NCERT Exemplar Solutions: Rational Numbers- 2 | Mathematics (Maths) Class 8 PDF Download

Q.1. Select those which can be written as a rational number with denominator 4 in their lowest form:
(7/8), (64/16), (36/-12), (-16/17), (5/-4), (140/28)
Solution: Rational number with denominator 4 in their lowest form are,
64/16 = 16/4, 36/-12 = 12/-4, 5/-4, 140/28 =20/4

Q.2. Using suitable rearrangement and find the sum:
(a) (4/7) + (-4/9) + (3/7) + (-13/9)
Solution: First rearrange the rational numbers and add the numbers with same denominator.
= (4/7) + (3/7) – (4/9) – (13/9)
= ((4 + 3)/7) – ((4 + 13)/9)
= (7/7) – (17/9)
= 1 – (17/9)
= (9 – 17)/9
= -8/9

(b) -5 + (7/10) + (3/7) + (-3) + (5/45) + (-4/5)
Solution:
= -5 + (-3) + (7/10) + (-4/5) + (3/7) + (5/14)
= 8 + [(7-8)/10] + [(6 + 5)/14]
= – 8 – (1/10) + (11/14)
LCM of 1, 10 and 14 is 70
= (-560 – 7 + 55)/70
= -512/70
= – 256/35

Q.3. Verify – (-x) = x for
(i) x = 3/5
Solution:
– x = -3/5
– (-x) = – (-3/5)
X = 3/5

(ii) x = -7/9
Solution:
– x = – (-7/9)
-x = 7/9
– (-x) = – (7/9)
X = – 7/9

(iii) x = 13/-15
Solution:
– x = – (-13/15)
-x = 13/15
– (-x) = – (13/15)
X = -13/15

Q.4. Give one example each to show that the rational numbers are closed under addition, subtraction and multiplication. Are rational numbers closed under division? Give two examples in support of your answer.
Solution:-
Rational numbers are closed under addition:-
Example:- 5/4 + 1/2
The LCM of the denominators 4 and 2 is 4
∴ (5/4) = [(5×1)/ (4×1)] = (5/4)
and (1/2) = [(1×2)/ (4×2)] = (2/4)
Then,
= 5/4 + 2/4
= (5 + 2)/ 4
= 7/4
= 7/4 is a rational number
Rational numbers are closed under subtraction:-
Example:- 5/4 – 1/2
The LCM of the denominators 4 and 2 is 4
∴ (5/4) = [(5×1)/ (4×1)] = (5/4)
and (1/2) = [(1×2)/ (4×2)] = (2/4)
Then,
= 5/4 –2/4
= (5 – 2)/ 4
= 3/4
= 3/4 is a rational number
Rational numbers are closed under addition:-
Example:- 5/4 × 1/2
= 5/8 is a rational number.
For any rational number x, x ÷ 0 is not defined,
Hence not all rational numbers are closed under division. We can say that except zero, all rational numbers are closed under division.
Example, ¾ ÷ 4/5
= ¾ × 5/4
= 15/16 is a rational number.

Q.5. Verify the property x + y = y + x of rational numbers by taking
(a) x = ½, y = ½
Solution:-
In the question is given to verify the property = x + y = y + x
Where, x = ½, y = ½
Then, ½ + ½ = ½ + ½
LHS = ½ + ½
= (1 + 1)/2
= 2/2
= 1
RHS = ½ + ½
= (1 + 1)/2
= 2/2
= 1
By comparing LHS and RHS
LHS = RHS
∴ 1 = 1
Hence x + y = y + x

(b) x = -2/3, y = -5/6
Solution:
In the question is given to verify the property = x + y = y + x
Where, x = -2/3, y = -5/6
Then, -2/3 + (-5/6) = -5/6 + (-2/3)
LHS = -2/3 + (-5/6)
= -2/3 – 5/6
The LCM of the denominators 3 and 6 is 6
(-2/3) = [(-2×2)/ (3×2)] = (-4/6)
and (-5/6) = [(-5×1)/ (6×1)] = (-5/6)
Then,
= – 4/6 – 5/6
= (- 4 – 5)/ 6
= – 9/6
RHS = -5/6 + (-2/3)
= -5/6 – 2/3
The LCM of the denominators 6 and 3 is 6
(-5/6) = [(-5×1)/ (6×1)] = (-5/6)
and (-2/3) = [(-2×2)/ (3×2)] = (-4/6)
Then,
= – 5/6 – 4/6
= (- 5 – 4)/ 6
= – 9/6
By comparing LHS and RHS
LHS = RHS
∴ -9/6 = -9/6
Hence x + y = y + x

