RS Aggarwal Solutions Class 6 Ex 5C

Find the sum:

Q1) \(\frac{ 5 }{ 8 } + \frac{ 1 }{ 8 }\)

Ans. 1) The given fractions are like fractions.

We know:

Sum of like fractions = \(\frac{ Sum \; of \; the \; numerators }{ Common \; denominator }\)

Thus, we have:

\(\frac{ 5 }{ 8 } \; + \; \frac{ 1 }{ 8 }\)

= \(\frac{ ( 5 + 1 ) }{ 8 } = \frac{ 6 }{ 8 } = \frac{ 3 }{ 4 }\)

Q2) \(\frac{ 4 }{ 9 } + \frac{ 8 }{ 9 }\)

Ans. 2) The given fractions are like fractions.

We know:

Sum of like fractions = \(\frac{ Sum \; of \; the \; numerators }{ Common \; denominator }\)

Thus, we have:

\(\frac{ 4 }{ 9 } \; + \; \frac{ 8 }{ 9 }\)

= \(\frac{ ( 4 + 8 ) }{ 9 } = \frac{ 12 }{ 9 } = \frac{ 4 }{ 3 }\)

= \(1\frac{ 1 }{ 3 }\)

Q3) \(1 \frac{ 3 }{ 5 } + 2 \frac{ 4 }{ 5 }\)

Ans. 3) The given fractions are like fractions.

We know:

Sum of like fractions = \(\frac{ Sum \; of \; the \; numerators }{ Common \; denominator }\)

Thus, we have:

\(1 \frac{ 3 }{ 5 } + 2 \frac{ 4 }{ 5 }\) = \(\frac{ 8 }{ 5 } \; + \; \frac{ 14 }{ 5 } = \frac{ ( 8 + 14 ) }{ 5 } = \frac{ 22 }{ 5 }\)

= \(4 \frac{ 2 }{ 5 }\)

Q4) \(\frac{ 2 }{ 9 } + \frac{ 5 }{ 6 }\)

Ans. 4) L.C.M. of 9 and 6 = ( 2 x 3 x 3 ) = 18

Now, we have:

\(\frac{ 2 }{ 9 } \; = \; \frac{ 2 \times 2 }{ 9 \times 2 } = \frac{ 4 }{ 18 }\)

\(\frac{ 5 }{ 6 } \; = \; \frac{ 5 \times 3 }{ 6 \times 3 } = \frac{ 15 }{ 18 }\)

Therefore, \(\frac{ 2 }{ 9 } + \frac{ 5 }{ 6 }\) = \(\frac{ 4 }{ 18 } + \frac{ 15 }{ 18 }\) = \(\frac{ ( 4 + 15 ) }{ 18 } \; = \frac{ 19 }{ 18 } = 1 \frac{ 1 }{ 18 }\)

Q5) \(\frac{ 7 }{ 12 } + \frac{ 9 }{ 16 }\)

Ans. 5) L.C.M. of 12 and 16 = ( 2 x 2 x 2 x 2 x 3 ) = 48

Now, we have:

\(\frac{ 7 }{ 12 } \; = \; \frac{ 7 \times 4 }{ 12 \times 4 } = \frac{ 28 }{ 48 }\)

\(\frac{ 9 }{ 16 } \; = \; \frac{ 9 \times 3 }{ 16 \times 3 } = \frac{ 27 }{ 48 }\)

Therefore, \(\frac{ 7 }{ 12 } + \frac{ 9 }{ 16 }\) = \(\frac{ 28 }{ 48 } + \frac{ 27 }{ 48 }\) = \(\frac{ ( 28 + 27 ) }{ 48 } \; = \frac{ 55 }{ 48 } = 1 \frac{ 7 }{ 48 }\)

Q6) \(\frac{ 4 }{ 15 } \; + \; \frac{ 17 }{ 20 }\)

Ans. 6)We have:

L.C.M. of 15 and 20 = ( 3 x 5 x 2 x 2 ) = 60

Therefore, \(\frac{ 4 }{ 15 } \; + \; \frac{ 17 }{ 20 } = \frac{ ( 16 +51 ) }{ 60 }\)

{ [ 60 / 15 = 4, 4 x 4 = 16 ] and [ 60 / 20 = 3, 17 x 3 = 51 ] }

= \(\frac{ 67 }{ 60 } \; = \; 1\frac{ 7 }{ 60 }\)

Q7) \(2 \frac{ 3 }{ 4 } + 5 \frac{ 5 }{ 6 }\)

Ans. 7) L.C.M. of 4 and 6 = ( 2 x 2 x 3 ) = 12

= \(2 \frac{ 3 }{ 4 } + 5 \frac{ 5 }{ 6 } = \frac{ 11 }{ 4 } + \frac{ 35 }{ 6 } = \frac{ ( 66 + 140 ) }{ 24 }\)

{ [ 24 / 4 = 6, 6 x 11 = 66 ] and [ 24 / 6 = 4, 4 x 35 = 140 ] }

= \(\frac{ 206 }{ 24 } = \frac{ 103 }{ 12 } = 8\frac{ 7 }{ 12 }\)

Q8) \(3 \frac{ 1 }{ 8 } + 1 \frac{ 5 }{ 12 }\)

