Question
The product of two numbers is 3600 and their HCF is 6.
Total number of such pairs of numbers is?  ÂSolution
As HCF is 6 so let numbers be 6x & 6y. It means x & y must be co-prime. Now 6x × 6y = 3600, So xy = 100. Now pairs of (x, y) can be (1,100) , (2,50) , (4, 25) , (5, 20) & (10, 10). Here only (1,100) & (4, 25) are co-prime so number of pairs are 2.Â
More HCF and LCM Questions
63.89% of 549.68 – (739.87 ÷ 5.34) = ? × 11.89Â
What approximate value should come in place of question mark (?) in the following equations?
39.9% of 1720 + 80.2% of 630 = 89.9% of 1280 + ?
416.021 ÷ 3.782 + 13.012 × 24.987 =?
...888.191 + 2.0001 X 7.961= ?
6940 ÷ 28 ÷ 7 =?
16(17/23)Â + 11(15/46)Â - 15(17/25) =? - 19(13/23)
32.12% of 2399.98 + 64.04% of 2499.95 = ? × 15.95
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
(67.2)2 – (8.9)2 – (22.02)2 =?
8 × 1099.95 ÷ 20.03 + 187.95 = ? × 3.96
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