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We need to convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 5 and 4 is 20.

3/5 = 3× 20 / 5×20 

= 60/100

3/4 = 3×25 / 4×25 

= 75/100

Here, we now know that 61, 62, 63,..74 are integers between numerators 60 and 75.

Hence, the rational numbers between 3/5 and 3/4 are 61/100, 62/100, 63/100, …., 74/100

A number of the form p/q is said to be a rational number if (b) p and q are integers and q ≠ 0

\(6 + \sqrt3\over \sqrt5 + \sqrt2\).

Multiply by √5 - √2 both numerator and denominator.

= \(6 + \sqrt3\over \sqrt5 + \sqrt2\) x \(\sqrt5 - \sqrt2\over\sqrt5 - \sqrt2\)

= \(6\sqrt5 - 6\sqrt2 + \sqrt15 - \sqrt6 \over 5 - 2\)

= \(6\sqrt5 - 6\sqrt2 + \sqrt15 - \sqrt6 \over 3\)

(c) 7

√49 = 7

Correct option is  A) 0

Correct option is (B) \(\frac{35}{100}\)

0.35 = \(\frac{35}{100}\)

Correct option is   B) \(\frac{35}{100}\)

It is given that,

The sum of the two numbers = \(\frac{-4}{3}\)

One of the number = -5

Suppose the other number is x

Since, the sum is \(\frac{-4}{3}\)

Therefore,

x - 5 = \(\frac{-4}{3}\)

3x – 15 = -4

3x = -4 + 15

3x = 11

x = \(\frac{11}{3}\)

Correct answer is (c) \(\frac{4}{6}\).

= \(\frac{4\, ÷\,2 }{6 \,÷ \,2}\)

= \(\frac{2}{3}\)

Hence, \(\frac{2}{3}\) is a rational number equivalent to \(\frac{4}{6}\).

It is given that,

The sum of the two numbers = \(\frac{5}{9}\)

One of the number = \(\frac{1}{3}\)

Since, the sum is \(\frac{5}{9}\)

Therefore,

The other number = Sum of numbers - given number

Other number = \(\frac{5}{9}-\frac{1}{3}\)

= \(\frac{5-3}{9}\)

= \(\frac{2}{9}\)

Hence, the other number is 2/9.

It is given that,

The sum of the two numbers = \(\frac{-1}{3}\)

One of the number = \(\frac{-12}{3}\)

Suppose the other number is x

Since, the sum is \(\frac{-1}{3}\)

Therefore,

x - \(\frac{12}{3}\) = \(\frac{-1}{3}\)

\(\frac{3x-12}{3}-\frac{-1}{3}\)

3x = 12-1

3x = 11

x = \(\frac{11}{3}\)

Correct answer is (c) 1/2.

Correct answer is (d) -1

(-4/3) – (-1/3)

= (-4 + 1)/3

= -3/3

= -1

Correct answer is (a) -1.

-3/2 + 1/2

= (-3 + 1)/2

= -2/2

= -1

Unlimited

There are unlimited number of rational numbers between two rational numbers.

We have given Class 8 Maths MCQ Questions of Rational numbers that will help you in getting ready for your Class 8 exam primarily based totally on the syllabus. The chapter-wise MCQ Questions for Class 8 Maths are given to students to lead them to apprehend every idea and assist them to attain good marks in exams.

Daily Maths practice will assist you to construct conceptual expertise approximately the concern and additionally lets you have a higher knowledge of the concepts. Multiple choice questions for Class 8 Rational numbers with every question which includes 4 answers, out of which one is correct. Students must solve the MCQ Questions of Rational numbers and select the right answer. They also can test their answers here.

Practice MCQ Question for Class 8 Maths chapter-wise

1. 0 is not

(a) a natural number
(b) a whole number
(c) an integer
(d) a rational number

2. The given property a+b = b+a is known as:

(a) Commutative property
(b) Distributive Property
(c) Associative Property
(d) Closure property

3. If a,b and c are whole numbers, then a+(b+c) = (a+b)+c. This property is called

(a) associative property
(b) distributive Property
(c) commutative property
(d) Closure property

4. The additive identity of any rational number is _____.

(a) 0
(b) 1
(c) -1
(d) 2

5. 1 is the multiplicative identity for ........

