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(i) Given (13/24) and (7/16)

To find the difference we have to make it equivalent fractions.

Taking LCM of 24 and 16 is 48.

Now converting the given fractions into equivalent fractions with denominator 48.

(13/24) – (7/16) = (26/48) – (21/48)

= (26 – 21)/48

= (5/48)

(ii) Given 6 and (23/3)

To find the difference we have to make it equivalent fractions.

Taking LCM of 3 and 1 is 3.

Now converting the given fractions into equivalent fractions with denominator 3.

(23/3) – 6 = (23/3) – (18/3)

= (23 – 18)/3

= (5/3)

(i) Given (21/25) and (18/20)

To find the difference we have to make it equivalent fractions.

Taking LCM of 25 and 20 is 100.

Now converting the given fractions into equivalent fractions with denominator 100.

(18/20) – (21/25) 

= (90/100) – (84/100)

= (90 – 84)/100

= 6/100

By converting it into its simplest form we get

= 3/50

(ii) Given 3 (3/10) and 2 (7/15)

First convert given mixed fractions into improper fractions.

(33/10) and (37/15)

To find the difference we have to make it equivalent fractions.

Taking LCM of 10 and 15 is 30.

Now converting the given fractions into equivalent fractions with denominator 30.

(33/10) – (37/15) 

= (99/30) – (74/30)

= (99 – 74)/30

= (25/30)

By converting it into simplest form we get

= (5/6)

From the question,

The cost of 1 kg of apples = [48(4/5)] = (244/5)

Therefore, the cost of [3(3/4)] kg of apples = (15/4)

Then,

= (244/5) × (15/4)

= (244 × 15) / (5 × 4)

On simplifying we get,

= (61 × 3) / (1 × 1)

= ₹ 183

Hence, the cost of [3(3/4)] kg is ₹ 183

(i) Given (6/7) – (9/11)

To find the difference we have to make it equivalent fractions.

Taking LCM of 7 and 11 is 77.

Now converting the given fractions into equivalent fractions with denominator 77.

Equivalent fractions are (66/77) and (63/77)

(6/7) – (9/11) 

= (66/77) – (63/77)

= (66 – 63)/77

= (3/77)

(ii) Given 8 – (5/9)

To find the difference we have to make it equivalent fractions.

Taking LCM of 1 and 9 is 9.

Now converting the given fractions into equivalent fractions with denominator 9.

Equivalent fractions are (72/9) and (5/9)

8 – (5/9) 

= (72/9) – (5/9)

= (72 -5)/9

= (67/9)

The cost of 6 (1/4) kg = (25/4) of apples is Rs 400

Cost of apple per kg is = (25/4) /400

= (4/25) × 400

= Rs 64

(i) Given 9 – 5 (2/3)

First convert the given mixed fraction into improper fraction.

We get 5 (2/3) = (17/3)

To find the difference we have to make it equivalent fractions.

Taking LCM of 1 and 3 is 3.

Now converting the given fractions into equivalent fractions with denominator 3.

Equivalent fractions are (27/3) and (17/3)

9 – 5 (2/3) = (27/3) – (17/3)

= (10/3)

(ii) Given 4 (3/10) – 1 (2/15)

First convert the given mixed fraction into improper fraction.

We get (43/10) and (17/15)

To find the difference we have to make it equivalent fractions.

Taking LCM of 10 and 15 is 30.

Now converting the given fractions into equivalent fractions with denominator 30.

Equivalent fractions are (129/30) and (34/30)

4 (3/10) – 1 (2/15) 

= (43/10) – (17/15)

= (129/30) – (34/30)

= (129 – 34)/30

= (95/30)

= (19/6)

Given product of two numbers is 25 (5/6) = (155/6)

One of the number is 6 (2/3) = (20/3)

Let the other number be x

(155/6) = x × (20/3)

x = (3/20) × (155/6)

x = (31/8)

x = 3 (7/8)

(i) Given (2/3) + (1/6) – (2/9)

Taking the LCM of 3, 6 and 9 is 18

Now we have to convert the given fraction into equivalent fraction with denominator 18

(2/3) + (1/6) – (2/9) 

= (12/18) + (3/18) – (4/18)

