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#### Question 1:

What do you mean by Euclid's division algorithm.

Euclid's division algorithm states that for any two positive integers a and b, there exist unique integers and r, such that a = bq + r, where 0 ≤ < b.

#### Question 2:

A number when divided by 61 gives 27 as quotient and 32 as remainder.
Find the number.

We know, Dividend = Divisor $×$ Quotient + Remainder
Given: Divisor = 61, Quotient = 27, Remainder = 32
Let the Dividend be x.
∴ x =
= 1679
Hence, the required number is 1679.

#### Question 3:

By what number should 1365 be divided to get 31 as quotient and 32 as remainder?

Given: Dividend = 1365, Quotient = 31, Remainder = 32
Let the divisor be x.
Dividend = Divisor $×$ Quotient + Remainder
1365 =  $×$ 31 + 32
⇒                1365 − 32 = 31x
⇒                      1333 = 31x

⇒                        x = $\frac{1333}{31}$ = 43
Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.

#### Question 4:

Using Euclid's algortihm, find the HCF of

(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575

(i)

On applying Euclid's algorithm, i.e. dividing 2520 by 405, we get:
Quotient = 6, Remainder = 90
∴ 2520 = 405 $×$ 6 + 90
Again on applying Euclid's algorithm, i.e. dividing 405 by 90, we get:
Quotient = 4, Remainder = 45
∴ 405 = 90 $×$ 4 + 45
Again on applying Euclid's algorithm, i.e. dividing 90 by 45, we get:
∴ 90 = 45 $×$ 2 + 0

Hence, the HCF of 2520 and 405 is 45.

(ii)

On applying Euclid's algorithm, i.e. dividing 1188 by 504, we get:
Quotient = 2, Remainder = 180
∴ 1188 = 504 $×$ 2 +180
Again on applying Euclid's algorithm, i.e. dividing 504 by 180, we get:
Quotient = 2, Remainder = 144
∴ 504 = 180$×$ 2 + 144
Again on applying Euclid's algorithm, i.e. dividing 180 by 144, we get:
Quotient = 1, Remainder = 36
∴ 180 = 144 $×$ 1 + 36
Again on applying Euclid's algorithm, i.e. dividing 144 by 36, we get:
∴ 144 = 36 $×$ 4 + 0

Hence, the HCF of 1188 and 504 is 36.

(iii)

On applying Euclid's algorithm, i.e. dividing 1575 by 960, we get:
Quotient = 1, Remainder = 615
∴ 1575 = 960 $×$ 1 + 615
Again on applying Euclid's algorithm, i.e. dividing 960 by 615, we get:
Quotient = 1, Remainder = 345
∴ 960 = 615 $×$ 1 + 345
Again on applying Euclid's algorithm, i.e. dividing 615 by 345, we get:
Quotient = 1, Remainder = 270
∴ 615 = 345 $×$ 1 + 270
Again on applying Euclid's algorithm, i.e. dividing 345 by 270, we get:
Quotient = 1, Remainder = 75
∴ 345 = 270 $×$ 1 + 75
Again on applying Euclid's algorithm, i.e. dividing 270 by 75, we get:
Quotient = 3, Remainder = 45
∴270 = 75 $×$ 3 + 45
Again on applying Euclid's algorithm, i.e. dividing 75 by 45, we get:
Quotient = 1,  Remainder = 30
∴ 75  = 45 $×$ 1 + 30
Again on applying Euclid's algorithm, i.e. dividing 45 by 30, we get:
Quotient = 1, Remainder = 15
∴ 45  = 30 $×$ 1 + 15
Again on applying Euclid's algorithm, i.e. dividing 30 by 15, we get:
Quotient = 2, Remainder = 0
∴ 30 = 15 $×$ 2 + 0

Hence, the HCF of 960 and 1575 is 15.

#### Question 5:

Show that every positive integer is either even or odd.

Let us assume that there exist a smallest positive integer that is neither odd nor even, say n. Since n is the least positive integer which is neither even nor odd, n − 1 must be either odd or even.

Case 1: If n − 1 is even, n − 1 = 2k for some k.
But this implies n = 2k + 1
this implies n is odd.

Case 2: If n − 1 is odd, n − 1 = 2k + 1 for some k.
But this implies n = 2k + 2 = 2(k + 1)
this implies n is even.

In both ways we have a contradiction.
Thus, every positive integer is either even or odd.

#### Question 6:

Show that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

Let n be any arbitrary positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder. So, by Euclid's division lemma, we have

n = 6m + r, where 0 ≤ r < 6.

As 0 ≤ r < 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5.
⇒ n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5

But n ≠ 6m or n ≠ 6m + 2 or n ≠ 6m + 4 (âˆµ 6m, 6m + 2, 6m + 4 are multiples of 2, so an even interger whereas n is an odd integer)

n = 6m + 1 or n = 6m + 3 or n = 6m + 5

Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

#### Question 7:

Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

Let n be any arbitrary positive odd integer.
On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid's division lemma, we have

n = 4m + r, where 0 ≤ r < 4.

As 0 ≤ r < 4 and r is an integer, r can take values 0, 1, 2, 3.
⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3

But n ≠ 4m or n ≠ 4m + 2 (âˆµ 4m, 4m + 2 are multiples of 2, so an even interger whereas n is an odd integer)

n = 4m + 1 or n = 4m + 3

Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

#### Question 8:

For any positive integer n, prove that n3n is divisible by 6.

Euclid's division lemma states that for given positive integers a and b, there exists unique integers q and r satisfying
Applying Euclid's division lemma om and 6, we have

Therefore, n can have six values, i.e.
$n=6q\phantom{\rule{0ex}{0ex}}n=6q+1\phantom{\rule{0ex}{0ex}}n=6q+2\phantom{\rule{0ex}{0ex}}n=6q+3\phantom{\rule{0ex}{0ex}}n=6q+4\phantom{\rule{0ex}{0ex}}n=6q+5$
Case I: When $n=6q$

Hence,  is divisible by 6

Case II:
When $n=6q+1$

Case III: When $n=6q+2$

Case IV: When $n=6q+3$
${n}^{3}=\left(6q+3{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q+3{\right)}^{3}-\left(6q+3\right)\phantom{\rule{0ex}{0ex}}=\left(6q+3\right)\left[\left(6q+3{\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+3\right)\left[36{q}^{2}+9+36q-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+3\right)\left[36{q}^{2}+36q+8\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+216{q}^{2}+48q+108{q}^{2}+108q+24\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+324{q}^{2}+156q+24\right]\phantom{\rule{0ex}{0ex}}=6\left[36{q}^{3}+54{q}^{2}+26q+4\right]\phantom{\rule{0ex}{0ex}}=6m$
Hence,  is divisible by 6.

Case V: When $n=6q+4$

Hence,  is divisible by 6.

Case VI: When $n=6q+5$

Hence,  is divisible by 6.

#### Question 9:

Prove that if x and y are both odd positive integers then x2 + y2 is even but not divisible by 4.

Let, be any positive odd integer and let .
So, ${x}^{2}+{y}^{2}=\left(n{\right)}^{2}+\left(n+2{\right)}^{2}$

$⇒{x}^{2}+{y}^{2}=2{n}^{2}+4+4n\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=2\left({n}^{2}+2+2n\right)$
$⇒{x}^{2}+{y}^{2}=2m$ (where $m={n}^{2}+2n+2$)
Because ${x}^{2}+{y}^{2}$ has 2 as a factor, so the value is an even number.
Also, because it does not have any multiple of 4 as a factor, therefore, it is not divisible by 4.

#### Question 10:

Use Euclid's algorithm to find HCF of 1190 and 1445. Express the HCF in the form 1190m + 1445n.

