Saturday, 21 February 2015

12. Aptitude Questions with Explanations HCF and LCM - Solved Problems

     Solved Problems on HCF and LCM

Aptitude questions with Explanations - HCF and LCM 
Express  $\frac{1168}{1095}$   in simple form?
a)$\frac{15}{16}$        b)$\frac{16}{15}$         c)$\frac{8}{9}$            d)$\frac{12}{17}$     e)None
Answer: B
Explanation:   To express any fraction in simple (small form), we have to find the HCF of numerator and denominator.

  1095)1168(1
           1095
               73)1095(15
                    1095
                   ———
                         0
 HCF of 1095 and 1168 is 73

Therefore  $\frac{\frac{1168}{73}}{\frac{1095}{73}}\,=\,\frac{16}{15}$

Quantitative Aptitude Questions- HCF and LCM 
Find the HCF of the numbers a and b, where 
a = 25 x 32 x 76 x 114 and n = 23x34x56x11x133?
a)396     b)408       c)792      d)1584    e)None of these
Answer: C
Explanation: We use prime factorization method, because already the numbers m and n written as product of prime factors
      We take only common factors with least powers 
                           HCF = 23 x 32 x 11 = 792 

Arithmetic  Questions with Answers - HCF and LCM 
Find the HCF of of 0.36, 0.48, 0.72 and 2.4
a)0.12     b)0.24       c)0.06       d)0.08     e)0.96
Answer: A
Explanation: Given numbers are decimal fractions.
  Make the number of decimal places in all the given numbers the same
   i.e., 0.36, 0.48, 0.72 and 2.40 this they become
36, 48, 72, 240 (multiply all by 100)
                HCF of 36,48,72,240,=1
Previously we multiplied the given numbers by 100. Now to get the answer divide the obtained HCF by 12
                    HCF of 0.36,0.48,0.72 and 2.4= 0.12

Aptitude Questions with Explanations -HCF and LCM 
What is the side of the largest possible square brick which can be paved on the floor of a room 4m 96cm long and 4m 3cm broad?
a)27       b)28       c)29             d)31        e)37
Answer: D
Explanation:  This is the problem on applications of HCF.
     Converting given measurements into centimeters
             4m 96 cm = 400+96 = 496 cm
              4m 3cm   = 400+3    =403 cm
To find the side of the largest possible brick , the side of brick must divide length as well as breadth exactly.
            Using division method
         403)496(1
                403
                  93) 403 (4
                        370
                           31) 93 (3
                                 93
                                   0
HCF = 31      
Side of the largest possible square brick =31cm 

Quantitative Aptitude Questions & Answers- HCF & LCM 
In a school, the number of students in Sections A , B and C of 10th class are 70,98 and 126 respectively. Due to overload of student in each class , the administration wants to increase the number rooms . What is the minimum number of rooms required, if in each room the same number of students are to seated and all of them being in the same section?
a)18       b)20          c)21         d)23          e)25.
Answer : C
Explanation:  This is the problem on real time applications of HCF 
  The number students in each room 
                      = HCF of 70,98 and 126 =14
            In each room , maximum 14 students can be seated.
                   Total number students = 70 + 98 +126 =294
                Number of rooms required  = $\frac{294}{14}$  =21

      Aptitude Questions with Explanations- HCF and LCM 
Find the greatest number that will divide 147,185 and 251 leaving the remainders 3,5 and 11 respectively.
a) 12             b)15       c)18         d)21          e)23
     Answer: A 
Explanation:  The greatest number when divide 147,185 and 251 leaving the remainders 3,5 and 11.Means the greatest number is not exactly dividing the given numbers. If we subtract the remainders from the respective numbers, then they are exactly divisible by the greatest numbers.
      The required greatest number = HCF of (147 - 3),(185-5) and ( 251-11)
                                                =HCF of 144,180 and 240
                                                =12

       Quantitative Aptitude For GMAT & GRE  - HCF & LCM 
     Find the greatest number that will divide 172, 205 and 304 so as to leave the same remainder in each case?
       a)19           b)21            c)33            d)45            e)49
     Answer : C
     Explanation: The greatest number that will divide 172,205 and 304 leaving the same remainder in each     case is       HCF ((205-172), (304-205), (304-172))
                = HCF of 33, 99, and 132
33=31x11             99=3 x 3 x 11      132=3 x 2 x 2 x11
         Required number =3 x 11=33

