Class 6 Playing with Numbers Exercise 2.10-2

\begin{array}{l} \text {Q6. In a morning walk three persons step off together. Their steps measure 80 cm, } \\ \text {85 cm and 90 cm respectively. What is the minimum distance each should walk so } \\ \text {that he can cover the distance in complete steps? } \\ \\\text {Sol. Prime factorization of 80} =2 \times 2 \times 2 \times 2 \times 5 \\ \text {Prime factorization of 85}=5 \times 17 \\ \text {Prime factorization of 90}=2 \times 3 \times 3 \times 5 \\\therefore \text {LCM of 80, 85 and 90 } \\ =2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 17\\ =12240 \\\text {Hence, the required minimum distance each should walk} \\ = \text {LCM of 80 cm, 85 cm and 90 cm } \\ =\text {12240 cm =122 m 40 cm} \\ \\\text {Q7. Determine the number nearest to 100000 but greater than 100000 which is } \\ \text {exactly divisible by each of 8,15 and 21 } \\ \\\text {Sol. In order to determine the number we must find LCM of 8,15 and 21.} \\\text {Prime factorization of 8}=2 \times 2 \times 2 \\ \text {Prime factorization of 15}=3 \times 5 \\ \text {Prime factorization of 21}=3 \times 7 \\\therefore \text {LCM of 8,15 and 21} \\ =2 \times 2 \times 2 \times 3 \times 5 \times 7 \\ =840 \\\therefore \text {840 is the number nearest to 100000 and divisible by 8,15 and 21.} \\\text {But we need a number greater than 100000 so to find that number we will divide 100000 by 840 } \\\text {On dividing we get remainder as 40.} \\\text {So the number greater than 100000 and exactly divisible by 840} \\ =100000 + 840-40 =100000+800= 100800 \\\text {Hence, the required number is 100800.} \\ \\\text {Q8. A school bus picking up children in a colony of flats stops at every sixth block of flats. } \\ \text {Another school bus starting from the same place stops at every eighth blocks of flats. } \\ \text {Which is the first bus stop at which both of them will stop? } \\ \\\text {Sol. We know that the first bus stop at which both buses will stop } \\ =\text {LCM of 6 and 8} \\ \text {Prime factorization of 6} =2 \times 3 \\ \text {Prime factorization of 8}=2 \times 2 \times 2 \\\therefore \text {LCM of 6 and 8} \\ =2 \times 2 \times 2 \times 3=24 \\\therefore \text {On 24th block bus stop is the first bus stop } \\ \text {where both the buses will stop.} \\ \\\text {Q9. Telegraph poles occur at equal distances of 220 m along a road and heaps of stones are } \\ \text {put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. } \\ \text {How far from it along the road is the next heap which lies at the foot of a pole? } \\\text {Sol. Prime factorization of 220}=2 \times 2 \times 5 \times 11 \\ \text {Prime factorization of 300}=2 \times 2 \times 3 \times 5 \times 5 \\\therefore \text {LCM of 220 and 300} \\ =2 \times 2 \times 3 \times 5 \times 5 \times 11 \\ =3300 \\\therefore \text {The next heap which lies at the foot of a pole is 3300 m far along the road.} \\ \\\text {Q10. Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively. } \\ \\\text {Sol. We know that LCM of 28 and 32 is the smallest number, } \\ \text {which when divided by 28 and 32 leaves remainder O.} \\\text {Prime factorization of 28}=2 \times 2 \times 7 \\ \text {Prime factorization of 32}=2 \times 2 \times 2 \times 2 \times 2 \\\therefore \text {LCM of 28 and 32}\\ =2 \times 2 \times 2 \times 2 \times 2 \times 7 \\ =224 \\\text {We know that Dividend = divisor } \times \text {Quotient + Remainder} \\ \text {Lets factorize 224 with 28 as one of the factor, we get } 224=28 \times 8+0 \\ \text {Now, we need 8 as remainder when divided by 28 } \\\Rightarrow 224=216+8 \\\Rightarrow 224=(28 \times 7 + 20 ) + 8 \\\Rightarrow 224-20=28 \times 7+8 \ldots \ldots \ldots \ldots (1) \\\text {Similarly, lets factorize 224 with 32 as one of the factor, we get } \\ 224=32 \times 7+0 \\ \text {Now, we need 12 as remainder when divided by 32 } \\ \Rightarrow 224=212+12 \\ \Rightarrow 224=(32 \times 6+20)+12 \\ \Rightarrow 224-20=32 \times 6+12 \ldots \ldots \ldots \ldots (2) \\\text {Now, we can see that LHS of both eq (1) and eq(2) are equal.} \\ \text {Divisor’s are 28 and 32, also required remainder’s are also 8 and 12 respectively.} \\ \text {Hence, the smallest dividend }=224-20=204 \\ \end{array}