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In the question is given to verify the property x × (y × z) = (x × y) × z

The arrangement of the given rational number is as per the rule of associative property for multiplication.

Then, 1 × (-½ × ¼) = (1 × -½) × ¼

LHS = 1 × (-½ × ¼)

= 1 × (-1/8)

= -1/8

RHS = (1 × -½) × ¼

= (-½) × ¼

= -1/8

By comparing LHS and RHS

LHS = RHS

∴ -1/8 = -1/8

Hence x × (y × z) = (x × y) × z

In the question is given to verify the property x × (y × z) = (x × y) × z

The arrangement of the given rational number is as per the rule of associative property for multiplication.

Then, (-2/7) × (-5/6 × ¼) = ((-2/7) × (-5/6)) × ¼

LHS = (-2/7) × (-5/6 × ¼)

= (-2/7) × (-5/24)

= 10/168

RHS = ((-2/7) × (-5/6)) × ¼

= (10/42) × ¼

= 10/168

By comparing LHS and RHS

LHS = RHS

∴ 10/168 = 10/168

Hence x × (y × z) = (x × y) × z

0 ÷ (2/3) = 0 × (3/2)

= 0/2

= 0

(i) Given (-5/7) and (3/7)

= (-5/7) + (3/7)

Here denominators are same so add the numerator

= ((-5+3)/7)

= (-2/7)

(ii) Given (-15/4) and (7/4)

= (-15/4) + (7/4)

Here denominators are same so add the numerator

= ((-15 + 7)/4)

= (-8/4)

On simplifying

= -2

(iii) Given (-8/11) and (-4/11)

= (-8/11) + (-4/11)

Here denominators are same so add the numerator

= (-8 + (-4))/11

= (-12/11)

(iv) Given (6/13) and (-9/13)

= (6/13) + (-9/13)

Here denominators are same so add the numerator

= (6 + (-9))/13

= (-3/13)

The LCM of the denominators 4, 16 and 8 is 16

∴ ¼ = [(1×4)/ (4×4)] = (4/16)

(13/16) = [(13×1)/ (16×1)] = (13/16)

(5/8) = [(5×2)/ (8×2)] = (10/16)

Now, 13 < 10 < 4

⇒ (13/16) > (10/16) > (4/16)

Hence, (13/16) > (5/8) > (¼)

Descending order 13/16, 5/8

(i) \(-2 = \frac{-2}{1}\)

And,

\(\frac{8}{-3} = \frac{8\times-1}{-3\times-1} = \frac{-8}{3}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 1, 6and 3 = 6

\(\frac{-2}{1}= \frac{-2\times6}{1\times6} = \frac{-12}{6}\)

\(\frac{-13}{6}= \frac{-13\times1}{6\times1} = \frac{-13}{6}\)

\(\frac{-8}{3}= \frac{-8\times2}{3\times2} = \frac{-16}{6}\)

\(\frac{1}{3}= \frac{1\times2}{3\times2} = \frac{-2}{6}\)

Clearly,

2 > -12 > -13 > -16

Therefore,

\(\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}\)

Hence,

\(\frac{1}{3}>\frac{-2}{1}>\frac{-13}{6}>\frac{-8}{3}\)

(ii) \(\frac{7}{-15}= \frac{7\times-1}{-30\times-1} = \frac{-17}{30}\)

And,

\(\frac{17}{-30}= \frac{17\times-1}{-30\times-1} = \frac{-17}{30}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 10, 15, 20 and 30 = 60

\(\frac{-3}{10}= \frac{-3\times6}{10\times6} = \frac{-18}{60}\)

\(\frac{-7}{15}= \frac{-7\times4}{15\times4} = \frac{-28}{60}\)

\(\frac{-11}{20}= \frac{-11\times3}{20\times3} = \frac{-33}{60}\)

\(\frac{-17}{30}= \frac{-17\times2}{30\times2} = \frac{-34}{60}\)

Clearly, 

-18>-28>-33>-34

Therefore,

\(\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}\)