Using Euclid's division algorithm, we have

Since 1445 > 1190, we apply Euclid's division lemma to 1445 and 1190 to get;
$1445=1190×1+255$
Since the remainder is not zero, we again apply division lemma to 1190 and 255 and get;
$1190=255×4+170$
Again, the remainder is not zero, so we apply division lemma to 255 and 170 to get;
$255=170×1+85$
Now we finally apply division lemma to 170 and 85 to get;
$170=85×2+0$
Since, in this step, 85 completely divides 170 leaving zero remainder, we stop the procedure.
Hence, the HCF is 85.
Now, using the above division, we have
$170×1+85=255\phantom{\rule{0ex}{0ex}}⇒85=255-170×1\phantom{\rule{0ex}{0ex}}⇒85=\left(1445-1190×1\right)-\left(1190-255×4\right)\phantom{\rule{0ex}{0ex}}⇒85=\left(1445-1190\right)-\left[1190-\left(1445-1190\right)×4\right]\phantom{\rule{0ex}{0ex}}⇒85=\left(1445-1190\right)-\left[1190-1445×4+1190×4\right]\phantom{\rule{0ex}{0ex}}⇒85=1445-1190-\left[1190×5-1445×4\right]\phantom{\rule{0ex}{0ex}}⇒85=1445-1190-1190×5+1445×4\phantom{\rule{0ex}{0ex}}⇒85=1445×5-1190×6$
Or, $85=1190\left(-6\right)+1445\left(5\right)$
Hence,

#### Question 1:

Using prime factorization, find the HCF and LCM of

(i) 36, 84
(ii) 23, 31
(iii) 96, 404
(iv) 144, 198
(v) 396, 1080
(vi) 1152, 1664

(i) 36, 84
Prime factorisation:
36 = 22 â¨¯ 32
84 = 22 â¨¯ 3 â¨¯ 7
â€‹ HCF = product of smallest power of each common prime factor in the numbers = 22 â¨¯ 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers = 22 â¨¯ 32 â¨¯ 7 = 252

(ii) 23, 31
Prime factorisation:
23 = 23
31 = 31
â€‹ HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers = 23 â¨¯ 31 = 713

(iii) 96, 404
Prime factorisation:
96 = 25 â¨¯ 3
404 = 22 â¨¯ 101
â€‹ HCF = product of smallest power of each common prime factor in the numbers = 22 = 4
LCM = product of greatest power of each prime factor involved in the numbers = 25 â¨¯ 3 â¨¯ 101 = 9696

(iv) 144, 198
Prime factorisation:
144 =
198 =
â€‹ HCF = product of smallest power of each common prime factor in the numbers =  = 18
LCM = product of greatest power of each prime factor involved in the numbers =  = 1584

(v) 396, 1080
Prime factorisation:
396 =
1080 =
â€‹ HCF = product of smallest power of each common prime factor in the numbers =  = 36
LCM = product of greatest power of each prime factor involved in the numbers =  = 11880

(vi) 1152 , 1664
Prime factorisation:
1152 =
1664 =
HCF = product of smallest power of each common prime factor involved in the numbers = ${2}^{7}$ = 128
LCM = product of greatest power of each prime factor involved in the numbers =  = 14976

#### Question 2:

Using prime factorization, find the HCF and LCM of

(i) 8, 9, 25
(ii) 12, 15, 21
(iii) 17, 23, 29
(iv) 24, 36, 40
(v) 30, 72, 432
(vi) 21, 28, 36, 45

(i) 8 = 2 â¨¯ 2 â¨¯ 2 = 23
9 = 3 â¨¯ 3 = 32
25 = 5 â¨¯ 5 = 52
HCF = Product of smallest power of each common prime factor in the numbers = 1
LCM = Product of the greatest power of each prime factor involved in the numbers = 2â¨¯ 32 â¨¯ 52 = 1800

(ii) 12 = 2 â¨¯ 2 â¨¯ 3 = 22 â¨¯ 3
15 = 3 â¨¯ 5
21 = 3 â¨¯7
HCF = Product of smallest power of each common prime factor in the numbers = 3
LCM = Product of the greatest power of each prime factor involved in the numbers = 22 â¨¯ 3 â¨¯ 5 â¨¯ 7 = 420

(iii) 17 = 17
23 = 23
29 = 29
HCF = Product of smallest power of each common prime factor in the numbers = 1
LCM = Product of the greatest power of each prime factor involved in the numbers = 17 â¨¯ 23 â¨¯ 29 = 11339

(iv) 24 = 2 =
36 =  = ${3}^{2}$
40 =  2  = ${2}^{3}$
∴ â€‹HCF = Product of smallest power of each common prime factor in the numbers = ${2}^{2}$ = 4
∴â€‹ LCM = Product of the greatest power of each prime factor involved in the numbers =

(v)
=
=
∴ HCF = Product of smallest power of each common prime factor in the numbers =
∴ â€‹LCM = Product of the greatest power of each prime factor involved in the numbers =

(vi)
28  =  =
=
=
∴â€‹ HCF = Product of smallest power of each common prime factor in the numbers = 1
∴â€‹ LCM = Product of the greatest power of each prime factor involved in the numbers =

#### Question 3:

The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Let the two numbers be and b.
â€‹Let the value of a be 161.
Given: HCF = 23 and LCM = 1449
we know,        × b = HCF â€‹× LCM
⇒     161 × b = 23 × 1449
⇒              ∴ b =   23 × 1449   =   33327  = 207
161                  161
Hence, the other number b is 207.

#### Question 4:

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other.

HCF of two numbers = 145
LCM of two numbers = 2175
Let one of the two numbers be 725 and other be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that

725 × x = 145 × 2175
x = $\frac{145×2175}{725}$
= 435

Hence, the other number is 435.

#### Question 5:

The HCF of two numbers is 18 and their product is 12960. Find their LCM.

â€‹HCF of two numbers = 18
Product of two numbers = 12960
Let their LCM be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that

12960 = 18 × x
x$\frac{12960}{18}$
= 720

Hence, their LCM is 720.

#### Question 6:

Is it possible to have two numbers whose HCF is 18 and LCM is 760? Give reason.

â€‹No, it is not possible to have two numbers whose HCF is 18 and LCM is 760.

Since, HCF must be a factor of LCM, but 18 is not a factor of 760.

#### Question 7:

Find the simplest form of:
(i) $\frac{69}{92}$
(ii) $\frac{473}{645}$
(iii) $\frac{1095}{1168}$
(iv) $\frac{368}{496}$

(i) Prime factorisation of 69 and 92 is:

69 = 3 × 23
92 = 22 × 23

Therefore, $\frac{69}{92}=\frac{3×23}{{2}^{2}×23}=\frac{3}{{2}^{2}}=\frac{3}{4}$
Thus, simplest form of $\frac{69}{92}$ is $\frac{3}{4}$.

(ii) Prime factorisation of 473 and 645 is:

473 = 11 × 43
645 = 3 × 5 × 43

Therefore, $\frac{473}{645}=\frac{11×43}{3×5×43}=\frac{11}{15}$
Thus, simplest form of $\frac{473}{645}$ is $\frac{11}{15}$.

(iii) Prime factorisation of 1095 and 1168 is:

1095 = 3 × 5 × 73
1168 = 24 × 73

Therefore, $\frac{1095}{1168}=\frac{3×5×73}{{2}^{4}×73}=\frac{15}{16}$
Thus, simplest form of $\frac{1095}{1168}$ is $\frac{15}{16}$.

(iv) Prime factorisation of 368 and 496 is:

368 = 24 × 23
496 = 24 × 31

Therefore, $\frac{368}{496}=\frac{{2}^{4}×23}{{2}^{4}×31}=\frac{23}{31}$
Thus, simplest form of $\frac{368}{496}$ is $\frac{23}{31}$.

#### Question 8:

Find the largest number which divides 438 and 606, leaving remainder 6 in each case.

Largest number which divides 438 and 606, leaving remainder 6 is actually the largest number which divides 438 − 6 = 432 and 606 − 6 = 600, leaving remainder 0.

Therefore, HCF of 432 and 600 gives the largest number.

Now, prime factors of 432 and 600 are:
432 = 24 × 33
600 = 2× 3 × 52

HCF = product of smallest power of each common prime factor in the numbers = 2× 3 = 24

Thus, the largest number which divides 438 and 606, leaving remainder 6 is 24.

#### Question 9:

Find the largest number which divides 320 and 457 leaving remainders 5 and 7 respectively.

