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32 by [1(3/5)]

The above question can be written as,

= 32 ÷ [1(3/5)]

Convert mixed fraction into improper fraction,

= [1(3/5)] = (8/5)

We have,

= (32/1) × (5/8)

(Because reciprocal of (8/5) is (5/8)

= (32 × 5) / (1 × 8)

= (4 × 5) / (1 × 1)

= 20

8 – [4(1/2)] – [2(1/4)]

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

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

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

LCM of 1, 2, 4 = 4

Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator.

= (9/2) × (2/2) = (18/4)

= (9/4) × (1/1) = (9/4)

= (8/1) × (4/4) = (32/4)

Then,

= (32/4) – (18/4) – (9/4)

= [(32-18-9)/4]

= [(32-27)/4]

= (5/4)

= [1(1/4)]

(3/5) × (7/11)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 7)/ (5 × 11)

= (21/55)

The total weight of fruits bought by Aneeta = [3(3/4)] + [4(1/2)]

We have,

First convert each mixed fraction into improper fraction

= [3(3/4)] = (15/4)

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

Then,

= (15/4) + (9/2)

LCM of 4, 2 = 4

Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator

= (15/4) × (1/1) = (15/4)

= (9/2) × (2/2) = (18/4)

= (15/4) + (18/4)

= (15 + 18)/ 4)

= (33/4)

= [8(1/4)]

The total weight of fruits purchased by Aneeta is [8(1/4)] kg

(2/3) + (5/6) – (1/9)

LCM of 3, 6, 9 = 18

Now, let us change each of the given fraction into an equivalent fraction having 18 as the denominator.

= (2/3) × (6/6) = (12/18)

= (5/6) × (3/3) = (15/18)

= (1/9) × (2/2) = (2/18)

Then,

= (12/18) + (15/18) – (2/18)

= (12+ 15 – 2)/ 18

= (27-2)/ 18

= (25/18)

= [1(7/18)]

(5/8) × (4/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 4)/ (8 × 7)

= (20/56) … [÷ by 4]

= (5/14)

(4/9) × (15/16)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4 × 15)/ (9 × 16)

= (60/144) … [ by 12]

= (5/12)

(2/5) × 15

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 15)/ (5 × 1)

= (30/5) … [÷ by 5]

= 6

(8/15) × 20

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (8 × 20)/ (15 × 1)

= (160/15) … [÷ by 5]

= (32/3)

= [10(2/3)]

(5/8) × 1000

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 1000)/ (8 × 1)

= (5000/8) … [÷ by 8]

= 625

[3(1/8)] × (16)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (25/8) × (16/1)

= (25 × 16)/ (8 × 1)

= (400/8) … [÷ by 8]

= 50

Fraction of students of class A who passed in 1^st class = 20/25 = 4/5

Fraction of student of class B Who passed in 1^st class = 24/30 = 4/5

From both classes, an equal fraction of student passed in first class

[2(4/15)] × (12)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (34/15) × (12/1)

= (34 × 12)/ (15 × 1)

= (408/15) … [÷ by 3]

= (136/5)

= [27(1/5)]

Yes, the fractions, where numerator is less than the denominator, are called proper fractions As : 4/9, 5/7

[3(6/7)] × [4(2/3)]

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (27/7) × (14/3)

= (27 × 14)/ (7 × 3)

= (378/21) … [÷ by 21]

= 18

[9(1/2)] × [1(9/19)]

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (19/2) × (28/19)

= (19 × 28)/ (2 × 19)

= (532/38)

= 14

(1) 20 rupees → \(\frac{1}{5}\) of 20 = 4, 20 ÷ 5 = 4 rupees.

(2) 30 km → \(\frac{1}{5}\) of 15 = 3, 15 ÷ 5 = 3 litres.

(3) 15 litres → \(\frac{1}{5}\) of 15 = 3, 15 ÷ 5 = 3 litres.

(4) 25 cm → \(\frac{1}{5}\) of 25 = 5, 25 ÷ 5 = 5 cm. 

i) One third of 15 pencils is:

1/3 x 15

= 1 x 5

= 5 pencils

ii) One third of 21 balloons is:

1/3 x 21

= 1 x 7

= 7 balloons

iii) One third of 9 children is:

1/3 x 9

= 1 x 3

= 3 children

iv) One third of 18 books is:

1/3 x 18

= 1 x 6

= 6 books

(1) 15 pencils → \(\frac{1}{3}\) of 15 = 5, 15 ÷ 3 = 5 pencils.

(2) 21 baloons → \(\frac{1}{3}\) of 21 = 7,21 ÷ 3 = 7 baloons.

(3) 9 children → \(\frac{1}{3}\) of 9 = 3, 9 ÷ 3 = 3 children.

(4) 18 books → \(\frac{1}{3}\) of 18 = 6, 18 ÷ 3 = 6 books.

