Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.12

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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 \(\frac{1}{x}\)
\(x-\frac{1}{x}=\frac{24}{5}\)
\(\frac{x^{2}-1}{x}=\frac{24}{5}\)
5x² – 5 -24x = 0 ⇒ 5x² – 24x – 5 = 0
5x² – 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 m². What is the width of the pathway?
Solution:
Area of ABCD = 16 × 12 ²
= 192 m²
Area of A’B’C’D’ (12 + 2w)(16 + 2w)
192 + 32 w + 24 w + 4 w² = 285

4w² + 56w – 93 = 0
4w² + 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
2y² + 5y+ 10y + 25 – 375 = 0
2y² + 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,
PX² + XQ² – 2PX . QX = 16 (∵ ∠Q × p = 90°)
PQ² – 2P × QX = 16
400 – 16 = 2PX × QX
384 = 2PX – QX
PX . QX = 192
∴ (PX + QX)² = PX² + QX² + 2PX . QX
= 400 + 2 × 192
= 784 = 28²
∴ 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 2x², 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 = 2x²

18x² – 9x – 16x² = 2 × 9
2x² – 9x – 18 = 0
(x – 6)(2x + 3) = 0
x = 6, x = \(\frac{-3}{2}\) (it is not possible)
No. of bees in total = 2x²
= 2 × 6² = 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.
∴ \(\frac{9}{4}=\frac{d^{2}}{(70-d)^{2}}\)
4d² = 9(70 – d)²
4d² = 9(70² – 140d + d²)
4d² = 9 × 70² – 9 × 140d + 9d²
∴ 5d² – 9 × 140d + 9 × 70² = 0
5d² = 1260d + 44100 = 0
d² – 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 = a²
Area of the gravel path = 100 – a²
Area of total garden =100
given cost of flower bed + gravelling = ₹ 364
3a² + 4 (100 – a²) = ₹ 364
3a² + 400 – 4a² = 364
∴ a² = 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
\(\frac { 15 }{ 100-x } \) = \(\frac { 20(100-x) }{ 3x } \)
45x² = 20(100 – x)²
(100 – x)² = \(\frac{45 x^{2}}{20}\) = \(\frac { 9 }{ 4 } \) x²
∴ 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
AB² + BC² = AC²
(AB + BC)² – 2AB . BC = AC2 [∵ a² + b² = (a + b)² – 2ab]
31² – 2AB . BC = 25²
-2AB . BC = 625 – 961


∴ The length of the smallest side is 7 cm.