Ans. 8) L.C.M. of 8 and 12 = ( 2 x 2 x 2 x 3 ) = 24

= \(3 \frac{ 1 }{ 8 } + 1 \frac{ 5 }{ 12 } = \frac{ 25 }{ 8 } + \frac{ 17 }{ 12 } = \frac{ ( 75 + 34 ) }{ 24 }\)

{ [ 24 / 8 = 3, 3 x 25 = 75 ] and [ 24 / 12 = 2, 2 x 17 = 34 ] }

= \(\frac{ 109 }{ 24 } = 4 \frac{ 13 }{ 24 }\)

Q9) \(2 \frac{ 7 }{ 10 } + 3 \frac{ 8 }{ 15 }\)

Ans. 9) We have:

L.C.M. of 10 and 15 = ( 2 x 3 x 5 ) = 30

= \(2 \frac{ 7 }{ 10 } + 3 \frac{ 8 }{ 15 } = \frac{ 27 }{ 10 } + \frac{ 53 }{ 15 } = \frac{ ( 81 + 106 ) }{ 30 }\)

{ [ 30 / 10 = 3, 3 x 27 = 81 ] and [ 30 / 15 = 2, 2 x 53 = 106 ] }

= \(\frac{ 187 }{ 30 } = 6 \frac{ 7 }{ 30 }\)

Q10) \(3 \frac{ 2 }{ 3 } + 1 \frac{ 5 }{ 6 } + 2\)

Ans. 10) We have:

L.C.M. of 3 and 6 = ( 2 x 3 ) = 6

\(3 \frac{ 2 }{ 3 } \; + \; 1 \frac{ 5 }{ 6 } \; + \; 2\)

= \(\frac{ 11 }{ 3 } \; + \; \frac{ 11 }{ 6 } \; + \; \frac{ 2 }{ 1 }\)

= \(\frac{ ( 22 + 11 + 12 ) }{ 6 }\)

{ [ 6 / 3 = 2, 2 x 11 = 22 ], [ 6 / 6 = 1, 1 x 11 = 11 ] and [ 6 / 1 = 6, 6 x 2 = 12 ] }

= \(\frac{ 45 }{ 6 } = \frac{ 15 }{ 2 } = 7 \frac{ 1 }{ 2 }\)

Q11) \(3 +1 \frac{ 4 }{ 15 } + 1 \frac{ 3 }{ 20 }\)

Ans. 10) We have:

L.C.M. of 15 and 20 = ( 2 x 2 x 3 x 5 ) = 60

\(3 \; + \; 1 \frac{ 4 }{ 15 } \; + \; 1 \frac{ 3 }{ 20 }\)

= \(\frac{ 3 }{ 1 } \; + \; \frac{ 19 }{ 15 } \; + \; \frac{ 23 }{ 20 }\)

= \(\frac{ ( 180 + 76 + 69 ) }{ 60 }\)

{ [ 60 / 1 = 60, 60 x 3 = 180 ], [ 60 / 15 = 4, 4 x 19 = 76 ] and [ 60 / 20 = 3, 3 x 23 = 69 ] }

= \(\frac{ 325 }{ 60 } = \frac{ 65 }{ 12 } = 5 \frac{ 5 }{ 12 }\)

Q12) \(3 \frac{ 1 }{ 3 } + 4 \frac{ 1 }{ 4 } + 6 \frac{ 1 }{ 6 }\)

Ans. 12) We have:

L.C.M. of 3, 4 and 6 = ( 2 x 2 x3 ) = 12

\(3 \frac{ 1 }{ 3 } + 4 \frac{ 1 }{ 4 } + 6 \frac{ 1 }{ 6 }\)

= \(\frac{ 10 }{ 3 } \; + \; \frac{ 17 }{ 4 } \; + \; \frac{ 37 }{ 6 }\)

= \(\frac{ ( 40 + 51 + 74 ) }{ 12 }\)

{ [ 12 / 3 = 4, 4 x 10 = 40 ], [ 12 / 4 = 3, 3 x 17 = 51 ] and [ 12 / 6 = 2, 2 x 37 = 74 ] }

= \(\frac{ 165 }{ 12 } = \frac{ 55 }{ 4 } = 13 \frac{ 3 }{ 4 }\)

Q13) \(\frac{ 2 }{ 3 } + 3 \frac{ 1 }{ 6 } + 4 \frac{ 2 }{ 9 } + 2 \frac{ 5 }{ 18 }\)

Ans. 13) We have:

L.C.M. of 3, 6 and 9 = ( 2 x 3 x 3 ) = 18

\(\frac{ 2 }{ 3 } \; + \; 3 \frac{ 1 }{ 6 } \; + \; 4 \frac{ 2 }{ 9 } \; + \; 2 \frac{ 5 }{ 18 }\)

= \(\frac{ 2 }{ 3 } \; + \; \frac{ 19 }{ 6 } \; + \; \frac{ 38 }{ 9 } \; + \; \frac{ 41 }{ 18 }\)

= \(\frac{ ( 12 + 57 + 76 + 41 ) }{ 18 }\)

{ [ 18 / 3 = 6, 6 x 2 = 12 ] , [ 18 / 6 = 3 , 3 x 19 = 57 ] , [ 18 / 9 = 2 , 2 x 38 = 76 ] and [ 18 / 18 = 1 , 1 x 41 = 41 ] }

= \(\frac{ 186 }{ 18 } = \frac{ 31 }{ 3 } = 10 \frac{ 1 }{ 3 }\)