= (12 + 3 – 4)/18

= 11/18

(ii) Given 12 – 3 (1/2)

First convert the given mixed fraction into improper fraction we get (7/2)

Taking the LCM of 2 and 1 is 2

Now we have to convert the given fraction into equivalent fraction with denominator 2

12 – 3 (1/2) 

= (24/2) – (7/2)

= (24 -7)/2

= (17/2)

(iii) Given 7 (5/6) – 4 (3/8) + 2 (7/12)

First convert the given mixed fraction into improper fraction we get (47/6), (35/8) and (31/12)

Taking the LCM of 12, 6 and 8 is 48

Now we have to convert the given fraction into equivalent fraction with denominator 48

7 (5/6) – 4 (3/8) + 2 (7/12) 

= (47/6) – (35/8) + (31/12)

= (376/48) – (210/48) + (124/48)

= (376 – 210 + 124)/48

= (290/48)

= (145/24)

Given 5 (3/7)

First convert the given mixed fraction into improper fraction we get (38/7)

Let x be the number added to (38/7) to get 12

Therefore x + (38/7) = 12

x = 12 – (38/7)

By taking LCM for 7 and 1 is 7

x = (12 × 7 – 38)/7

x = (84 – 38)/7

x = (46/7)

Hence (46/7) is the number which is added to 5 (3/7) to get 12.

Given 5 (4/15)

First convert the given mixed fraction into improper fraction we get (79/15)

Let x be the number added to (79/15) to get (63/5)

Therefore x + (79/15) = (63/5)

x = (63/5) – (79/15)

By taking LCM for 15 and 5 is 15

x = (63 × 3 – 79)/15

x = (189 – 79)/15

x = (110/15) 

= (22/3)

Hence (22/3) is the number which is added to 5 (4/15) to get 12 (3/5).

Let x be the number which needs to be multiplied by 6 (2/9) = (56/9)

x × (56/9) = 4 (4/9)

x × (56/9) = (40/9)

x = (40/9) × (9/56)

x = (40/56)

x = (5/7)

(A) 23.640

23.875 - 0.235 = 23.640

Correct option is  A) 23.640

Correct option is (D) 49/75

2\(\frac13\) \(\div\) 3\(\frac47\) = \(\frac{2\times3+1}{3}\div\frac{3\times7+4}7\) 

\(\frac73\div\frac{25}7\) = \(\frac73\times\frac7{25}\) = \(\frac{49}{75}\)

Correct option is   D) \(\frac{49}{75}\)

2/7 x \(\frac{3\times2+1}2\times\frac{1\times5+1}5\times\frac{2\times2+1}2\) 

\(=(\frac27\times\frac72)\times(\frac65\times\frac52)\) 

 = 1 x 3

 = 3

Correct option is  A) 3

Correct option is  A) 3/5

Correct option is  B) 2

Correct option is (C) 2

2/5 x 15 - 1/2 x 12

 = 2 x 3 - 4

 = 6 - 4

 = 2

Correct option is  C) 2

Correct option is (D) 13/2
\(=\frac{5\times6+7}6\times\frac{1\times37+2}{37}\) 

 = \(\frac{37}{6}\times\frac{35}{37} = \frac{39}6=\frac{13}2\)

Correct option is  D) \(\frac{13}{2}\)

Total time taken by Rohan in five races

= (3.20 +3.37 +3.29 +3.17 +3.32] minutes

= 16.35 minutes

Total number of races = 5

∴ The average time taken by Rohan

= 16.35/5  minutes = 3.27 minutes

The Value of 7.9 ÷ 1000 will be 3.0079.

Total participants of the function = 9

Total number of cakes = 3

∴ Each cake should be divided into 3 equal parts.

∴ Total number of equal parts of cake = 9

∴ Each one’s share may be 1/9 of total cakes or 1/3 of a cake.

Length of ribon = 111/2 cm = 23/2 cm

∴ Number of pieces to be cut = 23/2 ÷ 1/2

=23/2 x 2/1 = 23 pieces

Total amount = Rs. 12,0,000

Amount given to his wife = 2/5 of Rs. 1,20,000

= Rs. 2 x 24,000 = Rs. 48,000

Remaining amount = Rs. 120000 – Rs. 48000 = Rs. 72000

This amount is distributed among three children equally.