(c) x = -3/7, y = 20/21
Solution:
In the question is given to verify the property = x + y = y + x
Where, x = -3/7, y = 20/21
Then, -3/7 + 20/21 = 20/21 + (-3/7)
LHS = -3/7 + 20/21
The LCM of the denominators 7 and 21 is 21
(-3/7) = [(-3×3)/ (7×3)] = (-9/21)
and (20/21) = [(20×1)/ (21×1)] = (20/21)
Then,
= – 9/21 + 20/21
= (- 9 + 20)/ 21
= 11/21
RHS = 20/21 + (-3/7)
The LCM of the denominators 21 and 7 is 21
(20/21) = [(20×1)/ (21×1)] = (20/21)
and (-3/7) = [(-3×3)/ (7×3)] = (-9/21)
Then,
= 20/21 – 9/21
= (20 – 9)/ 21
= 11/21
By comparing LHS and RHS
LHS = RHS
∴ 11/21 = 11/21
Hence x + y = y + x

(d) x = -2/5, y = – 9/10
Solution:
In the question is given to verify the property = x + y = y + x
Where, x = -2/5, y = -9/10
Then, -2/5 + (-9/10) = -9/10 + (-2/5)
LHS = -2/5 + (-9/10)
= -2/5 – 9/10
The LCM of the denominators 5 and 10 is 10
(-2/5) = [(-2×2)/ (5×2)] = (-4/10)
and (-9/10) = [(-9×1)/ (10×1)] = (-9/10)
Then,
= – 4/10 – 9/10
= (- 4 – 9)/ 10
= – 13/10
RHS = -9/10 + (-2/5)
= -9/10 – 2/5
The LCM of the denominators 10 and 5 is 10
(-9/10) = [(-9×1)/ (10×1)] = (-9/10)
and (-2/5) = [(-2×2)/ (5×2)] = (-4/10)
Then,
= – 9/10 – 4/10
= (- 9 – 4)/ 10
= – 13/10
By comparing LHS and RHS
LHS = RHS
∴ -13/10 = -13/10
Hence x + y = y + x

Q.6. Simplify each of the following by using suitable property. Also name the property.
(a) [(½) × (¼)] + [(½) × 6]
Solution:
The arrangement of the given rational number is as per the rule of distributive law over addition.
Now take out ½ as common.
Then,
= ½ [¼ + 6]
= ½ [(1 + 24)/4]
= ½ [25/24]
= ½ × (25/24)
= 25/48

(b) [(1/5) × (2/15)] – [(1/5) × (2/5)]
Solution:
The arrangement of the given rational number is as per the rule of distributive law over subtraction.
Now take out 1/5 as common.
Then,
= 1/5 [(2/15) – (2/5)]
The LCM of the denominators 15 and 5 is 15
(2/15) = [(2×1)/ (15×1)] = (2/15)
and (2/5) = [(2×3)/ (5×3)] = (6/15)
= 1/5 [(2 – 6)/15]
= 1/5 [-4/15]
= (1/5) × (-4/15)
= -4/75

(c) (-3/5) × {(3/7) + (-5/6)}
Solution:
The arrangement of the given rational number is as per the rule of distributive law over addition.
= (-3/5) × {(3/7) + (-5/6)}
The LCM of the denominators 7 and 6 is 42
(3/7) = [(3×6)/ (7×6)] = (18/42)
and (-5/6) = [(-5×7)/ (6×7)] = (-35/42)
= -3/5 [(18 – 35)/42]
= -3/5 [-17/42]
= (-3/5) × (-17/42)
= 51/210 … [divide both denominator and numerator by 3]
= 17/30

Q.7. Tell which property allows you to compute
(1/5) × [(5/6) × (7/9)] as [(1/5) × (5/6)] × (7/9)
Solution: The arrangement of the given rational number is as per the rule of Associative property for Multiplication.