Q14) \(2 \frac{ 1 }{ 3 } + 1 \frac{ 1 }{ 4 } + 2 \frac{ 5 }{ 6 } + 3 \frac{ 7 }{ 12 }\)

Ans. 14) We have:

L.C.M. of 3, 4, 6 and 12 = ( 2 x 2 x 3 ) = 12

\(2 \frac{ 1 }{ 3 } + 1 \frac{ 1 }{ 4 } + 2 \frac{ 5 }{ 6 } + 3 \frac{ 7 }{ 12 }\)

= \(\frac{ 7 }{ 3 } \; + \; \frac{ 5 }{ 4 } \; + \; \frac{ 17 }{ 6 } \; + \; \frac{ 43 }{ 12 }\)

= \(\frac{ ( 28 + 15 + 34 + 43 ) }{ 12 }\)

{ [ 12 / 3 = 4, 4 x 7 = 28], [ 12 / 4 = 3, 3 x 5 = 15] , [ 12 / 6 = 2 , 2 x 17 = 34 ] and [ 12 / 12 = 1 , 1 x 43 = 43 ] }

= \(\frac{ 120 }{ 12 } \; = \; 10\)

Q15) \(2 + \frac{ 3 }{ 4 } + 1 \frac{ 5 }{ 8 } + 3 \frac{ 7 }{ 16 }\)

Ans. 15) We have:

L.C.M. of 4, 8 and 16 = ( 2 x 2 x 2 x 2 ) = 16

\(2 + \frac{ 3 }{ 4 } + 1 \frac{ 5 }{ 8 } + 3 \frac{ 7 }{ 16 }\)

= \(\frac{ 2 }{ 1 } + \frac{ 3 }{ 4 } + \frac{ 13 }{ 8 } + \frac{ 55 }{ 16 }\)

= \(\frac{ (32 + 12 + 26 + 55 ) }{ 16 }\)

{ [ 16 / 1 = 16, 16 x 2 = 32 ] , [ 16 / 4 = 4, 4 x 3 = 12 ] , [ 16 / 8 = 2, 2 x 13 = 26 ] and [ 16 / 16 = 1, 1 x 55 = 55 ] }

= \(\frac{ 125 }{ 16 } = 7 \frac{ 13 }{ 16 }\)

Q16) Rohit bought a pencil for Rs. \(3 \frac{ 2 }{ 5 }\) and an eraser for Rs. \(2 \frac{ 7 }{ 10 }\) . What is the total cost of both the articles?

Ans. 16) Total cost of both articles = Cost of pencil + cost of eraser

Thus, we have:

Rs.\(3 \frac{ 2 }{ 5 }\) + Rs. \(2 \frac{ 7 }{ 10 }\) = \(\frac{ 17 }{ 5 } + \frac{ 27 }{ 10 }\) = \(\frac{ ( 34 + 27 ) }{ 10 }\)

( L.C.M. of 5 and 10 = ( 5 x 2 ) = 10 ) = \(\frac{ 61 }{ 10 } = Rs. \; 6 \frac{ 1 }{ 10 }\)

Hence, the total cost of both the articles is Rs. \(6 \frac{ 1 }{ 10 }\)

Q17) Sohini bought \(4 \frac{ 1 }{ 2 }\) m of cloth for her kurta and \(2 \frac{ 2 }{ 3 }\) m of cloth for her pyjamas. How much cloth did she purchase in all?

Ans. 17) Total cloth purchased by Sohini = Cloth for kurta + pyjamas

Thus, we have:

( \(4 \frac{ 1 }{ 2 } + 2 \frac{ 2 }{ 3 }\) ) m = (\(\frac{ 9 }{ 2 } + \frac{ 8 }{ 3 }\) ) m

( L.C.M. of 2 and 3 = ( 2 x 3 ) = 6 ) = (\(\frac{ ( 27 + 16 ) }{ 6 }\)) m

{ [ 6 / 2 = 3, 3 x 9 = 27 ] and [ 6 / 3 = 2, 2 x 8 = 16 ] } = \(\frac{ 43 }{ 6 }\) m = \(7 \frac{ 1 }{ 6 }\) m

Therefore, total length of cloth purchased = \(7 \frac{ 1 }{ 6 }\) m

Q18) While coming back home from his school, Kishan covered \(4 \frac{ 3 }{ 4 }\) km by rickshaw and \(1 \frac{ 1 }{ 2 }\) km on foot. What is the distance of his house from the school?

Ans. 18) Distance from Kishan’s house to school = Distance covered by him by rickshaw + Distance covered by him on foot

Thus, we have:

( \(4 \frac{ 3 }{ 4 } + 1 \frac{ 1 }{ 2 }\) ) km

= \(\frac{ 19 }{ 4 } + \frac{ 3 }{ 2 }\)

= \(\frac{ ( 19 + 6 ) }{ 4 }\) km

\(\frac{ 25 }{ 4 }km = 6 \frac{ 1 }{ 4 }\) km

( L.C.M. of 2 and 4 = ( 2 x 2 ) = 4 )

Hence, the distance from Kishan’s house to school is \(6 \frac{ 1 }{ 4 }\) km

Q19) The weight of an empty gas cylinder is \(16 \frac{ 4 }{ 5 }\) kg and it contains \(16 \frac{ 2 }{ 3 }\) kg of gas. What is the weight of the cylinder filled with gas?

