# 10th Maths Chapter 3 Algebra Exercise 3.12

10th Standard Maths Chapter 3 Algebra Samacheer Kalvi Guide Exercise 3.12 Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

## 10th Maths Chapter 3 Algebra Exercise 3.12

### 1. If the difference between a number and its reciprocal is 24/5, find the number.

**Solution:
**Let a number be x.

Its reciprocal is 1x

x−1x=245

x2−1x=245

5×2 – 5 -24x = 0 ⇒ 5×2 – 24x – 5 = 0

5×2 – 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 2×2, 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 = 2×2

18×2 – 9x – 16×2 = 2 × 9

2×2 – 9x – 18 = 0

(x – 6)(2x + 3) = 0

x = 6, x = −32 (it is not possible)

No. of bees in total = 2×2

= 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. 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.

### 10. 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

45×2 = 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