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6.
The least multiple of 7, which leaves a remainder of 4. when divided by 6, 9, 15 and 18 is:
  • A.
    74
  • C.
    184
  • B.
    94
  • D.
    364
  • Answer & Explanation
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Answer : [D]
Explanation :
 
L.C.M. of 6, 9, 15 and 18 is 90.
 
Let required number be 90k + 4, which is a multiple of 7.
 
Least value of k for which (90k + 4) is divisible by 7 is k = 4
 
   Required number = 90 x 4 + 4 = 364  
 
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7.
L.C.M. of two prime numbers x and y (x > y) is 161 The value of 3y x is:
  • A.
    -2
  • C.
    1
  • B.
    -1
  • D.
    2
  • Answer & Explanation
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Answer : [A]
Explanation :
 
H.C.F. of two prime numbers is 1. Product of numbers = (1 x 161) = 161
 
Let the numbers be a and b. Then, ab = 161.
 
Now, co-primes with product 161 are (1, 161) and (7, 23).
 
Since x and y are prime numbers and x > y, we have x = 23 and y = 7.
 
    3y x = (3 x 7) - 23 = -2    
 
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8.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2:3. the sum of the numbers is:
  • A.
    28
  • C.
    40
  • B.
    32
  • D.
    64
  • Answer & Explanation
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Answer : [C]
Explanation :
 
Let the numbers be 2x and 3x. Then, their L.C.M. = 6x.
 
So, 6x = 48 or x = 8
 
The numbers are 16 and 24.
 
Hence, required sum = (16 + 24) = 40
 
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9.
The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
  • A.
    1677
  • C.
    2523
  • B.
    1683
  • D.
    3363
  • Answer & Explanation
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Answer : [B]
Explanation :
 
L.C.M. of 5, 6, 7, 8 = 840.
 
    Required number is of the form 840k + 3.  
 
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
 
    Required number = (840 x 2 + 3) = 1683  
 
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10.
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
  • A.
    276
  • C.
    322
  • B.
    299
  • D.
    345
  • Answer & Explanation
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Answer : [C]
Explanation :
 
Clearly, the numbers are (23 x 13) and (23 x 14).
 
    Larger number = (23 x 14) = 322  
 
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