Q.8. Verify the property x × y = y × z of rational numbers by using
(a) x = 7 and y = ½
Solution:
In the question is given to verify the property = x × y = y × x
Where, x = 7, y = ½
Then, 7 × ½ = ½ × 7
LHS = 7 × ½
= 7/2
RHS = ½ × 7
= 7/2
By comparing LHS and RHS
LHS = RHS
∴ 7/2 = 7/2
Hence x × y = y × x

(b) x = 2/3 and y = 9/4
Solution:
In the question is given to verify the property = x × y = y × x
Where, x = 2/3, y = 9/4
Then, (2/3) × (9/4) = (9/4) × (2/3)
LHS = (2/3) × (9/4)
= (1/1) × (3/2)
= 3/2
RHS = (9/4) × (2/3)
= (3/2) × (1/1)
= 3/2
By comparing LHS and RHS
LHS = RHS
∴ 3/2 = 3/2
Hence x × y = y × x

(c) x = -5/7 and y = 14/15
Solution:
In the question is given to verify the property = x × y = y × x
Where, x = -5/7, y = 14/15
Then, (-5/7) × (14/15) = (14/15) × (-5/7)
LHS = (-5/7) × (14/15)
= (-1/1) × (2/3)
= -2/3
RHS = (14/15) × (-5/7)
= (2/3) × (-1/1)
= -2/3
By comparing LHS and RHS
LHS = RHS
∴ -2/3 = -2/3
Hence x × y = y × x

(d) x = -3/8 and y = -4/9
Solution:
In the question is given to verify the property = x × y = y × x
Where, x = -3/8, y = -4/9
Then, (-3/8) × (-4/9) = (-4/9) × (-3/8)
LHS = (-3/8) × (-4/9)
= (-1/2) × (-1/3)
= 1/6
RHS = (-4/9) × (-3/8)
= (-1/3) × (-1/2)
= 1/6
By comparing LHS and RHS
LHS = RHS
∴ 1/6 = 1/6
Hence x × y = y × x

Q.9. Verify the property x × (y × z) = (x × y) × z of rational numbers by using
(a) x = 1, y = -½ and z = ¼
Solution:
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

(b) x = 2/3, y = -3/7 and z = ½
Solution:
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, (2/3) × (-3/7 × ½) = ((2/3) × (-3/7)) × ½
LHS = (2/3) × (-3/7 × ½)
= (2/3) × (-3/14)
= -6/42
RHS = ((2/3) × (-3/7)) × ½
= (-6/21) × ½
= -6/42
By comparing LHS and RHS
LHS = RHS
∴ -6/42 = -6/42
Hence x × (y × z) = (x × y) × z

(c) x = -2/7, y = -5/6 and z = ¼
Solution:
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

Q.10. Verify the property x × (y + z) = x × y + x × z of rational numbers by taking.
(a) x = -½, y = ¾, z = ¼
Solution:
In the question is given to verify the property x × (y + z) = x × y + x × z
The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.
Then, (-½) × (¾ + ¼) = (-½ × ¾) + (-½ × ¼)
LHS = (-½) × (¾ + ¼)
= (-½) × ((3 + 1)/4)
= -½ × (4/4)
= -½ × 1
= -½
RHS = (-½ × ¾) + (-½ × ¼)
= (-3/8) + (-1/8)
= (-3 – 1)/8
= -4/8
= -½
By comparing LHS and RHS
LHS = RHS
∴ -½ = -½
Hence x × (y + z) = x × y + x × z

(b) x = -½, y = 2/3, z = ¾
Solution:
In the question is given to verify the property x × (y + z) = x × y + x × z
The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.
Then, (-½) × ((2/3) + ¾) = (-½ × (2/3)) + (-½ × ¾)
LHS = (-½) × ((2/3) + ¾)
= (-½) × ((8 + 9)/12)
= -½ × (17/12)
= -17/24
RHS = (-½ × (2/3)) + (-½ × ¾)
= (-1/3) + (-3/8)
= (-8 – 9)/24
= -17/24
By comparing LHS and RHS
LHS = RHS
∴ -17/24 = -17/24
Hence x × (y + z) = x × y + x × z