[2(2/17)] × [7(2/9)] × [1(33/52)]

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

By Converting mixed fraction into improper fraction we get,

= (36/17) × (65/9) × (85/52)

= (36×65×85)/ (17×9×52)

On simplifying we get,

= (4×5×5) / (1×1×4)

Again simplifying we get,

= (1×5×5)/ (1×1×1)

= 25

[4(1/8)] × [2(10/11)]

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (33/8) × (32/11)

= (33 × 32)/ (8 × 11)

= (1056/88)

= 12

[5(5/6)] × [1(5/7)]

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (35/6) × (12/7)

= (35 × 12)/ (6 × 7)

= (420/42)

= 10

(1) 24 marbles → \(\frac{1}{3}\) of 24 = 8, 24 ÷ 3 = 8 marbles.

(1) 6 erasers → \(\frac{1}{3}\) of 6 = 2, 6 ÷ 3 = 2 erasers.

Diameter of Earth = 12756000 m

Diameter of a new planet = 5/86 x 12756000 m = 741627.90 m

i.e., 741627.90/1000 km = 741.6 km

From the question it is given that,

2/3 of a number is 10.

Let us assume the number be ‘P’.

Then,

2/3 of P = 10

2/3 × P = 10

By cross multiplication we get,

P = 10 × 3/2

P = 5 × 3

P = 15

So, the number is 15

Again it is given in question that, 1.75 times of that number = ?

= 1.75 of 15

= 1.75 × 15

= 26.25

(C) 0.749 and 0.75

0.7499 lies between 0.749 and 0.75

Total work done – work done by Neha = work done by Supriya

\(\frac{7}{9}-\frac{5}{9}=\frac{7-5}{9}=\frac{2}{9}\)

\(\frac{2}{9}\) work done by Supriya

Numbers of pages read by Lalita 

= 2/5 x 100 = 40

Number of pages read by Ila = 25

Hence, Ila has read less number of pages

(1) 35 gm → \(\frac{1}{5}\) of 35 = 7, 35 ÷ 5 = 7 gm.

(2) 40m → \(\frac{1}{5}\) of 40 = 8, 40 ÷ 5 = 8m.

(a) Diameter of the disc = 18.7 cm

∴ The distance around the disc

= 3.14 × 18.7 cm = 58.718 cm

(b) Radius of the disc = 6.45 cm

∴ Diameter of the disc = 2 × 6.45 cm = 12.9 cm

∴ The distance around the disc

= 3.14 × 12.9 cm = 40.506 cm

(i) (2/3) × [5/10] = (10/30)

(ii) (3/5) × [8/15] = (24/75)

(i) Given (4/7) × (14/25)

(4/7) × (14/25) 

= (4 × 14)/ (7 × 25)

= (56/175)

Converting above fractions into simplest form

= (8/25)

(ii) Given 7 (1/2) × 2 (4/15)

We have convert mixed fractions into improper fractions

Then we get (15/2) and (34/15)

= 7 (1/2) × 2 (4/15) 

= (15/2) × (34/15)

= (15 × 34)/ (2 × 15)

= (510/30)

= 17

(iii) Given 3 (6/7) × 4 (2/3)

We have convert mixed fractions into improper fractions

Then we get (27/7) and (14/3)

= 3 (6/7) × 4 (2/3) 

= (27/7) × (14/3)

On simplifying

= 9 × 2

= 18

(iv) Given 6 (11/14) × 3 (1/2)

We have convert mixed fractions into improper fractions

Then we get (95/14) and (7/2)

6 (11/14) × 3 (1/2) 

= (95/14) × (7/2)

= (95 × 7)/28

= (665/28)

= 23 (3/4)

(i) Given (7/11) by (3/5)

We have to multiply the given number

(7/11) × (3/5) 

= (21/55)

(ii) Given (3/5) by 25

(3/5) × 25 = 15 [dividing 25 by 5]

(iii) Given 3 (4/15) by 24

First convert the given mixed fraction to improper fraction.

(49/15) × 24 

= (1176/24)

= 78 (2/5)

(iv) Given 3 (1/8) by 4 (10/11)

First convert the given mixed fraction to improper fraction.

