**Q1 :** Three tankers contain 40 litres, 60 litres and 80 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times ?

**Solution : ** The required container has to measure the three tankers in a way that thecount is exact number of times. So, its capacity must be an exact divisor of the capacities of all the tankers.

Moreover, this capacity should be maximum.

Thus the maximum capacity of such container will be the HCF of 40 , 60, 80.

Factors of 40 = 1, 2, 4, 5

Factors of 60 = 1, 3, 4, 5.

Factors of 80 = 1, 2, 5, 8

The common factors of 40, 60, 80 are 1, 5

The HCF of 40, 60, 80 is 5.

Therefore maximum capacity of the required container is 5 litres.

It will fill first tanker in 8 , second 12 and third in 16 refills.

**Q2 :** Two cans contain 12 and 15 litres of milk respectively. Find a can of maximum capacity which can measure the milk in the two cans integral number of times ?

**Solution : ** To get maximum capacity of the container we need to find HCF of 12 and 15.

The factors of 12 are 1, 3, 4

The factors of 15 are 1, 3, 5

The common factors are 1, 3.

The HCF of 12 and 15 is 3.

Therefore maximum capacity of the required container is 3 litres.

**Q3 : **Find the greatest number which divides 39 and 79 leaving remainder 3 in each case ?

**Solution : **The numbers we get by subtracting 3 from 39 and 79 are 36 and 76 .

The HCF of 36 and 76 is the required answer.

The factors of 36 are 1, 4, 9

The factors of 76 are 1, 4, 19

The common factors are 1, 4

The HCF of 36 and 76 is 4.

Therefore greatest number which divides 39 and 79 leaving remainder 3 in each case is 4.

**Q4 : **What is the greatest number that divides 690 and 875 leaving remainders 10 and 25 respectively ?

**Solution : ** 690 - 10 = 680 ; 875 - 25 = 850.

The HCF of 680 and 850 is the required answer.

850 =680 x 1 + 170

680 = 170 x 4 + 0

Therefore HCF of 680 and 850 is 170 .

Hence the greatest number that divides 690 and 875 leaving remainders 10 and 25 respectively is 170 .

**Q5 :** A bookseller purchased 117 books out of which 45 are of Mathematics and 72 are of Physics . Each book has same size. Mathematics and Physics books are to be packed in separate bundles and each bundle must contain same number of books. Find the least number of bundles which can be made for these 117 books ?

**Solution :** Number of Mathematics books = 45

Number of Physics books = 72.

Let us find the HCF of 45 and 72.

72 = 45 x 1 + 27

45 = 27 x 1 + 18

27 = 18 x 1 + 9

18 = 9 x 2 + 0

Therefore HCF of 45 and 72 is 9 .

Now , each bundle has to contain 9 books so that the number of bundles made should be least.

The number of bundles of Physics books = 72/9 = 8

The number of bundles of Mathematics books = 45/9 = 5

Total number of bundles made = 8 + 5 = 13.