Hence,

\(\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}\)

(iii) \(\frac{23}{-24} = \frac{23\times-1}{-24\times-1} = \frac{-23}{24}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 6, 12, 18 and 24 = 72

 \(\frac{-5}{6}= \frac{-5\times12}{6\times12} = \frac{-60}{72}\)

\(\frac{-7}{12}= \frac{-7\times6}{12\times6} = \frac{-42}{72}\)

\(\frac{-13}{18}= \frac{-13\times4}{18\times4} = \frac{-52}{72}\)

\(\frac{-23}{24}= \frac{-23\times3}{24\times3} = \frac{-69}{72}\)

Clearly, -42>-52>-60>-69 

Therefore,

\(\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}\)

Hence,

\(\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}\)

(iv) Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 11, 22, 33 and 44 = 132

  \(\frac{-10}{11}= \frac{-10\times12}{11\times12} = \frac{-120}{132}\)

\(\frac{-19}{22}= \frac{-19\times6}{22\times6} = \frac{-144}{132}\)

\(\frac{-23}{33}= \frac{-23\times4}{33\times4} = \frac{-92}{132}\)

\(\frac{-39}{44}= \frac{-39\times3}{44\times3} = \frac{-117}{132}\)

Clearly, -92>-114>-117>-120 

Therefore,

\(\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}\)

Hence,

\(\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}\)

The LCM of the denominators 7, 8, 2 and 5 is 280

∴ 8/7 = [(8×40)/ (7×40)] = (320/280)

(-9/8) = [(-9×35)/ (8×35)] = (-315/280)

(-3/2) = [(-3×140)/ (2×140)] = (-420/280)

(2/5) = [(2×56)/ (56×56)] = (112/280)

Now, 320 > 112 > 0 > -315 > -420

Hence, ⇒ 8/7 > 2/5 > 0 > -9/8 > -3/2

Descending order 8/7, 2/5, 0, -9/8, -3/2

False

Let’s take 5/6 (first rational) and 15/18 (second rational)

As given both the rationals have different numerators.

Since, 15 and 18 have 3 in common hence, 

⇒ 15/18 = 5/6 

⇒ 5/6 (second rational) = 5/6 (first rational )

Hence, two rational with different numerators can be equal.

(i) False

Explanation:

It lies to the right of zero because it is a positive number.

(ii) False

Explanation:

It lies to the right of zero because it is a positive number.

(iii) True

Explanation:

Always positive number lie on the right of zero

(iv) True

Explanation:

Because they are of opposite sign

(v) False

Explanation:

Because they both are of same sign

(vi) True

Explanation:

They both are of opposite signs and positive number is greater than the negative number. Thus, it is on the right of the negative number.

(i) Given (3/5), (-17/-30), (8/-15), (-7/10)

The LCM of 5, 30, 15 and 10 is 30

Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 30

Consider (3/5)

Multiply both numerator and denominator by 6, then we get

(3/5) × (6/6) = (18/30) ….. (1)

Consider (8/-15)

Multiply both numerator and denominator by 2, then we get

(8/-15) × (2/2) = (16/-30) ….. (2)

Consider (-7/10)

Multiply both numerator and denominator by 3, then we get

(-7/10) × (3/3) = (-21/30) ….. (3)

In the above equation, denominators are same

Now on comparing the ascending order is:

(-7/10) < (8/-15) < (-17/30) < (3/5)

(ii) Given (-4/9), (5/-12), (7/-18), (2/-3)

The LCM of 9, 12, 18 and 3 is 36

Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 36

Consider (-4/9)

Multiply both numerator and denominator by 4, then we get

(-4/9) × (4/4) = (-16/36) ….. (1)

Consider (5/-12)

Multiply both numerator and denominator by 3, then we get

(5/-12) × (3/3) = (15/-36) ….. (2)

Consider (7/-18)

Multiply both numerator and denominator by 2, then we get

(7/-18) × (2/2) = (14/-36) ….. (3)

Consider (2/-3)

Multiply both numerator and denominator by 12, then we get

(2/-3) × (12/12) = (24/-36) ….. (4)

In the above equation, denominators are same

Now on comparing the ascending order is:

(2/-3) < ((-4/9) < (5/-12) < (7/-18)

(-5/-8), (37/53), 8 these are positive rational numbers. Because numerator and denominator are either both positive or both negative.

