LCM and HCF
 What least number must be subtracted from 1294 so that the remainder when divided by 9, 11,13 will leave in each case the same remainder 6 ?

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The number when divided by 9, 11, 13 leaving remainder 6 = (i.c.m. of 9, 11, 13) + 6 = 1293
Required least number = 1294  1293 = 1.Correct Option: B
The number when divided by 9, 11, 13 leaving remainder 6 = (i.c.m. of 9, 11, 13) + 6 = 1293
Required least number = 1294  1293 = 1.
 Find the number lying between 900 and 1000 which when divided by 38 and 57, leaves in each case a remainder 23 ?

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The least common multiple of 38 and 57 is 114 and the multiple which is between 900 and 1000 is 912.
Now, 912 + 23 i,e 935 lies between 900 and 1000 and when divided by 38 and 57 leaves in each case 23 as the remainder.
Therefore, 935 is number required.Correct Option: A
The least common multiple of 38 and 57 is 114 and the multiple which is between 900 and 1000 is 912.
Now, 912 + 23 i,e 935 lies between 900 and 1000 and when divided by 38 and 57 leaves in each case 23 as the remainder.
Therefore, 935 is number required.
 Find the sum of three numbers which are prime to one another such that the product of the first two is 437 and that of the last two is 551 ?

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From the question we see that the second number is a common factor of the two products and since the numbers are prime to one another. It is their . H. C. F . and is, therefore, 19.
∴ The first number = 437 / 19 = 23
and the third number = 551 / 19 = 29
Hence, the number are 23, 19 and 29
∴ Sum = 23 + 19 + 29
= 71.Correct Option: C
From the question we see that the second number is a common factor of the two products and since the numbers are prime to one another. It is their . H. C. F . and is, therefore, 19.
∴ The first number = 437 / 19 = 23
and the third number = 551 / 19 = 29
Hence, the number are 23, 19 and 29
∴ Sum = 23 + 19 + 29
= 71.
 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 7 ?

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The L.C.M of 9, 10, 15 = 90
On dividing 1936 by 90, the remainder = 46
But a part of this remainder = 7
Hence, the two numbers = 46  7 = 39.Correct Option: C
The L.C.M of 9, 10, 15 = 90
On dividing 1936 by 90, the remainder = 46
But a part of this remainder = 7
Hence, the two numbers = 46  7 = 39.
 An inspector of schools wishes to distribute 84 balls and 180 bats equally among a number of boys. Find the greatest number receiving the gift in this ways ?

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Find the H.C.F. of 84 and 180, which is 12 and this is the required answer .
Correct Option: D
Find the H.C.F. of 84 and 180, which is 12 and this is the required answer .