Each’s share = Rs. 72,000 x 1/3 = Rs. 24,000

(B) 0 and 1

A fraction whose numerator is less than the denominator is called a proper fraction.

So, 5/7 = 0.715

Therefore, 5/7 lies between 0 and 1.

Given Ravish takes 1 hour to read (1/3) part of the book

Then we have to calculate how much part he will read in 2 (1/5) hours

First convert the given mixed fraction into improper fractions is (11/5)

Now let x be the full part of book

1 hour = (1/3) x

Remaining part of the book, he will read in

= (11/5) × (1/3) x

= (11/5) part of the book

Given that the distance between two adjacent saplings is (3/4) m

There are 4 adjacent spacing for 5 sapling

Therefore, distance between the first and the last sapling is

= (3/4) × 4

= 3

The distance between them is 3m

(a) Here, 1 parts is shaded out of 2 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction 1/2

(b) Here, 4 parts is shaded out of 6 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction 4/6 = 2/3

(c) Here, 3 parts is shaded out of 9 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction 3/9 = 1/3

(d) Here, 2 parts is shaded out of 8 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction 2/8 = 1/4

(e) Here,3 parts is shaded out of 4 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction 3/4

(i) Here,6 parts are shaded out of 18 equal parts (i.e, triangles) Hence, this figure represents a fraction 6/18 = 1/3

(ii) Here, 4 parts are shaded out of 8 equal parts (i,e, rectangle) Hence, this figure represents a fraction 4/8 = 1/2

(iii) Here, 12 parts are shaded out o f 16 equal parts (i.e , squares) Hence, this figure 12/16 = 3/6

(iv) Here, 8 parts are shaded out of 12 equal parts (i,e . rectangle) Hence, this figure represents a fraction 8/12 = 2/3

(v) Here,4 parts are shaded out of 16 equal parts (i,e , triangles) Hence, this figure represents a fraction 4/16 = 1/4

Now these figures can be, matched correctly as (a)(ii), (b)(iv), (c)(i), (d)(v), (e)(iii).

i. Total quantity of rice = 9 kg 

Number of people = 5 

∴ Kilograms of rice received by each person = \(\cfrac{9}{5}\)

∴ Each person will get \(\cfrac{9}{5}\) kg of rice.

ii. Total meters of cloth = 11 meters 

Number of shirts to be made = 5

Meters of cloth needed to make 1 shirt = \(\cfrac{11}{5}\)
∴ Cloth needed to make 1 shirt is \(\cfrac{11}{5}\) meters.

Fraction as an operator ‘of ’

Fraction acts as an operator of 'of' represents multiplication.

Ex: 1/3 of 90 

= 1/3 × 90 

= 30.

True.

For example: consider two fractions 10/5 and 15/5.

Sum of two fractions = (10 + 15)/5

= 25/5

= 5

= 5/1

A fraction in which there is no common factor, except 1, in its numerator and denominator is called a fraction in the simplest or lowest form.

When 2 fractions are added, the result in most cases will be a fraction p/q form, but in some case if it does happen to be just a integer, it can always be written with denominator 1 (hence p/q form).

False.

Not necessarily a fraction. But can be written in fraction.

The normal body temperature = 98.6°F

Savitri’s temperature, when she was ill = 103.1°F

∴ Savitri’s temperature above normal

= 103.1°F – 98.6°F

= 4.5°F

∵ Price of 1 kg tomato = ₹16.50

∴ Price of 22.5 kg tomato = 6.50 x 22.5

= ₹371.25

In the given figure, total parts in which figure has been divided is 15 and out of which 4 parts are unshaded.

