-4 and -3
(ii) The decimal form of the rational number \(\frac{15}{-4}\) is _________ .
-3.75
(iii) The rational numbers \(\frac{-8}{3}\) and \(\frac{8}{3}\) are equidistant from _________.
0
(iv) The next rational number in the sequence \(\frac{-15}{24}, \frac{20}{-32}, \frac{-25}{40}\) is _________.
\(\frac{30}{-48}\)
(v) The standard form of \(\frac{58}{-78}\) is _________.
\(\frac{-29}{39}\)
-4 and -3
(ii) The decimal form of the rational number \(\frac{15}{-4}\) is _________ .
-3.75
(iii) The rational numbers \(\frac{-8}{3}\) and \(\frac{8}{3}\) are equidistant from _________.
0
(iv) The next rational number in the sequence \(\frac{-15}{24}, \frac{20}{-32}, \frac{-25}{40}\) is _________.
\(\frac{30}{-48}\)
(v) The standard form of \(\frac{58}{-78}\) is _________.
\(\frac{-29}{39}\)
False
(ii) \(\frac{-4}{5}\) lies to the left of \(\frac{-3}{4}\).
True
(iii) \(\frac{-19}{5}\) is greater than \(\frac{15}{-4}\).
False
(iv) The average of two rational numbers lies between them.
True
(v) There are an unlimited number of rational numbers between 10 and 11.
True
False
(ii) \(\frac{-4}{5}\) lies to the left of \(\frac{-3}{4}\).
True
(iii) \(\frac{-19}{5}\) is greater than \(\frac{15}{-4}\).
False
(iv) The average of two rational numbers lies between them.
True
(v) There are an unlimited number of rational numbers between 10 and 11.
True
v
(i) \(\frac{9}{4}\)
\(\frac{9}{4}=2 \frac{1}{4}\)
∴ \(\frac{9}{4}\) lies between 2 and 3
(ii) \(\frac{-8}{3}\)
\(\frac{-8}{3}=-2 \frac{2}{3}\)
\(-2 \frac{2}{3}\) lies between -2 and 3
(iii) \(\frac{-17}{-5}\)
\(\frac{-17}{-5}=3 \frac{2}{5}\)
\(3 \frac{2}{5}\) lies between 3 and 4 in the number line.
(iv) \(\frac{15}{-4}\)
\(\frac{15}{-4}=-3 \frac{3}{4}\)
\(-3 \frac{3}{4}\) lies between -3 and -4
(i) \(\frac{9}{4}\)
\(\frac{9}{4}=2 \frac{1}{4}\)
∴ \(\frac{9}{4}\) lies between 2 and 3
(ii) \(\frac{-8}{3}\)
\(\frac{-8}{3}=-2 \frac{2}{3}\)
\(-2 \frac{2}{3}\) lies between -2 and 3
(iii) \(\frac{-17}{-5}\)
\(\frac{-17}{-5}=3 \frac{2}{5}\)
\(3 \frac{2}{5}\) lies between 3 and 4 in the number line.
(iv) \(\frac{15}{-4}\)
\(\frac{15}{-4}=-3 \frac{3}{4}\)
\(-3 \frac{3}{4}\) lies between -3 and -4
(i) \(\frac{1}{11}\)
\(\frac{1}{11}\) = 0.0909….
(ii) \(\frac{13}{4}\)
\(\frac{13}{4}\) = 3.25
(iii) \(\frac{-18}{7}\)
\(\frac{-18}{7}\) = -2.571428571428….
(iv) \(1 \frac{2}{5}\)
\(1 \frac{2}{5}=\frac{7}{5}\) = 1.4
(v) \(-3 \frac{1}{2}\)
\(-3 \frac{1}{2}=-\frac{7}{2}=-3.5\)
(i) \(\frac{1}{11}\)
\(\frac{1}{11}\) = 0.0909….
(ii) \(\frac{13}{4}\)
\(\frac{13}{4}\) = 3.25
(iii) \(\frac{-18}{7}\)
\(\frac{-18}{7}\) = -2.571428571428….
(iv) \(1 \frac{2}{5}\)
\(1 \frac{2}{5}=\frac{7}{5}\) = 1.4
(v) \(-3 \frac{1}{2}\)
\(-3 \frac{1}{2}=-\frac{7}{2}=-3.5\)
(i) 2 and 0
i.e., \(\frac{-2}{1}\) and \(\frac{0}{1}\)
∴ Five rational number between \(\frac { -20 }{ 10 }\) (= -2) and \(\frac { 0 }{ 10 }\) (= 0) are
(ii) \(\frac{-1}{2}\) and \(\frac{3}{5}\)
LCM of 2 and 5 = 2 × 5 = 10
∴ Five rational number between
(iii) \(\frac{1}{4}\) and \(\frac{7}{20}\)
∴ Five rational number between
(iv) \(\frac{-6}{4}\) and \(\frac{-23}{10}\)
∴ Five rational number between
(i) 2 and 0
i.e., \(\frac{-2}{1}\) and \(\frac{0}{1}\)
∴ Five rational number between \(\frac { -20 }{ 10 }\) (= -2) and \(\frac { 0 }{ 10 }\) (= 0) are
(ii) \(\frac{-1}{2}\) and \(\frac{3}{5}\)
LCM of 2 and 5 = 2 × 5 = 10
∴ Five rational number between
(iii) \(\frac{1}{4}\) and \(\frac{7}{20}\)
∴ Five rational number between
(iv) \(\frac{-6}{4}\) and \(\frac{-23}{10}\)
∴ Five rational number between
The average of a and b is \(\frac { 1 }{ 2 }\)(a + b)
The average of a and b is \(\frac { 1 }{ 2 }\)(a + b)
(i) \(\frac{-11}{5}, \frac{-21}{8}\)
LCM of 5, 8 is 40
(ii) \(\frac{3}{-4}, \frac{-1}{2}\)
LCM of 4 and 2 = 4
(iii) \(\frac{2}{3}, \frac{4}{5}\)
LCM of 3 and 5 is 15.
(i) \(\frac{-11}{5}, \frac{-21}{8}\)
LCM of 5, 8 is 40
(ii) \(\frac{3}{-4}, \frac{-1}{2}\)
LCM of 4 and 2 = 4
(iii) \(\frac{2}{3}, \frac{4}{5}\)
LCM of 3 and 5 is 15.
(i) \(\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}\)
LCM of 12, 8, 24, 9, 36 is 4 × 3 × 2 × 3 = 72
Now comparing the numerators – 30, – 99, -45, 56, 24 we get 56 > 24 > – 30 > – 45 > – 99
(ii) \(\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}\)
LCM of 10, 5, 4, 20 is 5 × 2 × 2 = 20
Negative numbers are less than zero.
∴ Arranging the numerators we get
– 34 < – 28 < – 19 < – 10 < 0
Objective Type Questions:
(i) \(\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}\)
LCM of 12, 8, 24, 9, 36 is 4 × 3 × 2 × 3 = 72
Now comparing the numerators – 30, – 99, -45, 56, 24 we get 56 > 24 > – 30 > – 45 > – 99
(ii) \(\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}\)
LCM of 10, 5, 4, 20 is 5 × 2 × 2 = 20
Negative numbers are less than zero.
∴ Arranging the numerators we get
– 34 < – 28 < – 19 < – 10 < 0
Objective Type Questions:
- A. \(\frac{34}{99}\)
- B. \(\frac{-142}{99}\)
- C. \(\frac{142}{99}\)
- D. \(\frac{-34}{99}\)
(B) \(\frac{-142}{99}\)
Hint:
Let x be the number to be subtracted
\(\frac{-6}{11}-x\) = \(\frac{8}{9}\)
\(\frac{-6}{11}-\frac{8}{9}\) = x
(B) \(\frac{-142}{99}\)
Hint:
Let x be the number to be subtracted
\(\frac{-6}{11}-x\) = \(\frac{8}{9}\)
\(\frac{-6}{11}-\frac{8}{9}\) = x
- A. \(\frac{-20}{12}, \frac{5}{3}\)
- B. \(\frac{16}{-30}, \frac{-8}{15}\)
- C. \(\frac{-18}{36}, \frac{-20}{44}\)
- D. \(\frac{7}{-5}, \frac{-5}{7}\)
(B) \(\frac{16}{-30}, \frac{-8}{15}\)
Hint:
∴ \(\frac{16}{-30}\) and \(\frac{-8}{15}\)
(B) \(\frac{16}{-30}, \frac{-8}{15}\)
Hint:
∴ \(\frac{16}{-30}\) and \(\frac{-8}{15}\)
- A. 0 and \(\frac{-5}{4}\)
- B. -1 and 0
- C. -1 and -2
- D. -4 and -5
(C) -1 and -2
Hint:
\(\frac{-5}{4}\) = -1 \(\frac{1}{4}\)
∴ \(\frac{-5}{4}\) lies between -1 and -2.