(a) whole numbers
(b) integers
(c) rational numbers
(d) all of the above

6. The additive inverse of 23 is

(a) -23
(b) 32
(c) -32
(d) all of the above

7. How many rational numbers exist between any two distinct rational numbers?

(a) 2
(b) 3
(c) 11
(d) Infinite number of rational numbers

8. The rational number that does not have a reciprocal is

(a) 0
(b) 1
(c) 4
(d) -4

9. The reciprocal of 2 is

(a) 0
(b) 2^-1
(c) 2^-2
(d) -2

10. An integer can be:

(a) Only Positive
(b) Only Negative
(c) Both positive and negative
(d) None of the above

11. A rational number can be represented in the form of:

(a) p/q
(b) pq
(c) p+q
(d) p-q

12. The value of 1/2 x 3/5 is equal to:

(a) 1/2
(b) 3/10
(c) 3/5
(d) 2/5

13. The value of (1/2) ÷ (3/5) is equal to:

(a)3/5
(b) 3/10
(c) 6/5
(d) 5/6

14. The associative property is applicable to:

(a) Addition and subtraction
(b) Multiplication and division
(c) Addition and Multiplication
(d) Subtraction and Division

15. What is the sum of 2/3and 4/9?

(a) 6/3
(b) 6/9
(c) 10/9
(d) 10/3

16. What is the product of 2/9 and 3/4?

(a) 1/6
(b) 2/3
(c) 1/9
(d) 1/4

17. How many rational numbers are there in between 3/4 and 1?

(a) 0
(b) 1
(c) 2
(d) Countless

18. What should be subtracted from -2/3 to get -1?

(a) 1/3
(b) -1/3
(c) 2/3
(d) 2/3

19. The value of (-10/3) x (-15/2) x (17/19) x 0 is:

(a) 0
(b) 22.66
(c) 20
(d) 35

20. Which of the following numbers is the decimal form of 1/4

(a) 0.25
(b) -0.25
(c) 2.05
(d) 2.5

21. What number should be subtracted from both the terms of the ratio 15:19 in order to make it 3:4?

(a) 4
(b) 3
(c) 2
(d) 6

22.  ________ is not associative for rational numbers.

(a) Subtraction or Division
(b) Addition or Multiplication
(c) Addition or Division
(d) Multiplication or Division

23. Find the multiplicative inverse of 13.

(a) 13
(b) -13
(c) -1/13
(d) 1/13

24. The sum of the additive inverse and multiplicative inverse of 2 is ____________.

(a) 3/2
(b) -3/2
(c) 1/2
(d) -1/2

25. What is the product of 3/10 and 5/6?

(a) 1/6
(b) 1/3
(c) 2/9
(d) 1/4

Answer:

 1. Answer: (b) a whole number

Explanation: 0 is not a natural number. It is a whole number. Natural numbers only include positive integers.

2. Answer: (a) Commutative property

Explanation: Commutative property says that the numbers can be added in any order, and you will still get the same answer. a+b = b+a is a clear example of the commutative property.

3. Answer: (a) associative property

Explanation: a+(b+c) = (a+b)+c is associative property of whole numbers.

4. Answer: (a) 0

Explanation: The additive identity property says that if you add a real number to zero or add zero to a real number, then you get the same real number back. The number zero is known as the identity element. Then zero(0) is the additive identity of a real number and all rational numbers are real. Hence, 0 is the additive identity of rational numbers.

5.Answer: (d) all of the above

Explanation: We know that whole numbers are a subset of integers which in turn are a subset of rational numbers. Also, 1 is the multiplicative identity for rational numbers because the product of 1 and any rational number is the rational number itself. Thus, 1 is the multiplicative identity for whole numbers, integers, and rational numbers.

6. Answer: (a) -23

Explanation: Additive inverse of 23 will be -23.

7. Answer: (d) Infinite number of rational numbers

Explanation: Infinite number of rational numbers exist between any two distinct rational numbers. We know that a rational number is a number which can be written in the form of  p/q
where p and q are integers and \(q\neq0\)

8. Answer: (a) 0

Explanation: The rational number that does not have a reciprocal  0 because reciprocal of 0 is undefined.

