Time at which they meet at starting point
= LCM of 2, 4 and 5.5

Correct Option: B

Time at which they meet at starting point
= LCM of 2, 4 and 5.5
= 44 h

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.

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.

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

Correct 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

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 .