Chapter 1: Rational Numbers

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.

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

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.

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.

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

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.

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

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

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

Solution:-

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