(C) -1 and -2
Hint:
\(\frac{-5}{4}\) = -1 \(\frac{1}{4}\)
∴ \(\frac{-5}{4}\) lies between -1 and -2.
- A. \(\frac{-17}{24}\)
- B. \(\frac{-13}{16}\)
- C. \(\frac{7}{-8}\)
- D. \(\frac{-31}{32}\)
(A) \(\frac{-17}{24}\)
Hint:
LCM of 24, 16, 8, 32 = 8 × 2 × 3 × 2 = 96
∴ \(\frac{-17}{24}\) is the greatest number
(A) \(\frac{-17}{24}\)
Hint:
LCM of 24, 16, 8, 32 = 8 × 2 × 3 × 2 = 96
∴ \(\frac{-17}{24}\) is the greatest number
- A. 4
- B. 5
- C. 6
(C) 6
Hint:
Sum of digits in the denominator = 3 + 3 = 6
Posted in Class 8 on September 8, 2024 September 9, 2024
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(C) 6
Hint:
Sum of digits in the denominator = 3 + 3 = 6
Posted in Class 8 on September 8, 2024 September 9, 2024
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\(\frac{1}{20}\)
(ii) The value of \(\left(\frac{-3}{6}\right) \times\left(\frac{18}{-9}\right)\) is = ________ .
1
(iii) The value of \(\left(\frac{-15}{23}\right) \div\left(\frac{30}{-46}\right)\) is ________ .
1
(iv) The rational number ________ does not have a reciprocal.
0
(v) The multiplicative inverse of -1 is ________ .
-1
\(\frac{1}{20}\)
(ii) The value of \(\left(\frac{-3}{6}\right) \times\left(\frac{18}{-9}\right)\) is = ________ .
1
(iii) The value of \(\left(\frac{-15}{23}\right) \div\left(\frac{30}{-46}\right)\) is ________ .
1
(iv) The rational number ________ does not have a reciprocal.
0
(v) The multiplicative inverse of -1 is ________ .
-1
True
(ii) The rational numbers that are equal to their additive inverses are 0 and -1.
False
(iii) The additive inverse of \(\frac{-11}{-17}\) is \(\frac{11}{17}\)
False
(iv) The rational number which is its own reciprocal is -1.
True
(v) The multiplicative inverse exists for all rational numbers.
False
True
(ii) The rational numbers that are equal to their additive inverses are 0 and -1.
False
(iii) The additive inverse of \(\frac{-11}{-17}\) is \(\frac{11}{17}\)
False
(iv) The rational number which is its own reciprocal is -1.
True
(v) The multiplicative inverse exists for all rational numbers.
False
(i) \(\frac{7}{5}+\frac{3}{5}\)
(ii) \(\frac{7}{5}+\frac{5}{7}\)
(iii) \(\frac{6}{5}+\left(\frac{-14}{15}\right)\)
(iv) \(-4 \frac{2}{3}+7 \frac{5}{12}\)
(i) \(\frac{7}{5}+\frac{3}{5}\)
(ii) \(\frac{7}{5}+\frac{5}{7}\)
(iii) \(\frac{6}{5}+\left(\frac{-14}{15}\right)\)
(iv) \(-4 \frac{2}{3}+7 \frac{5}{12}\)
(i) \(\frac{-21}{5}\) by \(\frac{-7}{-10}\)
(ii) \(\frac{-3}{13}\) by -3
(iii) -2 by \(\frac{-6}{15}\)
(i) \(\frac{-21}{5}\) by \(\frac{-7}{-10}\)
(ii) \(\frac{-3}{13}\) by -3
(iii) -2 by \(\frac{-6}{15}\)
(i) a = \(\frac{1}{2}\), b = \(\frac{2}{3}\)
(ii) a = \(\frac{-3}{5}\), b = \(\frac{2}{15}\)
(i) a = \(\frac{1}{2}\), b = \(\frac{2}{3}\)
(ii) a = \(\frac{-3}{5}\), b = \(\frac{2}{15}\)
Let the number = a
Objective Type Questions
Let the number = a
Objective Type Questions
- A. 1
- B. \(\frac{-1}{2}\)
- C. \(\frac{1}{12}\)
- D. \(\frac{1}{22}\)
1
Hint:
1
Hint:
- A. \(\frac{5}{8}\)
- B. \(\frac{2}{3}\)
- C. \(\frac{15}{32}\)
- D. \(\frac{15}{16}\)
(D) \(\frac{15}{16}\)
Hint:
(D) \(\frac{15}{16}\)
Hint:
- A. 7
- B. \(\frac{-5}{7}\)
- C. 0
- D. all of these
(D) all of these
Hint:
Additive inverse of 7 is -7
Additive inverse of \(\frac{-5}{7}\) is \(\frac{5}{7}\)
Additive inverse of 0 is 0
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(D) all of these
Hint:
Additive inverse of 7 is -7
Additive inverse of \(\frac{-5}{7}\) is \(\frac{5}{7}\)
Additive inverse of 0 is 0
Posted in Class 8 on September 8, 2024 September 9, 2024
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closure property for addition
Let a = \(\frac{-5}{7}\) and b = \(\frac{8}{9}\)
∴ Closure property is true for addition of rational numbers.
Closure property for multiplication
∴ Closure property is true for rnultiplìcation of rational numbers.
closure property for addition
Let a = \(\frac{-5}{7}\) and b = \(\frac{8}{9}\)
∴ Closure property is true for addition of rational numbers.
Closure property for multiplication
∴ Closure property is true for rnultiplìcation of rational numbers.
Let a = \(\frac{-10}{11}\) and \(\frac{-8{33}\) be the given rational numbers.
From (1) and (2)
a + b = b + a and hence additionis commutative for rational numbers
From (3) and (4) a × b = b × a
Hence multiplication is commutative for rational numbers.
Let a = \(\frac{-10}{11}\) and \(\frac{-8{33}\) be the given rational numbers.
From (1) and (2)
a + b = b + a and hence additionis commutative for rational numbers
From (3) and (4) a × b = b × a
Hence multiplication is commutative for rational numbers.
From (1) and (2), (a + b) + c = a + (b + c) is true for rational numbers.
From (1) and (2) (a × b) × c = (a × b) × c is true for rational numbers.
Thus associative property.
From (1) and (2), (a + b) + c = a + (b + c) is true for rational numbers.
From (1) and (2) (a × b) × c = (a × b) × c is true for rational numbers.
Thus associative property.
From (1) and (2) we have a × (b + c) = (a × b) + (a × c) is true
Hence multiplication is distributive over addition for rational numbers Q.
From (1) and (2) we have a × (b + c) = (a × b) + (a × c) is true
Hence multiplication is distributive over addition for rational numbers Q.
Identify property for addition verified.
Identify property for multiplication verified.
Identify property for addition verified.
Identify property for multiplication verified.
Additive inverse for rational numbers verified.
Mulplicative inverse for rational numbers verified.
Objective Type Questions
Additive inverse for rational numbers verified.
Mulplicative inverse for rational numbers verified.