Ans. 19) Weight of the cylinder filled with gas = Weight of the empty cylinder + Weight of the gas inside the cylinder

Thus, we have:

( L.C.M. of 5 and 3 = ( 3 x 5 ) = 15 )

(\(16 \frac{ 4 }{ 5 } + 14 \frac{ 2 }{ 3 }\)) kg

= (\(\frac{ 84 }{ 5 } \; + \; \frac{ 44 }{ 3 }\) ) kg

= \(\frac{ ( 252 + 220 ) }{ 15 }\) kg

= \(\frac{ 472 }{ 15 }\)

= \(31 \frac{ 7 }{ 15 }\)

Hence, the weight of the cylinder filled with gas is \(31 \frac{ 7 }{ 15 }\) kg.

Exercise 5F

Find the difference:

Q1) \(\frac{ 5 }{ 8 } – \frac{ 1 }{ 8 }\)

Ans. 1) Difference of like fractions =

Difference of numerator \(\div\) Common denominator

\(\frac{ 5 }{ 8 } – \frac{ 1 }{ 8 }\)

\(\frac{ ( 5 -1 ) }{ 8 } = \frac{ 4 }{ 8 } = \frac{ 1 }{ 2 }\)

Q2) \(\frac{ 7 }{ 12 } – \frac{ 5 }{ 12 }\)

Ans. 2) Difference of like fractions =

Difference of numerator \(\div\) Common denominator

\(\frac{ 7 }{ 12 } – \frac{ 5 }{ 12 }\)

\(\frac{ ( 7 – 5 ) }{ 12 } = \frac{ 2 }{ 12 } = \frac{ 1 }{ 6 }\)

Q3) \(4 \frac{ 3 }{ 7 } – 2 \frac{ 4 }{ 7 }\)

Ans. 3) Difference of like fractions =

Difference of numerator \(\div\) Common denominator

\(4 \frac{ 3 }{ 7 } – 2 \frac{ 4 }{ 7 }\) =

\(\frac{ 31 }{ 7 } – \frac{ 18 }{ 7 }\)

= \(\frac{ ( 31 – 18 ) }{ 7 }\)

= \(\frac{ 13 }{ 7 }\)

Q4) \(\frac{ 5 }{ 6 } – \frac{ 4 }{ 9 }\)

Ans. 4)

L.C.M. of 6 and 9 = ( 3 x 2 x 3 ) = 18

Now, we have:

\(\frac{ 5 }{ 6 } = \frac{ 5 \times 3 }{ 6 \times 3 } = \frac{ 15 }{ 18 }\)

\(\frac{ 4 }{ 9 } = \frac{ 4 \times 2 }{ 9 \times 2 } = \frac{ 8 }{ 18 }\)

Therefore, \(\frac{ 5 }{ 6 } – \frac{ 4 }{ 9 } = \frac{ 15 }{ 18 } – \frac{ 8 }{ 18 }\)

= \(\frac{ ( 15 – 8 ) }{ 18 }\)

= \(\frac{ 7 }{ 18 }\)

Q5) \(\frac{ 1 }{ 2 } – \frac{ 3 }{ 8 }\)

Ans. 5)\(\frac{ 1 }{ 2 } – \frac{ 3 }{ 8 }\)

L.C.M. of 2 and 8 = ( 2 x 2 x 2 ) = 8

Now, we have:

\(\frac{ 1 }{ 2} = \frac{ 1 \times 4 }{ 2 \times 4 } = \frac{ 4 }{ 8 }\)

Therefore, \(\frac{ 1 }{ 2 } – \frac{ 3 }{ 8 }\) = \(\frac{ 4 }{ 8 } – \frac{ 3 }{ 8 }\) = \(\frac{ ( 4 – 3 ) }{ 8 } = \frac{ 1 }{ 8 }\)

Q6) \(\frac{ 5 }{ 8 } – \frac{ 7 }{ 12 }\)

Ans. 6) \(\frac{ 5 }{ 8 } – \frac{ 7 }{ 12 }\)

L.C.M. of 8 and 12 = ( 2 x 2 x 2 x 3 ) = 24

Now, we have:

\(\frac{ 5 }{ 8} = \frac{ 5 \times 3 }{ 8 \times 3 } = \frac{ 15 }{ 24 }\)

\(\frac{ 4 }{ 9 } = \frac{ 7 \times 2 }{ 12 \times 2 } = \frac{ 14 }{ 24 }\)

Therefore, \(\frac{ 5 }{ 8 } – \frac{ 7 }{ 12 } = \frac{ 15 }{ 24 } – \frac{ 14 }{ 24 }\)

= \(\frac{ ( 15 – 14 ) }{ 4 }\)

= \(\frac{ 1 }{ 24 }\)

Q7) \(2 \frac{ 7 }{ 9 } – 1 \frac{ 8 }{ 15 }\)

Ans. 7)\(2 \frac{ 7 }{ 9 } – 1 \frac{ 8 }{ 15 }\)

= \(\frac{ 25 }{ 9 } – \frac{ 23 }{ 15 }\)

L.C.M. of 9 and 15 = ( 3 x 3 x 5 ) = 45

\(\frac{ 25 }{ 9 } – \frac{ 23 }{ 15 } = \frac{ (125 – 69) }{ 45 } = \frac{ 56 }{ 45 } = 1 \frac{ 11 }{ 45 }\)