We know that the required number divides 315 (320 − 5) and 450 (457 − 7).
∴ Required number = HCF (315, 450)
On applying Euclid's lemma, we get:
315) 450 (1
−â€‹ 315
135) 315 (2
−   270
45) 135 (3
−â€‹ 135
0
Therefore, the HCF of 315 and 450 is 45.
Hence, the required number is 45.

#### Question 10:

Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case.

Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.

Prime factorization of 35, 56 and 91 is:

35 = 5 × 7
56 = 2× 7
91 = 7 × 13

LCM = product of greatest power of each prime factor involved in the numbers = 2× 5 × 7 × 13 = 3640

Least number which can be divided by 35, 56 and 91 is 3640.

Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.

Thus, the required number is 3647.

#### Question 11:

Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.

Let the required number be x.

Using Euclid's lemma,
x = 28p + 8 and x = 32q + 12, where p and q are the quotients
⇒ 28p + 8 = 32q + 12
⇒ 28p = 32q + 4
⇒ 7p = 8q + 1 ..... (1)

Here p = 8n − 1 and q = 7n − 1 satisfies (1), where n is a natural number
On putting n = 1, we get
p = 8 − 1 = 7 and q = 7 − 1 = 6

Thus, x = 28p + 8
= 28 × 7 + 8
= 204

Hence, the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 204.

#### Question 12:

Find the smallest number which when increased by 17 is exactly divisible by both 468 and 520.

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520.

Prime factorization of 468 and 520 is:
468 = 2× 32 × 13
520 = 23 × 5 × 13

LCM = product of greatest power of each prime factor involved in the numbers = 23 × 32 × 5 × 13 = 4680

The required number is 4680 − 17 = 4663.

Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

#### Question 13:

Find the greatest number of four digits which is exactly divisible by 15, 24 and 36.

Prime factorization:

15 = 3 × 5
24 = 23 × 3
36 = 22 × 32

LCM = product of greatest power of each prime factor involved in the numbers = 23 × 32 × 5 = 360

Now, the greatest four digit number is 9999.
On dividing 9999 by 360 we get 279 as remainder.
Thus, 9999 − 279 = 9720 is exactly divisible by 360.

Hence, the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

#### Question 14:

Find  the largest four-digits number which when divided by 4, 7 and 13 leaves a remainder of 3 in each case.

Largest 4 digit number is 9999
To find the largest 4 digit number divisible by 4, 7 and 13, we find the LCM of 4, 7 and 13 first.
LCM(4, 7, 13) = $4×7×13$ = 364
Now, to we divide 9999 by 364 and subtract the remainder from 9999 to get the number completely divisible by 4, 7 and 13.

$9999-171=9828$
Because the number leaves the remainder 3, so we add 3 to 9828.
Therefore, 9828 + 3 = 9831 is the required number.

#### Question 15:

Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3.

We find the LCM of 5, 6, 4 and 3 first.â€‹

So, $\mathrm{LCM}\left(5,6,4,3\right)=2×2×3×5=60$
Now, divide 2497 by 60, we get

To make the number completely divisible by 60, we must add a number that would make the remainder equal to 60.
Therefore, the number that must be added is 60 $-$ 37 = 23
Hence, 23 must be added to 2497.
So, the number exactly divisible by 5, 6, 4 and 3 is 2497 + 23 = 2520

#### Question 16:

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

We need to find the greatest number that would divide 43, 91 and 183 leaving the same remainder every time.
We first find the difference of the numbers and then find the HCF of the got numbers.
$183-91=92\phantom{\rule{0ex}{0ex}}183-43=140\phantom{\rule{0ex}{0ex}}91-43=48$
Now find HCF of 92, 140 and 48, we get
$92=2×2×23\phantom{\rule{0ex}{0ex}}140=2×2×5×7\phantom{\rule{0ex}{0ex}}48=2×2×2×2×3$
HCF(92, 140, 48) = 4
Therefore, 4 is the required number.

#### Question 17:

Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively.

First find the LCM of 20, 25, 35 and 40.

Now, we can see that
$20-14=6\phantom{\rule{0ex}{0ex}}25-19=6\phantom{\rule{0ex}{0ex}}35-29=6\phantom{\rule{0ex}{0ex}}40-34=6$
â€‹So, the required number would be
$=1400-6\phantom{\rule{0ex}{0ex}}=1394$

#### Question 18:

In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

Minimum number of rooms required =

Prime factorization of 60, 84 and 108 is:

60 = 2× 3 × 5
84 = 2× 3 × 7
108 = 2× 33

HCF = product of smallest power of each common prime factor in the numbers = 2× 3 = 12

Total number of paricipants = 60 + 84 + 108 = 252

Therefore, minimum number of rooms required = $\frac{252}{12}=21$

Thus, minimum number of rooms required is 21.

#### Question 19:

Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the higher of each stack is the same. How many stacks will be there?

Total number of English books = 336
Total number of mathematics books = 240
Total number of science books = 96
∴ Number of books stored in each stack = HCF (336, 240, 96)
Prime factorisation:
336 =
240 =
96 =
∴ HCF = Product of the smallest power of each common prime factor involved in the numbers =  = 48
Hence, we made stacks of 48 books each.

∴ Number of stacks =  = (7 + 5 + 2) = 14

#### Question 20:

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank? How many planks are formed?

The lengths of three pieces of timber are 42 m, 49 m and 63 m, respectively.
We have to divide the timber into equal length of planks.
∴ Greatest possible length of each plank = HCF(42, 49, 63)
Prime factorisation:
42 =
49 =
63 =
∴ HCF = Product of smallest power of each common prime factor in the numbers = 7
Therefore, the greatest possible length of each plank is 7 m.
Now, to find the total number of planks formed by each of the piece, we divide the length of each piece by the HCF, i.e. by 7.
We know that;
$7×6=42\phantom{\rule{0ex}{0ex}}7×7=49\phantom{\rule{0ex}{0ex}}7×9=63\phantom{\rule{0ex}{0ex}}$
Therefore, total number of planks formed$=6+7+9=22$
Hence, total 22 planks will be formed.

#### Question 21:

Find the greatest possible length which can be used to measure exactly the length 7 m, 3 m 85 cm and 12 m 95 cm.

The three given lengths are 7 m (700 cm), 3 m 85 cm  (385 cm) and 12 m 95 cm (1295 cm).   (âˆµ 1 m = 100 cm)
∴ Required length = HCF (700, 385, 1295)
Prime factorisation:
700 =
385 =
1295 =
∴ HCF =  = 35
Hence, the greatest possible length is 35 cm.

#### Question 22:

Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.

Total number of pens = 1001
Total number of pencils = 910
∴â€‹ Maximum number of students who get the same number of pens and pencils = HCF (1001, 910)
Prime factorisation:
1001 = $11×91$
910 = $10×91$
∴ HCF = 91
Hence, 91 students receive same number of pens and pencils.

#### Question 23:

Find the leash number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.

It is given that:
Length of a tile = 15 m 17 cm = 1517 cm                  [âˆµ 1 m = 100 cm]
Breadth of a tile = 9 m 2 cm = 902 cm
∴ Side of each square tile = HCF (1517, 902)
Prime factorisation:
1517 =
902 =
∴ HCF = Product of smallest power of each common prime factor in the numbers = 41
∴ Required number of tiles =  =  = 814

#### Question 24:

Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.

Length of the three measuring rods are 64 cm, 80 cm and 96 cm, respectively.
∴ Length of cloth that can be measured an exact number of times = LCM (64, 80, 96)
Prime factorisation:
64 = ${2}^{6}$
80 =
96 =
∴ LCM = Product of greatest power of  each prime factor involved in the numbers =  = 960 cm = 9.6 m
Hence, the required length of cloth is 9.6 m.

#### Question 25:

An electronic device makes a beep after every 60 seconds. Another device makes a beep after 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?

Beep duration of first device = 60 seconds
Beep duration of second device = 62 seconds
∴ Interval of beeping together = LCM (60, 62)
Prime factorisation:
60 =
62 =
∴ LCM =  seconds = $\frac{1860}{60}$ = 31 min
Hence, they will beep together again at 10 : 31 a.m.

#### Question 26:

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m. then at what time will they again change simultaneously?

We find the LCM of 48, 72 and 108 first to get the time after which they will blink together again.

