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I am a two-digit prime number and the sum of my digits is 10.1 am also one of the factors of 57. I am 19.

(D) The HCF of an even and an odd number is even.

The HCF of an even and an odd number is odd number.

(9 + 7) x 1000 = 16 x 1000 = 16,000

9 + 7 x 1000 = 9 + 7000 = 7,009 

16,000 > 7009 

∴ (9 + 7) x 1000 > [9 + 7 x 1000]

Smallest is 11

Biggest is 97

(A) 2

The smallest 5 – digit number = 10000

Prime factors of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

So, 10000 = 2⁴ × 5⁴

Therefore, distinct prime factors are = 2 and 5

Number of distinct prime factors of the smallest 5-digit number is = 2

80 ÷ [240 ÷ (-24)] + 7 

= 80 ÷ \([\frac{240}{-24}]\) + 7 

= 80 ÷ (-10) + 7 = - \([\frac{80}{10}]\) + 7 = (-8) + 7 = -1

2 is the smallest prime number.

A prime number is a positive integer having exactly two factors. If p is a prime, then it’s only factors are necessarily 1 and p itself. 

The first ten primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Given, to find 5 rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\)

We have, \(\frac{3}{5}\times\frac{6}{6} =\frac{18}{30}\) and \(\frac{4}{5}\times\frac{6}{6} =\frac{24}{30}\)

We know that,

18 < 19 < 20 < 21 < 22 < 23 < 24 < 

\(=\frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}\)

\(=\frac{3}{5}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{4}{5}\)

\(= \frac{3}{5}<\frac{19}{30}<\frac{2}{3}<\frac{7}{10}<\frac{11}{15}<\frac{23}{30}<\frac{4}{5}\)

Hence, 5 rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\) are:

\(\frac{19}{30},\frac{2}{3},\frac{7}{10},\frac{11}{15},\frac{23}{30}\) 

Note: You can multiply and divide with any number you want to find the rational numbers.

False.

If a number exactly divides the sum of three numbers, it must exactly divide the numbers separately.

50000 lakhs make 5 billions.

The total weight of a box containing 50 packets of biscuits each weighing 120 g

= 50 × 120 g = 6000 g

The capacity of a van = 900 kg = 900 × 1000 g = 900000 g

∴ The required number of boxes = 900000 + 6000 = 150

Therefore, 150 boxes can be loaded in the van.

Total tablets of Vitamin A = 180000

Number of tablets distributed among the students in a district = 18734

∴ The number of remaining vitamin tablets = 180000 – 18734 = 161266

Quantity of medicine in one capsule = 500 mg

∴ Quantity of medicine in 12 capsules or 1 strip = (500 × 12) mg = 6000 mg = 6 g

Quantity of medicine in 5 strips or 1 box = (6 × 5) g = 30 g

Quantity of medicine in 32 boxes = (30 × 32)g = 960 g

Weight of a tablet = 15 mg 

Weight of 3,00,000 tablets = 300000 × 15 

Weight of one box of tablets = 45,00,000 mg we know 1000 mg = 1 gram 

To convert mg into grams we have to divide grams by 1000 mg = 45,00,000 ÷ 1000 

Weight of one box of tablets in grams = 4500 grams. 

We know 1000g = 1kg

To convert ‘g’ into kilograms we have to divide kilograms by 1000g = \(\frac{4500}{1000}\) = 4.5 kg

Length covered in 1 rotation = 49.7 cm 

Length covered in 10 rotations = 49.7 x 10 cm 

= 497 cm

Cost of 1 chart = Rs 1.50 

Cost of 20 charts = Rs 1.50 x 20 

= Rs 30.00

Cost of 20 charts = Rs 30

Since it is in the form of p/q , and where q ≠ 0.

yes zero is a rational number. it can be written as 0/1,0/2 etc, in the form p/q where p and q are integers and q ≠0 ? 

Zero is a rational number. 

This can be written in the form of \(\frac{p}{q}\) because \(\frac{o}{q}\) is a rational number. 

E.g. \(\frac{0}{2}\)= 0, \(\frac{0}{5}\) = 0. etc. 

Zero belongs to set of rational number.

The decimal representation will be terminating, if the denominators have factors 2 or 5 or both. Therefore, p/q is a terminating decimal, when prime factorization of q must have only powers of 2 or 5 or both.

A rational number \(\frac{p}q\) is a terminating decimal only, when prime factors of q are 2 and 5 only. Therefore, \(\frac{p}q\) is a terminating decimal only, when prime factorisation of q must have only powers of 2 or 5 or both.

(a) We have, 4096

Since, the last two digits 96 is divisible by 4.

∴ 4096 must be divisible by 4.

(b) We have, 21084

Since, the last two digits 84 is divisible by 4.

∴ 21084 must be divisible by 4.

(c) We have, 31795012

Since, the last two digits 12 is divisible by 4.

∴ 31795012 must be divisible by 4.

25

Prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

So, there are 25 primes between 1 to 100.

If a number has 0 in ones place, then it is divisible by 10.

A number is divisible by 5, if it has 0 or 5 in its ones place.

True.

Consider the number 20, it is divisible by 4 but not divisible by 8.

A number is divisible by 2 if it has any of the digits 0, 2, 4, 6, or 8 in its ones place.

False.

As per the rule of divisibility test, the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.

The LCM of two or more given numbers is the lowest of their common multiple.

If the sum of the digits in a number is a multiple of 3, then the number is divisible by 3.

If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by 11, then the number is divisible by 11.