∴ The required fraction is 4/15.

i. \(\frac{9}{14}-\frac{3}{14}=\frac{9-3}{14}=\frac{6}{14}\)

ii. \(\frac{5}{6}-\frac{3}{6}=\frac{5-3}{6}=\frac{2}{6}\)

iii. \(\frac{9}{16}-\frac{5}{16}=\frac{9-5}{16}=\frac{4}{16}\)

iv. \(\frac{7}{8}-\frac{3}{8}-\frac{1}{8}=\frac{7-3-1}{8}=\frac{3}{8}\)

i. \(\frac{5}{7}-\frac{1}{7}=\frac{5-1}{7}=\frac{4}{7}\)

ii. \(\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}=\frac{2}{8}\)

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

iv. \(\frac{8}{11}-\frac{5}{11}=\frac{8-5}{11}=\frac{3}{11}\)

i. \(\frac{9}{13}-\frac{4}{13}=\frac{9-4}{13}=\frac{5}{13}\)

ii. \(\frac{7}{10}-\frac{3}{10}=\frac{7-3}{10}=\frac{4}{10}\)

iii. \(\frac{9}{12}-\frac{2}{12}=\frac{9-2}{12}=\frac{7}{12}\)

iv. \(\frac{10}{15}-\frac{3}{15}=\frac{10-3}{15}=\frac{7}{15}\)

i. \(\frac{3}{6}\)

ii. \(\frac{4}{4}\)

iii. \(\frac{10}{13}\)

iv. \(\frac{41}{63}\)

v. \(\frac{31}{33}\)

vi \(\frac{9}{10}\)

There are 10 letters from A to J

So, 2/5 of 10 = 2/5 × 10

= 2 × 2

= 4

The 4^th letter from A to J is D.

Therefore, D comes 2/5 of the way among A and J.

i. \(\frac{4}{6}\)

ii. \(\frac{1}{5}\)

iii. \(\frac{3}{16}\)

iv. \(\frac{3}{12}\)

v. \(\frac{3}{13}\)

vi. \(\frac{13}{90}\)

Devansh painted the wall = (2/3) part 

Janhvi painted the wall = (1/3) part 

∴ They both painted altogether = ((2/3) + (1/3)) part 

= ((2 + 1)/3) part = (3/3) part = 1 part

(a) When Nidhi is 4/6 of the way through the race, she will be at hurdle D.

(b) When Nidhi is 5/6 of the way through the race, she will be at hurdle E.

(c) When Nidhi is over hurdle C, she would finished the 3/6 or 1/2 or middle part of the race.

i. \(\frac{2}{3}\) x 30 So, we take \(\frac{1}{3}\) of 30, twice

\(\frac{1}{3}\) x 30 = 10, twice of 10 is 2 x 10 = 20

It means that \(\frac{2}{3}\) x 30 = 20

ii. \(\frac{7}{11}\) x 22 So, we take of 22, 7 times

\(\frac{7}{11}\) x 22 = 2, seven times of 2 is 2 x 7 = 14

(1) 5 marbles 

(2) 3 pens 

(3) 6 notebooks 

(4) 2 ladoos

Space painted by Shubhan = 2/3 of the room

Space painted by Madhavi = 1/3 of the room

Hence, together they painted = (2/3 + 1/3) of the room

= 1 = the complete wall.

(1) 3 bananas 

(2) 2 gms 

(3) 5 metres 

(4) 4 ₹

(2/3) × (5/44) × (33/35)

By the rule Multiplication of fraction,

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

Then,

= (2×5×33)/ (3×44×35)

On simplifying we get,

= (1×1×11) / (1×22×7)

= (11/154)

= (1/14)

(12/25) × (15/28) × (35/36)

By the rule Multiplication of fraction,

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

Then,

= (12×15×35)/ (25×28×36)

On simplifying we get,

= (1×3×5) / (5×4×3)

Again simplifying we get,

= (1×1×1)/ (1×4×1)

= (1/4)

(10/27) × (28/65) × (39/56)

By the rule Multiplication of fraction,

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

Then,

= (10×28×39)/ (27×65×56)

On simplifying we get,

= (10×1×3) / (27×5×2)

Again simplifying we get,

= (1×1×3)/ (27×1×1)

= (3/27)

=9

[5(5/9)] by [3(1/3)]

The above question can be written as,

= [5(5/9)] ÷ [3(1/3)]

Convert mixed fraction into improper fraction,

= [5(5/9)] = (50/9)

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

We have,

= (50/9) × (3/10)

(Because reciprocal of (10/3) is (3/10)

= (50 × 3) / (9 × 10)

= (5 × 1) / (3 × 1)

= (5/3)

= [1(2/3)]