Question

Sarfaraz and Khalid got a big packet of candy each as their birthday gifts. Sarfaraz’s packet has 90 gems and the numbers of yellow, pink, and blue gems are three prime numbers. The number of yellow gems is the least and the number of blue gems is highest. Khalid’s packet has 78 gems and the numbers of yellow, pink, and blue gems are three consecutive even numbers in the same order. The number of blue gems is the highest among the three even numbers. Find the LCM of the minimum number of blue gems in Sarfaraz’s packet and the number of yellow gems in Khalid’s packet.

• A. 1128
• B. 1356
• C. 1542
• D. 1824
• E. 1248

Answer 1

The Correct Option is A

Let the numbers of yellow, pink, and blue gems in Sarafaraz’s packet be \(‘a’, ‘b’\), and \(‘c’\), respectively, where \(a, b,\) and \(c\) are prime numbers.
So, \(a + b + c = 90\) such that \(a < b < c\).
Here, one or all the three numbers must be even as the sum is even
Also, as \(2\) is the only even prime number all the numbers cannot be even.
So, by hit and trial, the possible values of \(a, b\), and \(c\) are as follows:
\((2, 5, 83), (2, 17, 71), (2, 29, 59)\), or \((2, 41, 47)\)
So, the minimum number of blue gems in Sarfaraz’s packet = \(47\)
Let the numbers of yellow, pink, and blue gems in Khalid’s packet be ‘\(2x’, ‘2x + 2’\), and \(‘2x + 4’\), respectively.
So, \(2x + 2x + 2 + 2x + 4 = 78\)
\(⇒ 6x = 72\)
\(⇒ x = 12\)
Thus, the number of yellow gems in Khalid’s packet = \(24\)
So, the LCM of \(47\) and \(24\) is \(1128\).

Hence, option A is the correct answer.