Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Additional Questions

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Additional Questions

Question 1.
Find the least number by which 1100 must be multiplied so that the product becomes a perfect square. Also, in each case find the square root of the perfect square so obtained.
Solution:
We find 1100 = 2 × 2 × 5 × 5 × 11 =2² × 5² × 11
∴ The prime factor 11 has no pair.
∴ If we multiply 1100 by 11, then the product becomes a perfect square.
∴ New number = 1100 × 11 = 12100
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 add 1
12100 = 2² × 5² × 11²
\(\sqrt{12100}\) = 2 × 5 × 11 = 110
Square root of the new number = 110

Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Additional Questions

Question 2.
Find the square of 509 using (a + b)² = a² + 2ab + b²
Solution:
509² = (500 + 9)² = 500² + 2 x 500 x 9 + 9²
= 250000 + 9000 + 81
509² = 259081

Question 3.
Find the sum without adding
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Given sum is the sum of first 12 odd natural numbers.
Sum of first n odd natural numbers is n².
∴ 1 +3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144

Question 4.
Write a Pythagorean triplet whose one number is 110
Solution:
Here let 2m = 110
m = 55
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 add 3
m² – 1 = 55² – 1 = 3025 – 1 = 3024
m² + 1 = 55² + 1 = 3025 + 1 = 3026
∴ Pythagorean triplet is 110, 3024, 3026.

Question 5.
Find the square root of 10 \(\frac{2}{3}\) correct to three places of decimal.
Solution:
10 \(\frac{2}{3}\) = 10.6666…….
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 add 4
\(\sqrt{10 \frac{2}{3}}=3.2659 \Rightarrow \sqrt{10 \frac{2}{3}}\) = 3.266 correct to three places of decimal.

Question 6.
Find the square root of 0.053361
Solution:
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 add 5
\(\sqrt{0.053361}\) = 0.231

Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Additional Questions

Question 7.
Three numbers are in the ratio 2:3:4. The sum of their cubes is 33957. Find the numbers.
Solution:
Let the numbers be 2x, 3x and 4x
(2x)³ + (3x)³ + (4x)³ = 33957
8x³ + 27x³ + 64x³ = 33957
99x³ = 33957
x³ = \(\frac{33957}{99}\)
x³ = 343
x³ = 7 × 7 × 7
x³ = 7³
x = 7
∴ The numbers are 2x = 2 × 7 = 14
3x = 3 × 7 = 21
4x = 4 × 7 = 28

Question 8.
The volume of a cube is 9261000 m³. Find the side of the cube?
Solution:
Volume of the cube = side x side x side
side x side x side = 9261000
side = \(\sqrt[3]{9261×1000}\) = \(\sqrt[3]{9261}\) × \(\sqrt[3]{1000}\)
= \(\sqrt[3]{3^{3}×7^{3}}\) × \(\sqrt[3]{10×10×10}\) = 3 × 7 × 10 = 210
∴ Side of the cube = 210 m

Question 9.
If the diameters of the sun and Earth are 1.4 × 109 m and 1.275 × 107 m respectively. compare these two.
Solution:
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 add 6
So the diameter of the Sun is about 100 times the diameter of the Earth.

Question 10.
The size of a red blood cell is 0.000007 m and that of a plant, cell is 0.00001275 m. Compare these two.
Solution:
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 add 7
∴ RBC size if half the size of a plant cell.

Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Additional Questions