + b)^{2} – 2ab]

31^{2} – 2AB . BC = 25^{2}

-2AB . BC = 625 – 961

∴ The length of the smallest side is 7 cm.