Hence, LCM = $2×2×2×2×3×3×3=432$
So, they will blink again at 432 seconds past 8:00 am
or, $\frac{432}{60}=$7 minutes and 12 seconds past 8:00 am
So, the time will be 08:07:12 hrs

#### Question 27:

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12, minutes respectively. How many times do they toll together in 30 hours?

Six bells toll together at intervals of 2, 4, 6, 8, 10 and 12 minutes, respectively.
Prime factorisation:

∴ â€‹ =
Hence, after every 120 minutes (i.e. 2 hours), they will toll together.
∴ Required number of times =

#### Question 1:

Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form:

(i) $\frac{23}{\left({2}^{3}×{5}^{2}\right)}$

(ii) $\frac{24}{125}$

(iii)$\frac{171}{800}$

(iv) $\frac{15}{1600}$

(v) $\frac{17}{320}$

(vi) $\frac{19}{3125}$

(i)  =
We know either 2 or 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of $\left({2}^{m}×{5}^{n}\right)$.
Hence, the given rational is terminating.

(ii)
We know 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of .
Hence, the given rational is terminating.

(iii) $\frac{171}{800}$ =
We know either 2 or 5 is not a factor of 171, so it is in its simplest form.
Moreover, it is in the form of .
Hence, the given rational is terminating.

(iv)
We know either 2 or 5 is not a factor of 15, so it is in its simplest form.
Moreover, it is in the form of .
Hence, the given rational is terminating.

(v)   =  =
We know either 2 or 5 is not a factor of 17, so it is in its simplest form.
Moreover, it is in the form of .
Hence, the given rational is terminating.

(vi) $\frac{19}{3125}$ =
We know either 2 or 5 is not a factor of 19, so it is in its simplest form.
Moreover, it is in the form of .
Hence, the given rational is terminating.

#### Question 2:

Without actual division, show that each of the following rational numbers is a non-terminating repeating decimal:

(i) $\frac{11}{\left({2}^{3}×3\right)}$
(ii) $\frac{73}{\left({2}^{2}×{3}^{3}×5\right)}$
(iii)$\frac{129}{\left({2}^{2}×{5}^{7}×{7}^{5}\right)}$
(iv)$\frac{9}{35}$
(v)$\frac{77}{210}$
(vi) $\frac{32}{147}$
(vii) $\frac{29}{343}$
(viii) $\frac{64}{455}$

(i)
We know either 2 or 3 is not a factor of 11, so it is in its simplest form.
Moreover, ≠
Hence, the given rational is non-terminating repeating decimal.

(ii)
We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.
Moreover,  ≠
Hence, the given rational is non-terminating repeating decimal.

(iii)
We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.
Moreover,  ≠
Hence, the given rational is non-terminating repeating decimal.

(iv)
We know either 5 or 7 is not a factor of 9, so it is in its simplest form.
Moreover, () ≠
Hence, the given rational is non-terminating repeating decimal.

(v)
We know 2, 3 or 5 is not a factor of 11, so $\frac{11}{30}$ is in its simplest form.
Moreover,  ≠
Hence, the given rational is non-terminating repeating decimal.

(vi)
We know either 3 or 7 is not a factor of 32, so it is in its simplest form.
Moreover,  ≠
Hence, the given rational is non-terminating repeating decimal.

(vii)
We know 7 is not a factor of 29, so it is in its simplest form.
Moreover, ${7}^{3}$ ≠
Hence, the given rational is non-terminating repeating decimal.

(viii)
We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.
Moreover,  ≠
Hence, the given rational is non-terminating repeating decimal.

#### Question 3:

Express each of the following as a fraction in simplest form:

(i) $0.\overline{)8}$
(ii) $2.\overline{)4}$
(iii) $0.\overline{)24}$
(iv) $0.1\overline{)2}$
(v) $2.2\overline{)4}$
(vi) $0.\overline{)365}$

(i) Let x =  $0.\overline{)8}$
∴ x = 0.888                                             ...(1)
10x = 8.888                                            ...(2)
On subtracting equation (1) from (2), we get
9x = 8  ⇒ x = $\frac{8}{9}$

$0.\overline{)8}$ = $\frac{8}{9}$

(ii) Let x = $2.\overline{)4}$
∴ x = 2.444                                         ...(1)
10x24.444                                      ...(2)
On subtracting equation (1) from (2), we get
9x22 ⇒ x = $\frac{22}{9}$

∴ $2.\overline{)4}$$\frac{22}{9}$

(iii) Let x = $0.\overline{)24}$
∴â€‹ x = 0.2424                                   ...(1)
100x24.2424                              ...(2)
On subtracting equation (1) from (2), we get
99x24   ⇒ x$\frac{8}{33}$

∴ $0.\overline{)24}$  = $\frac{8}{33}$

(iv) Let $x=0.1\overline{2}$

On subtracting equation (1) from (2), we get
$100x-10x=\left(12.22222...\right)-\left(1.22222...\right)\phantom{\rule{0ex}{0ex}}⇒90x=11\phantom{\rule{0ex}{0ex}}⇒x=\frac{11}{90}$

(v) Let x = $2.2\overline{)4}$
∴  x = 2.2444                               ...(1)
10x = 22.444                              ...(2)
100x224.444                          ...(3)
On subtracting equation (2) from (3), we get
90x = 202  ⇒ x =$\frac{202}{90}$$\frac{101}{45}$

Hence, $2.2\overline{)4}$$\frac{101}{45}$

(vi) Let x = $0.\overline{)365}$
∴ x = 0.3656565                          ...(1)
10x3.656565                       ...(2)
1000x365.656565                 ...(3)
On subtracting (2) from (3), we get
990x362  ⇒ x =

Hence, $0.\overline{)365}$$\frac{181}{495}$

#### Question 1:

Define (i) rational numbers (ii) irrational numbers (iii) real numbers.

Rational numbers: The numbers of the form $\frac{p}{q}$ where  are integers and $q$ ≠ 0 are called rational numbers.
Example: $\frac{2}{3}$
Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.
Example: $\sqrt{2}$
Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational number or irrational number are called real numbers.
Example: 2, $\frac{1}{3}$$\sqrt{2}$, −3 etc.

#### Question 2:

Classify the following numbers as rational or irrational:

(i) $\frac{22}{7}$
(ii) 3.1416
(iii) π
(iv) $3.\overline{)142857}$
(v) 5.636363...
(vi) 2.040040004...
(vii) 1.535335333...
(viii) 3.121221222...
(ix) $\sqrt{21}$
(x) $\sqrt[3]{3}$

(i) $\frac{22}{7}$ is a rational number because it is of the form of ≠ 0.

(ii) 3.1416 is a rational number because it is a terminating decimal.

(iii) $\mathrm{\pi }$ is an irrational number because it is a non-repeating and non-terminating decimal.

(iv) 3.142857  is a rational number because it is a repeating decimal.

(v) 5.636363... is a rational number because it is a non-terminating, repeating decimal.

(vi) 2.040040004... is an irrational number because it is a non-terminating and non-repeating decimal.

(vii) 1.535335333... is an irrational number because it is a non-terminating and non-repeating decimal.

(viii) 3.121221222... is an irrational number because it is a non-terminating and non-repeating decimal.

(ix)  is an irrational number because  are irrational and prime numbers.

(x) $\sqrt[3]{3}$ is an irrational number because 3 is a prime number. So, $\sqrt{3}$ is an irrational number.

#### Question 3:

Prove that each of the following numbers is irrational:

(i) $\sqrt{6}$
(ii) $\left(2-\sqrt{3}\right)$
(iii) $\left(3+\sqrt{2}\right)$
(iv) $\left(2+\sqrt{5}\right)$
(v) $\left(5+3\sqrt{2}\right)$
(vi) $3\sqrt{7}$
(vii) $\frac{3}{\sqrt{5}}$
(viii) $\left(2-3\sqrt{5}\right)$
(ix) $\left(\sqrt{3}+\sqrt{5}\right)$

(i) Let  be rational.
Hence,   are both rational.
This contradicts the fact that  are irrational.
The contradiction arises by assuming $\sqrt{6}$ is rational.
Hence, $\sqrt{6}$ is irrational.

(ii) Let  be rational.
Hence, $2$ and  are rational.
∴  = rational   [âˆµ Difference of two rational is rational]
This contradicts  the fact that $\sqrt{3}$ is irrational.
The contradiction arises by assuming  is rational.
Hence,  is irrational.

(iii) Let  be rational.
Hence, $3$ and  are rational.
∴  = rational   [âˆµ Difference of two rational is rational]
This contradicts the fact that $\sqrt{2}$ is irrational.
The contradiction arises by assuming  is rational.
Hence,  is irrational.

(iv) Let  be rational.
Hence,  and $\sqrt{5}$ are rational.
∴  = rational         [âˆµ Difference of two rational is rational]
This contradicts the fact that $\sqrt{5}$ is irrational.
The contradiction arises by assuming  is rational.
Hence,  is irrational.

(v) Let,  be rational.
Hence, $5$ and  are rational.
∴  = rational       [âˆµ Difference of two rational is rational]
∴  = rational            [âˆµ Product of two rational is rational]
This contradicts the fact that $\sqrt{2}$ is irrational.
The contradiction arises by assuming  is rational.
Hence,  is irrational.

(vi) Let $3\sqrt{7}$ be rational.
∴  = rational           [âˆµ Product of two rational is rational]
This contradicts the fact that $\sqrt{7}$ is irrational.
The contradiction arises by assuming $3\sqrt{7}$ is rational.
Hence, $3\sqrt{7}$ is irrational.

(vii) Let $\frac{3}{\sqrt{5}}$ be rational.
∴  = rational         [âˆµ Product of two rational is rational]
This contradicts the fact that $\frac{1}{\sqrt{5}}$ is irrational.
∴
So, if $\frac{1}{\sqrt{5}}$ is rational, then $\frac{1}{5}\sqrt{5}$ is rational.
∴  = rational           [âˆµ Product of two rational is rational]
Hence, $\frac{1}{\sqrt{5}}$ is irrational.
The contradiction arises by assuming $\frac{3}{\sqrt{5}}$ is rational.
Hence, $\frac{3}{\sqrt{5}}$ is irrational.

(viii) Let  be rational.
Hence $2$ and  are rational.
∴  = rational   [âˆµ Difference of two rational is rational]
∴  = rational                                      [âˆµ Product of two rational is rational]
This contradicts the fact that $\sqrt{5}$ is irrational.
The contradiction arises by assuming  is rational.
Hence,  is irrational.

(ix) Let  be rational.
∴ , where $a$ is rational
∴                                  ... (1)
On squaring both sides of equation (1), we get

⇒
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
Hence,  is irrational.

#### Question 4:

Prove that $\frac{1}{\sqrt{3}}$ is irrational.

Let $\frac{1}{\sqrt{3}}$ be rational.
∴ $\frac{1}{\sqrt{3}}$ = $\frac{a}{b}$, where  are positive integers having no common factor other than 1
∴                            ...(1)
Since  are non-zero integers, $\frac{b}{a}$ is rational.
Thus, equation (1)  shows that $\sqrt{3}$ is rational.
This contradicts the fact that $\sqrt{3}$ is rational.
The contradiction arises by assuming $\sqrt{3}$ is rational.
Hence, $\frac{1}{\sqrt{3}}$ is irrational.

#### Question 5:

(i) Give an example of two irrationals whose sum is rational.
(ii) Give an examples of two irrationals whose product is rational.

(i) Let  be two irrationals.
∴  = rational number

(ii) Let  be two irrationals.
∴   = 18 = rational number

#### Question 6:

State whether the given statement is true of false:

(i) The sum of two rationals is always rational.
(ii) The product of two rationals is always rational.
(iii) The sum of two irrationals is an irrational.
(iv) The product of two irrationals is an irrational.
(v) The sum of a rational and and irrational is irrational.
(vi) The product of a rational and an irrational is irrational.

(i) True
(ii) True
(iii) False
Counter example:  are two irrational numbers. But their sum is 4, which is a rational number.
(iv) False
Counter example:
(v) True
(vi) True

#### Question 7:

Prove that ($2\sqrt{3}-1$) is an irrational number.

Let x$2\sqrt{3}-1$ be a rational number.

$x=2\sqrt{3}-1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\left(2\sqrt{3}-1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\left(2\sqrt{3}\right)}^{2}+{1}^{2}-2\left(2\sqrt{3}\right)\left(1\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=12+1-4\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-13=-4\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\frac{13-{x}^{2}}{4}=\sqrt{3}$
Since x is a rational number, x2 is also a rational number.
⇒ 13 − xis a rational number
⇒ $\frac{13-{x}^{2}}{4}$ is a rational number
⇒ $\sqrt{3}$ is a rational number
But $\sqrt{3}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus,  ($2\sqrt{3}-1$) is an irrational number.

#### Question 8:

Prove that ($4-5\sqrt{2}$) is an irrational number.

â€‹Let x = $4-5\sqrt{2}$ be a rational number.

$x=4-5\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\left(4-5\sqrt{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\left(4\right)}^{2}+{\left(5\sqrt{2}\right)}^{2}-2\left(4\right)\left(5\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=16+50-40\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-66=-40\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\frac{66-{x}^{2}}{40}=\sqrt{2}$
Since x is a rational number, x2 is also a rational number.
⇒ 66 − xis a rational number
⇒ $\frac{66-{x}^{2}}{40}$ is a rational number
⇒ $\sqrt{2}$ is a rational number
But $\sqrt{2}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus,  ($4-5\sqrt{2}$) is an irrational number.

#### Question 9:

Prove that ($5-2\sqrt{3}$) is an irrational number.

â€‹â€‹Let x = $5-2\sqrt{3}$ be a rational number.

$x=5-2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\left(5-2\sqrt{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\left(5\right)}^{2}+{\left(2\sqrt{3}\right)}^{2}-2\left(5\right)\left(2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=25+12-20\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-37=-20\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\frac{37-{x}^{2}}{20}=\sqrt{3}$
Since x is a rational number, x2 is also a rational number.
⇒ 37 − xis a rational number
⇒ $\frac{37-{x}^{2}}{20}$ is a rational number
⇒ $\sqrt{3}$ is a rational number
But $\sqrt{3}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus,  ($5-2\sqrt{3}$) is an irrational number.

#### Question 10:

Prove that $5\sqrt{2}$ is irrational.

Let $5\sqrt{2}$ is a rational number.

∴ $5\sqrt{2}=\frac{p}{q}$, where p and q are some integers and HCF(pq) = 1    ....(1)
$⇒5\sqrt{2}q=p\phantom{\rule{0ex}{0ex}}⇒{\left(5\sqrt{2}q\right)}^{2}={p}^{2}\phantom{\rule{0ex}{0ex}}⇒2\left(25{q}^{2}\right)={p}^{2}$
p2 is divisible by 2
p is divisible by 2  .....(2)

Let p = 2m, where m is some integer.

∴ $5\sqrt{2}q=2m$
$⇒{\left(5\sqrt{2}q\right)}^{2}={\left(2m\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2\left(25{q}^{2}\right)=4{m}^{2}\phantom{\rule{0ex}{0ex}}⇒25{q}^{2}=2{m}^{2}$
⇒ q2 is divisible by 2
⇒ q is divisible by 2   .....(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $5\sqrt{2}$ is irrational.

#### Question 11:

Prove that $\frac{2}{\sqrt{7}}$ is irrational.

$\frac{2}{\sqrt{7}}=\frac{2}{\sqrt{7}}×\frac{\sqrt{7}}{\sqrt{7}}=\frac{2}{7}\sqrt{7}$
â€‹Let $\frac{2}{7}\sqrt{7}$ is a rational number.

∴ $\frac{2}{7}\sqrt{7}=\frac{p}{q}$, where p and are some integers and HCF(pq) = 1    ....(1)
$⇒2\sqrt{7}q=7p\phantom{\rule{0ex}{0ex}}⇒{\left(2\sqrt{7}q\right)}^{2}={\left(7p\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒7\left(4{q}^{2}\right)=49{p}^{2}\phantom{\rule{0ex}{0ex}}⇒4{q}^{2}=7{p}^{2}$
⇒ q2 is divisible by 7
⇒ q is divisible by 7  .....(2)

Let q = 7m, where m is some integer.

∴ $2\sqrt{7}q=7p$
$⇒{\left[2\sqrt{7}\left(7m\right)\right]}^{2}={\left(7p\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒343\left(4{m}^{2}\right)=49{p}^{2}\phantom{\rule{0ex}{0ex}}⇒7\left(4{m}^{2}\right)={p}^{2}$
⇒ p2 is divisible by 7
⇒ p is divisible by 7   .....(3)

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\frac{2}{\sqrt{7}}$ is irrational.

#### Question 1:

State Euclid's division lemma.

Euclid's division lemma, states that for any two positive integers a and b, there exist unique whole numbers q and r, such that
a = b × q + r where 0 ≤ r < b

#### Question 2:

State fundamental theorem of arithmatic.

The fundamental theorem of arithmetic, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique.

#### Question 3:

Express 360 as product of its prime factors.

Prime factorization:

360 = 23 × 32 × 5

#### Question 4:

If a and b are two prime numbers then find HCF(a, b).

Prime factorization:
a = a
b = b

HCF = product of smallest power of each common prime factor in the numbers = 1

Thus, HCF(ab) = 1

#### Question 5:

If a and b are two prime numbers then find LCM(ab).

Prime factorization:
a = a
b = b

LCM = product of greatest power of each prime factor involved in the numbers = a × b

Thus, LCM(ab) = ab.

#### Question 6:

If the product of two numbers is 1050 and their HCF is 25, find their LCM.

HCF of two numbers = 25
Product of two numbers = 1050
Let their LCM be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that,

1050 = 25 × x
x$\frac{1050}{25}$
= 42

Hence, their LCM is 42.

#### Question 7:

What is a composite number?

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

#### Question 8:

If a and b are relatively prime then what is their HCF?

If two numbers are relatively prime then their greatest common factor will be 1.

Thus, HCF(a, b) = 1.

#### Question 9:

If the rational number $\frac{a}{b}$ has a terminating decimal expansion, what is the condition to be satisfied by b?

Let x be a rational number whose decimal expansion terminates.
Then, we can express x in the form $\frac{a}{b}$, where a and b are coprime, and prime factorization of b is of the form (2m × 5n), where m and n are non negative integers.

#### Question 10:

Simplify: $\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}$.

Thus, simplified form of $\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}$ is 6.

#### Question 11:

Write the decimal expansion of $\frac{73}{\left({2}^{4}×{5}^{3}\right)}$.

Decimal expansion:

Thus, the decimal expansion of $\frac{73}{\left({2}^{4}×{5}^{3}\right)}$ is 0.0365.

#### Question 12:

Show that there is no value of n for which (2n × 5n) ends in 5.

We can write:
(2n × 5n) = (2 × 5)n
= 10n

For any value of n, we get 0 in the end.

Thus, there is no value of n for which (2n × 5n) ends in 5.

#### Question 13:

Is it possible to have two numbers whose HCF is 25 and LCM is 520?

â€‹No, it is not possible to have two numbers whose HCF is 25 and LCM is 520.

Since, HCF must be a factor of LCM, but 25 is not a factor of 520.

#### Question 14:

Give an example of two irrationals whose sum is rational.

Let the two irrationals be .

$\left(4-\sqrt{5}\right)+\left(4+\sqrt{5}\right)=8$

â€‹Thus, sum (i.e., 8) is a rational number.

#### Question 15:

Give an example of two irrationals whose product is rational.

â€‹â€‹Let the two irrationals be .

$\left(4\sqrt{5}\right)×\left(3\sqrt{5}\right)=60$

â€‹Thus, product (i.e., 60) is a rational number.

#### Question 16:

If a and b are relatively prime, what is their LCM?

If two numbers are relatively prime then their greatest common factor will be 1.

∴ HCF(a, b) = 1

Using the formula, Product of two numbers = HCF × LCM
we conclude that,

a × b = 1 × LCM
∴ LCM = ab

Thus, LCM(ab) is ab.

#### Question 17:

The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500.

Since, HCF must be a factor of LCM, but 500 is not a factor of 1200.

#### Question 18:

Express $0.\overline{4}$ as a rational number in simplest form.

Let x be $0.\overline{4}$.

$x=0.\overline{4}$  .....(1)
Multiplying both sides by 10, we get
$10x=4.\overline{4}$  .....(2)

Subtracting (1) from (2), we get
$10x-x=4.\overline{4}-0.\overline{4}\phantom{\rule{0ex}{0ex}}⇒9x=4\phantom{\rule{0ex}{0ex}}⇒x=\frac{4}{9}$

Thus, simplest form of $0.\overline{4}$ as a rational number is $\frac{4}{9}$.

#### Question 19:

Express $0.\overline{23}$ as a rational number in simplest form.

â€‹Let be $0.\overline{23}$.

$x=0.\overline{23}$  .....(1)
Multiplying both sides by 100, we get
$100x=23.\overline{23}$  .....(2)

Subtracting (1) from (2), we get
$100x-x=23.\overline{23}-0.\overline{23}\phantom{\rule{0ex}{0ex}}⇒99x=23\phantom{\rule{0ex}{0ex}}⇒x=\frac{23}{99}$

Thus, simplest form of $0.\overline{23}$ as a rational number is $\frac{23}{99}$.

#### Question 20:

Explain why 0.15015001500015 ... is an irrational number.

Irrational numbers are non-terminating non-recurring decimals.

Thus, 0.15015001500015 ... is an irrational number.

#### Question 21:

Show that $\frac{\sqrt{2}}{3}$ is irrational.

Let $\frac{\sqrt{2}}{3}$ is a rational number.

∴ $\frac{\sqrt{2}}{3}=\frac{p}{q}$, where p and q are some integers and HCF(p, q) = 1   ....(1)

$⇒\sqrt{2}q=3p\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{2}q\right)}^{2}={\left(3p\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{q}^{2}=9{p}^{2}$
p2 is divisible by 2
⇒ p is divisible by 2   ....(2)

Let p = 2m, where m is some integer.

∴ $\sqrt{2}q=3p$
$⇒\sqrt{2}q=3\left(2m\right)\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{2}q\right)}^{2}={\left(3\left(2m\right)\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{q}^{2}=4\left(9{p}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒{q}^{2}=2\left(9{p}^{2}\right)$
⇒ q2 is divisible by 2
⇒ q is divisible by 2   ....(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\frac{\sqrt{2}}{3}$ is irrational.

#### Question 22:

Write a rational number between

Since, $\sqrt{3}$ = 1.732....
So, we may take 1.8 as the required rational number between

Thus, the required rational number is 1.8

#### Question 23:

Explain why $3.\overline{1416}$ is a rational number.

Since, $3.\overline{1416}$ is a non-terminating repeating decimal.

Hence, is a rational number.

#### Question 1:

Which of the following is a pair of co-primes?
(a) (14, 35)
(b) (18, 25)
(c) (31, 93)
(d) (32, 62)

The numbers that do not share any common factor other than 1 are called co-primes.
Clearly in option (b),
factors of 18 are: 1, 2, 3, 6, 9 and 18
factors of 25 are: 1, 5, 25
The two numbers do not share any common factor other than 1.
They are co-primes to each other.

#### Question 2:

If a = (22 × 33 × 54) and b = (23 × 32 × 5), then HCF (a, b) = ?

(a) 90
(b) 180
(c) 360
(d) 540

(b) 180
It is given that:
a and b = ()
∴ HCF (ab) = Product of smallest power of each common prime factor in the numbers
=
= 180

#### Question 3:

HCF of (23 × 32 × 5), (22 × 33 ×52) and (24 ×3 × 53 × 7) is

(a) 30
(b) 48
(c) 60
(d) 105

(c) 60

HCF = ()
HCF = Product of smallest power of each common prime factor in the numbers
=
= 60

#### Question 4:

LCM of (23 × 3 × 5) and (24 × 5 × 7) is

(a) 40
(b) 560
(c) 1680
(d) 1120

(c) 1680

LCM (
∴â€‹ LCM = Product of greatest power of each prime factor involved in the numbers
=
=
= 1680

#### Question 5:

The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number?

(a) 36
(b) 45
(c) 9
(d) 81

(d) 81
Let the two numbers be x and y.
It is given that:
x = 54
â€‹HCF = 27
LCM = 162
We know,
$×$ = HCF $×$ LCM
54 $×$ y = 27 $×$ 162
54y4374
∴â€‹ y = $\frac{4374}{54}$ = 81

#### Question 6:

The product of two numbers is 1600 and their HCF is 5. The LCM of the numbers is

(a) 8000
(b) 1600
(c) 320
(d) 1605

(c) 320
Let the two numbers be and y.
It is given that:
$×$ y = 1600
HCF = 5
We know,
HCF $×$ LCM = $×$ y
⇒       5 $×$ LCM = 1600
⇒              ∴  LCM = $\frac{1600}{5}$ = 320

#### Question 7:

What is the largest number that divides each one of 1152 and 1664 exactly?

(a) 32
(b) 64
(c) 128
(d) 256

(c) 128
Largest number that divides each one of 1152 and 1664 = HCF (1152, 1664)
We know,
1152 =
1164 =
∴ HCF = ${2}^{7}$ = 128

#### Question 8:

What is the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively?

(a) 13
(b) 9
(c) 3
(d) 585

(a) 13

We know the required number divides 65 (70 − 5) and 117 (125 − 8).
∴ Required number = HCF (65, 117)
we know,
65 =
117 =
∴ HCF = 13

#### Question 9:

What is the largest number that divides 245 and 1029, leaving remainder 5 in each case?

(a) 15
(b) 16
(c) 9
(d) 5

(b) 16

We know that the required number divides 240 (245 − 5) and 1024 (1029 − 5).
∴ Required number = HCF (240, 1024)
240 =
1024 =
∴ HCF =  = 16

#### Question 10:

The simplest form of $\frac{1095}{1168}$ is

(a) $\frac{17}{26}$
(b) $\frac{25}{26}$
(c) $\frac{13}{16}$
(d) $\frac{15}{16}$

(d) $\frac{15}{16}$

Hence, HCF of 1095 and 1168 is 73.

#### Question 11:

Euclid's division lemma sates that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy

(a) 1 < r < b
(b) 0 < rb
(c) 0 ≤ r < b
(d) 0 < r < b

(c) 0 ≤ r < b

Euclid's division lemma states that for any positive integers and b, there exist unique integers and such that a = bq + r,
where râ€‹ must satisfy 0 ≤ r < b

#### Question 12:

A number when divided by 143 leaves 31 as remainder. What will be the remainder when the same number is divided by 13?

(a) 0
(b) 1
(c) 3
(d) 5

(d) 5

We know,
Dividend = Divisor $×$ Quotient + Remainder.
It is given that:
Divisor = 143
Remainder = 13
So, the given number is in the form of 143x + 31, where x is the quotient.
∴ 143x + 31 = 13 (11x) + (13 ) + 5 = 13 (11x + 2) + 5
Thus, the remainder will be 5 when the same number is divided by 13.

#### Question 13:

Which of the following is an irrational number?

(a) $\frac{22}{7}$
(b) 3.1416
(c) $3.\overline{)1416}$
(d) 3.141141114...

(d) 3.141141114...

3.141141114 is an irrational number because it is a non-repeating and non-terminating decimal.

#### Question 14:

π is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

(c) an irrational number

$\mathrm{\pi }$ is an irrational number because it is a non-repeating and non-terminating decimal.

#### Question 15:

$2.\overline{)35}$ is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

(b) a rational number

2.35 is a rational number because it is a repeating decimal.

#### Question 16:

2.13113111311113...is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

(c) an irrational number

It is an irrational number because it is a non-terminating and non-repeating decimal.

#### Question 17:

3.24636363...is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

(b) a rational number

It is a rational number because it is a repeating decimal.

#### Question 18:

Which of the following rational numbers is expressible as a terminating decimal?

(a) $\frac{124}{165}$
(b) $\frac{131}{30}$
(c) $\frac{2027}{625}$
(d) $\frac{1625}{462}$

(c) $\frac{2027}{625}$

; we know 5 and 33 are not the factors of 124. It is in its simplest form and it cannot be expressed as the product of  for some non-negative integers .

So, it cannot be expressed as a terminating decimal.

$\frac{131}{30}$ = ; we know 5 and 6 are not the factors of 131. Its is in its simplest form and it cannot be expressed as the product of ( ) for some non-negative integers .

So, it cannot be expressed as a terminating decimal.

; as it is of the form , where  are non-negative integers.
So, it is a terminating decimal.

; we know 2, 7 and 33 are not the factors of 1625. It is in its simplest form and cannot be expressed as the product of  for some non-negative integers $m,n$.
So, it cannot be expressed as a terminating decimal.

#### Question 19:

The decimal expansion of the rational number $\frac{37}{{2}^{2}×5}$ will terminate after

(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

(b) two decimal places

So, the decimal expansion of the rational number will terminate after two decimal places.

#### Question 20:

The decimal expansion of the number $\frac{14753}{1250}$ will terminate after

(a) one decimal place
(b) two decimal place
(c) three decimal place
(d) four decimal place

(d) four decimal places

= $11.8024$

So, the decimal expansion of the number will terminate after four decimal places.

#### Question 21:

The number 1.732 is
(a) an irrational number
(b) a rational number
(c) an integer
(d) a whole number

â€‹Clearly, 1.732 is a terminating decimal.

Hence, a rational number.

Hence, the correct answer is option (b).

#### Question 22:

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then, the least prime factor of (a + b) is

(a) 2
(b) 3
(c) 5
(d) 8

(a) 2

Since 5 + 3 = 8, the least prime factor of a + b has to be 2, unless a + b is a prime number greater than 2.
If a + b is a prime number greater than 2, then a + must be an odd number. So, either a or b must be an even number. If a is even, then the least prime factor of is 2, which is not 3 or 5. So, neither a nor b can be an even number. Hence, a + b cannot be a prime number greater than 2 if the least prime factor of a is 3 or 5.

#### Question 23:

$\sqrt{2}$ is
(a) a rational number
(b) an irrational number
(c) a terminating decimal
(d) a nonterminating repeating decimal

Let $\sqrt{2}$ is a rational number.

∴ $\sqrt{2}=\frac{p}{q}$,  where p and q are some integers and HCF(p, q) = 1    .... (1)
$⇒\sqrt{2}q=p\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{2}q\right)}^{2}={p}^{2}\phantom{\rule{0ex}{0ex}}⇒2{q}^{2}={p}^{2}$
p2 is divisible by 2
⇒ p is divisible by 2  .... (2)

Let p = 2m, where m is some integer.

∴ $\sqrt{2}q=p$
$⇒\sqrt{2}q=2m\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{2}q\right)}^{2}={\left(2m\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{q}^{2}=4{m}^{2}\phantom{\rule{0ex}{0ex}}⇒{q}^{2}=2{m}^{2}$
⇒ q2 is divisible by 2
is divisible by 2  .... (3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\sqrt{2}$ is an irrational number.

Hence, the correct answer is option (b).

#### Question 24:

$\frac{1}{\sqrt{2}}$is

(a) a fraction
(b) a rational number
(c) an irrational number
(d) none of these

(c) an irrational number

$\frac{1}{\sqrt{2}}$ is an irrational number.

#### Question 25:

$\left(2+\sqrt{2}\right)$ is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

(c) an irrational number

is an irrational number.
if it is rational, then the difference of two rational is rational
∴â€‹  = $\sqrt{2}$ = irrational

#### Question 26:

What is the least number that divisible by all the natural numbers from 1 to 10 (both inclusive)?

(a) 100
(b) 1260
(c) 2520
(d) 5040

(c) 2520

We have to find the least number that is divisible by all numbers from 1 to 10.
∴ LCM (1 to 10) =
Thus, 2520 is the least number that is divisible by every element and is equal to the least common multiple.

#### Question 1:

The decimal representation of $\frac{71}{150}$is
(a) a terminating decimal
(b) a non-terminating, repeating decimal
(b) a non-terminating and non-repeating decimal
(d) none of these

(b) a non-terminating, repeating decimal

We know that 2, 3 or 5 are not factors of 71.
So, it is in its simplest form.
And,  ≠
∴
Hence, it is a non-terminating, repeating decimal.

#### Question 2:

Which of the following has terminating decimal expansion?

(a) $\frac{32}{91}$
(b) $\frac{19}{80}$
(c) $\frac{23}{45}$
(d) $\frac{25}{42}$

(b) $\frac{19}{80}$

We know 2 and 5 are not factors of 19, so it is in its simplest form.
And
Hence, $\frac{19}{80}$ is a terminating decimal.

#### Question 3:

On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n − 1) is divided by 9?

(a) 1
(b) 2
(c) 3
(d) 4

(b) 2

Let be the quotient.
It is given that:
remainder = 7
On applying Euclid's algorithm, i.e. dividing by 9, we have
n = 9q + 7
⇒    3n27q21
⇒ 3n − 1 = 27q20
⇒ 3n − 1 = 9 $×$ 3q$×$ 2 + 2
⇒ 3n − 1 = 9 $×$ (3q + 2) + 2
So, when (3n − 1) is divided by 9, we get the remainder 2.

#### Question 4:

$0.\overline{)68}+0.\overline{)73}=?$
(a) $1.\overline{)41}$
(b) $1.\overline{)42}$
(c) $0.\overline{)141}$
(d) None of these

(b) $1.\overline{)42}$

#### Question 5:

Show that any number of the form 4n, nN can never end with the digit 0.

If 4n ends with 0, then it must have 5 as a factor.
But we know the only prime factor of 4n is 2.
Also we know from the fundamental theorem of arithmetic that prime factorisation of each number is unique.
Hence, 4n can never end with the digit 0.

#### Question 6:

The HCF  of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.

Let the two numbers be .
It is given that:
x = 81
HCF = 27 and  LCM = 162
We know,    Product of two numbers = HCF $×$ LCM
⇒                           = 27 $×$ 162
⇒                        81 $×$ $y$  = 4374
⇒                                   = 54
Hence, the other number y is 54.

#### Question 7:

Examine whether $\frac{17}{30}$ is a terminating decimal.

We know that 2, 3 and 5 are not the factors of 17.
So, $\frac{17}{30}$ is in its simplest form.
Also, 30 =  ≠
Hence, $\frac{17}{30}$ is a non-terminating decimal.

#### Question 8:

Find the simplest form of $\frac{148}{185}$.

(âˆµ HCF of 148 and 185 is 37)

Hence, the simplest form is $\frac{4}{5}$.

#### Question 9:

Which of the following numbers are irrational?

(a) $\sqrt{2}$
(b) $\sqrt[3]{6}$
(c) 3.142857
(d) $2.\overline{)3}$
(e) π
(f) $\frac{22}{7}$
(g) 0.232332333...
(h) $5.27\overline{)41}$

(a) $\sqrt{2}$ is irrational (âˆµ if $p$ is prime, then $\sqrt{p}$ is irrational).

(b)  is irrational.

(c) 3.142857 is rational because it is a terminating decimal.

(d) 2.3 is rational because it is a non-terminating, repeating decimal.

(e) $\mathrm{\pi }$ is irrational because it is a non-repeating, non-terminating decimal.

(f) $\frac{22}{7}$ is rational because it is in the form of  ≠ 0.

(g) 0.232332333...  is irrational because it is a non-terminating, non-repeating decimal.

(h) 5.2741 is rational because it is a non-terminating, repeating decimal.

#### Question 10:

Prove that $\left(2+\sqrt{3}\right)$ is irrational.

Let  be rational.
Then, both  and 2 are rational.
∴  is rational [âˆµ Difference of two rational is rational]
⇒ $\sqrt{3}$ is rational.
This contradicts the fact that $\sqrt{3}$ is irrational.
The contradiction arises by assuming  is rational.
Hence,  is irrational.

#### Question 11:

Find the HCF and LCM of 12, 15, 18, 27.

Prime factorisation:
12 =  =
15 =
18 =  =
27 =  = ${3}^{3}$
Now,
HCF = Product of smallest power of each common prime factor in the number
= 3
LCM = Product of greatest power of each prime factor involved in the number
=   = 540

#### Question 12:

Give an example of two irrationals whose sum is rational.

Let  be two irrational numbers.
Sum = , which is a rational number.

#### Question 13:

Give prime factorisation of 4620.

Prime factorisation:
4620 =

#### Question 14:

Find the HCF of 1008 and 1080 by prime factorization method.

Prime factorisation:
1008 =
1080 =
HCF = Product of  smallest power of each common prime factor in the number
=  = 72

#### Question 15:

Find the HCF and LCM of $\frac{8}{9},\frac{10}{27}$ and $\frac{16}{81}$.

Prime factorisation of the numbers given in the numerators are as follows:

= $80$

Prime factorisation of numbers given in the denominators are as follows:

∴

∴ $\frac{80}{9}$

#### Question 16:

Find the largest number which divides 546 and 764, leaving remainders 6 and 8 respectively.

We know the required number divides 540 (546 − 6) and 756 (764 − 8), respectively.
∴ Required largest number = HCF (540, 756)
Prime factorisation:
540 =
756 =
∴ HCF = ${2}^{2}×{3}^{3}=108$
Hence, the largest number is 108.

#### Question 17:

Prove that $\sqrt{3}$ is an irrational number.

Let $\sqrt{3}$ be rational and its simplest form be $\frac{a}{b}$.
Then,  are integers with no common factors other than 1 and $b$ ≠ 0.
Now  ⇒                      [on squaring both sides]
⇒            ... (1)

⇒ $3$ divides ${a}^{2}$                    [since 3 divides 3${b}^{2}$]
⇒ $3$ divides $a$                     [since 3 is prime, 3 divides ${a}^{2}$ ⇒ 3 divides $a$]
Let  for some integer $c$.
Putting  in equation (1), we get
⇒
⇒ 3 divides ${b}^{2}$               [since 3 divides 3${c}^{2}$]
⇒ 3 divides $b$                 [since 3 is prime, 3 divides ${b}^{2}$ ⇒ 3 divides $b$]
Thus, 3 is a common factor of both
But this contradicts the fact that  have no common factor other than 1.
The contradiction arises by assuming $\sqrt{3}$ is rational.
Hence, $\sqrt{3}$ is rational.

#### Question 18:

Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.

Let be the given positive odd integer.
On dividing a by 4,let q be the quotient and r the remainder.
Therefore,by Euclid's algorithm we have
a = 4q + r           0 ≤ < 4
⇒      a = 4q + r             râ€‹ = 0,1,2,3
⇒      a = 4qa = 4q1,  a = 4q2,  a = 4q + 3
But, 4q  and  4q + 2 = 2 (2q1) = even
Thus, when is odd, it is of the form (4q + 1) or (4q3) for some integer q.

#### Question 19:

Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.

Let be quotient and be the remainder.
On applying Euclid's algorithm, i.e. dividing by 3, we have
n = 3q r       0 ≤ < 3
⇒  n = 3q + r       r = 0, 1 or 2
⇒  n = 3q  or  n = (3q1) or n = (3q2)
Case 1â€‹: If n = 3q, then is divisible by 3.
Case 2: If n = (3q1), then (n + 2) = 3q3 = 3(3q1), which is clearly divisible by 3.
In this case, (n + 2) is divisible by 3.
Case 3 : If n = (3q2), then (n + 4) = 3q + 6 = 3(q + 2), which is clearly divisible by 3.
In this case, (n + 4) is divisible by 3.
Hence, one and only one out of n, (+ 1) and (n + 2) is divisible by 3.

#### Question 20:

Show that $\left(4+3\sqrt{2}\right)$ is irrational.

Then both (4 + $3\sqrt{2}$) and 4 are rational.
⇒ ( 4 + $3\sqrt{2}$ − 4) = $3\sqrt{2}$ = rational   [âˆµ Difference of two rational numbers is rational]
⇒ $3\sqrt{2}$ is rational.
⇒ $\sqrt{2}$ is rational.
This contradicts the fact that $\sqrt{2}$ is irrational (when 2 is prime, $\sqrt{2}$ is irrational).