Objective Type Questions
- A. 1
- B. 1
- C. 0
- D. \(\frac { 1 }{ 2 }\)
(C) 0
(C) 0
- A. commutative
- B. closure
- C. distributive
- D. associative
(D) associative
(D) associative
- A. \(\frac{1}{8}-\frac{1}{8}=0\)
- B. \(\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\)
- C. \(\frac{1}{8}+0=\frac{1}{8}\)
- D. \(\frac{1}{8}-0=\frac{1}{8}\)
(A) \(\frac{1}{8}-\frac{1}{8}=0\)
(A) \(\frac{1}{8}-\frac{1}{8}=0\)
- A. addition
- B. subtraction
- C. multiplication
- D. division
(B) subtraction
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(B) subtraction
Posted in Class 8 on September 8, 2024 September 9, 2024
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9
(ii) The number of non-square numbers between 242 and 252 is ________ .
48
(iii) The number of perfect square numbers between 300 and 500 is ________ .
5
(iv) If a number has 5 or 6 digits in it, then its square root will have ________ digits.
3
(v) The value of Jii lies between integers ______ and ________ .
13, 14
9
(ii) The number of non-square numbers between 242 and 252 is ________ .
48
(iii) The number of perfect square numbers between 300 and 500 is ________ .
5
(iv) If a number has 5 or 6 digits in it, then its square root will have ________ digits.
3
(v) The value of Jii lies between integers ______ and ________ .
13, 14
True
(ii) A square number will not have odd number of zeros at the end.
True
(iii) The number of zeros in the square of 91000 is 9.
False
(iv) The square of 75 is 4925.
False
(v) The square root of 225 is 15.
True
True
(ii) A square number will not have odd number of zeros at the end.
True
(iii) The number of zeros in the square of 91000 is 9.
False
(iv) The square of 75 is 4925.
False
(v) The square root of 225 is 15.
True
(i) 17
(ii) 203
(iii) 1098
(i) 17
(ii) 203
(iii) 1098
(i) 725
725 = 5 × 5 × 29 = 5 2 × 29
Here the second prime factor 29 does not have a pair.
Hence 725 is not a perfect square number.
(ii) 190
190 = 2 × 5 × 19
Here the factors 2, 5 and 9 does not have pairs.
Hence 190 is not a perfect square number.
(iii) 841
841 = 29 × 29
Hence 841 is a perfect square
(vi) 1089
1089 = 3 × 3 × 11 × 11 = 33 × 33
Hence 1089 is a perfect square
The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.
(i) 725
725 = 5 × 5 × 29 = 5 2 × 29
Here the second prime factor 29 does not have a pair.
Hence 725 is not a perfect square number.
(ii) 190
190 = 2 × 5 × 19
Here the factors 2, 5 and 9 does not have pairs.
Hence 190 is not a perfect square number.
(iii) 841
841 = 29 × 29
Hence 841 is a perfect square
(vi) 1089
1089 = 3 × 3 × 11 × 11 = 33 × 33
Hence 1089 is a perfect square
The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.
(i) 144
144 = 2 × 2 × 2 × 2 × 3 × 3
√144 = 2 × 2 × 3 = 12
(ii) 256
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√256 = 2 × 2 × 2 × 2 = 16
(iii) 784
784 = 2 × 2 × 2 × 2 × 7 × 7
√784 = 2 × 2 × 2 × 2 × 7 × 7 = 28
(iv) 1156
1156 = 2 × 2 × 17 × 17
1156 = 2 2 × 17 2
1156 = (2 × 17) 2
∴ \(\sqrt{1156}\) = \(\sqrt{(2 \times 17)^{2}}\) = 2 × 17 = 34
∴ \(\sqrt{1156}\) = 34
(v) 4761
4761 = 3 × 3 × 23 × 23
4761 = 3 2 × 23 2
4761 = (3 × 23) 2
√4761 = \(\sqrt{(3 \times 23)^{2}}\)
√4761 = 3 × 23
√4761 = 69
(vi) 9025
9025 = 5 × 5 × 19 × 19
9025 = 5 2 × 19 2
9025 = (5 × 19) 2
√925 = \(\sqrt{(5 \times 19)^{2}}\) = 5 × 19 = 95
(i) 144
144 = 2 × 2 × 2 × 2 × 3 × 3
√144 = 2 × 2 × 3 = 12
(ii) 256
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√256 = 2 × 2 × 2 × 2 = 16
(iii) 784
784 = 2 × 2 × 2 × 2 × 7 × 7
√784 = 2 × 2 × 2 × 2 × 7 × 7 = 28
(iv) 1156
1156 = 2 × 2 × 17 × 17
1156 = 2 2 × 17 2
1156 = (2 × 17) 2
∴ \(\sqrt{1156}\) = \(\sqrt{(2 \times 17)^{2}}\) = 2 × 17 = 34
∴ \(\sqrt{1156}\) = 34
(v) 4761
4761 = 3 × 3 × 23 × 23
4761 = 3 2 × 23 2
4761 = (3 × 23) 2
√4761 = \(\sqrt{(3 \times 23)^{2}}\)
√4761 = 3 × 23
√4761 = 69
(vi) 9025
9025 = 5 × 5 × 19 × 19
9025 = 5 2 × 19 2
9025 = (5 × 19) 2
√925 = \(\sqrt{(5 \times 19)^{2}}\) = 5 × 19 = 95
(i) 1764
√1764 = 42
(ii) 6889
√6889 = 83
(iii) 11025
√11025 = 105
(iv) 17956
√17956 = 134
(v) 418609
√418609 = 647
Roots Calculator is a free online tool that displays the roots of the given quadratic equation.
(i) 1764
√1764 = 42
(ii) 6889
√6889 = 83
(iii) 11025
√11025 = 105
(iv) 17956
√17956 = 134
(v) 418609
√418609 = 647
Roots Calculator is a free online tool that displays the roots of the given quadratic equation.
(i) √440
we have 20 2 = 400
21 2 = 441
∴ √440 ≃ 21
(ii) √800
we have 28 2 = 784
29 2 = 841
∴ √800 ≃ 28
(iii) √1020
we have 31 2 = 961
32 2 = 1024
∴ √1020 ≃ 32
(i) √440
we have 20 2 = 400
21 2 = 441
∴ √440 ≃ 21
(ii) √800
we have 28 2 = 784
29 2 = 841
∴ √800 ≃ 28
(iii) √1020
we have 31 2 = 961
32 2 = 1024
∴ √1020 ≃ 32
(i) 2.89
√2.89 = 1.7
(ii) 67.24
√67.24 = 8.2
(iii) 2.0164
√2.0164 = 1.42
(iv) \(\frac{144}{225}\)
(v) \(7 \frac{18}{49}\)
\(\sqrt{7 \frac{18}{49}}=2 \frac{5}{7}\)
(i) 2.89
√2.89 = 1.7
(ii) 67.24
√67.24 = 8.2
(iii) 2.0164
√2.0164 = 1.42
(iv) \(\frac{144}{225}\)
(v) \(7 \frac{18}{49}\)
\(\sqrt{7 \frac{18}{49}}=2 \frac{5}{7}\)
We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 2 2 × 3 2 × 5 2 × 2
Here the last factor 2 has no pair. So if we multiply 1800 by 2, then the number becomes a perfect square.
∴ 1800 × 2 = 3600 is the required perfect square number.
∴ 3600 = 1800 × 2
3600 = 2 2 × 3 2 × 5 2 × 2 × 2
3600 = 2 2 × 3 2 × 5 2 × 2 2
= (2 × 3 × 5 × 2) 2
\(\sqrt{3600}=\sqrt{(2 \times 3 \times 5 \times 2)^{2}}\)
= 2 × 3 × 5 × 2 = 60
∴ √3600 = 60
Objective Type Questions
We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 2 2 × 3 2 × 5 2 × 2
Here the last factor 2 has no pair. So if we multiply 1800 by 2, then the number becomes a perfect square.
∴ 1800 × 2 = 3600 is the required perfect square number.