9. Answer: (b) 2^-1

Explanation: Therefore, the reciprocal of the number 2 is 1/2or 2⁻¹

10. Answer: (c) Both positive and negative

Explanation:  An integer can be both positive and negative as well as zero. i.e. …-3, -2, -1, 0, 1, 2, 3,…

11. Answer: (a) p/q

Explanation:  A rational number can be represented in the form p/q where p and q are integers and q is not equal to zero.

12. Answer: (b) 3/10

Explanation:  1/2 x3/5 

= (1 x 3)/(2 x 5) 

= 3/10

13. Answer: (d) 5/6

Explanation: (1/2) ÷ (3/5)

 = (1/2) x (5/3) 

= (1 x 5)/(2 x 3) 

= 5/6

14. Answer: (c) Addition and Multiplication

Explanation: As per associative property:

A + (B + C) = (A + B) + C

A × (B × C) = (A × B) × C

15. Answer: (c) 10/9

Explanation: 2/3+ 4/9

⇒ 2/3 x (3/3) + 4/9

⇒ 6/9 + 4/9

⇒ 10/9

16. Answer: (a) 1/6

Explanation: The product of 2/9 and 3/4:

⇒ 2/9 x 3/4

⇒ (2 x 3)/(9 x 4)

⇒ (2 x 3)/(3 x 3 x 2 x 2)

By canceling the common terms from numerator and denominator, we get;

⇒ 1/(3×2)

⇒ 1/6

17. Answer: (d) Countless

Explanation: Let us write 3/4 as 30/40 and 1 as 40/40.

Hence the rational numbers between them are:

31/40, 32/40, 33/40, 34/40,35/40,36/40, 37/40, 38/40, 39/40.

There are countless rational numbers between any two rational numbers.

18. Answer: (a) 1/3

Explanation: Let x be subtracted from -2/3.

-2/3 – x = -1

-x = -1 + 2/3

-x = -1/3

x = 1/3

19. Answer: (a) 0

Explanation:  Any number multiplied by zero is equal to zero.

20. Answer: (a) 0.25

Explanation: \(\frac{1}{4}=\frac{1\times25}{4\times25}\)

= 25/100

= 0.25

21. Answer: (b) 3

Explanation: Let the required number be x. 

\(\frac{15-x}{19-x}=\frac{3}{4}\)

60−4x = 57−3x

x = 3

22. Answer: (a) Subtraction or Division

Explanation: subtraction and division are not associative for rational numbers.

23. Answer:(d) 1/13

Explanation: The multiplicative inverse of 13 is (13)¹ = \(\frac{1}{13}\)

24.Answer: (b) -3/2

Explanation:  the additive inverse of 2 is −2

the multiplicative inverse of 2 is 1/2

the sum of the additive and the multiplicative inverse is

= -2 + 1/2

= -3/2

25. Answer: (d) 1/4

Explanation: The product of 3 /10 and 5/6:

⇒ 3/10 x 5/6

⇒ (3 x 5)/(10 x 6)

⇒ 15/60

⇒ 1/4

Click here Practice MCQ Question for Rational Numbers Class 8

(a) Compare the distance between Himgaon and Rawalpur to Sonapur and Ramgarh?

From the table the distance between Himgaon and Rawalpur = 98¾ km = 395/4 km

The distance between Sonapur and Ramgarh =
40 2/3 = 122/3 km

Then,

Difference of the distance between Himgaon and Rawalpur to Sonapur and Ramgarh,

= ((395/4) – (122/3))

= (1185 – 488)/ 12

= 697/12

=58 1/2 km

(b) If you drove from Himgaon to Sonapur and then from Sonapur to Rawalpur, how far would you drive?

From the table,

Distance between Himgaon and Sonapur = 100 5/6 km = 605/6 km

Distance between Sonapur and Rawalpur = 16 ½ km = 33/2

Then,

Total distance that he would drive,

= 605/6 + 33/2

= (605 + 99)/6

= 704/6

= 352/3

=117 1/3 km

From the question,

The overall width in cm of several wide screen television are,

97.28 cm = 9728/100 … [∵by the decimal removing method]

By dividing both numerator and denominator by 4 we get,

= 2432/25 cm

98 4/9 cm = by converting mixed fraction into improper fraction we get,

= 886/9 cm

98 1/25 cm = by converting mixed fraction into improper fraction we get,

= 2451/25 cm

97.94 cm = 9794/100 … [∵by the decimal removing method]

By dividing both numerator and denominator by 2 we get,

= 4897/50 cm

Now, we have to take the LCM of denominators to arrange them in ascending order.