(25/8) × (54/11) 

= (1350/88) 

= (675/44)

= 15 (15/44)

Mrs. Sharma’s height = \(\frac{14}{9}-\frac{4}{9}=\frac{14-4}{9}\) = \(\frac{10}{9}\)

Mrs. Sharma’s height = \(\frac{10}{9}\)

On making three equivalent fractions of given fractions – 

(i) First equivalent fraction of 3/4 = (3/4) × (2/2) = 6/8

Second equivalent fraction of 3/4 = (3/4) × (3/3) = 9/12 

Third equivalent fraction of 3/4 = (3/4) × (4/4) = 12/16

Thus three equivalent fractions of 3/4 are 6/8, 9/12 and 12/16

(ii) First equivalent fraction of 1/3 = (1/3) × (2/2) = 2/6

Second equivalent fraction of 1/3 = (1/3) × (3/3) = 3/9

Third equivalent fraction of 1/3 = (1/3) × (4/4) = 4/12 

Thus, three equivalent fraction of 1/3 are 2/6, 3/9 and 4/12

(iii) First equivalent fraction of 2/7 = (2/7) × (2/2) = 4/14

Second equivalent fraction of 2/7 = (2/7) × (3/3) = 6/21

Third equivalent fraction of 2/7 = (2/7) × (4/4) = 8/28 

Thus, three equivalent fractions of 2/7 are 4/14, 6/21 and 8/28

Yes, unit fraction always has 1 as numerator and denominator are different. Thus, the rule is that the fraction having smaller denominator will be greater fraction.

(i) (b),

(ii) (c),

(iii) (a)

(i) 2/3

(ii) 1/2 x 2 = 1

(iii) 3/4 x 3 = 9/4

Correct option is  C) 16/5 

Correct option is B) 7 \(\frac{2}{3}\)

Correct option is  B) 1/3 

(a) 7/8 - 5/8 = 1/4

(b) 7/10 - 1/5 = 1/2

(c) 1/2 - 1/3 = 1/6

Atomic weight of Helium = 4.0030

Atomic weight of Hydrogen = 1.0080

Atomic weight of Oxygen – 16.0000

(a) Difference between the atomic weights of Oxygen and Hydrogen

= 16.0000 – 1.0080

= 14.9920

(b) Difference between atomic weights of Oxygen and Helium = 16.0000 – 4.0030 = 11.9970

(c) Difference between atomic weights of Helium and Hydrogen = 4.0030 – 1.0080 = 2.9950

The correct length of the iron rod = 19.33 cm

The length of the rod measured by Ravi = 19.34 cm

∴ Error made by Ravi = (19.34 – 19.33)cm = +0.01 cm

The length of the rod measured by Kamal = 19.25 cm

∴ Error made by Kamal = (19.25 – 19.33) cm = -0.08 cm

The length of the rod measured by Tabish = 19.27 cm

∴ Error made by Tabish = (19.27 – 19.33) cm = -0.06 cm.

The reciprocal of 5/129 = 129/5

5/129 x 129/5 = 1

Let the number be x.

According to question,

1/8 x x = 2/5 ÷ 1/20

x/8 = 2/5 x 20/1

x/8 = 8

⇒ x = 64

Hence, the required number is 64.

Given (3/5)

By multiplying or dividing both the numerator and denominator so that it keeps the same value by this we can get the equivalent fractions.

(3 × 2)/ (5 × 2), (3 × 3)/ (5 × 3), (3 × 4)/ (5 × 4), (3 × 5)/ (5 × 5), (3 × 6)/ (5 × 6)

Equivalent fractions are

(6/10), (9/15), (12/20), (15/25), (18/30)

(i) Given (5/8) + (3/10)

Taking LCM for 8 and 10 we get 40

Now we have to convert the given fractions into equivalent fractions with denominator 40

(5/8) + (3/10) = (5 × 5)/ (8 × 5) + (3 × 4)/ (10 × 4)

= (25/40) + (12/40)

= (37/40)

(ii) Given 4 (3/4) + 9 (2/5)

First convert given mixed fractions into improper fractions.

4 (3/4) + 9 (2/5) = (19/4) + (47/5)

Taking LCM for 4 and 5 we get 20

Now we have to convert the given fractions into equivalent fractions with denominator 20

4 (3/4) + 9 (2/5) = (19/4) + (47/5) = (19 × 5)/ (4 × 5) + (47 × 4)/ (5 × 4)

= (95/20) + (188/20)

= (283/20)

(i) Given (5/6) + 3 + (3/4)

Taking LCM for 6, 1 and 4 we get 12

Now we have to convert the given fractions into equivalent fractions with denominator 12

(5/6) + 3 + (3/4) 

= (5 × 2)/ (6 × 2) + (3 × 12)/ (1 × 12) + (3 × 3)/ (4 × 3)

= (10/12) + (36/12) + (9/12)

= (55/12)

(ii) Given 2 (3/5) + 4 (7/10) + 2 (4/15)

First convert given mixed fractions into improper fractions

2 (3/5) + 4 (7/10) + 2 (4/15) 

= (13/5) + (47/10) + (34/15)

Taking LCM for 5, 10 and 15 we get 30

Now we have to convert the given fractions into equivalent fractions with denominator 30

2 (3/5) + 4 (7/10) + 2 (4/15) 

= (13/5) + (47/10) + (34/15) 

= (13 × 6)/ (5 × 6) + (47 × 3)/ (10 × 3) + (34 × 2)/ (15 × 2)

= (78/30) + (141/30) + (68/30)

= (287/30)