LCM and HCF
 Three runners running around a circular track can complete one revolution in 2, 4 and 5.5 respectively. When will they meet at starting point ?

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Time at which they meet at starting point
= LCM of 2, 4 and 5.5Correct Option: B
Time at which they meet at starting point
= LCM of 2, 4 and 5.5
= 44 h
 Find the two numbers whose L.C.M. is 1188 and H.C.F. is 9 ?

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Let the two number be 9a and 9b where a and b are two numbers prime to each other. The L.C.M. of 9a and 9b is 9ab.
∴ 9ab = 1188
∴ ab = 132
Now, the possible pairs of factors of 132 are 1 x 132, 2 x 66, 3 x 44, 6 x 22, 11 x 12 of these pairs (2, 66) and (6, 22) are not prime to each other and, therefore, not admissible.
Hence the admissible pairs are
(1, 132), (3, 44), (4, 33), (11, 12)
∴ a = 1, b = 132; or
a=3, b=44 or
a=4, b=33 or
a=11, b = 12
Hence, the possible number are 9, 9 x 13, 9 x 3, 9 x 44, 9 x 4, 9 x 33, 9 x 11, 9 x 12
So option A is correct 27, 396.Correct Option: A
Let the two number be 9a and 9b where a and b are two numbers prime to each other. The L.C.M. of 9a and 9b is 9ab.
∴ 9ab = 1188
∴ ab = 132
Now, the possible pairs of factors of 132 are 1 x 132, 2 x 66, 3 x 44, 6 x 22, 11 x 12 of these pairs (2, 66) and (6, 22) are not prime to each other and, therefore, not admissible.
Hence the admissible pairs are
(1, 132), (3, 44), (4, 33), (11, 12)
∴ a = 1, b = 132; or
a=3, b=44 or
a=4, b=33 or
a=11, b = 12
Hence, the possible number are 9, 9 x 13, 9 x 3, 9 x 44, 9 x 4, 9 x 33, 9 x 11, 9 x 12
So option A is correct 27, 396.
 In a school 391 boys and 323 girls have been divided into the largest possible equal classes. So that there are equal number of boys and girls in each class. What is the number of classes ?

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The largest possible number of persons in a class is given by the H.C.F. of 391 and 323 i,e. 17
∴ No of classes of boys = 391/17 = 23 and
No. of classes of girls = 323 / 17 = 19.Correct Option: B
The largest possible number of persons in a class is given by the H.C.F. of 391 and 323 i,e. 17
∴ No of classes of boys = 391/17 = 23 and
No. of classes of girls = 323 / 17 = 19.
 The product of two number is 7168 and their H.C.F. is 16. find the sum of all possible numbers ?

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Let the number be 16a and 16b, where a and be are two number prime to each other.
∴ 16a x 16b = 7168
∴ ab = 28
Now, the pairs of number whose products is 28, are (28, 1) : (14, 2), (7,4)
14 and 2 which are not prime to each other should be rejected.
Hence, the required answer
= (448 + 16) + (112 + 64) = 640Correct Option: A
Let the number be 16a and 16b, where a and be are two number prime to each other.
∴ 16a x 16b = 7168
∴ ab = 28
Now, the pairs of number whose products is 28, are (28, 1) : (14, 2), (7,4)
14 and 2 which are not prime to each other should be rejected.
Hence, the required answer
= (448 + 16) + (112 + 64) = 640
 The Sum of two number is 1215 and their H.C.F is 81, how many pairs of such number can be formed ? Find them ?

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Let the number be 81a and 81b where a and b are two numbers prime to each other.
∴ 81a + 81b = 1215
⇒ a + b = 1215/81 = 15
Now, find two numbers, whose sum is 15, the possible pairs are (14, 1) , (13, 2), (12, 3), (11, 4), (10, 5), (9, 6 ), (8, 7) of these the only pairs of numbers that are prime to each other are (14, 1) , (13, 2), (11, 4), and (8, 7).
Hence, the required number are
(14 x 81, 1 x 81), (13 x 81, 2 x 81), (11 x 81, 4 x 81), (8 x 81, 7 x 81)
or (1134, 81), (1053, 162), (891, 324), (648, 567)
So, there are four such pairs .Correct Option: D
Let the number be 81a and 81b where a and b are two numbers prime to each other.
∴ 81a + 81b = 1215
⇒ a + b = 1215/81 = 15
Now, find two numbers, whose sum is 15, the possible pairs are (14, 1) , (13, 2), (12, 3), (11, 4), (10, 5), (9, 6 ), (8, 7) of these the only pairs of numbers that are prime to each other are (14, 1) , (13, 2), (11, 4), and (8, 7).
Hence, the required number are
(14 x 81, 1 x 81), (13 x 81, 2 x 81), (11 x 81, 4 x 81), (8 x 81, 7 x 81)
or (1134, 81), (1053, 162), (891, 324), (648, 567)
So, there are four such pairs .