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(i) Given (3/10) ÷ (10/3)

= (3 × 3)/ (10 × 10)

= (9/100)

(ii) Given 4 (3/5) ÷ (4/5)

First convert the given mixed fractions into improper fraction

4 (3/5) = (23/5)

(23/5) ÷ (4/5) = (23 × 5)/ (5 × 4)

= (23/4)

= 5 (3/4)

(iii) Given 5 (4/7) ÷ 1 (3/10)

First convert the given mixed fractions into improper fraction

(39/7) and (13/10)

(39/7) ÷ (13/10) = (39 × 10)/ (7 × 13)

= (390/91)

= (30/7)

= 4 (2/7)

(iv) Given 4 ÷ 2 (2/5)

First convert the given mixed fractions into improper fraction

2 (2/5) = (12/5)

4 ÷ (12/5) = (4 × 5)/12

= (20/12)

= 1 (2/3)

Given area of rectangular plot is 65 (1/3) m² = (196/3) m²

Length of the same plot is 12 (1/4) m = (49/4) m

Width of the plot is

Area = length × breadth

(196/3) = (25/2) × breadth

Breadth = (196/3)/ (25/2)

= (196 × 4)/ (49 × 3)

= (184/147)

= 5 (3/4)

[2(5/9)] – [1(7/15)]

Convert mixed fraction into improper fraction, and then find the difference.

= [2(5/9)] = (23/9)

= [1(7/15)] = (22/15)

We get,

= (23/9) – (22/15)

LCM of 9, 15 = 45

Now, let us change each of the given fraction into an equivalent fraction having 45 as the denominator.

= [(23/9) × (5/5)] = (115/45)

= [(22/15) × (3/3)] = (66/45)

Then,

= (115/45) – (66/45)

= (115 -66)/45

= (49/45)

= [1(4/45)]

[8(5/6)] – [3(3/8)] + [1(7/12)]

First convert each mixed fraction into improper fraction.

We get,

= [8(5/6)] = (53/6)

= [3(3/8)] = (27/8)

= [1(7/12)] = (19/12)

Then,

(53/6)- (27/8) – (19/12)

LCM of 6, 8, 12 = 24

Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator.

= [(53/6) × (4/4)] = (212/24)

= [(27/8) × (3/3)] = (81/ 24)

= [(19/12) × (2/2)] = (38/24)

Now,

= (212/24) – (81/ 24) – (38/24)

= [(212- 81+ 38)/24]

= [(250-81/24)]

= (169/24)

= [7(1/24)]

(9) ÷ (1/3)

We have,

= (9/1) ÷ (1/3)

= (9/1) × (3/1)

(Because reciprocal of (1/3) is (3/1)

= (9 × 3) / (1 × 1)

= 27

[6(7/8)] by (11/16)

The above question can be written as,

= [6(7/8)] ÷ (11/16)

Convert mixed fraction into improper fraction,

= [6(7/8)] = (55/8)

We have,

= (55/8) × (16/11)

(Because reciprocal of (11/16) is (16/11)

= (55 × 16) / (8 × 11)

= (5 × 2) / (1 × 1)

= 10

63 by [2(1/4)]

The above question can be written as,

= 63 ÷ [2(1/4)]

Convert mixed fraction into improper fraction,

= [2(1/4)] = (9/4)

We have,

= (63/1) × (4/9)

(Because reciprocal of (9/4) is (4/9)

= (63 × 4) / (1 × 9)

= (7 × 4) / (1 × 1)

= 28

Given total length of the wire is = 12 (1/2) = (25/2) m

It is cut into 10 pieces, so length of one piece is

= (25/2)/10

= (25/20)

= (5/4)

= 1 (1/4) m

(24) ÷ (6/7)

We have,

= (24/1) ÷ (6/7)

= (24/1) × (7/6)

(Because reciprocal of (6/7) is (7/6)

= (24 × 7) / (1 × 6)

= (4 × 7) / (1 × 1)

= 28

(11/24) by (7/8)

The above question can be written as,

= (11/24) ÷ (7/8)

We have,

= (11/24) × (8/7)

(Because reciprocal of (7/8) is (8/7)

= (11 × 8) / (24 × 7)

= (11 × 1) / (3 × 9)

= (11/21)

From the question,

Cost for 1 orange = ₹ [6(3/4)] = (27/4)

Man gets = ₹ 378

Then we have,

= (378/1) ÷ (27/4)

= (378/1) × (4/27)

(Because reciprocal of (27/4) is (4/27)

= (378 × 4) / (1 × 27)

= (42×4) / (1×3)

= (14×4) / (1×1)

= 56

Hence, the man sold 56 orange.