(i)

\(-2=\frac{-2}{1}\)

And,

\(\frac{8}{-3}=\frac{8\times-1}{-3\times-1}=\frac{-8}{3}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. LCM of 1, 6 and 3 = 6

\(\frac{-2}{1}=\frac{-2\times6}{1\times6}=\frac{-12}{6}\)

\(\frac{-13}{6}=\frac{-13\times1}{6\times1}=\frac{-13}{6}\)

\(\frac{-8}{3}=\frac{-8\times2}{3\times2}=\frac{-16}{6}\)

\(\frac{1}{3}=\frac{1\times2}{3\times2}=\frac{2}{6}\)

Clearly, 2 > -12 > -13 > -16

Therefore,

\(\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}\)

Hence,

\(\frac{1}{3}>\frac{-2}{1}>\frac{-13}{6}>\frac{-8}{6}\)

(ii)

\(\frac{7}{-15}=\frac{7\times-1}{-15\times-1}=\frac{-7}{15}\)

 And,

\(\frac{17}{-30}=\frac{17\times-1}{-30\times-1}=\frac{-17}{30}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. LCM of 10, 15, 20 and 30 = 60

 \(\frac{-3}{10}=\frac{-3\times6}{10\times6}=\frac{-18}{60}\)

 \(\frac{-7}{15}=\frac{-7\times4}{15\times4}=\frac{-28}{60}\)

\(\frac{-11}{20}=\frac{-11\times3}{20\times3}=\frac{-33}{60}\)

\(\frac{-17}{30}=\frac{-17\times2}{30\times2}=\frac{-34}{60}\)

Clearly, -18>-28>-33>-34

 Therefore,

 \(\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}\)

Hence,

\(\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}\)

(iii)

\(\frac{23}{-24}=\frac{23\times-1}{-24\times-1}=\frac{-23}{24}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. LCM of 6, 12, 18 and 24 = 72

 \(\frac{-5}{6}=\frac{-5\times12}{6\times12}=\frac{-60}{72}\)

\(\frac{-7}{12}=\frac{-7\times6}{12\times6}=\frac{-42}{72}\)

\(\frac{-13}{18}=\frac{-13\times4}{18\times4}=\frac{-52}{72}\)

\(\frac{-23}{24}=\frac{-23\times3}{24\times3}=\frac{-69}{72}\)

Clearly, -42>-52>-60>-69 

Therefore,

  \(\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}\)

Hence,

\(\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}\)

(iv)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. LCM of 11, 22, 33 and 44 = 132

 \(\frac{-10}{11}=\frac{-10\times12}{11\times12}=\frac{-120}{132}\)

\(\frac{-19}{22}=\frac{-19\times6}{22\times6}=\frac{-114}{132}\)

\(\frac{-23}{33}=\frac{-23\times4}{33\times4}=\frac{-92}{132}\)

\(\frac{-39}{44}=\frac{-39\times3}{44\times3}=\frac{-117}{132}\)

Clearly, -92>-114>-117>-120

 Therefore,

  \(\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}\)

Hence,

\(\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}\)

(i) 5

Rational number= (5/1)

5 = Numerator

1 = Denominator

(ii) -3

Rational number= (-3/1)

-3 = Numerator

1= Denominator

(iii) 1

Rational number= (1/1)

1 = Numerator

1= Denominator

(iv) 0

Rational number= (0/1)

0 = Numerator

1= Denominator

(v) -23

Rational number= (-23/1)

-23= Numerator

1= Denominator

(i) Given (7/8), (64/16), (39/-12), (5/-4), (140/28)