∴ 3600 = 1800 × 2
3600 = 2 2 × 3 2 × 5 2 × 2 × 2
3600 = 2 2 × 3 2 × 5 2 × 2 2
= (2 × 3 × 5 × 2) 2
\(\sqrt{3600}=\sqrt{(2 \times 3 \times 5 \times 2)^{2}}\)
= 2 × 3 × 5 × 2 = 60
∴ √3600 = 60
Objective Type Questions
- A. 9
- B. 6
- C. 4
- D. 3
(A) 9
Hint:
Ones digit = 3 × 3 = 9
(A) 9
Hint:
Ones digit = 3 × 3 = 9
- A. 4 2
- B. 5 2
- C. 6 2
- D. 7 2
(D) 7 2
Hint:
25 2 = 25 × 25 = 625
24 2 = 24 × 24 = 576
(D) 7 2
Hint:
25 2 = 25 × 25 = 625
24 2 = 24 × 24 = 576
- A. 5
- B. 6
- C. 7
- D. 8
(C) 7
Hint:
√49 = 7
(C) 7
Hint:
√49 = 7
- A. √2
- B. √8
- C. √48
- D. √32
(D) √32
(D) √32
- A. 4
- B. 5
- C. 6
- D. 7
(B) 5
Hint:
\(\frac{n+1}{2}=\frac{10}{2}=5\)
Posted in Class 8 on January 3, 2025 January 4, 2025
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(B) 5
Hint:
\(\frac{n+1}{2}=\frac{10}{2}=5\)
Posted in Class 8 on January 3, 2025 January 4, 2025
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7
(ii) The maximum number of digits in the cube of a two digit number is __________ .
6
(iii) The smallest number to be added to 3333 to make it a perfect cube is __________ .
42
(iv) The cube root of 540×50 is __________ .
30
(v) The cube root of 0.000004913 is __________ .
0.017
7
(ii) The maximum number of digits in the cube of a two digit number is __________ .
6
(iii) The smallest number to be added to 3333 to make it a perfect cube is __________ .
42
(iv) The cube root of 540×50 is __________ .
30
(v) The cube root of 0.000004913 is __________ .
0.017
True
(ii) Subtracting 103 from 1729 gives 93.
True
(iii) The cube of 0.0012 is 0.000001728.
False
(iv) 79570 is not a perfect cube.
True
(v) The cube root of 250047 is 63.
True
True
(ii) Subtracting 103 from 1729 gives 93.
True
(iii) The cube of 0.0012 is 0.000001728.
False
(iv) 79570 is not a perfect cube.
True
(v) The cube root of 250047 is 63.
True
1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2 3 × 3 3 × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.
1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2 3 × 3 3 × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.
We have 10985 = 5 × 13 ×13 × 13
= 5 × 13 ×13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.
We have 10985 = 5 × 13 ×13 × 13
= 5 × 13 ×13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.
Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.
1000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.
Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.
1000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.
(i)
= 3 × 3
\(\sqrt[3]{729}\) = 9
(ii) \(\sqrt[3]{6859}\) = \(\sqrt[3]{19 \times 19 \times 19}\)
\(\sqrt[3]{6859}\) = 19
(i)
= 3 × 3
\(\sqrt[3]{729}\) = 9
(ii) \(\sqrt[3]{6859}\) = \(\sqrt[3]{19 \times 19 \times 19}\)
\(\sqrt[3]{6859}\) = 19
We have to find out \(\sqrt{(\sqrt[3]{46656})}\)
First we will find \(\sqrt[3]{46656}\)
∴ The required number is 6.
We have to find out \(\sqrt{(\sqrt[3]{46656})}\)
First we will find \(\sqrt[3]{46656}\)
∴ The required number is 6.
(729) 1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]
We have to find out √3,
√3 = 1.732
(729) 1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]
We have to find out √3,
√3 = 1.732
Consider the numbers 2 2 and 4 2
The numbers are 4 and 16.
Their procluct 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16
Posted in Class 8 on January 3, 2025 January 4, 2025
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Consider the numbers 2 2 and 4 2
The numbers are 4 and 16.
Their procluct 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16
Posted in Class 8 on January 3, 2025 January 4, 2025
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1
(ii) For a ≠ 0, a 0 is __________ .
1
(iii) 4 -3 × 5 -3 = __________ .
20 -3
(iv) (-2) -7 is = __________ .
\(\frac{-1}{128}\)
(v) \(\left(-\frac{1}{3}\right)^{-5}\) = _________ .
– 243
1
(ii) For a ≠ 0, a 0 is __________ .
1
(iii) 4 -3 × 5 -3 = __________ .
20 -3
(iv) (-2) -7 is = __________ .
\(\frac{-1}{128}\)
(v) \(\left(-\frac{1}{3}\right)^{-5}\) = _________ .
– 243
True
(ii) The simplified form of \((256)^{\frac{-1}{4}} \times 4^{2}\) is \(\frac{1}{4}\).
True
(iii) Using the power rule, \(\left(3^{7}\right)^{-2}\) = 3 5
True
(iv) The standard form of 2 × 10 -4 is 0.0002.
False
(v) The scientific form of 123.456 is 1.23456 × 10 -2 .
True
True
(ii) The simplified form of \((256)^{\frac{-1}{4}} \times 4^{2}\) is \(\frac{1}{4}\).
True
(iii) Using the power rule, \(\left(3^{7}\right)^{-2}\) = 3 5
True
(iv) The standard form of 2 × 10 -4 is 0.0002.
False
(v) The scientific form of 123.456 is 1.23456 × 10 -2 .
True
(i) (3 2 ) 3 × (2 × 3 5 ) -2 × (18) 2
(ii) \(\frac{9^{2} \times 7^{3} \times 2^{5}}{84^{3}}\)
(iii) \(\frac{2^{8} \times 2187}{3^{5} \times 3^{2}}\)
= 2 8-5 × 3 7-5
= 2 3 × 3 2
= 8 × 9
= 72
(i) (3 2 ) 3 × (2 × 3 5 ) -2 × (18) 2
(ii) \(\frac{9^{2} \times 7^{3} \times 2^{5}}{84^{3}}\)
(iii) \(\frac{2^{8} \times 2187}{3^{5} \times 3^{2}}\)
= 2 8-5 × 3 7-5
= 2 3 × 3 2
= 8 × 9
= 72
(i) 6054.321
6054.321 = (6 × 1000) + (0 × 100) + (5 × 10) + (4 × 10 0 ) + \(\frac{3}{10}+\frac{2}{100}+\frac{1}{1000}\)
= (6 × 10 3 ) + (5 × 10 1 ) + (4 × 10 0 ) + \(\frac{3}{10}+\frac{2}{100}+\frac{1}{1000}\)
= (6 × 10 3 ) + (5 × 10 1 ) + (4 × 10 0 ) + (3 × 10 -1 ) + (2 × 10 -2 ) + (1 × 10 -3 )
(ii) 897.14
= (8 × 100) + (9 × 10) + (7 × 10 0 ) + \(\frac{1}{10}+\frac{4}{100}\)
= (8 × 1o 2 ) +( 9 × 10 1 ) + (7 × 10 0 ) + \(\left(1 \times \frac{1}{10}\right)+\left(4 \times \frac{1}{100}\right)\)
= (8 × 10 3 ) + (9 × 10 3 ) + (7 × 10 0 ) + (1 × 10 -1 ) + (4 × 10 -2 )
(i) 6054.321
6054.321 = (6 × 1000) + (0 × 100) + (5 × 10) + (4 × 10 0 ) + \(\frac{3}{10}+\frac{2}{100}+\frac{1}{1000}\)
= (6 × 10 3 ) + (5 × 10 1 ) + (4 × 10 0 ) + \(\frac{3}{10}+\frac{2}{100}+\frac{1}{1000}\)
= (6 × 10 3 ) + (5 × 10 1 ) + (4 × 10 0 ) + (3 × 10 -1 ) + (2 × 10 -2 ) + (1 × 10 -3 )
(ii) 897.14
= (8 × 100) + (9 × 10) + (7 × 10 0 ) + \(\frac{1}{10}+\frac{4}{100}\)