(c) x = -2/3, y = -4/6, z = -7/9
Solution:
In the question is given to verify the property x × (y + z) = x × y + x × z
The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.
Then, (-2/3) × ((-4/6) + (-7/9)) = ((-2/3) × (-4/6)) + ((-2/3) × (-7/9))
LHS = (-2/3) × ((-4/6) + (-7/9))
= (-2/3) × ((-12 – 14)/18)
= – (2/3) × (-26/18)
= – (1/3) × (-26/9)
= 26/27
RHS = ((-2/3) × (-4/6)) + ((-2/3) × (-7/9))
= ((-1/3) × (-4/3)) + ((-2/3) × (-7/9))
= (4/9) + (14/27)
= (12 + 14)/27
= 26/27
By comparing LHS and RHS
LHS = RHS
∴ 26/27 = 26/27
Hence x × (y + z) = x × y + x × z

(d) x = -1/5, y = 2/15, z = -3/10
Solution: 
In the question is given to verify the property x × (y + z) = x × y + x × z
The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.
Then, (-1/5) × ((2/15) + (-3/10)) = ((-1/5) × (2/15)) + ((-1/5) × (-3/10))
LHS = (-1/5) × ((2/15) + (-3/10))
= (-1/5) × ((4 – 9)/30)
= (-1/5) × (-5/30)
= (-1/1) × (-1/30)
= 1/30
RHS = ((-1/5) × (2/15)) + ((-1/5) × (-3/10))
= (-2/75) + (3/50)
= (-4 + 9)/150
= 5/150
= 1/30
By comparing LHS and RHS
LHS = RHS
∴ 1/30 = 1/30
Hence x × (y + z) = x × y + x × z

Q.11. Use the distributivity of multiplication of rational numbers over addition to simplify.
(a) (3/5) × [(35/24) + (10/1)]
Solution:
We know that the distributivity of multiplication of rational numbers over addition, a × (b + c) = a × b + a × c
Where, a =3/5, b =35/24, c = 10/1
Then, (3/5) × [(35/24) + (10/1)] = ((3/5) × (35/24)) + ((3/5) × (10/1))
= ((1/1) × (7/8)) + ((3/1) × (2/1))
= (7/8) + (6/1)
= (7 + 48)/8
= 55/8
=

(b) (-5/4) × [(8/5) + (16/15)]
Solution:
We know that the distributivity of multiplication of rational numbers over addition, a × (b + c) = a × b + a × c
Where, a =-5/4, b =8/5, c = 16/15
Then, (-5/4) × [(8/5) + (16/15)] = ((-5/4) × (8/5)) + ((-5/4) × (16/15))
= ((-1/1) × (2/1)) + ((-1/1) × (4/3))
= (-2/1) + (-4/3)
= (-6 – 4)/3
= -10/3
=-3 1/3

(c) (2/7) × [(7/16) – (21/4)]
Solution:
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, (2/7) × [(7/16) – (21/4)] = ((2/7) × (7/16)) – ((2/7) × (21/4))
= ((1/1) × (1/8)) – ((1/1) × (3/2))
= (1/8) – (3/2)
= (1 – 12)/8
= -11/8

(d) ¾ × [(8/9) – 40]
Solution:
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

Q.12. Simplify
(a) (32/5) + (23/11) × (22/15)
Solution:
= (32/5) + (23/1) × (2/15)
= (32/5) + (46/15)
= (96 + 46)/15
= 142/15

(b) (3/7) × (28/15) ÷ (14/5)
Solution:-
= (3/7) × (28/15) ÷ (14/5)
= (1/1) × (4/5) ÷ (14/5)
= (4/5) ÷ (14/5)
= (4/5) × (5/14)
= (2/1) × (1/7)
= 2/7

(c) (3/7) + (-2/21) × (-5/6)
Solution:
= (3/7) – (2/21) × (-5/6)
= (3/7) – (1/21) × (-5/3)
= (3/7) – (-5/63)
= (3/7) + (5/63)
= (27 + 5)/63
= 32/63

(d) (7/8) + (1/6) – (1/12)
Solution:
= (7/8) + (1/6) – (1/12)
= ((14 + 1)/16) – (1/12)
= (15/16) – (1/12)
= (45-4)/48
= 41/48

Q.13. Identify the rational number that does not belong with the other three. Explain your reasoning (-5/11), (-1/2), (-4/9), (-7/3)
Solution: The rational number that does not belong with the other three is -7/3 as it is smaller than –1 whereas rest of the numbers are greater than –1.

