Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.5

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.5

Miscellaneous and Practice Problems

Question 1.
A square carpet covers an area of 1024 m² of a big hall. It is placed in the middle of the hall. What is the length of a side of the carpet?
Solution:
Area of the carpet = 1024 m²
side × side = 1024 m²
(side)² = 1024 m²

(side)² = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2²
= (2 × 2 × 2 × 2 × 2)²
(side)² = 32²
side = 32
Length of a side of the carpet = 32 m

Question 2.
There is a large square portrait of a leader that covers an area of 4489 cm². If each side has a 2 cm liner, what would be its area?
Solution:
Area of the square = 4489 cm²
(side)² = 4489 cm²

(side)² = 67 x 67
side² = 67²
Length of a side = 67

Length of a side Length of a side with liner = 67 + 2 + 2 cm = 71 cm
Area of the larger square = 71 x 71 cm²
= 5041 cm²
Area of the liner = Area of big square-Area of small square
= (5041 – 4489) cm² = 552 cm²

Question 3.
2401 plants are planted in a garden such that each contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Given number of plants in a row = Number of rows.
Number of rows × number of plants in a row = Total plants
Total plants = 2401

= 7 × 7 × 7 × 7 = 7² × 7²
= 49 × 49
∴ number of rows = 49
number of plants in a row = 49

Question 4.
If \(\sqrt[3]{1906624} \times \sqrt{x}\) = 3100, find x.
Solution:

\(\sqrt{x}\) = 25
Squaring on both sides \((\sqrt{x})^{2}\) = 25²
x = 625

Question 5.
If (625)^x = 15625, find x² and x³
Solution:
(625)^x = 15625

(5 x 5 x 5 x 5)^x = 5 x 5 x 5 x 5 x 5 x 5
(5⁴)^x = 5⁶
5^4x = 5⁶
5^4x = 5⁶
Comparing the powers of 5 both sides
4x = 6
x = \(\frac{6}{4}\)
x = \(\frac{3}{2}\)

Question 6.
If 2^m-1 + 2^m+1 = 640, then find ‘m’
Solution:
Given 2^m-1 + 2^m+1 = 640
2^m-1 + 2^m+1 = 128 + 512 [consecutive powers of 2]
2^m-1 + 2^m+1 = 2⁷+ 2⁹ [powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, 512, …..]
m – 1 = 7
m = 7 + 1
m = 8

Question 7.
Simplify \(\frac{16 \times 10^{2} \times 64}{4^{2} \times 2^{4}}\)
Solution:

\(\frac{16 \times 10^{2} \times 64}{4^{2} \times 2^{4}}\) = \(\frac{2^{4} \times 10^{2} \times 2^{6}}{\left(2^{2}\right)^{2} \times 2^{4}}=\frac{2^{4+6} \times 10^{2}}{2^{4} \times 2^{4}}=\frac{2^{10} \times 100}{2^{8}}\)
= 2^10-8 × 100= 2² × 100 = 400

Question 8.
Give the answer in scientific notation:
A human heart beats at an average of 80 beats per minute. How many times does it beat in
(i) an hour?
(ii) a day?
(iii) a year?
(iv) 100 years?
Solution:
Heart beat per minute = 80 beats
(i) an hour One hour = 60 minutes
Heart beat in an hour = 60 x 80 = 4800 = 4.8 x 10³

(ii) In a day
One day = 24 hours = 24 x 60 minutes
∴ Heart beat in one day = 24 x 60 x 80 = 24 x 4800 = 115200 = 1.152 x 10⁵

(iii) a year
One year = 365 days = 365 x 24 hours = 365 x 24 x 60 minutes
∴ Heart beats in a year = 365 x 24 x 60 x 80 = 42048000 = 4.2048 x 10⁷

(iv) 100 years
Heart beats in one year = 4.2048 x 10⁷
Heart beats in 100 years = 4.2048 x 10⁷ x 100 = 4.2048 x 10⁷ x 10²
= 4.2048 x 10⁹

Challenging Problems

Question 9.
A greeting card has an area 90 cm². Between what two whole numbers is the length of its side?
Solution:
Area of the greeting card = 90 cm²
(side)² = 90 cm²

(side)² = 2 x 5 x 3 x 3 = 2 x 5 x 3²
\(\sqrt{({side})^{2}}=\sqrt{2 \times 5 \times 3^{2}}\)

side = 3 \(\sqrt{2 × 5}\)
side = \(\sqrt{10}\) cm
side = 3 × 3.2 cm
side = 9.6 cm
∴ Side lies between the whole numbers 9 and 10.

Question 10.
225 square shaped mosaic tiles, each of area 1 square decimetre exactly cover a square shaped verandah. How long is each side of the square shaped verandah?
Solution:
Area of one tile = 1 sq. decimeter
Area of 225 tiles = 225 sq.decimeter
225 square tiles exactly covers the square shaped verandah.
∴ Area of 225 tiles = Area of the verandah
Area of the verandah = 225 sq.decimeter
side x side = 15 x 15 sq.decimeter
side = 15 decimeters
Length of each side of verandah = 15 decimeters.

Question 11.
A group of 1536 cadets wanted to have a parade forming a square design. Is it possible? If it is not possible how many more cadets would be required?
Solution:
Number of cadets to form square design

1536 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
The numbers 2 and 3 are unpaired
It is impossible to have the parade forming square design with 1536 cadets.

39 x 39 = 1521
Also 40 x 40 = 1600
∴ We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
∴ 64 more cadets would be required to form the square design.

Question 12.
Find the decimal fraction which when multiplied by itself gives 176.252176.
Solution:
We will find the square root of 176.252176

176.252176 = 13.276 x 13.276
∴ The required number is 13.276

Question 13.
Evaluate \(\sqrt{286225}\) and use it to compute \(\sqrt{2862.25}\) + \(\sqrt{28.6225}\)
Solution:

Question 14.
The speed of light in glass is about 2 x 10⁸ m/sec. Use the formula, time = \(\frac{distence}{speed}\) to find the time (in hours) for a pulse of light to travel 7200 km in glass.
Solution:

Required time = 10^-5 hours

Question 15.
Simplify : (3.769 x 10⁵) + (4.21 x 10⁵)
Solution:

(3.769 x 10⁵) + (4.21 x 10⁵) = 3,76,900 + 4,21,00
= 7,97,900 = 7.979 x 10⁵

Question 16.
Order the following from the least to the greatest: 1625, 8100, 3500, 4400, 2600
Solution:
16²⁵ = (2⁴)²⁵ = 2¹⁰⁰
8¹⁰⁰ = (2³)¹⁰⁰ = 2³⁰⁰
4⁴⁰⁰ = (2²)⁴⁰⁰ = 2⁸⁰⁰
2⁶⁰⁰ = 2⁶⁰⁰
Comparing the powers we have, 2¹⁰⁰ < 2³⁰⁰ < 2⁶⁰⁰< 2⁸⁰⁰
∴ The required order : 16²⁵, 8¹⁰⁰, 2⁶⁰⁰, 3⁵⁰⁰, 4⁴⁰⁰