TYPES OF NUMBERS

I. Natural Numbers : Counting numbers I, 2. 3, 4, 5, ..... are called natural artillery.

2. Whole Numbers : All counting numbers together with zero form the set of whole number. Thus,

   (A) 0 is the only whole number which is not a natural number.

   (B) Every natural number is a whole number

3. Integers : All natural numbers. 0 and negatives of counting numbers ie., - 3, - 2, - 1. 0. 

I. 2, 3...... together form the set of integers

(A) Positive Integers t 11. 2. 3, 4, .....I is the set of all positive integers.

(B) Negative Integers : (- I, - 2, - 3......I is the set of all negative integers.

(C) Non-Positive and Non-Negative integers : 0 is neither positive nor negative. 

    So, 0, 1, 2, 3. ..... I represents the set of non-negative integers, while (0, - I, - 2, - 3,  ) 

    represents the set of non-positive integers.

4. Even Numbers:   A number divisible by 2 is called an over, number. e.g.. 2. 4. 6, 8. 10. etc

5. Odd Numbers :   A number not divisible by 2 is called an odd number. es, 1. 3. 5, 7, 9, 11, etc.

6. Prime Numbers : A number greater than 1 is called a prime number, if it has exactly 

   two factors namely 1 and the number itself.

Prime lumbers upto 100 are . 2, 3, 5, 7. II, 13. 17. 19. 23. 29. 31, 37. 41, 43, 47, 53, 59, 81, 87, 71. 73, 79, 83, 89, 97.

7. Composite Numbers : Numbers greater than 1 which are not prime, are known OS composite numbers.  eg., 4. 6. 8. 9, 10. 12.

Note : (1) 1 as neither prime nor composite.

      (ii) 2 is the only even number which is prime.

      (iii)There are 25 prime numbers between 1 and 100.

8. Co-primes prime numbers a and b are said to be co primes. if their H.C.F. is 1.

 e,g., (2, 3), (4. 5). (7, 9), (8. 11). etc. are co-primes.

sbi po preparation continues.....

LCM and HCF - Solved Examples

11.The total number of prime factors of the product (8)20×(15)24×(7)15$(8{)}^{20}\times (15{)}^{24}\times (7{)}^{15}$ is

a. 59

b. 98

c. 123

d. 4

Correct Option: D

Explanation:

The prime numbers are 2,3,5,17 in the expression.  The expression can be written as (23)20×(3×5)24×(17)15⇒260×324×524×1715$(2^3{)}^{{}^{20}}\times (3\times 5{)}^{24}\times (17{)}^{15}\Rightarrow 2^{60}\times 3^{24}\times 5^{24}\times {17}^{15}$

So number of prime factors are 4. i.e., 2, 3, 5, 17

12. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divisible by 3, then the first number is

a. 264

b. 132

c. Both a and b

d. 33

Correct Option: C

Explanation:

Let the numbers are ah, bh respectively.  Here h is HCF of two numbers.  (obviously a, b are coprimes i.e., HCF (a, b) = 1)

Given that HCF = h = 44 and LCM = abh = 264

Dividing LCM by HCF we get ab = 6.

ab can be written as 1 x 6, 2 x 3, 3 x 2, 6 x 1.

But given that the first number is divisible by 3. So only two options possible for A. 3 x 44, 6 x 44. So option C is correct

13. What least number must be subtracted from 1294 so that the remainder when divided 9, 11, 13 will leave in each case the same remainder 6 ?

a. 0

b. 1
c. 2
d. 3

Correct Option: B

Explanation:

LCM of 9,11,13 is 1287. Dividing 1294 with 1287, the remainder will be 7, to get remainder 6, 1 is to be deducted from 1294 so that 1293 when divided by 9,11,13 leaves 6 as remainder.

14. The least number which is divisible by 12, 15, 20 and is a perfect square, is

a. 400

b. 900
c. 1600

d. 3600

Correct Option: b

Explanation: 

LCM = 5 × 3 × 2² = 60
To make this number as a perfect square, we have to multiply this number by 5 × 3

The number is 60 × 15= 900

15. The least perfect square number which is divisible by 3,4,5,6 and 8 is

a. 900

b. 1200

c. 2500

d. 3600

Correct Option: D

Explanation: 

LCM = 2×2×2×3×5=23×3×5$2\times 2\times 2\times 3\times 5=2^3\times 3\times 5$

But the least perfect square is = 23×3×5×(2×3×5)=24×32×52=3600$2^3\times 3\times 5\times (2\times 3\times 5)=2^4\times 3^2\times 5^2=3600$  as the perfect squares have their powers even.

Mensuration II

The following results are very important to solve various mensuration problems.

1.  The largest possible sphere that can be chiseled out from a cube of side "a" cm. 

 Diagonal of the sphere is a, so radius = a/2.
Remaining empty space in the cube = a3−πa36$a^3-{\displaystyle \frac{\pi a^3}{6}}$
${\displaystyle \frac{\mathrm{<span\; style="color:\; blue;\; font-size:\; 13.65px;\; line-height:\; 27.3px;\; white-space:\; normal;">2.\; The\; largest\; possible\; cube\; that\; can\; be\; chiseled\; out\; from\; a\; sphere\; of\; radius\; "a"\; cm</span>}}{}}$

Here OA = radius of the sphere. So diameter of the sphere = 2a.

Let the side of the square = x$x$, then the diagonal of the cube = 3√x$\sqrt{3}x$

⇒3√x=2a$\Rightarrow \sqrt{3}x=2a$ ⇒x=2a3√$\Rightarrow x={\displaystyle \frac{2a}{\sqrt{3}}}$

${\displaystyle \frac{\mathrm{}}{}}$

Therefore side of the square = 2a3√${\displaystyle \frac{2a}{\sqrt{3}}}$ 

3.  The largest possible cube that can be chiseled out from a hemisphere of radius 'a' cm. 

Sol:

Given, the radius of the hemi sphere AC = a$a$.  Let the side of the cube is x$x$.
From the above diagram, BE2+ED2=BD2$B{E}^2+E{D}^2=B{D}^2$
⇒x2+x2=BD2$\Rightarrow x^2+x^2=B{D}^2$
⇒BD=2√x$\Rightarrow BD=\sqrt{2}x$
⇒BC=2√x2=x2√$\Rightarrow BC={\displaystyle \frac{\sqrt{2}x}{2}}={\displaystyle \frac{x}{\sqrt{2}}}$
From Δ$\mathrm{\Delta }$ABC, AC2=AB2+BC2$A{C}^2=A{B}^2+B{C}^2$
⇒a2=x2+(x2√)2$\Rightarrow a^2=x^2+{\left({\displaystyle \frac{x}{\sqrt{2}}}\right)}^2$
⇒a2=3x22$\Rightarrow a^2={\displaystyle \frac{3x^2}{2}}$
⇒x=23−−√a$\Rightarrow x=\sqrt{{\displaystyle \frac{2}{3}}}a$
The edge of the cube = a23−−√