Find the least number which when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7.

Given that,
x = 130, y = 305, z = 245
a = 6, b = 9, c = 17

According to the formula,
Required number = HCF of [(130 - 6), (305 - 9), (245 - 17)]
= HCF of 124, 296, 228 = 4.

Let HCF = N
According to the question, LCM = 20N
Given that, HCF + LCM = 2520
? N + 20N = 2520
? N = 2520 / 21 = 120
Now, LCM = 20N = 20 x 120 = 2400

We know that,
1st number x 2nd number = HCF x LCM
? 2nd number = (LCM x HCF) / (1st number)
= (120 x 2400) / 480 = 600
? Required answer = 600 x 3 = 1800

Let LCM = m, HCF = n,

According to the question,
m = 12n, .......(i)
and m + n = 403.......(ii)
? 12n + n = 403 [from Eq.(i)]
? 13n = 403
? n = 403/13 = 31
? m = 12 x 31 = 372

Let the another number = k
? 93 x k = 372 x 31
[as product of two numbers HCF x LCM]
? k = (372 x 31)/93 = 124

By Hit and Trial
From option (d), we can say that the HCF of 36 and 24 is 12 and it is also satisfies the given condition
a > b > 12
? a = 36 and b = 24

Let numbers be 10N, 12N, 15N and 18N.
Then, LCM = 180N and HCF = N
Hence, required LCM = 180 x 3 = 540

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.

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

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 .