AnswersRelated What least number must be subtracted from 1936 so that the remainder when divided by 9.10,15 will leave in each case the same remainder unseen 7?a)30b)39c)23d)29Correct answer is option 'B'. Can you explain this answer?
The L.C.M. of 9.10 and 15 = 90.On dividing 1936 by 90. the remainder = 46. But 7 is also a pait of this remainder. Therefore, the required number = 46 - 7 = 39.
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LCM of 9, 10 and 15 = 90⇒ The multiple of 90 are also divisible by 9, 10 or 15.∴ 21 × 90 = 1890 will be divisible by them.∴ Now, 1897 will be the number that will give remainder 7. ∴ Required number = 1936 – 1897 = 39
The answer for this question is 39. Explanation : Step 1 - Take LCM (9,10,15) = 90 Step 2 - Divide 1936 by 90. 1936÷90 = 21 (quotient) and 46(remainder) Step 3 - To get 7 as remainder you must subtract 39. As 46–39 = 7 Result : 1936–39 = 1897 1897÷9 = 210(quotient) + 7(remainder) 1897÷10 = 189(quotient) + 7(remainder) 1897÷15 = 126(quotient) + 7(remainder |
Which least number must be subtracted from 1936 so that the remainder when divided by 9 10 and 15 will leave in each case the same remainder?
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