Understanding the word statements of the problem and writing the word statements in the solution along with the method of solution are explained in a lucid way.

Solved Example 1 of Fraction Word Problems

Three boxes weigh 10 ^{1}⁄_{2} kg,15 ^{1}⁄_{5} kg and20 ^{3}⁄_{4} kg respectively.A porter carries all the three boxes. What is the total weightcarried by the porter ?

Solution : Total weight = Sum of the weights of all the three boxes = 10 ^{1}⁄_{2} +15 ^{1}⁄_{5} +20 ^{3}⁄_{4}

= (10 x 2 + 1)⁄2 + (15 x 5 + 1)⁄5 + (20 x 4 + 3)⁄4= 21⁄2 + 76⁄5 + 83⁄4

Let us find the L.C.M. of the denominators.

2| 2 4 5
-----------------------
1 2 5

L.C.M. = 2 x 2 x 5 = 20

Total weight = 21⁄2 + 76⁄5 + 83⁄4

Now 120 is taken as the common denominator. The number with which we have to multiply each denominator to get 120, is taken and that number multiplies each numerator as shown below.

= (21 x 10 + 76 x 4 + 83 x 5)⁄20 = (210 + 304 + 415)⁄20 = 929⁄20 = 46 ^{9}⁄_{20} kg. Ans.

Solved Example 2 of Fraction Word Problems

From a rope of 15 ^{3}⁄_{4} mlong, a piece of length 6 ^{5}⁄_{6} mhas been cut off. What is the length of the remaining piece ?

Solution : Length of the remaining piece = Total length of the rope - Length of cut piece = 15 ^{3}⁄_{4} m- 6 ^{5}⁄_{6} m = (15 x 4 + 3)⁄4 - (6 x 6 + 5)⁄6 = 63⁄4 - 41⁄6

2| 4 6
----------------
2 3

L.C.M. = 2 x 2 x 3 = 12

Length of the remaining piece = 63⁄4 - 41⁄6 = (63 x 3 - 41 x 2)⁄12 = (189 - 82)⁄12 = 107⁄12 = 8 ^{11}⁄_{12} m. Ans.

Solved Example 3 of Fraction Word Problems

The weight of 15 packets of sweets is 7 ^{1}⁄_{2} kg.What is the weight of each packet.

Solution : Weight of each packet = weight of 15 packets of sweets ÷ 15 = (7 ^{1}⁄_{2}) ÷ 15 = [(7 x 2 + 1)⁄2] ÷ 15 = (15⁄2) ÷ 15 = (15⁄2) x (1⁄15) [ since dividing is same as multiplying with reciprocal ] = 1⁄2 kg. Ans.

Solved Example 4 of Fraction Word Problems

An iron rod is 9 ^{1}⁄_{2}metres long and it is cut into 19⁄20 metre long pieces.How many pieces are made out of the long rod ?

Solution : Number of pieces of the rod = (Total length of the rod) ÷ (length of each piece of rod) = (9 ^{1}⁄_{2}) ÷ (19⁄20) = [(9 x 2 + 1)⁄2] ÷ (19⁄20) = (19⁄2) ÷ (19⁄20) = (19⁄2) x (20⁄19) [ since dividing is same as multiplying with reciprocal ] = 10. Ans.

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A mango weighs 1⁄3 kilo. If a basket of mangoesweighs 10 ^{2}⁄_{3}kg., find the number of mangoes in the basket (weight of basket is ignored).

Solution : Number of mangoes in the basket = (Total weight of mangoes in the basket) ÷ (weight of each mango) = (10 ^{2}⁄_{3}) ÷ (1⁄3) = [(10 x 3 + 2)⁄3] ÷ (1⁄3) = (32⁄3) ÷ (1⁄3) = (32⁄3) x (3⁄1) [ since dividing is same as multiplying with reciprocal] = 32. Ans.

The Examples 6, 7, 8 and 10 need the knowledge of solving

A student was asked to multiply a given number by 9⁄16.Instead he divided the given number by 9⁄16. His answer was175 more than the correct answer. What was the given number.

Solution to fraction word problem involving linear equation :

Let x be the given number. Correct answer is multiplying x with 9⁄16 = x x (9⁄16) = 9x⁄16.

By data, dividing x with 9⁄16 gave 175 more than correct answer which is 9x⁄16. ⇒ x ÷ (9⁄16) = 175 + 9x⁄16 ⇒ x x (16⁄9) = 175 + 9x⁄16 ⇒ 16x⁄9 = 175 + 9x⁄16 ⇒ 16x⁄9 - 9x⁄16 = 175

L.C.M. of the denominators (9, 16) = 9 x 16 = 144

∴ (16 x 16 - 9 x 9)x⁄144 = 175 ⇒ 175x⁄144 = 175 ⇒ x = 175 x (144⁄175) = 144. Ans.

Solved Example 7 of Fraction Word Problems

A pole has half of its length in the mud, 1⁄3 of its length in waterand 1 ^{2}⁄_{3} m above thewater. Find the whole length of the pole.

Solution to fraction word problem involving linear equation :

Let x m be the length of the pole. Then, length of the pole in the mud = x x (1⁄2) = x⁄2 Length of the pole in water = x x (1⁄3) = x⁄3 Length of the pole above water = 1 ^{2}⁄_{3} = (1 x 3 + 2)⁄3 = 5⁄3

We have, length of the pole = x = x⁄2 + x⁄3 + 5⁄3 Let us find the L.C.M. of the denominators (2, 3, 3)

3| 2 3 3
-------------------
2 1 1

L.C.M. of the denominators = 3 x 2 = 6

x = x⁄2 + x⁄3 + 5⁄3 ⇒ x = (3x + 2x + 5 x 2)⁄6 ⇒ 6x = 5x + 10 ⇒ 6x - 5x = 10 ⇒ x = 10.

Thus, length of the pole = 10 m. Ans.

Solved Example 8 of Fraction Word Problems

A drum of water is 3⁄7 full. When 28 litres are drawn from it, it is just5⁄14 full. Find the total capacity of the drum, in litres.

Solution to fraction word problem involving linear equation :

Let x litres be the total capacity of the drum. When the drum is 3⁄7 full, the volume of water in the drum = x x 3⁄7 = 3x⁄7 When 28 litres are drawn from it, the volume of water in the drum = 3x⁄7 - 28 By data this is 5⁄14 full. ⇒ volume = x x 5⁄14 = 5x⁄14 ∴ 5x⁄14 = 3x⁄7 - 28 ⇒ 3x⁄7 - 5x⁄14 = 28 Let us find the L.C.M. of the denominators.

7| 14 7
----------------
2 1

L.C.M. = 7 x 2 = 14

∴ (3x x 2 - 5x)⁄14 = 28 ⇒ (6x - 5x)⁄14 = 28 ⇒ x⁄14 = 28⇒ x = 14 x 28 = 392

Thus, total capacity of the drum = 392 litres. Ans.

Solved Example 9 of Fraction Word Problems

From a class of 44 students, 1⁄5th of the girls and 1⁄8th of the boys took part in a social camp. What fraction of the total strength took part in the camp?

Solution : Number of girls in the class is a multiple of 5 and Number of boys in the class is a multiple of 8.

A trian starts with a number of passengers. At the first station, it dropsone-third of the passengers and takes 280 more. At the second station, it drops one-half of the new total and takes 12 more. On arriving at the third station, it is found to have 248 passengers. find the number of passengers in the beginning.

Solution to fraction word problem involving linear equation :

let x be the number of passengers in the beginning.

At the first station, number of passengers dropped = 1⁄3rd of x = x⁄3 number of passengers added = 280. So, total number of passengers while leaving first platform = x - x⁄3 + 280. = (3x - x + 280 x 3)⁄3 = (2x + 840)⁄3

At the second station, number of passengers dropped = 1⁄2 of [(2x + 840)⁄3] = (x + 420)⁄3 number of passengers added = 12. So, total number of passengers while leaving second platform = (2x + 840)⁄3 - (x + 420)⁄3 + 12. = (2x + 840 -x - 420 + 12 x 3)⁄3 = (x + 456)⁄3

By data, this = 248 ⇒ (x + 456)⁄3 = 248 ⇒ x + 456 = 3 x 248 = 744 ⇒ x = 744 - 456 = 288.

Thus, the number of passengers in the beginning = 288. Ans.

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2⁄3 students in a class are boys. If there are 19 girlsin the class, how many are boys ?

A water tank can hold 56 ^{1}⁄_{4}litres of water. How much water is contained in the tank,when it is 8⁄15 full ?

From a string 15 metres long, how many pieces of length 2 ^{1}⁄_{2} metres can be cut ?

A packet of toffees weighs 3⁄4 kg. and a tin containing thepackets weighs 22 ^{1}⁄_{2}kg. How many packets of toffees arethere in the tin ? (weight of tin ignored.)

A person takes 4 ^{1}⁄_{2}minutes to read a page. How many pages can the person read in36 minutes ?

A tumbler holding 1 ^{1}⁄_{2}litres is used to fill a drum which can hold 30 litres. How many timesshould the water be poured using the tumbler to fill the drum ?

Find the fraction which is as much greater than 5⁄8as is less than 3⁄4.

If 1⁄8 of a pencil is black, 1⁄2 of the remaining is whiteand the remaining 3 ^{1}⁄_{2} cmis blue, find the total length of the pencil.

On a certain day, 810 people visited the school exhibition.Out of this, 7⁄15 were men and 11⁄30 were ladies and the rest were children. How many children visited the exhibition?

A cake weighs 2 kg. If 2⁄7 of its weight is flour, 1⁄8 of its weight is sugar, 3⁄14 of its weight is milkand the rest is nuts and plums, find the weight of nuts and plumsin the cake.

For Answers see at the bottom of the page.

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