= (8 × 1o 2 ) +( 9 × 10 1 ) + (7 × 10 0 ) + \(\left(1 \times \frac{1}{10}\right)+\left(4 \times \frac{1}{100}\right)\)
= (8 × 10 3 ) + (9 × 10 3 ) + (7 × 10 0 ) + (1 × 10 -1 ) + (4 × 10 -2 )
(i) 8 × 10 4 + 7 × 10 3 + 6 × 10 2 + 5 × 10 1 + 2 × 1 + 4 × 10 -2 + 7 × 10 -4
= 8 × 10 4 + 7 × 10 3 + 6 × 10 2 + 5 × 10 1 + 2 × 1 + 4 × 10 -2 + 7 × 10 -4
= 8 × 10000 + 7 × 1000 + 6 × 100 + 5 × 10 + 2 × 1 + 4 × \(\frac{1}{100}\) + 7 × \(\frac{1}{10000}\)
= 80000 + 7000 + 600 + 50 + 2 + \(\frac{4}{100}\) + \(\frac{7}{10000}\)
= 87652.0407
(ii) 5 × 10 3 + 5 × 10 1 + 5 × 10 -1 + 5 × 10 -3
= 5 × 10 3 + 5 × 10 1 + 5 × 10 -1 + 5 × 10 -3
= 5 × 1000 + 5 × 10 + 5 × \(\frac{1}{10}\) + 5 × \(\frac{1}{1000}\)
= 5000 + 50 + \(\frac{5}{10}+\frac{5}{1000}\) = 5050.505
(iii) The radius of a hydrogen atom is 2.5 10 -11 m
Radiys of a hydrogen atom = 2.5 × 10 -11 m
= \(2.5 \times \frac{1}{10^{11}} \mathrm{m}=\frac{2.5}{10^{11}} \mathrm{m}\)
= 0.000000000025 m
(i) 8 × 10 4 + 7 × 10 3 + 6 × 10 2 + 5 × 10 1 + 2 × 1 + 4 × 10 -2 + 7 × 10 -4
= 8 × 10 4 + 7 × 10 3 + 6 × 10 2 + 5 × 10 1 + 2 × 1 + 4 × 10 -2 + 7 × 10 -4
= 8 × 10000 + 7 × 1000 + 6 × 100 + 5 × 10 + 2 × 1 + 4 × \(\frac{1}{100}\) + 7 × \(\frac{1}{10000}\)
= 80000 + 7000 + 600 + 50 + 2 + \(\frac{4}{100}\) + \(\frac{7}{10000}\)
= 87652.0407
(ii) 5 × 10 3 + 5 × 10 1 + 5 × 10 -1 + 5 × 10 -3
= 5 × 10 3 + 5 × 10 1 + 5 × 10 -1 + 5 × 10 -3
= 5 × 1000 + 5 × 10 + 5 × \(\frac{1}{10}\) + 5 × \(\frac{1}{1000}\)
= 5000 + 50 + \(\frac{5}{10}+\frac{5}{1000}\) = 5050.505
(iii) The radius of a hydrogen atom is 2.5 10 -11 m
Radiys of a hydrogen atom = 2.5 × 10 -11 m
= \(2.5 \times \frac{1}{10^{11}} \mathrm{m}=\frac{2.5}{10^{11}} \mathrm{m}\)
= 0.000000000025 m
467800000000 = 4.678 × 10 11
(ii) 0.000001972
0.000001972 = 1.972 × 10 -6
(iii) 1642.398
1642.398 = 1.642398 × 10 3
(iv) Earth’s volume is about 1,083,000,000,000 cubic kilometres
1,083,000,000,000
Earth’s volume = 1.083 110 × 10 2 cubic kilometres
(v) If you fill a bucket with dirt, the portion of the whole Earth that is in the bucket will be 0.00000000000000000000000 16 kg
Portion of earth in the bucket = 0.00000000000000000000000 16 kg
= 1.6 10 × 10 24 kg.
Objective Type Questions
467800000000 = 4.678 × 10 11
(ii) 0.000001972
0.000001972 = 1.972 × 10 -6
(iii) 1642.398
1642.398 = 1.642398 × 10 3
(iv) Earth’s volume is about 1,083,000,000,000 cubic kilometres
1,083,000,000,000
Earth’s volume = 1.083 110 × 10 2 cubic kilometres
(v) If you fill a bucket with dirt, the portion of the whole Earth that is in the bucket will be 0.00000000000000000000000 16 kg
Portion of earth in the bucket = 0.00000000000000000000000 16 kg
= 1.6 10 × 10 24 kg.
Objective Type Questions
- A. \(\frac{2}{3}\)
- B. \(\frac{-2}{5}\)
- C. \(\frac{5}{2}\)
- D. \(\frac{-5}{2}\)
(B) \(\frac{-2}{5}\)
Hint:
(-4) -1 = \(\left(-\frac{1}{4}\right)^{1}=\frac{-1}{4}\)
(B) \(\frac{-2}{5}\)
Hint:
(-4) -1 = \(\left(-\frac{1}{4}\right)^{1}=\frac{-1}{4}\)
- A. \(\left(\frac{-1}{4}\right)^{2}\) = 4 -2
- B. \(\left(\frac{-1}{4}\right)^{2}=\left(\frac{1}{2}\right)^{4}\)
- C. \(\left(\frac{-1}{4}\right)^{2}\) = 16 -1
- D. \(-\left(\frac{1}{4}\right)^{2}\) = 16 -1
\(-\left(\frac{1}{4}\right)^{2}\) = 16 -1
Hint:
(-2) – 3 x (- 2) – 2 = (-2) – 3 – 2 = (-2) – 5 (\(-\frac { 1 }{ 2 }\))5 = \(-\frac { 1 }{ 32 }\)
\(-\left(\frac{1}{4}\right)^{2}\) = 16 -1
Hint:
(-2) – 3 x (- 2) – 2 = (-2) – 3 – 2 = (-2) – 5 (\(-\frac { 1 }{ 2 }\))5 = \(-\frac { 1 }{ 32 }\)
- A. 2.02 × 10 9
- B. 2.02 × 10 -9
- C. 2.02 × 10 -8
- D. 2.02 × 10 -10
(D) 2.02 × 10 -10
Hint:
0.0000000002020
Posted in Class 8 on September 9, 2024 September 10, 2024
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(D) 2.02 × 10 -10
Hint:
0.0000000002020
Posted in Class 8 on September 9, 2024 September 10, 2024
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Capacity of the small waterug = 3\(\frac{4}{5}\) litres.
Capacity of the big jug = \(2 \frac{2}{3}\) times the small one.
= \(2 \frac{2}{3} \times 3 \frac{4}{5}=\frac{8}{3} \times \frac{19}{5}=\frac{152}{15}\)
= \(\frac{2}{15}\) litres
Capacity of the large jug = \(\frac{2}{15}\) litres.
Capacity of the small waterug = 3\(\frac{4}{5}\) litres.
Capacity of the big jug = \(2 \frac{2}{3}\) times the small one.
= \(2 \frac{2}{3} \times 3 \frac{4}{5}=\frac{8}{3} \times \frac{19}{5}=\frac{152}{15}\)
= \(\frac{2}{15}\) litres
Capacity of the large jug = \(\frac{2}{15}\) litres.
∴ The product is \(\frac{400}{120}\) and its simplest form improper fraction is \(\frac{10}{3}\)
And mixed fraction is \(3 \frac{1}{3}\)
∴ Both are correct
∴ The product is \(\frac{400}{120}\) and its simplest form improper fraction is \(\frac{10}{3}\)
And mixed fraction is \(3 \frac{1}{3}\)
∴ Both are correct
Length of the room × Breadth = Area of the room
Breadth of the room = \(2 \frac{11}{20}\) m
Area of the room = \(\frac{153}{10}\) sq.m
Length x \(2\frac{11}{20}\) = \(\frac{153}{10}\)
Length of the room = 6 m
Length of the room × Breadth = Area of the room
Breadth of the room = \(2 \frac{11}{20}\) m
Area of the room = \(\frac{153}{10}\) sq.m
Length x \(2\frac{11}{20}\) = \(\frac{153}{10}\)
Length of the room = 6 m
Area of the square = 4489 cm 2
(side)2 = 4489 cm 2
(side)2 = 67 × 67
side = 67 2
Length of a side = 67
Length of a side with liner = 67 + 2 + 2 cm
= 71 cm
Area of the larger square = 71 × 71 cm 2
= 5041 cm 2
Area of the liner = Area of big square – Area of small square
= (5041 – 4489) cm 2
= 552 cm 2
Area of the square = 4489 cm 2
(side)2 = 4489 cm 2
(side)2 = 67 × 67
side = 67 2
Length of a side = 67
Length of a side with liner = 67 + 2 + 2 cm
= 71 cm
Area of the larger square = 71 × 71 cm 2
= 5041 cm 2
Area of the liner = Area of big square – Area of small square
= (5041 – 4489) cm 2
= 552 cm 2
Area of the greeting card = 90 cm 2
(side) 2 = 90 cm 2
(side) 2 = 2 × 5 × 3 × 3 = 2 × 5 × 3 2
Side = 3\(\sqrt{2 \times 5}\)
side = 3√10 cm
side = 3 × 3.2cm
side = 9.6 cm
∴ Side lies between the whole numbers 9 and 10.
