Class 6 Playing with Numbers Exercise 2.8-2

\begin{array}{l} \text {Q6. Determine the longest tape which can be used to measure exactly } \\ \text {the lengths 7 m, 3 m 85 cm and 12 m 95 cm } \\ \\\text {Sol. Given that first length =7 m =700 cm} \quad \quad [\text { 1 m = 100 cm}]\\ \text {Second length =3 m 85 cm=385 cm} \quad \quad [\text { 1 m = 100 cm}]\\ \text {Third length =12 m 95 cm=1295 cm} \quad \quad [\text { 1 m = 100 cm}]\\\therefore \text {The length of the longest tape will be the HCF of 700,385 and 1295} \\\text {Prime factorization of 700}=2 \times 2 \times 5 \times 5 \times 7 \\\text {Prime factorization of 385}=5 \times 7 \times 11 \\ \text {Prime factorization of 1295}=5 \times 7 \times 37 \\\Rightarrow \text {HCF of 700,385, and 1295}=5 \times 7=35 \\\text {Hence, the longest tape length which can be used to measure given lengths is 35 cm} \\ \\\text {Q7. 105 goats, 140 donkeys and 175 cows have to be taken across a river. } \\ \text {There is only one boat which will have to make many trips in order to do so. } \\ \text {The lazy boatman has his own conditions for transporting them. He insists } \\ \text {that he will take the same number of animals in every trip and they have to be } \\ \text {of the same kind. He will naturally like to take the largest possible number } \\ \text {each time. Can you tell how many animals went in each trip?} \\ \\ \ \text {Sol. We have to find the largest possible number of animals means } \\ \text {we have to find the HCF of 105,140, and 175 } \\\text {Prime factorization of 105 }=3 \times 5 \times 7 \\\text {Prime factorization of 140}=2 \times 2 \times 5 \times 7 \\\text {Prime factorization of 175}=5 \times 5 \times 7 \\\Rightarrow \text {HCF of 105,140, and 175}=5 \times 7=35 \\\text {Hence, 35 animals went in each trip.} \\ \\\text {Q8. Two brands of chocolates are available in packs of 24 and 15 respectively. } \\ \text {If I need to buy an equal number of chocolates of both kinds, what is } \\ \text { the least number of boxes of each kind would need to buy? } \\ \\\text {Sol. Let the brand ‘A’ contain 24 chocolates in one packet } \\ \text {and brand ‘B’ contain 14 chocolates in one packet.} \\\text {Equal number of chocolates of each kind can be found out by } \\ \text {taking LCM of the number of chocolates in each packet.} \\ \end{array}\begin{array}{l|l} 2 & 15,24 \\ \hline 2 & 15,12 \\ \hline 2 & 15,6 \\ \hline 3 & 15,3 \\ \hline 5 & 5,1 \\ \hline & 1,1 \end{array}\begin{array}{l} \Rightarrow \text {LCM of 15 and 24 }=2 \times 2 \times 2 \times 3 \times 5=120 \\\therefore \text {Minimum 120 chocolates of each kind should be purchased.} \\\text {Number of boxes of brand ‘A’ which needs to be purchased } = {\large \frac {120}{24}}=5 \\ \text {Number of boxes of brand ‘B’ which needs to be purchased } = {\large \frac {120}{15}}=8 \\\text {Q9. During a sale, colour pencils were being sold in packs of 24 each and } \\ \text {crayons in packs of 32 each. If you want full packs of both and the same } \\ \text {number of pencils and crayons, how many of each would you need to buy?} \\ \\\text {Sol. Required number of pencils and crayons } = \text {LCM of 24 and 32} \\\text {Prime factorization of 24}=2 \times 2 \times 2 \times 3 \\ \text {Prime factorization of 32}=2 \times 2 \times 2 \times 2 \times 2 \\ \therefore \text {LCM of 24 and 32}=2 \times 2 \times 2 \times 2 \times 2 \times 3=96 \\\therefore \text {96 pencils and 96 crayons needed to be bought.} \\\text {Number of pencils pack which needs to be bought } = {\large \frac {96}{24}}=4 \\ \text {Number of crayons pack which needs to be bought } = {\large \frac {96}{32}}=3 \\ \\\text {Q10. Reduce each of the following fractions to the lowest terms:} \\ (i) \quad {\large \frac{161}{207}}\\ (ii) \quad {\large \frac{296}{481}} \\ \\\text {Sol. For reducing any fraction to the lowest terms, we have } \\ \text {to divide its numerator and denominator by their HCF.} \\ \\(i) \quad {\large \frac{161}{207}} \\\text {Prime factorization of 161}=7 \times 23 \\ \text {Prime factorization of 207}=3 \times 3 \times 23 \\ \therefore \text {HCF of 161 and 207}=23 \\\Rightarrow {\large \frac{161}{207}} = {\large \frac {\frac {161}{23}}{\frac {207}{23}} } \\ ={\large \frac {7}{9}} \\(ii) \quad {\large \frac{296}{481}} \\\text {Prime factorization of 296}=2 \times 2 \times 2 \times 37 \\ \text {Prime factorization of 481}=13 \times 37 \\ \therefore \text {HCF of 296 and 481}=37 \\ \Rightarrow {\large \frac{296}{481}} = {\large \frac{\frac {296}{37}}{\frac {481}{37}}}\\ ={\large \frac {8}{13}} \\ \\\text {Q11. A merchant has 120 litres of oil of one kind, 180 litres of another kind } \\ \text {and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil } \\ \text {in tins of equal capacity. What should be the greatest capacity of such a tin? } \\ \\\text {Sol. The maximum capacity of the required tin is the HCF of 120, 180 and 240.} \\ \text {Prime factorization of 120}=2 \times 2 \times 2 \times 3 \times 5 \\\text {Prime factorization of 180}=2 \times 2 \times 3 \times 3 \times 5 \\\text {Prime factorization of 240}=2 \times 2 \times 2 \times 2 \times 3 \times 5 \\\therefore \text {HCF of 120,180, and 240}=2 \times 3 \times 5=60 \\\text {Hence, the required greatest capacity of the tin must be 60 liters.} \end{array}