The LCM of the denominators 25, 9, 25 and 50 is 450

∴ 2432/25= [(2432×18)/ (25×18)] = (43776/450)

(886/9) = [(886×50)/ (9×50)] = (44300/450)

(2451/25) = [(2451×18)/ (25×18)] = (44118/450)

(4897/50) = [(4897×9)/ (50×9)] = (44073/450)

Then,

Now, 43776 < 44073 < 44118 < 44300

Hence, in ascending order = (2432/25) < (4897/50) < (2451/25) < (886/9)

∴97.28 < 97.94 < 98 1/25 cm < 98 4/9 cm.

From the question it is given that,

Height of the roller coaster at an amusement park = 2/3 m

Height of the new roller coaster is about to build = 3/5 times the height of the existing

Coaster

= (2/3) × (3/5)

= (2/1) × (1/5)

= (2/5) m

(2/3) × [¾ × (5/7)] and [(2/3) × (5/7)] × ¾ this can be compared with associative property and commutative property.

(i) May

2.6924 cm = 26924/10000 … [∵by the decimal removing method]

By dividing both numerator and denominator by 4 we get,

= 6731/2500 cm

(ii) June

0.6096 cm = 0.6096/10000 … [∵by the decimal removing method]

By dividing both numerator and denominator by 16 we get,

= 381/625 cm

(iii) July

-6.9088 cm = -69088/10000 … [∵by the decimal removing method]

By dividing both numerator and denominator by 4 we get,

= -4318/625 cm

(iv) August

-8.636 cm = -8636/1000 … [∵by the decimal removing method]

By dividing both numerator and denominator by 4 we get,

= -2159/250 cm

From the question it is given that,

A mother and her two daughters got a room constructed for = ₹ 62,000

Let us assume mother’s share be x,

Then,

The elder daughter’s contribute = 3/8 of her mother’s share

= 3/8 x

The younger daughter’s contribute = ½ of her mother’s share

= ½ x

So, mother’s share + elder daughter’s share + younger daughter’s share = ₹ 62,000

x + (3/8) x + ½ x = ₹ 62,000

The LCM of the denominators 1, 8 and 2 is 8

(1/1) = [(1×8)/ (1×8)] = (8/8)

(3/8) = [(3×1)/ (8×1)] = (3/8)

(1/2) = [(1×4)/ (2×4)] = (4/8)

Then,

(8/8) x + (3/8) x + (4/8) x = 62,000

(8x + 3x + 4x)/8 = 62,000

(15x/8) = 62,000

15x = 62,000 × 8

X = 496000/15

X = ₹ 33,066.6

∴Mother’s share = ₹ 33,066.6

Elder daughter’s share = 3/8 of her mother’s share

= 3/8 x

= 3/8 × 33066.6

= ₹ 12,400

Younger daughter’s share = ½ of her mother’s share

= ½ x

= ½ × 33066.6

= ₹ 16,533.3

From the question it is given that,

Length of the skirt = 35 7/8 cm = 287/8 cm

Dimension of hem = 3 1/8 cm = 25/8 cm

Length of skirt, if hem is let down = ((287/8) + (25/8)) cm

= 312/8 cm

= 39 cm

From the question it is given that,

Total monthly allowance received by Huma, Hubna and Seem = ₹ 2,016

from their mother

Seema gets allowance = ½ of Huma’s share

Hubna gets allowance = 1 2/3 of Seema’s share

= 5/3 of Seema’s share

= 5/3 of ½ of Huma’s share … [∵ given]