Area of the greeting card = 90 cm 2
(side) 2 = 90 cm 2
(side) 2 = 2 × 5 × 3 × 3 = 2 × 5 × 3 2
Side = 3\(\sqrt{2 \times 5}\)
side = 3√10 cm
side = 3 × 3.2cm
side = 9.6 cm
∴ Side lies between the whole numbers 9 and 10.
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 × side = 15 × 15 sq.decimeter
side = 15 decimeters
Length of each side of verandah = 15 decimeters.
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 × side = 15 × 15 sq.decimeter
side = 15 decimeters
Length of each side of verandah = 15 decimeters.
Heart beat per minute = 80 beats
(i) an hour
One hour = 60 minutes
Heart beat in an hour = 60 × 80
= 4800
= 4.8 × 10 3
(ii) In a day
One day = 24 hours = 24 × 60 minutes
∴ Heart beat in one day = 24 × 60 × 80 = 24 × 4800 = 115200
= 1.152 × 10 5
(iii) a year
One year = 365 days = 365 × 24 hours = 365 × 24 × 60 minutes
∴ Heart beats in a year = 365 × 24 × 60 × 80
= 42048000
= 4.2048 × 10 7
(iv) 100 years
Heart beats in one year = 4.2048 × 10 7
heart beats in 100 years = 4.2048 × 10 7 × 100 = 4.2048 × 10 7 × 10 2
= 4.2048 × 10 9
Challenging Problems:
Heart beat per minute = 80 beats
(i) an hour
One hour = 60 minutes
Heart beat in an hour = 60 × 80
= 4800
= 4.8 × 10 3
(ii) In a day
One day = 24 hours = 24 × 60 minutes
∴ Heart beat in one day = 24 × 60 × 80 = 24 × 4800 = 115200
= 1.152 × 10 5
(iii) a year
One year = 365 days = 365 × 24 hours = 365 × 24 × 60 minutes
∴ Heart beats in a year = 365 × 24 × 60 × 80
= 42048000
= 4.2048 × 10 7
(iv) 100 years
Heart beats in one year = 4.2048 × 10 7
heart beats in 100 years = 4.2048 × 10 7 × 100 = 4.2048 × 10 7 × 10 2
= 4.2048 × 10 9
Challenging Problems:
1 inch = 120 km
Distance between A and B = \(4\frac{1}{6}\)
Distance between A and C = \(3\frac{1}{3}\)
∴ Distance between B and C = \(4 \frac{1}{6}+3 \frac{1}{3}\) inches
1 inch = 120km
∴ \(\frac{45}{6}\) inches = \(\frac{45}{6}\) × 120 km = 900 km
Distance between B and C = 900 km
1 inch = 120 km
Distance between A and B = \(4\frac{1}{6}\)
Distance between A and C = \(3\frac{1}{3}\)
∴ Distance between B and C = \(4 \frac{1}{6}+3 \frac{1}{3}\) inches
1 inch = 120km
∴ \(\frac{45}{6}\) inches = \(\frac{45}{6}\) × 120 km = 900 km
Distance between B and C = 900 km
let a = \(\frac{5}{6}\) and b = \(\frac{-4}{3}\) be two non zero rational numbers.
∴ Collection of non-zero rational numbers are closed under division.
(ii) Subtraction is not commutative for rational numbers.
let a = \(\frac{1}{2}\) and b = \(-\frac{5}{6}\) be two rational numbers.
a – b ≠ b – a
∴ Subtraction is not commutative for rational numbers.
(iii) Division is not associative for rational numbers.
Let a = \(\frac{2}{5}\), b = \(\frac{6}{5}\), c = \(\frac{3}{5}\) be three rational numbers.
a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c
∴ Division is not associative for rational numbers.
(iv) Distributive property of multiplication over subtraction is true for rational numbers. That is, a (b – c) = ab – ac.
Let a = \(\frac{2}{9}\), b = \(\frac{3}{6}\), c = \(\frac{1}{3}\) be three rational numbers.
To prove a × (b – c) = ab – bc
∴ From (1) and (2)
a × (b – c) = ab – bc
∴ Distributivity of multiplication over subtraction is true for rational numbers.
(v) The mean of two rational numbers is rational and lies between them.
Let a = \(\frac{2}{11}\) and b = \(\frac{5}{6}\) be two rational numbers
∴ The mean lies between the given rational numbers \(\frac{2}{11}\) and \(\frac{5}{6}\)
let a = \(\frac{5}{6}\) and b = \(\frac{-4}{3}\) be two non zero rational numbers.
∴ Collection of non-zero rational numbers are closed under division.
(ii) Subtraction is not commutative for rational numbers.
let a = \(\frac{1}{2}\) and b = \(-\frac{5}{6}\) be two rational numbers.
a – b ≠ b – a
∴ Subtraction is not commutative for rational numbers.
(iii) Division is not associative for rational numbers.
Let a = \(\frac{2}{5}\), b = \(\frac{6}{5}\), c = \(\frac{3}{5}\) be three rational numbers.
a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c
∴ Division is not associative for rational numbers.
(iv) Distributive property of multiplication over subtraction is true for rational numbers. That is, a (b – c) = ab – ac.
Let a = \(\frac{2}{9}\), b = \(\frac{3}{6}\), c = \(\frac{1}{3}\) be three rational numbers.
To prove a × (b – c) = ab – bc
∴ From (1) and (2)
a × (b – c) = ab – bc
∴ Distributivity of multiplication over subtraction is true for rational numbers.
(v) The mean of two rational numbers is rational and lies between them.
Let a = \(\frac{2}{11}\) and b = \(\frac{5}{6}\) be two rational numbers
∴ The mean lies between the given rational numbers \(\frac{2}{11}\) and \(\frac{5}{6}\)
Let the weight of 1 ragi adai = x grams
given \(\frac { 1 }{ 4 }\) of x = 120gm
\(\frac { 1 }{ 4 }\) × x = 120
x = 120 × 4
x = 480gm
∴ \(\frac { 2 }{ 3 }\) of the adai = \(\frac { 2 }{ 3 }\) × 480 gm = 2 × 160 gm = 320gm
\(\frac { 2 }{ 3 }\) of the weight of adai = 320gm
Let the weight of 1 ragi adai = x grams
given \(\frac { 1 }{ 4 }\) of x = 120gm
\(\frac { 1 }{ 4 }\) × x = 120
x = 120 × 4
x = 480gm
∴ \(\frac { 2 }{ 3 }\) of the adai = \(\frac { 2 }{ 3 }\) × 480 gm = 2 × 160 gm = 320gm
\(\frac { 2 }{ 3 }\) of the weight of adai = 320gm
Given p + 2q = 18 ……… (1)
pq = 40 ……… (2)
Given p + 2q = 18 ……… (1)
pq = 40 ……… (2)
Number of cadets to form square design
The numbers 2 and 3 are unpaired
∴ It is impossible to have the parade forming square design with 1536 cadets.
39 × 39 = 1521
Also 40 × 40 = 1600
∴ We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
∴ 64 more cadets would be required to form the square design.
Number of cadets to form square design
The numbers 2 and 3 are unpaired
∴ It is impossible to have the parade forming square design with 1536 cadets.
39 × 39 = 1521
Also 40 × 40 = 1600
∴ We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
∴ 64 more cadets would be required to form the square design.