From the question,

Rope length = [13(1/2)] m = (27/2)

Number of equal pieces divided into = 9

Then we have,

= (27/2) ÷ (9/1)

= (27/2) × (1/9)

(Because reciprocal of (9/1) is (1/9)

= (27 × 1) / (2 × 9)

= (3×1) / (2×1)

= (3 / 2)

= [1(1/2)] m

Hence, the length of 9 pieces of rope is [1(1/2)] m

1/16 kg chocolates to be filled in 1 box.

∴ 1 kg chocolates to be filled in 1 ÷ 1/16 = 1 × 16 = 16 boxes.

Now, 1 1/2 kg chocolates to be filled in 16 x 1 1/2 = 16 x 3/2 = 24 boxes.

From the question,

Total weight of boxes= [49(1/2)] kg = (99/2)

Number of boxes = 18

Then we have,

= (99/2) ÷ (18/1)

= (99/2) × (1/18)

(Because reciprocal of (18/1) is (1/18)

= (99 × 1) / (2 × 18)

= (11×1) / (2×2)

= (11 / 4)

= [2(3/4)] kg

Hence, the weight of each box is [2(3/4)] kg

(7/10) ÷ (3/5)

We have,

= (7/10) ÷ (3/5)

= (7/10) × (5/3)

(Because reciprocal of (3/5) is (5/3)

= (7 × 5) / (10 × 3)

= (7 × 1) / (2 × 3)

= (7/6)

= [1(1/6)]

[5(4/7)] ÷ [1(3/10)]

Convert mixed fraction into improper fraction,

= [5(4/7)] = (39/7)

= [1(3/10)] = (13/10)

We have,

= (39/7) ÷ (13/10)

= (39/7) × (10/13)

(Because reciprocal of (13/10) is (10/13)

= (39 × 10) / (7 × 13)

= (390) / (91)

= (30 / 7)

= [4(2/7)]

[15(3/7)] ÷ [1(23/49)]

Convert mixed fraction into improper fraction,

= [15(3/7)] = (108/7)

= [1(23/49)] = (72/49)

We have,

= (108/7) ÷ (72/49)

= (108/7) × (49/72)

(Because reciprocal of (72/49) is (49/72)

= (108 × 49) / (7 × 72)

= (9×7) / (1 × 6)

= (3×7) / (1×2)

= (21/2)

= [10(1 /2)]

[3(3/7)] ÷ (8/21)

Convert mixed fraction into improper fraction,

= [3(3/7)] = (24/7)

We have,

= (24/7) ÷ (8/21)

= (24/7) × (21/8)

(Because reciprocal of (8/21) is (21/8)

= (24 × 21) / (7 × 8)

= (3 × 3) / (1 × 1)

= 9

(4/7) ÷ (9/14)

We have,

= (4/7) ÷ (9/14)

= (4/7) × (14/9)

(Because reciprocal of (9/14) is (14/9)

= (4 × 14) / (7 × 9)

= (4 × 2) / (1×9)

= (8/9)

(i) True

(ii) False

(iii) True

(iv) True

(v) False

(2/3), (3/5), (7/10), (8/15)

LCM of 3, 5, 10, 15 = 3 × 5 × 2 = 30

Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator.

[(2/3) × (10/10)] = (20/30)

[(3/5) × (6/6)]= (18/30)

[(7/10) × (3/3) = (21/30)

[(8/15) × (2/2)] = (16/30)

Clearly,

(21/30) > (20/30) > (18/30) > (16/30)

Hence,

(7/10) > (2/3) > (3/5) > (8/15)

Hence, the given fractions in descending order are (7/10), (2/3), (3/5), (8/15)

(3/4), (7/8), (7/12), (17/24)

LCM of 4, 8, 12, 24 = 4 × 2 × 3 = 24

Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator.