{ [ 45 / 9 = 5, 5 x 25 = 125 ] and [ 45 / 15 = 3 , 3 x 23 = 69 ] }

Q8) \(3 \frac{ 5 }{ 8 } – 2 \frac{ 5 }{ 12 }\)

Ans. 8) \(3 \frac{ 5 }{ 8 } – 2 \frac{ 5 }{ 12 }\)

= \(\frac{ 29 }{ 8 } – \frac{ 29 }{ 12 }\)

L.C.M. of 8 and 12 = ( 2 x 2 x 2 x 3 ) = 24

Therefore, \(\frac{ 29 }{ 8 } – \frac{ 29 }{ 12 } = \frac{ ( 87 – 58) }{ 24 } = \frac{ 29 }{ 24 } = 1 \frac{ 5 }{ 24 }\)

{ [ 24 / 8 = 2 , 3 x 29 = 87 ] and [ 24 / 12 = 2 , 2 x 29 = 58 ] }

Q9) \(2 \frac{ 3 }{ 10 } – 1 \frac{ 7 }{ 15 }\)

Ans. 9) \(2 \frac{ 3 }{ 10 } – 1 \frac{ 7 }{ 15 }\)

= \(\frac{ 23 }{ 10 } – \frac{ 22 }{ 15 }\)

L.C.M. of 10 and 15 = ( 2 x 3 x 5 ) = 30

= \(\frac{ ( 69 – 44 ) }{ 30 } = \frac{ 25 }{ 30 } = \frac{ 5 }{ 6 }\)

{ [ 30 / 10 = 3 , 3 x 23 = 69 ] and [ 30 / 15 = 2 , 2 x 22 = 44 ] }

Q10) \(6 \frac{ 2 }{ 3 } – 3 \frac{ 3 }{ 4 }\)

Ans. 10) \(6 \frac{ 2 }{ 3 } – 3 \frac{ 3 }{ 4 }\)

= \(\frac{ 20 }{ 3 } – \frac{ 15 }{ 4 }\)

L.C.M. of 3 and 4 = ( 2 x 2 x 3 ) = 12

= \(\frac{ ( 80 – 45 ) }{ 12 } = \frac{ 35 }{ 12 } = 2 \frac{ 11 }{ 12 }\)

{ [12 / 3 = 4 , 4 x 20 = 80 ] and [ 12 / 4 = 3 , 3 x 15 = 45 ] }

Q11) \(7 \; – \; 5\frac{ 2 }{ 3 }\)

Ans. 11) \(7 \; – \; 5\frac{ 2 }{ 3 }\)

= \(\frac{ 7 }{ 1 } – \frac{ 17 }{ 3 }\)

L.C.M. of 1 and 3 = 3

= \(\frac{ ( 21 – 17 ) }{ 3 } = \frac{ 4 }{ 3 } = 1 \frac{ 1 }{ 3 }\)

{ [ 3 / 1 = 3 , 3 x 7 = 21 ] and [ 3 / 3 = 1 , 1 x 17 = 17 ] }

Q12) \(10 \; – \; 6 \frac{ 3 }{ 8 }\)

Ans. 12) \(10 \; – \; 6 \frac{ 3 }{ 8 }\)

= \(\frac{ 10 }{ 1 } – \frac{ 51 }{ 8 }\)

L.C.M. of 1 and 8 = 8

= \(\frac{ ( 80 – 51) }{ 8 } = \frac{ 29 }{ 8 } = 3 \frac{ 5 }{ 8 }\)

{ [ 8 / 1 = 8 , 8 x 10 = 80 ] and [ 8 / 8 = 1 , 1 x 51 = 51 ] }

Simplify:

Q13) \(\frac{ 5 }{ 6 } – \frac{ 4 }{ 9 } + \frac{ 2 }{ 3 }\)

Ans. 13) We have:

\(\frac{ 5 }{ 6 } – \frac{ 4 }{ 9 } + \frac{ 2 }{ 3 }\)

L.C.M. of 3, 6 and 9 = ( 2 x 3 x 3 ) = 18

= \(\frac{ ( 15 – 8 + 12 ) }{ 18 }\)

{ [ 18 / 6 = 3 , 3 x 5 = 15 ] , [ 18 / 9 = 2 , 2 x 4 = 8 ] and [ 18 / 3 = 6 , 6 x 2 = 12 ] }

= \(\frac{ ( 27 – 8 ) }{ 18 }\)

= \(\frac{ 19 }{ 18 }\)

= \(1 \frac{ 1 }{ 18 }\)

Q14) \(\frac{ 5 }{ 8 } + \frac{ 3 }{ 4 } – \frac{ 7 }{ 12 }\)

Ans. 14) We have:

\(\frac{ 5 }{ 8 } + \frac{ 3 }{ 4 } – \frac{ 7 }{ 12 }\)

L.C.M. of 4, 8, and 12 = ( 2 x 2 x 2 x 3 ) = 24

\(\frac{ ( 15 + 18 – 14 ) }{ 24 }\)

{ [ 24 / 8 = 3 , 3 x 5 = 15 ] , [ 24 / 4 = 6 , 6 x 3 = 18 ] and [ 24 / 12 = 2 , 2 x 7 = 14]}