Three numbers whos decimal expansions are non-terminating, nonrecurring are 

(i) 0.123123312333 

(ii) 0.20200200020000 

(iii) 0.56566566656666

Answer is (D) √9

Answer is (A) 4/11

Answer is (B) a rational number

(i) √23 = 4.7958………. 

This is not terminating or non-recurring decimal. 

∴ √23 is an irrational number. 

(ii) √23 = 15 this is a rational number. 

(iii) 0.3796 

This is rational number, because trminating decimal has expansion. 

(iv) 0.478478… 

This is rational number, because decimal expansion is recurring.

 (v) 0.101001000100001… 

This is an irrational number because termianting or recruring decimal has no expansion.

Answer is (D) non-terminating non-repeating

(i) 2 – √5 is an irrational number.

(ii) (3 + √23) – √23 

= 3 + √23 – √23 = 3

Hence, it is a rational number.

\(\frac { 2\sqrt { 11 } }{ 7\sqrt { 11 } } =\frac { 2 }{ 7 } \)

Hence, it a rational number.

(iv) \(\frac { 1 }{ \sqrt { 3 } } \) is an irrational number.

(v) 2π is an irrational number.

(i) √23

Here, 23 is not a perfect square number, 

hence, √23 is an irrational number.

(ii) √225

Here 225 is a perfect square of 15, 

hence √225 is a rational number.

(iii) 0.3796 can be written in \(\frac { p }{ q }\) form,

where q ≠ 0 i.e \(\frac { 3796 }{ 10000 }\)

Hence, it is a rational number.

(iv) 7.478478 i.e. \(7.\bar { 478 } \)

Here, pair of digits (478) are repeating as it is non-terminating but recurring.

(v) 1.101001000100001…. is an irrational number is its decimal expansion is non-terminating and non-recurring.

We know that LCM × HCF = Product of two numbers

Also, we know that HCF of two co-primes = 1

LCM × 1 = 117

LCM = 117

(i) –1 and –2

Three rational numbers between –1 and –2 are –1.1, –1.2 and –1.3.

(ii) 0.1 and 0.11

Three rational numbers between 0.1 and 0.11 are 0.101, 0.102 and 0.103.

(iii) 5/7 and 6/7

5/7 can be written as (5 × 10)/(7 × 10) = 50/70

Similarly,

6/7 can be written as (6 × 10)/(7 × 10) = 60/70

Three rational numbers between 5/7 and 6/7 = three rational numbers between 16/80 and 20/80.

Three rational numbers between 5/7 and 6/7 are 51/70, 52/70, 53/70.

(iv) 1/4 and 1/5

Here, according to the question,

LCM of 4 and 5 is 20.

Let us make the denominators common, 80.

(4 × 20) = 80 and (5 × 16) = 80

Hence,

1/4 can be written as (1 × 20)/(4 × 20) = 20/80

Similarly,

1/5 can be written as (1 × 16)/(5 × 16) = 16/80

Three rational numbers between 1/4 and 1/5 = three rational numbers between 16/80 and 20/80.

Therefore, the three rational numbers are 17/80, 18/80 and 19/80.

Given (-8/7), (9/8), (-19/-13), (-21/13)

A rational number is said to be positive if its numerator and denominator are either positive integers or both negative integers.

Therefore the positive rational numbers are (9/8) and (-19/-13).

1. (d) Closed under addition. 

2. (a) Distributive property of multiplication over addition.

3. (e) Additive identity.

4. (c) Commutative property under multiplication.

5. (b) Multiplicative identity.

Given that HCF of 2 numbers is 5

The numbers may like 5x and 5y

Also given their product = 300

5x × 5y = 300

⇒ 25xy = 300

⇒ xy = 300/25

⇒ xy = 12

The possible values of x and y be (1, 12) (2, 6) (3, 4)

The numbers will be (5x, 5y)

⇒ (5 × 1, 5 × 12) = (5, 60)

⇒ (5 × 2, 5 × 6) = (10, 30)

⇒ (5 × 3, 5 × 4) = (15, 20)

(5, 60) is impossible because the given the numbers are two digit numbers.

The remaining numbers are (10, 30) and (15, 20)
But given that HCF is 5

(10, 30) is impossible, because its HCF = 10

The numbers are 15, 20

Let a = 0.5 = 0.50 

And, b = 0.55 

We observe that in the second decimal place a has digit 0 and b has digit 5. Therefore a < b. So, if we consider irrational numbers 

x = 0.51051005100051… 

y = 0.5305343055353530… 

We find that, 

a < x < y < b 

Hence, x and y are required irrational numbers.

(i) True. Because set of natural number belongs to set of whole numbers. 

∴ W = {0, 1, 2, 3 …………………….} 

(ii)  False. Because zero belongs to set of integers. But -2, -3, -1 are not whole numbers. 

(iii)  False. Because \(\frac{1}{2}\) is a rational number but not a whole number.

(i) 45293 or 45427

We observe that both the numbers are of 5-digits.

And at the digits at leftmost and second place from the left are same.

But the digits at the third place from the left are different, the first number has 2 and the second number has 4.

Since 2 < 4

45427 is greater

(ii) 380362 or 381007

We observe that both the numbers are of 6-digits.

And at the digits at leftmost and second place from the left are same.

But the digits at the third place from the left are different, the first number has 0 and the second number has 1.

Since 0 < 1

381007 is greater

(iii) 63520 or 63250

We observe that both the numbers are of 5-digits.

And at the digits at left most and second place from the left are same.

But the digits at the third place from the left are different, the first number has 5 and the second number has 2.

Since 5 < 2

63520 is greater