Aptitude Questions for CRT - HCF and LCM 
What is the least number of that is exactly divisible by 4, 12, 15, and 18?
a) 120       b)108        c)180       d)240     e)360
Answer: C     
Explanation:    LCM of (4, 12, 15, and 18)
                                      2|4   12   15   18
                                      3|2    6    15     9
                                      2|2    2      5     3
                                         1    1     5     3
 Least number which is exactly divisible by 4,12,15 and 18 is =180

Aptitude Questions with Explanations- HCF and LCM 
LCM of 0.12, 0.15, 0.2 and 0.54?
     a) 5.4       b)10.8    c)16.2       d)2.7   e)None of these
Answer: A
Explanation : Make the number of decimal places in all the given numbers the same i.e., 0.12, 0.15, 0.2 and 0.54
LCM of 12, 15, 20 and 54
2|12   15   20   54
2|6     15    10  27
3|3     15    5    27
5|1      5     5     9
    1      1      1    9
LCM=22×3×5×9=540
                           LCM=$ \frac{540}{10} $ =5.4

Aptitude Questions with Answers  - HCF and LCM 
A, B and C start running around a circular stadium and complete one round in 27 s, 9 s and 36 s, respectively.  In how much time will they meet again at the starting point?
a)1 hour                      b) 1 minute                          c) 1 minute 48 seconds      
d) 2 minutes               e)3 minutes
Answer: C
Explanation :  This problem is on real time applications of LCM
 LCM of 27, 9 and 36 = 108
     So they will meet again at the starting point after 108 s. i.e., 1 min 48 s.

Aptitude Questions and Answers - HCF and LCM 
Three friends Raju , Ramesh and Sunil start running around a circular stadium and complete a single round in 24 s, 36 s and 40 s, respectively. After how many minutes will they meet against at the starting point?
a)1 hour      b) 30 minutes            c)15 minutes              d)6 minutes        e)10 minutes
Answer: D
Explanation : This is the problem on appliations of LCM in real life.
24 = 3 × 2 × 2 × 2 = 3 × 2³
              36 = 3 × 3 × 2 × 2 = 3² × 2²
            and 40 = 2 × 2 × 2 × 5 = 5¹ × 23
            LCM of 24, 36 and 40 = 3² × 2³ × 5
                                         = 9 × 8 × 5 = 360
Hence, they will meet again at the starting point after 360 s, i.e., 6 min

Aptitude Questions for CAT - HCF and LCM 
What is the least number when divided by 4, 5, 6, 8 and 10 leaves 3 as remainder?
a)63      b)117        c)123       d)243       e)363
Answer: C
Explanation:  The least number which when divided by 4,5,6,8 and 10 is the LCM of these numbers. But each time  to get 3 as remainder, we have to add the remainder 3 to the obtained LCM.
LCM of (4, 5, 6, 8, 10)+ 3
=120+3
=123

Quantitative Aptitude Questions - HCF and LCM 
The least number which should be added to 2497 so that sum is divisible by 5, 6, 4, 3?
a)    43       b)23         c)45         d)37     e)46
Answer: B
Explanation:  LCM of 5,6,4,3 is 60.
On dividing 2497 by 60 we get 37 as remainder.
Therefore number to be added is 60 - 37 =23.
Answer is 23.

Aptitude Questions -HCF and LCM 
The HCF of 2 numbers is 11 and LCM is 693.If one of numbers is 77.find the other number ?
a)99     b)198         c)66      d)143       e)None of these
Answer: A
Explanation : Product of two numbers= HCF x  LCM
                        77 x a = 11 x 693
Other number a  = $\frac{11\times693}{77}$ =99.

Quantitative Aptitude - HCF and LCM 
Find largest number of four digits divisible by 12, 15, 18, 27 ?
a)    9820       b)9720   c)9960       d)9760   e)9620
Answer: B
Explanation : The largest 4 digit number is 9999.
We know that , if LCM of given numbers divide a number N , then N is exactly divisible by all the given numbers .
LCM of 12,15,18,27 is 540.
On dividing 9999 by 540 we get 279 as remainder.
Therefore number =9999 - 279 =9720.