= 5/3 × ½ of Huma’s share

= 5/6 of Huma’s share

So,

Huma’s share + Hubna’s share + Seema’s share = ₹ 2,016

Let Huma’s share be 1,

1 + (5/6) Huma’s share + ½ Huma’s share = ₹ 2,016

(1 + (5/6) + ½) = ₹ 2,016

The LCM of the denominators 1, 6 and 2 is 6

(1/1) = [(1×6)/ (1×6)] = (6/6)

(5/6) = [(5×1)/ (6×1)] = (5/6)

(1/2) = [(1×3)/ (2×3)] = (3/6)

Then,

(6/6) + (5/6) + (3/6) = ₹ 2,016

(6 + 5 + 3)/ 6 = ₹ 2,016

(14/6) = ₹ 2,016

So, Huma’s share = ₹ 2,016 ÷ (14/6)

= 2,016 × (6/14)

= 144 × 6

∴Huma’s Share is = ₹ 864

Seema’s share = ½ Huma’s share

= ½ × 864

= ₹ 432

Hubna’s share = 5/6 of Huma’s share

= 5/6 × 864

= 5 × 144

= ₹ 720

From the question,

Manavi and Kuber each receives and equal allowance = ₹ 1260

Let us assume total cost be ₹ 1

For Manavi, left over = Total cost – Total spends

= 1 – (½ + ¼)

= 1 – (2 + 1)/4

= 1 – (3/4)

= (4 – 3)/4

= ¼

So, Amount = 1260 × ¼ = ₹ 315

For Kuber, left over = Total cost – Total spends

= 1 – (1/3 + 3/5)

= 1 – (5 + 9)/15

= 1 – (14/15)

= (15 – 14)/15

= 1/15

So, Amount = 1260 × (1/15) = ₹ 84

The LCM of the denominators 25, 32, 40 and 20 is 800

∴ 1/25 = [(1×32)/ (25×32)] = (32/800)

(1/32) = [(1×25)/ (32×25)] = (25/800)

(1/40) = [(1×20)/ (40×20)] = (20/800)

(1/20) = [(1×40)/ (20×40)] = (40/800)

Then,

(a) Soni hop more than Nancy = (40/800) – (25/800)

= (40 – 25)/800

= (15/800)

= 3/160 km

(b) The total distance covered by Seema and Megha = (32/800) + (20/800)

= (32 + 20)/800

= (52/800)

= 13/200 km

(c) Nancy walked farther.

3/5 and 3/4 can be represented as 48/80 and 60/80 respectively.

Therefore, ten rational numbers between 3/5 and 3/4 are

49/80, 50/80, 51/50, 52/80, 53/80, 54/80, 55/80, 56/80, 57/80

Correct option is A) 2

D) can’t be determine

Correct option is  B) 1

Correct option is (D) 5/2

(A) \(\frac{1}{7}\) < 1 < 2

(B) \(\frac{1}{2}\) < 1 < 2

(C) \(\frac{2}{5}\) < 1 < 2

(D) \(\frac{5}{2}\) = 2.5 < 3

& \(\frac94\) = 2.25

\(\because\) 2.25 < 2.5 < 3

\(\Rightarrow\) \(\frac94\) < \(\frac{5}{2}\) < 3 which satisfies \(\frac{9}{4}<k<3\)

Therefore \(k=\frac{5}{2}.\)

Correct option is  D) 5/2

Correct option is  C) 0

(i) (-5/7) = (-25/35) = (-35/49)

Explanation:

Given (-5/7) = (…/35) = (…/49)

Here (-5/7) × (5/5) = (-25/35)

And also (-5/7) × (7/7) = (-35/49)

(ii) (-4/-9) = (8/18) = (12/27)

Explanation:

Given (-4/-9) = (…/18) = (12/…)

On multiplying by -2 we get

(-4/-9) × (-2/-2) = (8/18)

Also on multiplying by -3

(-4/-9) × (-3/-3) = (12/27)

(iii) (6/-13) = (-12/26) = (24/-52)

Explanation:

Given (6/-13) = (-12/…) = (24/…)

On multiplying by -2

(6/-13) × (-2/-2) = (-12/26)

Also multiplying by 4

And also (6/-13) × (4/4) = (24/-52)

(iv) (-6/-22) = (3/11) = (-15/-55)

Explanation:

Given (-6/…) = (3/11) = (…/-55)