Q.14. The cost of 19/4 metres of wire is ₹ 171/2. Find the cost of one metre of the wire.
Solution:-
From the question it is given that,
The cost of 19/4 meters of wire is = ₹ 171/2
Then, cost of one meter of wire = (171/2) ÷ (19/4)
= (171/2) × (4/19)
= (9/1) × (2/1)
= 18/1
= ₹ 18
∴ The cost of one meter of wire is ₹ 18.

Q.15. A train travels 1445/2 km in 17/2 hours. Find the speed of the train in km/h.
Solution:-
From the question it is given that,
Distance travelled by train = 1445/2 km
Time taken by the train to cover distance 1445/2 = 17/2 hours
The speed of the train = (1445/2) ÷ (17/2)
= (1445/2) × (2/17)
= (85/1) × (1/1)
= 85 km/h
∴ The speed of the train is 85 km/h.

Q.16. If 16 shirts of equal size can be made out of 24m of cloth, how much cloth is needed for making one shirt?
Solution:
From the question it is given that,
The 16 shirts are made out of= 24m of cloth.
Cloth needed for making one shirt = 24/16 m of cloth
= 3/2m of cloth i.e. 1.5m
So, Cloth is needed for making one shirt is 1.5m.

Q.17. 7/11 of all the money in Hamid’s bank account is ₹ 77,000. How much money does Hamid have in his bank account?
Solution:
From the question, it is given that
7/11 of all the money in Hamid’s bank account = ₹ 77,000
Now, let us assume money in Hamid’s bank account be ₹ x.
Then,
(7/11) × (x) = 77,000
x = 77,000/ (7/11)
x = 77000 × (11/7)
x = 11000 × (11/1)
x = 121000
∴ The total money in Hamid’s bank account is ₹ 121000.

Q.18. A  m long rope is cut into equal pieces measuring  m each. How many such small pieces are these?
Solution: 
From the question it is given that,
The length of the rope = m
= (117 × 3 + 1)/3
= 352/3m
Then length of each piece measures =m
= 22/3 m
So, the number of pieces of the rope = total length of the rope/ length of each piece
= (352/3)/ (22/3)
= (352/3) × (3/22)
= (16/1) × (1/1)
= 16
Hence, number of small pieces cut from the  m long rope is 16.

Q.19. 1/6 of the class students are above average, ¼ are average and rest are below average. If there are 48 students in all, how many students are below average in the class?
Solution:
From the question it is given that,
Number of students in the class are above average = 1/6
Number of students in the class are average = ¼
Number of students in the class are below average = 1 – ((1/6) + (¼))
= 1 – ((2 + 3)/12)
= 1 – (5/12)
= (12 – 5)/12
= 7/12 students.
So, the number of students in the class = 48
Then,
Number of students below average = (7/12) × 48
= (7/1) × 4
= 28 students
∴ The number of students in the class are below average are 28.

Q.20. 2/5 of total number of students of a school come by car while ¼ of students come by bus to school. All the other students walk to school of which 1/3 walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in that school?
Solution:
Let us assume total number of students in the school be x.
From the question it is given that,
The number of students come by car = (2/5) × x
The number of students come by bus = (¼) × x
Remaining students walk to school = x – ((2x/5) + (¼x))
= x – ((8x – 5x)/20)
= x – (13x/20)
= (20x – 13x)/20
= 7x/20
Then, number of students walk to school on their own = (1/3) of (7x/20)
= 7x/60
Since, 224 students come to school on their own.
As per the data given in the question,
= (7x/60) = 224
x = (224 × 60)/7
x = 32 × 60
x = 1920
∴ The total number of students in that school is 1920.