Aptitude Questions  - HCF and LCM 
Find the smallest number of two digits which on being divided by 2, 3, 4 and 5 leaves 0, 1, 2 and 3 as remainder respectively.
a)    40           b)52        c)54          d)67           e)58
Answer : E
Explanation: LCM of 2, 3, 4 and 5 is 60.
The smallest number of two digits is 10.
But we also see that the smallest number divisible by 2, 3, 4 and 5 is 60. Therefore, 10 has no significance here.
The difference between divisor and remainder is 2 in each case. So, we subtract 2 from 60.
60 - 2 = 58 Ans.

Aptitude Questions with Solutions- HCF and LCM 
The least positive integer which is a perfect square and also divisible by each of 6,12, 18 and 24?
a)36       b)100        c)144         d)196        e)108
Answer: C
Explanation : Perfect Square : A number when written as product of prime factors , if each prime factor has even number as its power , then the number is perfect square.
    Writing the given number as product of prime factors
              6 = 2 x 3
             12 = 23 x 3
             18 = 2 x 32
             24= 23 x 3
        LCM = 23 x 32 = 72
     But 72 is not the perfect square. But multiples of 72 are also exactly divisible by given numbers.
 72 x 2 =144 is the perfect square.  So answer is 144


Aptitude Questions with Answers - HCF and LCM 
What is the greatest number that will divide 99, 147 and 219 leaving the same reminder in each case?
   a) 396      b) 12      c) 369      d) 3       e) 24
Answer: E
Explanation : Given that, a = 99, b = 147,  c = 219
     Required number = [H.C.F. of (a-b), (b-c), (c-a)]
               Now,
                            (a-b)  = (99 – 147) = 48
                            (b-c)  = (147 – 219) = 72
                            (c-a) =   (219 – 99) = 120
             Also              
           48 = 24 x 3¹
           72 = 23 x 32
           120 = 23 x 3¹ x 5¹
           HCF of 48,72 and 120 = 23 x 3¹ = 8 x 3  = 24


Aptitude Questions and Answers  - HCF and LCM 
What is the least number, which when divided by 5,6,8,10 and 12 leaves the remainder 2 in each case , but when divided by 13 leaves no remainder ?
a)262       b)362     c)960     d)962    e)None of these
Answer : D
Explanation :  LCM of 5,6,8,10 and 12 = 120
But in each case , we have a remainder 2  .So the required number is 120n+2 which is exactly divisible by 13.
              120 n + 2 = 13 x 9 n + 2
       So clearly 3n+2 must be divisible by 13.
  For n=8,  3n +2 is exactly divisible by 13.
     Therefore , the required number = 120n + 2
                                                 = 120 x 8 + 2   = 962

Aptitude Questions with Explanations - HCF and LCM 
HCF of two numbers is 27 and sum of the two numbers is 216. How many such pairs of numbers are possible?
a)0         b)1         c)2             d)3          e)4
Answer : C
Explanation : HCF of two numbers =27 
           Let two numbers are in the ratio a:b , then numbers are 27a  and 27b.
             Given that , sum of two numbers is 216
                                  27a + 27 b = 216
 a+b = 216/27= 8
  Writing the possibilities of a and b to be sum is 8
                 ( 1,7 ), (2,6),(3,5) and (4,4).
              Of these pairs, (1,7) and (3,5) are relatively primes .
 Therefore, 2 such pairs are possible.
And those required numbers are ( 1 x 27 , 7 x 27) and ( 3 x 27, 5 x 27) 
                               = (27 , 189) and (81,135)
  

Quantitative Aptitude for Bank Exams - HCF and LCM 
Find the smallest number which when increased by 5 is divisible by 9, 21, 25 and 30?
a)3145         b)6305    c)3155      d)1605      e)6295
Answer:
Explanation :
LCM of the given numbers.
3 |9, 21, 25, 30
5 |3, 7, 25, 10
   | 3, 7, 5, 2
LCM = 3 × 5 × 3 × 7 × 5 × 2 = 3150
By definition, LCM of a given set of numbers is their smallest common multiple. In other words, it is the smallest numbers which is divisible by all the given numbers.
So, to get the answer , we have to  deduct 5 from the LCM.
3150 – 5 = 3145 Ans.


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