The LCM of 8, 16, 12, 4 and 28 is 336

Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 336

Consider (7/8)

Multiply both numerator and denominator by 42, then we get

(7/8) × (42/42) = (294/336) ….. (1)

Consider (64/16)

Multiply both numerator and denominator by 21, then we get

(64/16) × (21/21) = (1344/336) ….. (2)

Consider (39/-12)

Multiply both numerator and denominator by 28, then we get

(39/-12) × (28/28) = (-1008/336) ….. (3)

Consider (5/-4)

Multiply both numerator and denominator by 84, then we get

(5/-4) × (84/84) = (-420/336) ….. (4)

In the above equation, denominators are same

Now on comparing the descending order is:

(140/28) > (64/16) > (7/8) > (5/-4) > (36/-12)

(ii) Given (-3/10), (17/-30), (7/-15), (-11/20)

The LCM of 10, 30, 15 and 20 is 60

Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 60

Consider (-3/10)

Multiply both numerator and denominator by 6, then we get

(-3/10) × (6/6) = (-18/60) ….. (1)

Consider (17/-30)

Multiply both numerator and denominator by 2, then we get

(17/-30) × (2/2) = (34/-60) ….. (2)

Consider (7/-15)

Multiply both numerator and denominator by 4, then we get

(7/-15) × (4/4) = (28/-60) ….. (3)

In the above equation, denominators are same

Now on comparing the descending order is:

(-3/10) > (7/-15) > (-11/20) > (17/20)

False

8 can be written as a rational number with any integer as denominator, it is false because 8 can be written as a rational number with 1 as denominator i.e.8/1.

Correct option is  B) 138.71

True.

In any rational number p/q,denominator is always a nonzero integer.

(b) 1

A rational number is said to be in the standard form, if its denominator is a positive integer and the numerator and denominator have no common factor other than 1.

Given rational numbers that can be written as a rational number with denominator 4 are:

(7/8) = (3.5/4) (On dividing both denominator and denominator by 2)

(64/16) = (16/4) (On dividing both denominator and numerator by 4)

(36/-12) = (-12/4) (On dividing both denominator and numerator by -3)

(5/- 4) = (- 5/4) (On multiplying both denominator and numerator by -1)

(140/28) = (20/4) (On dividing both numerator and denominator by 7)

Length of cord \(=71\frac{1}{2}\,m\)

No of pieces \(=26\)

Length of each piece = Length of cord \(\div\) No of pieces

\(=71\frac{1}{2}\,m\div26\)

\(=\frac{143}{2}\,m\div26\)

\(=\frac{143}{2}\,m\times\frac{1}{26}\)

\(=\frac{143}{2\times26}\,m\)

\(=\frac{143}{52}\,m\)

\(=\frac{11}{4}\,m\)

\(=2\frac{3}{4}\,m\)

Length of each piece \(=2\frac{3}{4}\,m\)

i) 7/11 by 5/4
\(=\frac{7}{11}\times\frac{5}{4}\\=\frac{7\times5}{11\times4}\\=\frac{35}{44}\)

ii)  5/7 by -3/4
\(=\frac{5}{7}\times\frac{-3}{4}\\=\frac{(5\times-3)}{7\times4}\\=\frac{-15}{28}\)

iii)  -2/9 by 5/11
\(=\frac{-2}{9}\times\frac{5}{11}\\=\frac{-2\times5}{9\times11}\\=\frac{-10}{99}\)

iv)  -3/5 by -4
\(=\frac{-3}{5}\times\frac{-4}{7}\\=\frac{-3\times-4}{5\times7}\\=\frac{12}{35}\)

v) -15/11 by 7
\(=\frac{-15}{11}\times7\\=\frac{-15}{11}\times7\\=\frac{-15\times7}{11}\\=\frac{-105}{11}\)

(i) \(\frac{-5}{17}\times \frac{51}{-60}\)

= \(\frac{-1\times 3}{1\times -12}\)