Q.21. Huma, Hubna and Seema received a total of ₹ 2,016 as monthly allowance from their mother such that Seema gets ½ of what Huma gets and Hubna gets times Seema’s share. How much money do the three sisters get individually
Solution:
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 =  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

Q.22. A mother and her two daughters got a room constructed for ₹ 62,000. The elder daughter contributes 3/8 of her mother’s contribution while the younger daughter contributes ½ of her mother’s share. How much do the three contribute individually?
Solution:
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

Q.23. Tell which property allows you to compare
(2/3) × [¾ × (5/7)] and [(2/3) × (5/7)] × ¾
Solution: (2/3) × [¾ × (5/7)] and [(2/3) × (5/7)] × ¾ this can be compared with associative property and commutative property.

Q.24. Name the property used in each of the following.
(i) (-7/4) × (-3/4) = (-3/5) × (-7/11)
Solution: The above rational number is in the form of Commutative property over multiplication.

(ii) (-2/3) × [(3/4) + (-½)] = [(-2/3) × (3/4)] + [(-2/3) × (½)]
Solution: The above rational number is in the form of Distributive property over addition

(iii) (1/3) + [(4/9) + (-4/3)] = [(1/3) + (4/9)] + [-4/3]
Solution: The above rational number is in the form of Associative property over addition

(iv) (-2/7) + 0 = 0 + (-2/7) = (-2/7)
Solution: The above rational number is in the form of Additive identity of rational number

(v) (3/8) × 1 = 1 × (3/8) = (3/8)
Solution: The above rational number is in the form of Multiplicative identity of rational number.

Q.25. Find the multiplicative inverse of
(i)
Solution: The given number  can be written as = -9/8
The multiplicative inverse = -8/9

(ii)  
Solution: can be written as = 10/3
The multiplicative inverse = 3/10

Q.26. Arrange the numbers ¼, 13/16, 5/8 in the descending order.
Solution:
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, 1/4

Q.27. The product of two rational numbers is -14/27. If one of the numbers be 7/9, find the other.
Solution:
Let us assume the other number be y.
Given, product of two rational number = -14/27
One number = 7/9
Then,
= y × (7/9) = -14/27
= y = (-14/27)/ (7/9)
= y = (-14/27) × (9/7)
= y = (-2/3) × (1/1)
= y = -2/3
So, the other number is -2/3

Q28. By what numbers should we multiply -15/20 so that the product may be -5/7
Solution:
Let us assume the other number be y.
Given, product of two rational number = -5/7
One number = -15/20
Then,
= y × (-15/20) = -5/7
= y = (-5/7)/ (-15/20)
= y = (-5/7) × (-20/15)
= y = (-1/7) × (-20/3)
= y = -20/21
So, the other number is -20/21

Q.29. By what number should we multiply -8/13 so that the product may be 24
Solution:
Let us assume the other number be y.
Given, product of two rational number = 24
One number = -8/13
Then,
= y × (-8/13) = 24
= y = 24/ (-8/13)
= y = (24/1) × (-13/8)
= y = (3/1) × (-13/1)
= y = -39
So, the other number is -39

Q.30. The product of two rational numbers is –7. If one of the number is –5, find the other?
Solution:
Let us assume the other number be y.
Given, product of two rational number = -7
One number = -5
Then,
= y × (-5) = -7
= y = -7/ (-5)
= y = 7/5
So, the other number is 7/5

Q.31. Can you find a rational number whose multiplicative inverse is –1?
Solution: No, we cannot find a rational number whose multiplicative inverse is –1.

Q.32. Find five rational numbers between 0 and 1.
Solution: The five rational numbers between 0 and 1 are, 1/6, 2/6, 3/6, 4/6, 5/6.

Q.33. Find two rational numbers whose absolute value is 1/5.
Solution: 1/5 and -1/5 are the rational number whose absolute value is 1/5.

Q.34. From a rope 40 metres long, pieces of equal size are cut. If the length of one piece is 10/3 metre, find the number of such pieces.
Solution:
From the question it is given that,
The length of rope = 40 m
The length of one piece of rope = 10/3
Let us assume the total number of pieces be y.
So,
(10/3) y = 40
y = (40 × 3)/10
y = 120/10
y = 12 pieces
∴ The number of pieces cut from the rope are 12.

Q.35. 5½ metres long rope is cut into 12 equal pieces. What is the length of each piece?
Solution:
From the question it is given that,
The length of rope = 5½ m = 11/2 m
The total number of pieces = 12
Let us assume the length of one piece of rope be y.
So,
12y = 11/2 m
y = (11/2) × (1/12)
y = 11/24
∴ The length of one piece of rope 11/24.

