Q) A trader has three different types of oils of volume 870 L, 812 L and 638 L. Find the least number of containers of equal size required to store all the oil without getting mixed.

(Q 26 – 30/4/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s examine & understand the question calmly:

Since all containers need to be of equal size

∴ this size will be common factor of all 3 volumes

we need least number of containers, then the size of each container has to be largest

∴ this will be highest factor

∴ we need to calculate the container size as highest common factor

Step 2: By prime factorisation:

870 = 2 x 3 x 5 x 29

812 = 2 x 2 x 7 x 29

638 = 2 x 11 x 29

∴ HCF of 870, 812, 638 = 2 x 29 = 58

Step 3: To store 870 L, Minimum no. of 29 L containers required = = 15

To store 812 L, Minimum no. of 29 L containers required = = 14

To store 638 L, Minimum no. of 29 L containers required = = 11

Step 4: Minimum no. of 29 L containers required to keep all 3 types of oil

= 15 + 14 + 11 = 40.

Therefore, the least number of containers to store all the oils is 40.

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