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Exercise:

Exercise 1

01. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: 
A. 504 B. 536
C. 544 D. 548

Answer and Explanation

Answer: 548

Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.

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02. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is: 
A. 75 B. 81
C. 85 D. 89

Answer and Explanation

Answer: 85

Explanation:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = 551 / 29   = 19; Third number = 1073/ 29 = 37.

Required sum = (19 + 29 + 37) = 85.

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03. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ? 
A. 26 minutes and 18 seconds B. 42 minutes and 36 seconds
C. 45 minutes D. 46 minutes and 12 seconds

Answer and Explanation

Answer: 46 minutes and 12 seconds

Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

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04. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: 
A. 101 B. 107
C. 111 D. 185

Answer and Explanation

Answer: 111

Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.

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05. The least number which is a multiple of 11 and when divided by 3, 5 and 9 leaves 2 as remainder:
A. 297 B. 319
C. 407 D. None of these

Answer and Explanation

Answer: 407

Explanation:
The L.C.M. of 3, 5 and 9 is 45.
Hence, the number is of the form (45a + 2).
The least value of ‘a’ for which (45a + 2) is multiple of 11 is 9.
Therefore, the number is 407.

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06. 3 bells beep at an interval of 12, 20, and 35 minutes. If they beep together at 10 a.m., then they will again beep together at:
A. 12 p.m. B. 1 p.m.
C. 4 p.m. D. 5 p.m

Answer and Explanation

Answer: 5 p.m

Explanation:
The L.C.M. of 12, 20 and 35 is 420. Hence, all 3 bells beep together after 420 minutes = 7 hours
Hence, the 3 bells will beep together 7 hours after 10 a.m. i.e. at 5 p.m.

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07. A man decides to pave with square tiles his hall which is 4.8 metres long and 5.38 metres wide. Find the largest size of the tile that he could use.
A. 21 cm B. 2 cm
C. 269 cm D. None of these

Answer and Explanation

Answer: 2 cm

Explanation:
The largest size of the tile is H.C.F. of 480 cm and 538 cm is 2 cm.

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08. Find the least number that is a perfect cube and can be divisible by 8, 14, and 18.
A. 5832 B. 74088
C. 504 D. Cannot be determined

Answer and Explanation

Answer: 74088

Explanation:
A number that is divisible by 8, 14 and 18 = L.C.M. of 8, 14 and 18 = 504
504 = 2*2*2*3*3*7
Hence, to get a perfect cube 504 should be multiplied by 3*7*7 = 147
The number is 74088.

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09. If the product of the L.C.M. and H.C.F. of two numbers is 60 and the difference between the two numbers is 4, find the numbers.
A. 6 and 10 B. 8 and 12
C. 4 and 8 D. Cannot be determined

Answer and Explanation

Answer: 6 and 10

Explanation:
H.C.F. * L.C.M = product of the two numbers
H.C.F.*L.C.M = 60 = product of the two numbers
Let one of the numbers be x. The second number is (x+4)
60 = product of the two numbers = x*(x+4)
x2 + 4x – 60 = 0 = (x + 10)*(x – 6)
x = 6
Therefore, the numbers are 6 and 10.

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