Factors, Multiples, GCF & LCM: Part: 2 Exercise problem and solution

Question: 7 What is the minimum number of apples that must be added to the existing stock of 264 apples so that the total stock can be equally distributed among 6, 7 or 8 persons?
Explore: Since the apples need to be divided equally among 6, 7 or 8 people, first we calculate their LCM.
2/(6,7,8) = 3, 7, 4
:. LCM of 6, 7 or 8 = 2x3x7x4 = 168

 The division produces a reminder 96. To make this remainder 0, the minimum no. of apples that must be added
= 168 - 96
= 72
Answer : 72

Question: 8 It takes Russel 20 minutes to inspect a car and Sohel needs 18 minutes to inspect a car. They both start inspecting cars separately at 8.00 am. At certain points of time, both of them will finish inspecting a car at the same time. When will this occur for the first time?
Option: (a) 9.30 am (b) 9.42 am (c) 10.00 am (d) 11.00 am

Explore: Since we are looking for a time that is common to both the inspection, the time must be longer than both 20 & 18 min. We calculate their LCM.

LCM of 20 & 18 = 2x10x9 =180
180 minutes = 180/6 hours = 3 hours
:. After 3 hours they will finish inspecting at the same time i.e. at (8.00 am + 3 hrs) or 11.00 am.
Answer: (d)

Question: 9 What is the smallest number of apples that can be distributed equally among 4, 6, 9 or 15 students having a surplus of two apples each time?
Option: (a) 422 (b) 362 (c) 182 (d) 62 (e) None

Explore: First, let us find the number of apples that may be distributed equally among 4, 6, 9 or 15 students without any remaining. For that we have to calculate the LCM of these numbers.
4 = 2x2
6 = 2x3
9 = 3x3
15 = 3x5
:. LCM = 2x2x3x3x5 = 180
Now, according to the problem, we always have a surplus of 2 apples. Hence, the required number = 180 + 2 = 182
Answer: (c)

Question: 10 What is the largest number of apples not exceeding 440 that can be distributed among three persons in the proportions of 5:6:7?
Option: (a) 440 (b) 430 (c) 432 (d) 420

Explore: Since we want to divide the apples in the proportion 5:6:7, we need a number that is a common multiple of all of them. That is, we need that LCM to find out the least number of apples that may be distributed according to that proportion.
LCM of 5, 6, 7 = 5x6x7 = 210
Multiples of 210 are 210, 420,630 ....
:. Largest number not exceeding 440 is 420.
Answer: (d)

Question: 11 The greatest common factor of two positive integers is A. The least common multiple of the two numbers is B. If one of the number is C, then what is the other one?
Option: (a) ab/c (b) bc/a (c) a/c + b (d) a + b/c (e) None

Explore: We know,
Product of two numbers = (GCL x LCM) of the two numbers.
:. C x the other number = AxB
Or, the other number = AB/C
Answer: (A)

Question: 12 3 & 5 are factors of F. We can conclude that
Option: (a) 3x5 = F (b) 8 is a factor of F (c) F is a multiple of 15 (d) 3 & 5 are the only factors of F (e) 15 is a multiple of F

Explore: Since 3 & 5 are both factors F, we can conclude that,
f/(3x5) = N, where N is an integer.
Or, F = N x 15
So, (c) is definitely true. (A) is true only when N = 1. We cannot say anything about (b). D is false since 1 & F are also factors of itself. And again (e) is true only when N = 1
Answer: (c)

Question: 13 Which of the following must be an integer if x is a positive integer and (4/x) + (5/x) + (6/x) is also an integer?
Option: (a) x/5 (b) 5/x (c) x/30 (d) 30/x (e) None of these.

Explore: (4/x) + (5/x) + (6/x) = (4+5+6)/x = (15/x)
So, if 15/x is an integer, (15/x) x 2 will also be an integer.
(15/x) x 2 = 30/x
Answer: (d)

Please also check first and last Post:

Factors, Multiples, GCF & LCM: Exercise problem and solution

Factors, Multiples, GCF & LCM: Part: 3 Exercise problem and solution

Posted in: Exercise Problem, Factors, GCF, LCM, Multiples, Solutions to Problems