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1. Find the HCF of 1365, 1560 and 1755.
  A.  215
  B.  205
  C.  195
  D.  190
     
   
View Answer

HCF can be find with two methods: (1) Method of prime factors (2) Method of Division
Method 1: In this method, we break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers. The product will be the required HCF.
1365 = (3) x (5) x 7 x (13)
1560 = 2 x 2 x 2 x (3) x (5) x (13)
1755 = 3 x 3 x (3) x (5) x (13)
∴ HCF = 3 x 5x 13 = 195
Method 2: In this method divide the greater number by the smaller number, divide the divisor by the remainder, divide the remainder by the next remainder, and so on until no remainder is left.
The last divisor is the required HCF.
1365 )1560( 1  
  1365    
  195 )1365( 7
    1365  
    0  

∴ 195 is the HCF of 1365 and 1560
Again,
195 )1755( 9
1755
0
Hence, the required HCF = 195

2. A girl bought a certain number of apples Rs 14.40P, he gained 44P by selling some of them for Rs 8. Find at least how much applies he had left with.
  A.  19
  B.  21
  C.  17
  D.  25
     
   
View Answer

Cost price of all apples = 1440 P
Cost price of the apples sold = 800 P - 44 P = 756 P (Since Rs 8 = 800 P)
Now, the HCF of 1440 P and 756 P = 36 P
∴ Highest possible cost price of each apple = 36 P
Again the C.P of the apples left = 1440 P - 756 P = 684 P
∴ The minimum number of apples left = 684 ÷ 36 = 19

3. Find the HCF of 16.5, 0.45 amd 15.
  A.  15
  B.  0.15
  C.  1.5
  D.  None of the Above
     
   
View Answer

The given numbers are equivalent to 16.50, 0.45 and 15.00
Multiply 100 to each term to remove decimal, we get
1650, 45, 1500
The HCF of 1650, 45 and 1500 is 15
Now, divide the HCF by 100, we get 15/100 = 0.15
∴ 0.15 is the required HCF

4. Find the LCM of 18, 24, 60 and 150
  A.  1700
  B.  1820
  C.  1780
  D.  1800
     
   
View Answer

There are two methods for it.
Method 1.
Resolve the given numbers into their prime factors and then find the product of the highest power of all the factors that occur in the given numbers. This product will be the LCM.
18 = 2 x 3 x 3 = 2 x 32
24 = 2 x 2 x 2 x 3 = 23 x 3
60 = 2 x 2 x 3 x 5 = 22 x 3 x 5
150 = 2 x 3 x 5 x 5 = 2 x 3 x 52
Hence, the required LCM = 23 x 32 x 52 = 1800
Method 2:
  2  18, 24, 60,150
  3  9, 12, 30, 75
  5  3, 4, 10, 25
  3  3, 4, 2, 5
  2  1, 4, 2, 5
  2  1, 2, 1, 5
  5  1, 1, 1, 5
    1, 1, 1, 1
LCM = 2 x 3 x 5 x 3 x 2 x 2 x 5 = 1800

5. Find the LCM of 0.6, 9.6 and 3.6
  A.  29.80
  B.  28.6
  C.  28.80
  D.  27.8
     
   
View Answer

The given numbers are equivalent to 0.60, 9.60 and 0.36
Now, find the LCM of 60, 960 and 36
  2  60, 960, 36
  2  30,480,18
  3  15, 240, 9
  5  5, 80, 3
  3  1,16,1
  16  1,1,1
Now LCM = 2 x 2 x 3 x 5 x 3 x 16 = 2880
∴ the required LCM = 28.80

6. Find the HCF of
54 / 9
, 3
9 / 17
and
36 / 51
  A.  
6 / 17
  B.  
7 / 14
  C.  
5 / 17
  D.  None of the Above
     
   
View Answer

Here,
54 / 6
=
6 / 1
, 3
9 / 17
=
60 / 17
and
36 / 51
=
12 / 17

Thus the fractions are
6 / 1
,
60 / 17
and
12 / 17

∴ HCF=
HCF of 6,60,12 / LCM of 1,17,17
=
6 / 17

7. Find the LCM of 4
1 / 2
, 3 and 10
1 / 2
  A.  
63 / 2
  B.  
63 / 1
  C.  
65 / 2
  D.  
17 / 2
     
   
View Answer

4
1 / 2
=
9 / 2
, 10
1 / 2
=
21 / 2

∴ the required LCM =
LCM of 9, 3 and 21 / HCF of 2, 1 and 2
=
63 / 1
= 63

8. The circumference of the front and hind-wheels of cart are 2
2 / 5
and 3
3 / 7
meters respectively. A chalk mark is put one point of contact of each wheel with the ground at any given moment. How far will the cart have travelled so that their chalk marks may be again on the ground at the same time?
  A.  20 meters
  B.  27 meters
  C.  22 meters
  D.  24 meters
     
   
View Answer

A little reflection will show that chalk marks will touch the ground together for the first time after the wheels have passed over a distance which is the LCM of 2
2 / 5
meteres and 3
3 / 7
metres
LCM of
12 / 5
and
24 / 7

LCM of 12 and 24 / HCF of 5 and 7

∴ LCM = 24 metres

9. The circumferences of the wheels of a cart are 6
3 / 14
dm and 8
1 / 18
dm. What is the least distance in which both wheels simultaneously complete an integral number of revolutions? How often will the points of the two wheels which were lowest at the time of starting touch the ground together in 1 km? (without including the touch at the start)
  A.  50 times
  B.  49 times
  C.  45 times
  D.  44 times
     
   
View Answer

LCM of 6
3 / 14
dm and 8
1 / 18
dm
LCM of
87 / 14
and
145 / 18

LCM =
LCM of 87 and 145 / HCF of 14 and 18
=
435 / 2

∴ the required number of revolutions =
2 x 10000 / 435
= 45.9
Hence, 45 times (without including the touch at the start)

10. The LCM of two number is 2079 and their HCF is 27. If one number is 189 then find the other number.
  A.  287
  B.  297
  C.  295
  D.  290
     
   
View Answer

Let the other no. be y then,
LCM x HCF = 1st Number x 2nd number
2079 x 27 = 189 x y
∴ y =
2079 x 27 / 189
= 297

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