Chapter 1: Rational Numbers

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

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

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.

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.

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.

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.

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.

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.

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.

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.