0n multiplying by -2

(3/11) × (-2/-2) = (-6/-22)

And also on multiplying by -5

(3/11) × (-5/-5) = (-15/-55)

Correct option is  D) Infinite

Correct option is (D) none

\(\frac{1}{2}\times(\frac{-2}{3}+\frac{1}{4})\) \(=\frac{1}{2}\times(\frac{-2\times4+3}{12})=\frac{1}{2}\times(\frac{-8+3}{12})\)

\(=\frac{1}{2}\times\frac{-5}{12}=\frac{-5}{2\times12}=\frac{-5}{24}.\)

Correct option is  D) none

Correct option is (C) 0

\(\because\) \(\frac{1}{2}-\frac{1}{2}=0\)

\(\therefore\) \(\frac{7}{9}\times 1\frac{1}{2}\times8\frac{1}{17}\times(\frac{1}{2}-\frac{1}{2})\) \(=\frac{7}{9}\times 1\frac{1}{2}\times8\frac{1}{17}\times0=0.\)

Correct option is  C) 0

Correct option is (B) -3/14

\(\frac{5}{12}\div X=\frac{-35}{18}\)

\(\Rightarrow\) \(\frac{5}{12}\times\frac1X=\frac{-35}{18}\)

\(\Rightarrow\) \(X=\frac{5}{12}\times\frac{-18}{35}\) \(=\frac{-3}{2\times7}=\frac{-3}{14}.\)

Correct option is  B) -3/14

Correct option is  C) -2

Correct option is (B)  −16/15

\(\frac{-5}{6}\times x\) = \(\frac{8}{9}\)

\(\Rightarrow\) x = \(\frac{8}{9}\) \(\times\frac{-6}{5}\) \(=\frac{8\times-2}{3\times5}\) \(=\frac{-16}{15}.\)

Correct option is   B) \(\frac{-16}{15}\)

Correct option is (A) 5/9

\(\left((\frac{5}{9})^{-1}\right)^{-1}\) \(=(\frac{5}{9})^{-1\times-1}\) \(=(\frac{5}{9})^1\) \(=\frac{5}{9}.\)

Correct option is  A) 5/9

(16/-21) × (-14/5)

First we write (16/-21) in standard form

= (16×-1)/ (-21×-1)

= (-16/21)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (-16×-14)/ (21×5)

On simplifying,

= (-16×-2)/ (3×5)

= 32/15

Correct option is  D) xp – xq

Correct option is (D) -8

\((\frac{1}{-8})^{-1}\) \(=\left((-8)^{-1}\right)^{-1}\) \(=(-8)^{-1\times-1}=(-8)^1\) = -8.

Correct option is  D) -8

Correct option is  D) Division

Let the other number be x.

Then,

\(\frac{-15}{4}\times\text{x}=\frac{-16}{35}\)

\(\Rightarrow\) \(\text{x}=\frac{-16}{35}\div\frac{-15}{14}\)

\(\Rightarrow\) \(\text{x}=\frac{-16}{35}\times\frac{14}{-15}\)

\(\Rightarrow\) \(\text{x}=\frac{-224}{-525}=\frac{-224\times-1}{-525\times-1}=\frac{224}{525}\)

\(\Rightarrow\) \(\text{x}=\frac{224}{525}=\frac{224\div7}{525\div7}=\frac{32}{75}\)

Correct option is  B) 1

Correct option is (A) 4/7

\(\frac{-16}{21}\div\frac{-4}{3}\) \(=\frac{-16}{21}\times\frac{-3}{4}\) \(=-(-\frac{4}{7})\) \(=\frac{4}{7}.\)

Correct option is  A) 4/7

Correct option is (C) -7/36

\(\frac{5}{9}-\frac{3}{4}\) \(=\frac{5\times4-3\times9}{36}=\frac{20-27}{36}\) \(=\frac{-7}{36}.\)

Correct option is  C) -7/36

Correct option is  A) 0

Correct option is  D) -2/35

Correct option is (D)  16/15

\(\frac{16}{35}\div\frac{3}{7}\) \(=\frac{16}{35}\times\frac73=\frac{16}{5\times3}\) \(=\frac{16}{15}.\)

Correct option is   D) \(\frac{16}{15}\)