## LCM and HCF Practice Problems – Set 2

LCM and HCF Practice Problems – Set 2
1. A student can divide her books into groups of 5, 9 and 13. what is the smallest possible number of the books ?
a) 487
b) 585
c) 635
d) 705
Explanation:
The smallest possible number of books = L.C.M of 5, 9 and 13.
Therefore, L.C.M of 5,9 and 13 is = 5 x 9 x 13 = 585.

2. A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is  ?
a) 63
b) 31
c) 16
d) 27
Explanation:
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31

3. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to ?
a) 13/125
b) 14/57
c) 11/120
d) 6/41
Explanation:
Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

4. A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?
a) 47
b) 46
c) 37
d) 35
Explanation:
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.

5. In a palace, three different types of coins are there namely gold, silver and bronze. The number of gold, silver and bronze coins is 18000, 9600 and 3600 respectively. Find the minimum number of rooms required if in each room should give the same number of coins of the same type ?
a) 26
b) 24
c) 18
d) 12
Explanation:
Gold coins = 18000 , Silver coins = 9600 , Bronze coins = 3600
Find a number which exactly divide all these numbers
That is HCF of 18000, 9600& 3600
All the value has 00 at end so the factor will also have 00.
HCF for 180, 96 & 36.
Factors of
180 = 3 x 3 x 5 x 2 x 2
96 = 2 x 2 x 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
Common factors are 2x2×3=12
Actual HCF is 1200
Gold Coins (18000/1200) will be 15 rooms
Silver Coins (9600/1200) will be in 8 rooms
Bronze Coins (3600/1200) will be in 3 rooms
Total rooms will be (15+8+3)  =  26 rooms.

6. If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to
a) 55/601
b) 601/55
c) 11/120
d) 120/11
Explanation:
Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.
Therefore, Required sum = 1/a + 1/b = (a + b)/ab = 55/600 = 11/120

7. The largest measuring cylinder that can accurately fill 3 tanks of capacity 98, 182 and 266 litres each, is of capacity ?
a) 7 lit
b) 14 lit
c) 98 lit
d) 42 lit
Explanation:
To know the the measuring cylinder that can fill all the given capacities , they must be divisible by the required number.
98,182,266 all are divisible by 14
So  14 litres  is the largest cylinder that can fill all the given cylinders.
(or)
The other method is take HCF of all given capacities i.e 98, 182 and 266.

8. A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.
a) 107
b) 117
c) 127
d) 137
Explanation:
Let us calculate both the length and width of the room in centimeters.
Length = 6 meters and 24 centimeters = 624 cm
width = 4 meters and 32 centimeters = 432 cm
As we want the least number of square tiles required, it means the length of each square tile should be as large as possible. Further, the length of each square tile should be a factor of both the length and width of the room.
Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48
Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

9. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 1 hour, how many times do they toll together ?
a) 26
b) 25
c) 30
d) 31
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 30 minutes, they will together (60/2)+1=31 times

10. The H.C.F and L.C.M of two numbers are  84 and 21 respectively.  If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is
a) 12
b) 48
c) 84
d) 108