(3.769 × 10 5 ) + (4.21 × 10 5 ) = 3,76,900 + 4,21,000
= 7,97,000
= 7.979 × 10 5
(3.769 × 10 5 ) + (4.21 × 10 5 ) = 3,76,900 + 4,21,000
= 7,97,000
= 7.979 × 10 5
16 25 = (2 4 ) 25 = 2 100
8 100 = (2 3 ) 100 = 2 300
4 400 = (2 2 ) 400 = 2 800
2 600 = 2 600
Comparing the powers we have.
2 100 < 2 300 < 2 600 < 2 800
∴ The required order: 16 25 , 8 100 , 3 500 , 4 400 , 2 600
Posted in Class 8 on September 9, 2024 September 10, 2024
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16 25 = (2 4 ) 25 = 2 100
8 100 = (2 3 ) 100 = 2 300
4 400 = (2 2 ) 400 = 2 800
2 600 = 2 600
Comparing the powers we have.
2 100 < 2 300 < 2 600 < 2 800
∴ The required order: 16 25 , 8 100 , 3 500 , 4 400 , 2 600
Posted in Class 8 on September 9, 2024 September 10, 2024
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\(\frac{125}{200}=\frac{125 \div 25}{200 \div 25}=\frac{5}{8}\)
= \(\frac{5}{8}\)
\(\frac{125}{200}=\frac{125 \div 25}{200 \div 25}=\frac{5}{8}\)
= \(\frac{5}{8}\)
- A. \(\frac{2}{3}\)
- B. \(\frac{16}{24}\)
- C. \(\frac{32}{60}\)
- D. \(\frac{24}{36}\)
(C) \(\frac{32}{60}\)
\(\frac{8}{12}=\frac{8+4}{12 \div 4}=\frac{2}{3}\)
\(\frac{8}{12}=\frac{8 \times 2}{12 \times 2}=\frac{16}{24}\)
\(\frac{8}{12}=\frac{8 \times 3}{12 \times 3}=\frac{24}{36}\)
But \(\frac{32}{60}=\frac{32 \div 5}{60 \div 5}=\frac{6.4}{12}\)
∴ \(\frac{32}{60}\) is not equivalent fraction of \(\frac{8}{12}\)
(C) \(\frac{32}{60}\)
\(\frac{8}{12}=\frac{8+4}{12 \div 4}=\frac{2}{3}\)
\(\frac{8}{12}=\frac{8 \times 2}{12 \times 2}=\frac{16}{24}\)
\(\frac{8}{12}=\frac{8 \times 3}{12 \times 3}=\frac{24}{36}\)
But \(\frac{32}{60}=\frac{32 \div 5}{60 \div 5}=\frac{6.4}{12}\)
∴ \(\frac{32}{60}\) is not equivalent fraction of \(\frac{8}{12}\)
LCM of 5, 8, 10 = 5 × 2 × 4
= 40
LCM of 5, 8, 10 = 5 × 2 × 4
= 40
\(\frac{7}{36}+\frac{35}{81}=\frac{7}{36} \times \frac{81}{35}=\frac{9}{20}\)
\(\frac{7}{36}+\frac{35}{81}=\frac{7}{36} \times \frac{81}{35}=\frac{9}{20}\)
Let the total population = 1
Population of men = Total population – Women – Children
∴ Population of men = \(\frac{2}{5}\)
Let the total population = 1
Population of men = Total population – Women – Children
∴ Population of men = \(\frac{2}{5}\)
Try These (Text Book Page No. 3)
Try These (Text Book Page No. 3)
A rational number, Because – 7 = \(\frac{-14}{2}=\frac{p}{q}\)
A rational number, Because – 7 = \(\frac{-14}{2}=\frac{p}{q}\)
\(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}\)
Try These (Text Book Page No. 4)
That’s literally all there is to it! 1/8 as a decimal is 0.125.
Write the decimal forms of the following rational numbers:
\(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}\)
Try These (Text Book Page No. 4)
That’s literally all there is to it! 1/8 as a decimal is 0.125.
Write the decimal forms of the following rational numbers:
\(\frac{486}{1000}\) = 0.486
\(\frac{486}{1000}\) = 0.486
(i) \(\frac{-6}{4}, \frac{18}{-12}\)
\(\frac{-6}{4}=\frac{-6 \times 3}{4 \times 3}=\frac{-18}{12}\)
∴ \(\frac{-6}{4}\) equivalent to \(\frac{-18}{12}\)
(ii) \(\frac{-4}{-20}, \frac{1}{-5}\)
\(\frac{-4}{-20}=\frac{-4 \div(-4)}{-20 \div(-4)}=\frac{1}{5} \neq-\frac{1}{5}\)
∴ \(\frac{-4}{-20}\) equivalent to \(\frac{1}{-5}\)
(iii) \(\frac{-12}{-17}, \frac{60}{85}\)
\(\frac{-12}{-17}=\frac{-12 x-5}{-17 x-5}=\frac{60}{85}\)
∴ \(\frac{-12}{-17}\) equivalent to \(\frac{60}{85}\)
(i) \(\frac{-6}{4}, \frac{18}{-12}\)
\(\frac{-6}{4}=\frac{-6 \times 3}{4 \times 3}=\frac{-18}{12}\)
∴ \(\frac{-6}{4}\) equivalent to \(\frac{-18}{12}\)
(ii) \(\frac{-4}{-20}, \frac{1}{-5}\)
\(\frac{-4}{-20}=\frac{-4 \div(-4)}{-20 \div(-4)}=\frac{1}{5} \neq-\frac{1}{5}\)
∴ \(\frac{-4}{-20}\) equivalent to \(\frac{1}{-5}\)
(iii) \(\frac{-12}{-17}, \frac{60}{85}\)
\(\frac{-12}{-17}=\frac{-12 x-5}{-17 x-5}=\frac{60}{85}\)
∴ \(\frac{-12}{-17}\) equivalent to \(\frac{60}{85}\)
(i) \(\frac{36}{-96}\)
= \(\frac{-36 \div 12}{96 \div 12}=\frac{-3}{8}\)
(ii) \(\frac{-56}{-72}\)
= \(\frac{-56 \div(-8)}{-72 \div(-8)}=\frac{7}{9}\)
(iii) \(\frac{27}{18}\)
= \(1 \frac{9}{18}=1 \frac{1}{2}\)
(i) \(\frac{36}{-96}\)
= \(\frac{-36 \div 12}{96 \div 12}=\frac{-3}{8}\)
(ii) \(\frac{-56}{-72}\)
= \(\frac{-56 \div(-8)}{-72 \div(-8)}=\frac{7}{9}\)
(iii) \(\frac{27}{18}\)
= \(1 \frac{9}{18}=1 \frac{1}{2}\)
\(\frac{-2}{3}\) lies betveen 0 and -1.
T?ìe unit part between O and —lis divided into 3 equal parts and second part is taken.
(ii) \(\frac{-8}{-5}\)
\(\frac{-8}{-5}\) = \(1 \frac{3}{5}\)
\(1 \frac{3}{5}\) lies between I and 2, The unit part between I and 2 is divided into 5 equal parts and the third part is taken.
(iii) \(\frac{5}{-4}\)
\(\frac{5}{-4}\) = \(-\frac{5}{4}\) = \(-1 \frac{1}{4}\)
\(-1 \frac{1}{4}\) lies between -1 and -2. The unit part between -1 and -2 is divided into four equal parts and the first part is taken.
Think (Text Book Page No. 15)
Is zero a rational number? If so, what is its additive inverse
Yes zero a rational number Additive inverse of zero is zero.
Think (Text Book Page No. 16)
What is the multiplicative inverse of 1 and -1?
Multiplicative inverse of 1 is 1 and -1 is -1.
Try These (Text Book Page No. 16)
Divide
(i) \(\frac{-7}{3}\) by 5
(ii) 5 by \(\frac{-7}{3}\)
(iii) \(\frac{-7}{3}\) by \(\frac{35}{6}\)
(i) \(\frac{-7}{3}\) by 5
(ii) 5 by \(\frac{-7}{3}\)
(iii) \(\frac{-7}{3}\) by \(\frac{35}{6}\)
Try These (Text Book Page No. 20)
The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Let 0 and \(\frac{1}{2}\) te two rational numbers 0 – \(\frac{1}{2}\) = –\(\frac{1}{2}\) is a rational numter
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number \(\frac{5}{2}\) and 0.