= \(\frac{ ( 33 – 14 ) }{ 24 }\)

= \(\frac{ 19 }{ 24 }\)

Q15) 2 + \(\frac{ 11 }{ 15 } – \frac{ 5 }{ 9 }\)

Ans. 15) We have :

\(\frac{ 2 }{ 1 } + \frac{ 11 }{ 15 } – \frac{ 5 }{ 9 }\)

L.C.M. of 15 and 9 = ( 3 x 3 x 5 ) = 45

= \(\frac{ ( 90 + 33 – 25 ) }{ 45 }\)

{ [ 45 / 1 = 45 , 45 x 2 = 90 ] , [ 45 / 15 = 3 , 3 x 11 = 33 ] and [ 45 / 9 = 5 , 5 x 5 = 25 ]}

= \(\frac{ ( 90 + 8 ) }{ 45 }\)

= \(\frac{98 }{ 45 }\)

= \(2 \frac{ 8 }{ 45 }\)

Q16) \(5 \frac{ 3 }{ 4 } – 4 \frac{ 5 }{ 12 } + 3 \frac{ 1 }{ 6 }\)

Ans. 16) We have:

\(5 \frac{ 3 }{ 4 } – 4 \frac{ 5 }{ 12 } + 3 \frac{ 1 }{ 6 }\)

= \(\frac{ 23 }{ 4 } – \frac{ 53 }{ 12 } + \frac{ 19 }{ 6 }\)

L.C.M. of 4, 12 and 6 = ( 2 x 2 x 3 ) = 12

= \(\frac{ ( 69 – 53 + 38 ) }{ 12 }\)

{ [ 12 / 4 = 3, 3 x 23 = 69 ], [ 12 / 12 = 1, 1 x 53 = 53 ] and [ 12 / 6 = 2 , 2 x 19 = 38 ]}

= \(\frac{ ( 107 – 53 ) }{ 12 }\)

= \(\frac{54 }{ 12 }\)

= \(\frac{9 }{ 2 }\)

= \(4 \frac{ 1 }{ 2 }\)

Q17) \(2 + 5 \frac{ 7 }{ 10 } – 3 \frac{ 14 }{ 15 }\)

Ans. 17) We have:

\(2 + 5 \frac{ 7 }{ 10 } – 3 \frac{ 14 }{ 15 }\)

= \(\frac{ 2 }{ 1 } + \frac{ 57 }{ 10 } – \frac{ 59 }{ 15 }\)

L.C.M. of 10 and 15 = ( 2 x 5 x 3 ) = 30

= \(\frac{ ( 60 + 171 – 118 ) }{ 30 }\)

{ [ 30 / 1 = 30, 30 x 2 = 60 ], [ 30 / 10 = 3, 3 x 57 = 171 ] and [ 30 / 15 = 2 , 2 x 59 = 118 ]}

= \(\frac{ ( 231 – 118 ) }{ 30 }\)

= \(\frac{113 }{ 30 }\)

= \(3 \frac{ 23 }{ 30 }\)

Q18) \(8 – 3 \frac{ 1 }{ 2 } – 2 \frac{ 1 }{ 4 }\)

Ans. 18) We have:

\(8 – 3 \frac{ 1 }{ 2 } – 2 \frac{ 1 }{ 4 }\)

= \(\frac{ 8 }{ 1 } + \frac{ 7 }{ 2 } – \frac{ 9 }{ 4 }\)

L.C.M. of 1, 2 and 4 = ( 2 x 2 ) = 4

= \(\frac{ ( 32 – 14 – 9 ) }{ 4 }\)

{ [ 4 / 1 = 4, 4 x 8 = 32 ], [ 4 / 2 = 2, 2 x 7 = 14 ] and [ 4 / 4 = 1 , 1 x 9 = 9 ]}

= \(\frac{ ( 32 – 23 ) }{ 34 }\)

= \(\frac{9 }{ 4 }\)

= \(2 \frac{ 1 }{ 4 }\)

Q19) \(8 \frac{ 5 }{ 6 } – 3 \frac{ 3 }{ 8 } + 2 \frac{ 7 }{ 12 }\)

Ans. 19) We have:

\(8 \frac{ 5 }{ 6 } – 3 \frac{ 3 }{ 8 } + 2 \frac{ 7 }{ 12 }\)

= \(\frac{ 53 }{ 6 } – \frac{ 27 }{ 8 } + \frac{ 31 }{ 12 }\)

L.C.M. of 6, 8 and 12 = ( 2 x 2 x 2 x 3 ) = 24

= \(\frac{ ( 212 – 81 + 62 ) }{ 24 }\)

{ [24 / 6 = 4, 4 x 53 = 212 ], [ 24 / 8 = 23, 3 x 7 = 81 ] and [ 24 / 12 = 2 , 2 x 31 = 62 ]}

= \(\frac{ ( 274 – 81 ) }{ 24 }\)

= \(\frac{193 }{ 24 }\)

= \(8 \frac{ 1 }{ 24 }\)

Q20) \(6 \frac{ 1 }{ 6 } – 5 \frac{ 1 }{ 5 } + 3 \frac{ 1 }{ 3 }\)

Ans. 19) We have:

\(6 \frac{ 1 }{ 6 } – 5 \frac{ 1 }{ 5 } + 3 \frac{ 1 }{ 3 }\)