= \(\frac{3}{12}\)

= \(\frac{1}{4}\)

 (ii) \(\frac{-6}{11}\times \frac{-55}{36}\)

= \(\frac{-1\times -5}{1\times 6}\)

= \(\frac{5}{6}\)

 (iii) \(\frac{-8}{25}\times \frac{-5}{16}\)

= \(\frac{-1\times -1}{5\times 2}\)

= \(\frac{1}{10}\)

 (iv) \(\frac{6}{7}\times \frac{-49}{36}\)

= \(\frac{1\times -7}{1\times 6}\)

= \(\frac{-7}{6}\)

 (v) \(\frac{8}{-9}\times \frac{-7}{-16}\)

= \(\frac{1\times -7}{-9\times -2}\)

= \(\frac{-7}{18}\)

 (vi) \(\frac{8}{-9}\times \frac{3}{64}\)

= \(\frac{1\times 1}{-3\times 8}\)

= \(\frac{-1}{24}\)

(c) positive integer

A rational number is said to be in the standard form, if its denominator is a positive integer.

(12/-17)

The denominator of rational number is negative then we multiply its numerator and denominator by -1 to get an equivalent rational number with positive denominator.

= [(12×-1)/ (-17/-1)]

= (-12/17)

\(\frac{-12}{26}\) is not in standard form since 12 and 26 have a common divisor 2.

\(\frac{28}{-105}\) is not in standard form since its denominator is negative.

Therefore,

only \(\frac{-49}{71}\) and \(\frac{-9}{16}\) are in standard forms as their numerator and denominator have no common divisor and their denominators are positive.

i)  3/5 × 35/24 + 3/5 × 10

 = 1/1 × 7/8 + 6/1

= 7/8 + 6 = (7×1 + 6×8)/8

= (7+48)/8

= 55/8

ii) 2/7 × 7/16 – 2/7 × 21/4

1/1 × 1/8 – 1/1 × 3/2

= 1/8 – 3/2

= 1/8 – 3/2 = (1×1 – 3×4)/8

= (1 – 12)/8

= -11/8

(i) \(\frac{7}{11}\times \frac{5}{4}\)

= \(\frac{7\times 5}{11\times 4}\)

= \(\frac{35}{44}\)

 (ii) \(\frac{5}{7}\times \frac{-3}{4}\)

= \(\frac{5\times -3}{7\times 4}\)

= \(\frac{-15}{28}\)

 (iii) \(\frac{-2}{9}\times \frac{5}{11}\)

= \(\frac{-2\times 5}{9\times 11}\)

= \(\frac{-10}{99}\)

 (iv) \(\frac{-3}{17}\times \frac{-5}{-4}\)

= \(\frac{-3\times -5}{17\times 4}\)

= \(\frac{15}{68}\)

 (v) \(\frac{9}{-7}\times \frac{36}{-11}\)

= \(\frac{9\times 36}{-7\times -11}\)

= \(\frac{324}{77}\)

 (vi) \(\frac{-11}{13}\times \frac{-21}{7}\)

= \(\frac{-11\times -21}{13\times 7}\)

= \(\frac{231}{91}\)

= \(\frac{33}{13}\)

 (vii) \(\frac{-3}{5}\times \frac{-4}{7}\)

= \(\frac{-3\times -4}{5\times 7}\)

= \(\frac{12}{35}\)

 (viii) \(\frac{15}{11}\times 7\)

= \(\frac{-15\times 7}{11}\)

= \(\frac{-105}{11}\)

(i) We have,

\(\frac{-3}{2}+\frac{5}{4}-\frac{7}{4}\) = \(\frac{-3\times 2+5\times 1-7\times 1}{4}\) (L.C.M of 2, 4 and 4 is 4)

= \(\frac{-6+5-7}{4}\)

= \(\frac{-13+5}{4}\)

= \(\frac{-8}{4}\)