[(3/4) × (6/6)] = (18/24)

[(7/8) × (3/3)] = (21/24)

[(7/12) × (2/2)] = (14/24)

[(17/24) × (1/1)] = (17/24)

Clearly,

(21/24) > (18/24) > (17/24) > (14/24)

Hence,

(7/8) > (3/4) > (17/24) > (7/12)

Hence, the given fractions in descending order are (7/8), (3/4), (17/24), (7/12)

(1) 3/5 > 3/7

(2) 9/15 < 9/11

(3) 4/7 < 4/5

(4) 5/12 < 7/12

(5) 6/17 > 3/17

(6) 5/19 < 11/19

(7) 12/21 < 12/15

(8) 11/17 < 11/15

(9) 6/11< 6/15

(10) 14/23 < 5/23

(11) 17/20 > 12/20

(12) 11/15 > 8/15

(3/4), (5/6), (7/9), (11/12)

LCM of 4, 6, 9, 12 = 2× 2× 3× 3=36

Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator.

[(3/4) × (9/9)] = (27/36)

[(5/6) × (6/6)] = (30/36)

[(7/9) × (4/4)] = (28/360

[(11/12) × (3/3)] = (33/36)

Clearly,

(27/36) < (28/36) < (30/36) < (33/36)

Hence,

(3/4) < (7/9) < (5/6) < (11/12)

Hence, the given fractions in ascending order are (3/4), (7/9), (5/6), (11/12)

(i) 6/7< 8/7

(ii) 5/8 > 3/8

(iii) 6/13 > 6/17

(iv) 5/22 > 3/22

(v) 9/47 < 9/42

(11/12) and (15/16)

By cross multiplication, we have:

11 × 16 = 176 and 12 × 15 = 180

But,

176 < 180

∴ (11/12) < (15/16)

(5/9) and (11/15)

By cross multiplication, we have:

5 × 15 = 75 and 9 × 11 = 99

But,

75 < 99

∴ (5/9) < (11/15)

(5/8)and (7/12)

By cross multiplication, we have:

5 × 12 = 60 and 8 × 7 = 56

But,

60 > 56

∴ (5/8) > (7/12)

The part of the work finished by Ramu in 1 hour = 1/3

So, the part of the work finished by Ramu in 2 1/5 hours 

= 2 1/5 x 1/3 = 11/5 x 1/3 = (11 x 1)/(5 x 

3) = 11/15

Ramu will finish 11/15 part of the work in 2 1/5 hours

False

Reciprocal of an improper fraction is an improper fraction.

From the question it is given that,

Renu completed 2/3 part of her home work in 2 hours.

Let us assume total part of homework be ‘P’.

Then,

2/3 of P = 2

2/3 × P = 2

By cross multiplication we get,

P = 2 × 3/2

P = 3 home work

So, part of her home work had she completed in 1¼ hours i.e. 5/4 hours

Let us assume part of home work she had completed in 5/4 hours be = Q

Then,

Q of 3 = 5/4

Q × 3 = 5/4

By cross multiplication we get,

Q = (5/4) × (1/3)

Q = 5/12

Therefore, Reenu completed 5/12 part of her home work in 5/4 hours.

Side of a square = 6 (2/3) = (20/3) m

Area of square = side × side

= (20/3) × (20/3)

= (400/9)

=44 (4/9) m²

Given (3/10) liter of milk is given to each student

Number of student given (3/10) liter of milk = 1

Number of students giving 1 liter of milk = (10/3)

Numbers of students giving 30 liters of milk = (10/3) × 30 = 100 students

(c) (9/7). 

Because, its numerator is more than its denominator.