= \(\frac{ 37 }{ 6 } – \frac{ 26 }{ 5 } + \frac{ 10 }{ 3 }\)

L.C.M. of 6, 5 and 3 = ( 2 x 5 x 3 ) = 30

= \(\frac{ ( 185 – 156 + 100 ) }{ 30 }\)

{ [30 / 6 = 5, 5 x 37 = 185 ], [ 30 / 5 = 6, 6 x 26 = 156 ] and [ 30 / 3 = 10 , 10 x 10 = 100 ] }

= \(\frac{ ( 285 – 156 ) }{ 30 }\)

= \(\frac{ 129 }{ 30 }\)

= \(\frac{43 }{ 10 }\)

= \(4 \frac{ 3 }{ 10 }\)

Q21) \(3 + 1 \frac{ 1 }{ 5 } + \frac{ 2 }{ 3 } – \frac{ 7 }{ 15 }\)

Ans. 21) We have:

\(3 + 1 \frac{ 1 }{ 5 } + \frac{ 2 }{ 3 } – \frac{ 7 }{ 15 }\)

= \(\frac{ 3 }{ 1 } + \frac{ 6 }{ 5 } + \frac{ 2 }{ 3 } – \frac{ 7 }{ 15 }\)

L.C.M. of 1, 5, 3 and 15 = ( 5 x 3 ) = 15

= \(\frac{ ( 45 + 18 + 10 – 7 ) }{ 15 }\)

{ [ 15 / 1 = 15, 15 x 3 = 45 ], [ 15 / 5 = 3, 3 x 6 = 18 ] , [ 15 / 3 = 5 , 5 x 2 = 10 ] and [ 15 / 15 = 1, 1 x 7 = 7 ] }

= \(\frac{ ( 73 – 7 ) }{ 15 }\)

= \(\frac{66 }{ 15 }\)

= \(\frac{22 }{ 5 }\)

= \(4 \frac{ 2 }{ 5 }\)

Q22) What should be added to \(9 \frac{ 2 }{ 3 }\) to get 19 ?

Ans. 22) Let x be added to \(9 \frac{ 2 }{ 3 }\) to get 19.

Therefore, \(9 \frac{ 2 }{ 3 }\) + x = 19

Thus, we have:

x = 19 – \(9 \frac{ 2 }{ 3 }\)

= \(\frac{ 19 }{ 1 } – \frac{ 29 }{ 3 }\)

L.C.M. of 1 and 3 is 3.

= \(\frac{ ( 57 – 29 ) }{ 3 }\)

{ [ 3 / 1 = 3 , 3 x 19 = 57 ] and [ 3 / 3 = 1 , 1 x 29 = 29 ] }

= \(\frac{28 }{ 3 }\)

= \(9 \frac{ 1 }{ 3 }\)

Q23) What should be added to \(6 \frac{ 7 }{ 15 }\) to get \(8 \frac{ 1 }{ 5 }\) ?

Ans. 23) Let x be added to \(6 \frac{ 7 }{ 15 }\) to get \(8 \frac{ 1 }{ 5 }\)

Therefore, \(6 \frac{ 7 }{ 15 }\) + x = \(6 \frac{ 1 }{ 5 }\)

Therefore, we have:

x = \(8 \frac{ 1 }{ 5 }\)\(6 \frac{ 7 }{ 15 }\)

= \(\frac{ 41 }{ 5 } – \frac{ 97 }{ 15 }\)

= L.C.M. of 5 and 15 = ( 5 x 3 ) = 15

= \(\frac{ ( 57 – 29 ) }{ 3 }\)

{ [ 15 / 5 = 3 , 3 x 41 = 123 ] and [ 15 / 15 = 1, 1 x 97 = 97 ] }

= \(\frac{26 }{ 15 }\)

= \(1 \frac{ 11 }{ 15 }\)

Q24) Subtract the sum of \(3 \frac{ 5 }{ 9 }\) and \(3 \frac{ 1 }{ 3 }\) from the sum of \(5 \frac{ 5 }{ 6 }\) and \(4 \frac{ 1 }{ 9 }\)

Ans. 24) \(\left (5 \frac{ 5 }{ 6 } + 4 \frac{ 1 }{ 9 } \right ) – \left ( 3 \frac{ 5 }{ 9 } + 3 \frac{ 1 }{ 3 } \right )\)

\(\left ( \frac{ 35 }{ 6 } + \frac{ 37 }{ 9 } \right ) – \left ( \frac{ 32 }{ 9 } + \frac{ 10 }{ 3 } \right )\)

L.C.M. of 3, 6, 9 = ( 2 x 3 x 3 ) = 18

= \(\frac{ [ 105 + 74 ] – [ 64 + 60 ] }{ 18 }\)

{ [ 18 / 6 = 3 , 3 x 35 = 105 ] and [ 18 / 9 = 2 , 2 x 37 = 74 ] }

{ [ 18 / 9 = 2 , 2 x 32 = 64 ] and [ 18 / 3 = 6 , 6 x 10 = 60 ] }

= \(\frac{ [ 179 ] – [ 124 ] }{ 18 } = \frac{ 55 }{ 18 } = 3 \frac{ 1 }{ 18 }\)

Q25) Of \(\frac{ 3 }{ 4 }\) and \(\frac{ 5 }{ 7 }\) , which is greater and by how much ?