\(\frac{5}{2}\) + 0 = \(\frac{5}{2 \times 0}=\frac{5}{0}\)
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.
Try These (Text Book Page No. 22)
(i) Is \(\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}\) ?
LHS ≠ RHS
∴ \(\frac{3}{5} \div \frac{7}{8}\) ≠ \(\frac{7}{8}-\frac{3}{5}\)
∴ Subtraction of rational numbers is not commutative.
(ii) \(\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}\)? So, what do you conclude?
LHS ≠ RHS
∴ \(\frac{3}{5} \div \frac{7}{8}\) ≠ \(\frac{7}{8} \div \frac{5}{3}\)
∴ Commutative property not hold good br division of rational numbers.
Try This (Text Book Page No. 22)
Check whether associative property holds for subtraction and division.
Consider for rational numbers \(\frac{2}{3}, \frac{1}{2}\) and \(\frac{3}{4}\)
∴ Associative property not holds for subraction of rational numbers
∴ Associative property not holds for division of rational numbers
Try This (Text Book Page No. 25)
\(\frac{-2}{3}\) lies betveen 0 and -1.
T?ìe unit part between O and —lis divided into 3 equal parts and second part is taken.
(ii) \(\frac{-8}{-5}\)
\(\frac{-8}{-5}\) = \(1 \frac{3}{5}\)
\(1 \frac{3}{5}\) lies between I and 2, The unit part between I and 2 is divided into 5 equal parts and the third part is taken.
(iii) \(\frac{5}{-4}\)
\(\frac{5}{-4}\) = \(-\frac{5}{4}\) = \(-1 \frac{1}{4}\)
\(-1 \frac{1}{4}\) lies between -1 and -2. The unit part between -1 and -2 is divided into four equal parts and the first part is taken.
Think (Text Book Page No. 15)
Is zero a rational number? If so, what is its additive inverse
Yes zero a rational number Additive inverse of zero is zero.
Think (Text Book Page No. 16)
What is the multiplicative inverse of 1 and -1?
Multiplicative inverse of 1 is 1 and -1 is -1.
Try These (Text Book Page No. 16)
Divide
(i) \(\frac{-7}{3}\) by 5
(ii) 5 by \(\frac{-7}{3}\)
(iii) \(\frac{-7}{3}\) by \(\frac{35}{6}\)
(i) \(\frac{-7}{3}\) by 5
(ii) 5 by \(\frac{-7}{3}\)
(iii) \(\frac{-7}{3}\) by \(\frac{35}{6}\)
Try These (Text Book Page No. 20)
The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Let 0 and \(\frac{1}{2}\) te two rational numbers 0 – \(\frac{1}{2}\) = –\(\frac{1}{2}\) is a rational numter
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number \(\frac{5}{2}\) and 0.
\(\frac{5}{2}\) + 0 = \(\frac{5}{2 \times 0}=\frac{5}{0}\)
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.
Try These (Text Book Page No. 22)
(i) Is \(\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}\) ?
LHS ≠ RHS
∴ \(\frac{3}{5} \div \frac{7}{8}\) ≠ \(\frac{7}{8}-\frac{3}{5}\)
∴ Subtraction of rational numbers is not commutative.
(ii) \(\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}\)? So, what do you conclude?
LHS ≠ RHS
∴ \(\frac{3}{5} \div \frac{7}{8}\) ≠ \(\frac{7}{8} \div \frac{5}{3}\)
∴ Commutative property not hold good br division of rational numbers.
Try This (Text Book Page No. 22)
Check whether associative property holds for subtraction and division.
Consider for rational numbers \(\frac{2}{3}, \frac{1}{2}\) and \(\frac{3}{4}\)
∴ Associative property not holds for subraction of rational numbers
∴ Associative property not holds for division of rational numbers
Try This (Text Book Page No. 25)
Think (Text Book Page No. 26)
Think (Text Book Page No. 26)
No, the square of a prime number ‘P’ has at Least 3 divisors 1, P and P 2 . But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.
No, the square of a prime number ‘P’ has at Least 3 divisors 1, P and P 2 . But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.
The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. Bu the product of two perfect square is a perfect square.
Try These (Text Book Page No. 26)
The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. Bu the product of two perfect square is a perfect square.
Try These (Text Book Page No. 26)
256 = 16 2
576 = 24 2
4096 = 64 2
∴ 256, 576, and 4096 are perfect squares
256 = 16 2
576 = 24 2
4096 = 64 2
∴ 256, 576, and 4096 are perfect squares
Because the unit digit ola perfect square will be 0, 1,4, 5, 6, 9. But the given numbers have unit digits 2, 3, 7, 8. So they are not perfect squares.
Think (Text Book Page No. 27)
Consider the claim: “Between the squares of the consecutive numbers n and (n + 1), there are 2n non-square numbers’ Can it be true? FInd how many non-square numbers are there
(i) between 4 and 9 ?
(ii) between 49 and 64? and Verify the claim.
Therefore we conclude that there are 2n non-square numbers between two consecutive square numbers.
Think (Text Book Page No. 30)
In this quick guide we’ll describe what the factors of 96 are, how you find them and list out the factor pairs of 96 for you to prove the calculation works.
In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?
108 = 2 × 2 × 3 × 3 = 2 2 × 3 2 × 3
If we multiply the factors by 2, then we get
2 2 × 3 2 × 3 × 3 = 2 2 × 3 2 × 3 2 = (2 × 3 × 3) 2
Which is perfect square.
∴ Again if we divide by 3 then we get 2 2 × 3 2 ⇒ (2 × 3) 2 , a perfect square.
∴ We have to multiply or divide 108 by 3 to get a perfect square.
Try These (Text Book Page No. 32)
Find the square root by long division method.
Because the unit digit ola perfect square will be 0, 1,4, 5, 6, 9. But the given numbers have unit digits 2, 3, 7, 8. So they are not perfect squares.
Think (Text Book Page No. 27)
Consider the claim: “Between the squares of the consecutive numbers n and (n + 1), there are 2n non-square numbers’ Can it be true? FInd how many non-square numbers are there
(i) between 4 and 9 ?
(ii) between 49 and 64? and Verify the claim.
Therefore we conclude that there are 2n non-square numbers between two consecutive square numbers.
Think (Text Book Page No. 30)
In this quick guide we’ll describe what the factors of 96 are, how you find them and list out the factor pairs of 96 for you to prove the calculation works.
In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?
108 = 2 × 2 × 3 × 3 = 2 2 × 3 2 × 3
If we multiply the factors by 2, then we get
2 2 × 3 2 × 3 × 3 = 2 2 × 3 2 × 3 2 = (2 × 3 × 3) 2
Which is perfect square.
∴ Again if we divide by 3 then we get 2 2 × 3 2 ⇒ (2 × 3) 2 , a perfect square.
∴ We have to multiply or divide 108 by 3 to get a perfect square.
Try These (Text Book Page No. 32)
Find the square root by long division method.
Mass of Planet Uranus = 86800000000000000000000000 kg
[23 zeros after 88]
Mass of Planet Uranus = 86800000000000000000000000 kg
[23 zeros after 88]
0.000012005 = 1.2005 × 10 -5
(ii) 43 12.345
43 12.345 = 4.312345 × 10 3
(iii) 0.10524
0.10524 = 1.0524 × 10 -1
(iv) The distance between the Sun and the planet Saturn 1.4335 × 10 12 miles.
Posted in Class 8 on January 5, 2025 January 6, 2025
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0.000012005 = 1.2005 × 10 -5
(ii) 43 12.345
43 12.345 = 4.312345 × 10 3
(iii) 0.10524
0.10524 = 1.0524 × 10 -1
(iv) The distance between the Sun and the planet Saturn 1.4335 × 10 12 miles.
Posted in Class 8 on January 5, 2025 January 6, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi