10th Maths Book Back Question and Answers – Chapter 3 Exercise 3.12:
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Samacheer Kalvi 10th Maths Book Back Answers – Ex 3.12 Algebra
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Chapter 3
Exercise 3.12 Algebra
1. If the difference between a number and its reciprocal is 245, find the number.
Solution:
Let a number be x.
Its reciprocal is 1x
x−1x=245
x2−1x=245
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 = −15, 5
∴ The number is −15 or 5.
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 −312 = 15.5
w = – 15.5 cannot possible 3
∴ w = 32 = 1.5 m
(w cannot be (-ve))
The width of the pathway = 1.5 m.
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 = 90x hrs.
When the speed is increased bus 15 km/hr.
= 90x+15
It is given that the time to cover 90 km is reduced by 12 hrs.
The speed of the bus cannot be -ve value.
∴ The original speed of the bus is 45 km/hr.
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.
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 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.
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 = −32 (it is not possible)
No. of bees in total = 2x2
= 2 × 62 = 72
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.
∴ 94=d2(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
= 4202 or 842
= 120 or 42
∴ The person stand at a distance 28m from the first and 42 m from second gallery.
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 =10−62=42 = 2 cm
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 23”. 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 = y100−x
By the given condition
(100 – x) yx = 15 (for first women)
y = 15100−x ……(1)
x × y(100−x) = 203 [For second women]
y = 20(100−x)3x ……..(2)
From (1) and (2) We get
15100−x = 20(100−x)3x
45x2 = 20(100 – x)2
(100 – x)2 = 45×220 = 94 x2
∴ 100 – x = 94×2−−−√
100 – x = 3×2
3x = 2(100 – x)
3x = 200 – 2x
3x + 2x = 200 ⇒ 5x = 200
x = 2005 ⇒ x = 40
Number of eggs with the first women = 40
Number of eggs with the second women = (100 – 40) = 60
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.
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