Ans. 25) Let us compare \(\frac{ 3 }{ 4 }\) and \(\frac{ 5 }{ 7 }\)

3 x 7 = 21 and 4 x 5 = 20

Clearly, 21 > 20

Therefore, \(\frac{ 3 }{ 4 } > \frac{ 5 }{ 7 }\)

Required difference:

= \(\frac{ 3 }{ 4 } – \frac{ 5 }{ 7 }\)

L.C.M. of 4 and 7 = ( 2 x 2 x 7 ) = 28

= \(\frac{ 21 – 20 }{ 28 }\)

{ [ 28 / 4 = 7 , 7 x 3 = 21 ] and [ 28 / 7 = 4 , 4 x 5 = 20 ] }

= \(\frac{ 1 }{ 28 }\)

Hence, \(\frac{ 3 }{ 4 }\) is greater than \(\frac{ 5 }{ 7 }\) by \(\frac{ 1 }{ 28 }\) .

Q26) Mrs Soni bought \(7 \frac{ 1 }{ 2 }\) litres of milk . Out of his milk, \(5 \frac{ 3}{ 4 }\) litres was consumed. How much milk is left with her ?

Ans. 26) Amount of milk left with Mrs. Soni = Total amount of milk bought by her – Amount of milk consumed

Therefore, Amount of milk left with Mrs. Soni

= \(7 \frac{ 1 }{ 2 }\)\(5 \frac{ 3 }{ 4 }\)

= \(\frac{ 15 }{ 2 } – \frac{ 23 }{ 4 }\)

L.C.M. of 2 and 4 = ( 2 x 2 ) = 4

= \(\frac{ 30 – 23 }{ 4 }\)

{ [ 4 / 2 = 2 , 2 x 15 = 30 ] and [ 4 / 4 = 1 , 1 x 23 = 23 ]}

= \(\frac{ 7 }{ 4 } = 1 \frac{ 3 }{ 4 }\) litres

Therefore, Milk left with Mrs. Soni = \(1 \frac{ 3 }{ 4 }\) litres

Q27) A film show lasted for \(3 \frac{ 1 }{ 3 }\) hours. Out of this time, \(1 \frac{ 3 }{ 4 }\) hours was spent on advertisements. What was the actual duration of the film ?

Ans. 27) Actual duration of the film = Total duration of the show – Time spent on advertisements

= \(\left (3 \frac{ 1 }{ 3 } – 1 \frac{ 3 }{ 4 } \right )\) hours

= \(\left ( \frac{ 10 }{ 3 } – \frac{ 7 }{ 4 } \right )\) hours

L.C.M. of 3 and 4 = ( 2 x 2 x 3 ) = 12

\(\left ( \frac{ 40 – 21 }{ 12 } \right )\) hours

{ [ 12 / 3 = 4 , 4 x 10 = 40 ] and [ 12 / 4 = 3 , 3 x 7 = 21 ] }

= \(\left ( \frac{ 19 }{ 12 } \right )\) hours = \(1 \frac{ 7 }{ 12 }\) hours

Thus, the actual duration of the film was \(1 \frac{ 7 }{ 12 }\) hours .

Q28) In one day, rickshaw puller earned Rs. \(137 \frac{ 1 }{ 2 }\) . Out of his money, he spent Rs. \(56 \frac{ 3 }{ 4 }\) on food. How much money is left with him?

Ans. 28) Money left with the rickshaw puller =

Money earned by him in a day – Money spent by him on food

= Rs. \(\left (137 \frac{ 1 }{ 2 } – 56 \frac{ 3 }{ 4 } \right )\)

L.C.M. of 2 and 4 = ( 2 x 2 ) = 4

= Rs. \(\left ( \frac{ 275 }{ 2 } – \frac{ 227 }{ 4 } \right )\)

{ [ 4 / 2 = 2 , 2 x 275 = 550 ] and [ 4 / 4 = 1 , 1 x 227 = 227 ] }

= Rs. \(\left (\frac{ 550 – 227 }{ 4 } \right )\)

= Rs. \(\left (\frac{ 323 }{ 4 } \right )\)

= Rs. \(80 \frac{ 3 }{ 4 }\)

Hence, Rs. \(80 \frac{ 3 }{ 4 }\) is left with the rickshaw puller .

Q29) A piece of wire , \(2 \frac{ 3 }{ 4 }\) meters long , broke into pieces. One piece is \(\frac{ 5 }{ 8 }\) meters long. How long is the other piece ?

Ans. 29) The length of the other piece =

( Length of the wire – Length of one piece )

= \(\left (3 \frac{ 3 }{ 4 } – \frac{ 5 }{ 8 } \right )\) m

= \(\left ( \frac{ 11 }{ 4 } – \frac{ 5 }{ 8 } \right )\)m

L.C.M. of 4 and 8 = ( 2 x 2 x 2 ) = 8

= \(\left (\frac{ 22 – 5 }{ 8 } \right )\) m

{ [ 8 / 4 = 2 , 2 x 11 = 22 ] and [ 8 / 8 = 1 , 1 x 5 = 5 ] }

= \(\left (\frac{ 17 }{ 8 } \right )\) m

= \(2 \frac{ 1 }{ 8 }\) m

Hence, the other piece is \(2 \frac{ 1 }{ 8 }\) m long .


Practise This Question

Which organism has a streamlined body?