HCF and LCM Problems and Solutions

• 8. HCF of
  \begin{aligned}
  2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
  \end{aligned} is

  1. \begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
  2. \begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
  3. \begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
  4. \begin{aligned} 2 \times 3 \times 5 \end{aligned}

  Answer :

  Option C

  Explanation:

  As in HCF we will choose the minimum common factors among the given.. So answer will be third option

• 9. Find the HCF of
  \begin{aligned}
  2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
  \end{aligned}

  1. 128
  2. 126
  3. 146
  4. 434

  Answer :

  Option B

  Explanation:

  HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
  2 * 3*3 * 7

• 10. Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

  1. 15
  2. 16
  3. 30
  4. 31

  Answer :

  Option D

  Explanation:

  LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.
  
  Now 60/2 = 30
  Adding one bell at the starting it will 30+1 = 31

• 11. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is

  1. 10:28 am
  2. 10:30 am
  3. 10:31 am
  4. None of above

  Answer :

  Option C

  Explanation:

  L.C.M. of 60 and 62 seconds is 1860 seconds
  1860/60 = 31 minutes
  
  They will beep together at 10:31 a.m.
  
  Sometimes questions on red lights blinking comes in exam, which can be solved in the same way

• 12. If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.

  1. 2574
  2. 2500
  3. 1365
  4. 1574

  Answer :

  Option A

  Explanation:

  HCF * LCM = 84942, because we know
  
  Product of two numbers = Product of HCF and LCM
  
  LCM = 84942/33 = 2574

• 13. Find the HCF of 54, 288, 360
  

  1. 18
  2. 36
  3. 54
  4. 108

  Answer :

  Option A

  Explanation:

  Lets solve this question by factorization method.
  
  \begin{aligned}
  18 = 2 \times 3^2, 288 = 2^5 \times 3^2, 360 = 2^3 \times 3^2 \times 5
  \end{aligned}
  
  So HCF will be minimum term present in all three, i.e.
  \begin{aligned}
  2 \times 3^2 = 18
  \end{aligned}
  

• 14. Reduce \begin{aligned}
  \frac{803}{876}
  \end{aligned} to the lowest terms.

  1. \begin{aligned} \frac{11}{12} \end{aligned}
  2. \begin{aligned} \frac{23}{24} \end{aligned}
  3. \begin{aligned} \frac{26}{27} \end{aligned}
  4. \begin{aligned} \frac{4}{7} \end{aligned}

  Answer :

  Option A

  Explanation:

  HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12