Q.36. Write the following rational numbers in the descending order.
(8/7), (-9/8), (-3/2), 0, (2/5)
Solution:-
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

Q.37. Find (i) 0 ÷ (2/3)
Solution:
0 ÷ (2/3) = 0 × (3/2)
= 0/2
= 0

Q.38. On a winter day the temperature at a place in Himachal Pradesh was –16°C. Convert it in degree Fahrenheit (^oF) by using the formula.
(C/5) = (F – 32)/9
Solution:
Given, a winter day the temperature at a place in Himachal Pradesh was –16°C.
Formula, (C/5) = (F – 32)/9
Where, C = -16^o
Then,
(-16^o/5) = (F – 32)/9
(-16^o/5) × 9 = F -32
(-144/5) = F – 32
F = 32 – (144/5)
F = (160 – 144)/5
F = 16/5
F = 3.2 ^oF

Q.39. Find the sum of additive inverse and multiplicative inverse of 7.
Solution:
Additive inverse of 7 = – 7
Multiplicative inverse of 7 = 1/7
Then,
Sum of additive inverse and multiplicative inverse of 7 = -7 + (1/7)
= (-49 + 1)/7
= – 48/7

Q.40. Find the product of additive inverse and multiplicative inverse of 1/3.
Solution:
Additive inverse of -1/3 = 1/3
Multiplicative inverse of -1/3 = -3/1
Then,
The product of additive inverse and multiplicative inverse of 1/3 = 1/3 × (-3)
= -1

Q.41. The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:

(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull?
Solution: We have to find out the difference of wingspan of an Albatross and wingspan of a Sea gull.
Length of wingspan of an Albatross =
 = 18/5 m 

Length of wingspan of a Sea gull =
 = 17/10 m
Difference of both = (18/5) – (17/10)
= (36 – 17)/ 10
= 19/10 m
∴ The wingspan of an Albatross is 19/10 m longer than the wingspan of a Sea gull.

(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay?
Solution: We have to find out the difference of wingspan of a Golden eagle and wingspan of a Blue jay.
Length of wingspan of a Golden eagle = 2½ = 5/2 m
Length of wingspan of a Blue jay = 41/100 m
Difference of both = (5/2) – (41/100)
= (250 – 41)/ 100
= 209/100 m
∴ The wingspan of a Golden eagle is 209/100 m longer than the wingspan of a Blue jay.

Q.42. Shalini has to cut out circles of diameter 1¼ cm from an aluminum strip of dimensions 8¾ cm by 1¼ cm. How many full circles can Shalini cut? Also calculate the wastage of the aluminum strip.

Solution:
From the question it is given that,
Diameter of the circle = Breadth of the aluminium strip 1¼ cm = 5/4 cm
Length of aluminium strip = 8¾ cm = 35/4 cm
∴ The number of full circles cut from the aluminum strip = (35/4) ÷ (5/4)
= (35/4) × (4/5)
= (7/1) × (1/1)
= 7 circles
Radius of circle = (5/ (4 × 2)) = 5/8 cm
Area to be cut by one circle = πr²
= (22/7) × (5/8)²
= (22/7) × (25/64) cm²
Now, area to be cut by 7 full circles = 7 × (22/7) × (25/64)
= (22 × 25)/64
= 550/64 cm²
Area of the aluminum strip = length × breadth
= (35/4) × (5/4) cm²
= (175/16) cm²
∴ The wastage of aluminum strip = (175/16) – (550/64)
= (700 – 550)/64
= 150/64
= 75/32 cm²

Q.43. One fruit salad recipe requires ½ cup of sugar. Another recipe for the same fruit salad requires 2 tablespoons of sugar. If 1 tablespoon is equivalent to 1/16 cup, how much more sugar does the first recipe require?
Solution:
From the question it is given that,
One fruit salad recipe requires = ½ cup of sugar
Sugar required for another salad = 2 × (1/16) = 2/16 cup
Hence, the required sugar = ½ – (2/16)
= (8 – 2)/16
= 6/16
= 3/8 cup of sugar.