= -2

(ii) We have,

 \(\frac{5}{3}-\frac{7}{6}+\frac{2}{3}\) = \(\frac{5\times 2-7\times 1-2\times 2}{6}\) (L.C.M of 3, 6 and 3 is 6)

 = \(\frac{10-7-4}{6}\)

= \(\frac{-1}{6}\)

(iii) We have,

 \(\frac{5}{4}-\frac{7}{6}+\frac{2}{3}\) = \(\frac{5\times 6-7\times 4+2\times 8}{24}\) (L.C.M of 4, 6 and 3 is 24)

 = \(\frac{30-28+16}{24}\)

= \(\frac{2+16}{24}\)

= \(\frac{18}{24}\)

 = \(\frac{3}{4}\)

(iv) We have,

 \(\frac{-2}{5}+\frac{3}{10}+\frac{4}{7}\) = \(\frac{-2\times 14+3\times 7+4\times 10}{70}\) (L.C.M of 5, 10 and 7 is 70)

  = \(\frac{-28+21+40}{70}\)

= \(\frac{-28+61}{70}\)

= \(\frac{33}{70}\)

(v) We have,

 \(\frac{5}{6}-\frac{2}{5}+\frac{2}{15}\) = \(\frac{5\times 5-2\times 6+2\times 2}{30}\) (L.C.M of 6, 5 and 15 is 30)

 = \(\frac{25-12+4}{30}\)

= \(\frac{13+4}{30}\)

 = \(\frac{17}{30}\)

 (vi) We have,

 \(\frac{3}{8}+\frac{2}{9}-\frac{5}{36}\) = \(\frac{3\times 9+2\times 8-5\times 2}{72}\) (L.C.M of 8, 9 and 36 is 72)

 = \(\frac{27+16-10}{72}\)

= \(\frac{43-10}{72}\)

= \(\frac{33}{72}\)

 = \(\frac{11}{24}\)

(-8/-19)

The denominator of rational number is negative then we multiply its numerator and denominator by -1 to get an equivalent rational number with positive denominator.

= [(-8×-1)/ (-19/-1)]

= (8/1)

Let the other number be x.

Then,

\(\frac{-10}{3}+\text{x}=\frac{-3}{1}\)

\(\Rightarrow\) \(\text{x}=\frac{-3}{1}-\frac{-10}{3}\)

\(\Rightarrow\) \(\text{x}=\frac{-3\times3-(-10)\times1}{3}\)

\(\Rightarrow\) \(\text{x}=\frac{-9+10}{3}\)

\(\Rightarrow\) \(\text{x}=\frac{1}{3}\)

(c) (-3/-4)

When the numerator and denominator both are positive integers or both are negative integers, it is a positive rational number.

i) 

\(\frac{3}{4}+\frac{5}{6}-\frac{7}{8}\\\frac{((3\times6)+(5\times4)-(7\times3))}{24}\\=\frac{(18+20-21)}{24}\\=\frac{(38-21)}{24}\\=\frac{17}{24}\)

ii) 

\(\frac{2}{3}+\frac{-5}{6}+\frac{-7}{9}\\=\frac{((2\times3)+(-5\times3)+(-7\times2))}{18}\\=\frac{(12-15-14)}{18}\\=\frac{-17}{18}\)

iii) 

\(\frac{-11}{2}+\frac{7}{6}+\frac{-5}{8}\\=\frac{((-11\times12)+(7\times4)+(-5\times3))}{24}\\=\frac{(-132+28-15)}{24}\\=\frac{-119}{24}\)

iv) 

\(\frac{5}{3}+\frac{3}{-2}+\frac{-7}{3}+3\\=\frac{((5\times2)+(-3\times2)+(-7\times2)+(3\times6))}{6}\\=\frac{(10-9-14+18)}{6}\\=\frac{5}{6}\)

We know that the distributivity of multiplication of rational numbers over subtraction, a × (b – c) = a × b – a × c

Where, a = -2/7, b = 7/16, c = 21/4

Then, (¾) × [(8/9) – (40)] = ((¾) × (8/9)) – ((¾) × (40))

= ((1/1) × (2/3)) – ((3/1) × (10))

= (2/3) – (30)

= (2 – 90)/3

= -88/3

(i) Largest number is 65

Given largest number is 65.