From the question,

The total distance covered by a car in 1 hour = [66(2/3)] km = (200/3)

Therefore, the distance covered by a car in 9 hour = (200/3) × 9

Then,

= (200/3) × (9/1)

= (200 × 9) / (3 × 1)

On simplifying we get,

= (200× 3) / (1 × 1)

= 600 km

Hence, the distance covered by a car in 9 hour is 600 km.

(c) (10/3). 

Because, its denominator is a whole number, other than 10, 100, 10000 etc.

A sugar bag contains 30kg of sugar.

After consuming, the left sugar in the bag is = 30- (2/3) × 30

= 30 – 2 × 10

= 30 – 20

= 10kg

Given the cost of milk per liter is = 17(3/4) = Rs (71/4)

And the cost of 7 (2/5) = (37/5) is

= (37/ 5) × (71/4)

= (37 × 71)/20

= (2327/20)

= Rs 131 (7/20)

Given distance covered by Sharada in one hour = 8 (1/3) = (25/3) km

Distance covered by her in 2 (2/5) hours = (12/5) is

= (25/3) × (12/5)

= (25 × 12)/15

= (300/15)

= 20 km

From the question,

Distance covered by vikas in [7(3/4)] hours on foot = [20(2/3)] km = (62/3)

Distance covered by vikas in 1 hour = (62/3) ÷ (31/4)

Then we have,

= (62/3) × (4/31)

(Because reciprocal of (31/4) is (4/31)

= (62 × 4) / (3 × 31)

= (2×4) / (3×1)

= (8) / (3)

= [2(2/3)] km

Hence, Distance covered by vikas in 1 hour is [2(2/3)] km

Let money earned by Kajol be ₹ x.

According to question,

3/8 x x = 75

x = 75 x 8/3

⇒ x = 200

∴ The amount earned by Kajol is ₹ 200.

The estimated value of 75.195 to the nearest hundredths is 75.20.

2/7 is closer to 1/4, 3/7​ is closer to 1/2

5/7 is closer to 3/4

2/9 is closer to 1/4, 4/9 ​​​​ is closer to 1/2

5/9 is closer to 1/2, 7/9​​​​ is closer to 3/4

Total quantity of cotton wool = 6 1/4 kg = 25/4 kg

Quantity of cotton wool required for one pillow = 1 1/4 kg = 5/4 kg

∴ Required number of pillows

= 25/4 ÷ 5/4

= 25/4 x 4/5

= 5

Given Suman studies for 5 (2/3) hours i.e. (17/3) hours

And she devotes 2 (4/5) hours i.e. (14/5) hours for Science and mathematics..

Let x be the hours she devotes for other subjects.

(17/3) = x + (14/5)

x = (17/3) – (14/5)

By taking LCM of 3 and 5 is 15

x = (17 × 5 – 14× 3)/15

x = (85 – 42)/15

x = (43/15) 

= 2 (13/15) hours

∵ Quantity of ghee sold by Raju in one month = 2.500 kg

∴ Quantity sold in 10 months

= 10 x 2.500 kg

= 25.00 kg

(c) between 40 cups and 50 cups

From the question it is given that,

One packet of biscuits requires 2½ cups of flour = 5/2

One packet of biscuits requires 1 2/3 cups of sugar. = 5/3

Total ingradiants for one packet of biscuits = (5/2) + (5/3)

= (15 + 10)/6

= 25/6

Then, total quantity of both ingredients used in 10 such packets of biscuits = 10 × (25/6)

= 5 × (25/3)

= 125/3

= 41 2/3

(a) In the given circles, 1 out of 2, 2 out of 4, 3 out of 6, and 4 out of 8 equal part of shaded respectively.

Therefore these circles represent.

Also, all these fractions are equivalent.

(b) In the given rectangles, 4 out of 12, 3 out of 9, 2 out of 6, 1 out of 3, and 6 out of 15 equal parts (i.e, circles) are shaded respectively 

Therefore these rectangles represent, Not, not all of these fractions are equivalent.

Let a number x be divided by 520.

According to question,

x ÷ 520 = 85 ÷ 0.625

x/520 = 85/0.625

x = (85 x 520)/0.625 = 44200/625 x 1000

= 70.72 × 1000 = 70720

Thus, the required number is 70720.