Q.44. Four friends had a competition to see how far could they hop on one foot. The table given shows the distance covered by each.

(a) How farther did Soni hop than Nancy?
(b) What is the total distance covered by Seema and Megha?
(c) Who walked farther, Nancy or Megha?
Solution:
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.

Q.45. The table given below shows the distances, in kilometers, between four villages of a state. To find the distance between two villages, locate the square where the row for one village and the column for the other village intersect.

(a) Compare the distance between Himgaon and Rawalpur to Sonapur and Ramgarh?
Solution: From the table the distance between Himgaon and Rawalpur = 98¾ km = 395/4 km
The distance between Sonapur and Ramgarh =
 = 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
=  km

(b) If you drove from Himgaon to Sonapur and then from Sonapur to Rawalpur, how far would you drive?
Solution: From the table,
Distance between Himgaon and Sonapur =  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
=  km

Q.46. The table shows the portion of some common materials that are recycled.

(a) Is the rational number expressing the amount of paper recycled more than ½ or less than ½?
Solution: The rational number expressing the amount of paper recycled is less than ½.

(b) Which items have a recycled amount less than ½?
Solution: Paper and Glass have a recycled amount less than ½.

(c) Is the quantity of aluminium cans recycled more (or less) than half of the quantity of aluminium cans?
Solution: The quantity of aluminium cans recycled is more than half of the quantity of aluminium cans.

(d) Arrange the rate of recycling the materials from the greatest to the smallest.
Solution:-
The LCM of the denominators 11, 8, 5 and 4 is 440
∴ 5/11 = [(5 × 40)/ (11 × 40)] = (200/440)
(5/8) = [(5 × 55)/ (8 × 55)] = (275/440)
(2/5) = [(2 × 88)/ (5 × 88)] = (176/440)
(3/4) = [(3 × 110)/ (4 × 110)] = (330/440)
Then,
Now, 330 > 275 > 200 > 176
Hence, ⇒ 3/4 > 5/8 > 5/11 > 2/5
∴ Scrap > Aluminium > Cans > Paper > Glass.

Q.47. The overall width in cm of several wide-screen televisions are 97.28 cm,  cm,  cm and 97.94 cm. Express these numbers as rational numbers in the form p/q and arrange the widths in ascending order.
Solution: 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,
= 2432/25cm
 cm = by converting mixed fraction into improper fraction we get,
= 886/9 cm
 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,
= 4897/50cm
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 <  cm <  cm

Q.48. Roller Coaster at an amusement park is 2/3m high. If a new roller coaster is built that is 3/5 times the height of the existing coaster, what will be the height of the new roller coaster?
Solution:
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

Q.49. Here is a table which gives the information about the total rainfall for several months compared to the average monthly rains of a town. Write each decimal in the form of rational number p/q.

Solution:
(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

Q.50. The average life expectancies of males for several states are shown in the table. Express each decimal in the form p/q and arrange the states from the least to the greatest male life expectancy. State-wise data are included below; more indicators can be found in the “FACTFILE” section on the homepage for each state.

Source: Registrar General of India (2003) SRS Based Abridged Lefe Tables. SRS Analytical Studies, Report No. 3 of 2003, New Delhi: Registrar General of India. The data are for the 1995-99 period; states subsequently divided are therefore included in their pre-partition states (Chhatisgarh in MP, Uttaranchal in UP and Jharkhand in Bihar)
Solution:

Kerala; Punjab; HP; Maharashtra; Haryana; Tamil Nadu; West Bengal; Karnataka; Gujarat; Andhra Pradesh; Bihar; Rajasthan; UP; Orissa; Assam; MP

Q.51. A skirt that is  cm long has a hem of  cm. How long will the skirt be if the hem is let down?
Solution: From the question it is given that,
Length of the skirt =
 cm = 287/8 cm
Dimension of hem =  cm = 25/8 cm
Length of skirt, if hem is let down = ((287/8) + (25/8)) cm
= 312/8 cm
= 39 cm

Q.52. Manavi and Kuber each receives an equal allowance. The table shows the fraction of their allowance each deposits into his/her saving account and the fraction each spends at the mall. If allowance of each is ₹ 1260 find the amount left with each.
Solution:

Solution:
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