We know that 2m, m² – 1, m² + 1 form a

Pythagorean triplet.

let m² + 1 = 65

m² = 65 – 1

m² = 64

m² = 8 × 8

m = 8

∴ 2m = 2 × 8 = 16

m² – 1 = 64 – 1 = 63

∴ The required Pythagorean triplet is (16, 63, 65)

(ii) Smallest number is 10

We know that (2m, m² – 1, m² + 1) form a

Pythagorean triplet.

Given smallest number is 10

let 2m = 10

m = 102

m = 5

m² + 1 = 5² + 1 = 25 + 1 = 26

m² – 1 = 5² – 1 = 25 – 1 = 24

∴ The required Pythagorean triplet is (10, 24, 26)

For any natural number m, we have 2m, m² – 1, m² + 1 is a Pythagorean triplet.

(i) Consider 2m = 16 ⇒ m = 8

m² – 1 = 64 – 1 = 63
m² + 1 = 64 + 1 = 65

So Pythagorean triplet is 16, 63, 65.

(ii) Consider 2m = 18 ⇒ m = 9

m² – 1 = 81 – 1 = 80
m² + 1 = 81 + 1 = 82

So Pythagorean triplet is 18, 80, 82.

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

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 10

(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

(i) 441 = 220 + 221

(ii) 225 = 112 + 113

(iii) 289 = 144 + 145

(iv) 1089 = 544 + 545

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.

Let the weight of 1 ragi adai = x grams given 1/4 of x = 120 gm

1/4 × x = 120

x = 120 × 4

x = 480 gm

∴ 2/3 of the adai

= 2/3 × 480 gm = 2 × 160 gm = 320 gm

2/3 of the weight of adai = 320 gm

Total money = Rs 300

Fraction spent on notebooks \(=\frac{1}{3}\)

Amount spent on notebooks \(=\frac{1}{3}\times300=Rs\,100\)

Amount left = Rs 300 – Rs 100 = 200

Fraction spent on stationary \(=\frac{1}{4}\)

Amount spent on stationary \(=\frac{1}{2}\times200=Rs\,50\)

Money left = Rs 300 – Rs 150 \(=Rs \,150\)

Let the total weight of a box of apple = x kg.

Weight of 3/4 of a box apples = 3 kg 225 gm. = 3.225 kg

34 × x = 3225

x = 3.225 × 43 kg = 1.075 × 4 kg 

= 4.3 kg = 4 kg 300 gm

Weight of the box of apples = 4 kg 300 gm.

256 = 16²

576 = 24²

4096 = 64²

∴ 256, 576 and 4096 are perfect squares

We know that the numbers end with odd number of zeros, 7 and 8 not perfect squares.

∴ 1000, 34567 and 408 cannot be perfect squares.

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

True

Every natural number is a rational number.

Rational number is the quotient of two integers such that the denominator is a nonzero inter, i.e. Rational number =p/q , where p and q are integers and q ≠ 0.

1 is the multiplicative identity for rational numbers.

The given rational numbers are 2/3 and 3/4

\(\frac23\,\times\,\frac44=\frac8{12}\), \(\frac34\,\times\,\frac33=\frac9{12}\)

The rational numbers between 8/12 , 9/12 is

\(\cfrac{(\cfrac{8}{12}\,+\,\cfrac{9}{12})}{2} = \cfrac{\cfrac{17}{12}}{2}=\frac{17}{24}\)

(∵ the rational number between a, b is (a+b) / 2)

∴ the rational number between 2/3 and 3/4 is 17/24

Five rational numbers less than two are:

0, \(\frac{1}{5},\) \(\frac{2}{5},\) \(\frac{3}{5},\) \(\frac{4}{5},\)