{"id":7310853,"date":"2020-07-30T11:59:58","date_gmt":"2020-07-30T06:29:58","guid":{"rendered":"https:\/\/entri.app\/blog\/?p=7310853"},"modified":"2020-07-30T16:35:31","modified_gmt":"2020-07-30T11:05:31","slug":"lcm-and-hcf-questions-for-bank-exam","status":"publish","type":"post","link":"https:\/\/entri.app\/blog\/lcm-and-hcf-questions-for-bank-exam\/","title":{"rendered":"LCM And HCF Questions for Bank, SSC, RRB Exam 2020"},"content":{"rendered":"

HCF and LCM is one of the most frequently asked topics in most of the exams. HCF stands for Highest Common Factor.<\/strong> It\u00a0is the greatest number that divides each of the two or more numbers. <\/strong>HCF\u00a0is also called the Greatest Common Measure (GCM) and Greatest Common Divisor (GCD). <\/strong>LCM stands for Least Common Multiple. <\/strong>It is the smallest common multiple of any two or more numbers. Here are more details on them and some of the LCM and HCF questions for Bank exam<\/a> 2020.<\/p>\n

HCF of numbers can be found out using two methods<\/strong>:<\/p>\n

    \n
  1. Prime factorization method<\/li>\n
  2. Division method<\/li>\n<\/ol>\n

    HCF by Prime factorization Method:<\/strong><\/h3>\n

    Step 1: Write each number as the product of its prime factors (Hence the name prime factorization method)<\/p>\n

    2: List the common factors of each number.<\/p>\n

    3: The product of all the common prime factors in the smallest power is the HCF.<\/p>\n

    For example<\/strong>,<\/h4>\n

    Find the HCF of 42,63 and 140. (L.I.C.A.D.O 2008)<\/p>\n

    42 = 2x3x7<\/p>\n

    63 = 3x3x7<\/p>\n

    140 = 2x2x5x7<\/p>\n

    Therefore, the common factor with the least power is 7. Hence, HCF is 7.<\/p>\n

    HCF by Division Method:<\/strong><\/h3>\n

    Step 1: Divide the larger number by smaller one.<\/p>\n

    2: Divide the divisor by the remainder.<\/p>\n

    3: Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as the remainder.<\/p>\n

    Finding HCF of three or more numbers:<\/strong><\/h3>\n

    Suppose we have to find the HCF of three numbers.<\/p>\n

    Step 1: Find the HCF of any two numbers.<\/p>\n

    2: Now, repeat the process with this new HCF and the third number to get the HCF of all the three numbers.<\/p>\n

    For example<\/strong>,<\/h4>\n

    Find the HCF of 513,1134,1215.<\/p>\n

    Step 1: HCF of 1134 and 1215<\/p>\n

    2: Required HCF = HCF of (HCF of 1134,1215) and 513.<\/p>\n

    \"hcff\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"hcff<\/p>\n

    Hence, required HCF is 27.<\/p>\n

    LCM can be found using two methods<\/strong>:<\/p>\n

      \n
    1. Factorization method<\/li>\n
    2. Common division method (short cut method)<\/li>\n<\/ol>\n

      LCM by Factorization Method<\/strong><\/h3>\n

      Step 1: Resolve each one of the given numbers into products of prime factors.<\/p>\n

      2: LCM is the highest powers of all the factors.<\/p>\n

      For example<\/strong>,<\/h4>\n

      Find the LCM of 87 and 145.<\/p>\n

      87 = 3×29<\/p>\n

      145 = 5×29<\/p>\n

      Therefore, LCM = 3x5x29<\/p>\n

      = 435.<\/p>\n

      LCM by Common Division Method<\/strong><\/h3>\n

      Step 1: Arrange the given numbers in a row in any order.<\/p>\n

      2: Divide by a number which divides exactly at least two of the given numbers and carry forward the\u00a0 numbers which are not divisible.<\/p>\n

      3: Repeat the above process till no two of the numbers are divisible by the same except 1.<\/p>\n

      4: The product of the divisors and the undivided numbers is the LCM.<\/p>\n

      For example<\/strong>,<\/h4>\n

      Find the LCM of 16,24,36 and 54.<\/p>\n

      \"lcm\"<\/p>\n

      Hence, LCM = 2x2x2x3x3x2x1x1x3 = 432.<\/p>\n

      Some of the important points to keep in mind:<\/h3>\n

      Product of two numbers = Product of their HCF and LCM<\/p>\n

      HCF of fractions = HCF of numerators\/ LCM of denominators<\/p>\n

      LCM of fractions = LCM of numerators \/ HCF of denominators<\/p>\n

      Prepare for all exams using Entri app<\/a><\/p>\n

      LCM and HCF Questions for Bank Exam 2020<\/h2>\n

      Here are some questions that appeared in various examinations:<\/p>\n

      1.<\/strong>Find the LCM of 2x2x3x3x3x5x7x7, 2x2x2x3x3x5x5x7x7x7x7, 2x3x5x5x5x7x11\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (L.I.C.A.D.O ,2008)<\/strong><\/p>\n

      Answer:<\/strong> LCM = Product of highest powers of 2,3,5,7 and 11<\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 7 x 7 x 7 x 7 x 11<\/p>\n

      2.<\/strong>Two numbers are in the ratio 3:4. Their LCM is 84, Find the numbers\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (SSC, 2010)<\/strong><\/p>\n

      Answer<\/strong>: Let the numbers be 3x and 4x. Then, their LCM = 12x<\/p>\n

      So, 12x = 84 or x=7<\/p>\n

      Therefore, the numbers are 21 and 28.<\/p>\n

      3. <\/strong>The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.\u00a0 \u00a0 \u00a0 (PSC,2009)<\/strong><\/p>\n

      Answer:<\/strong>\u00a0 Product of numbers = Product of their HCF and LCM<\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 77x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 11*693<\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0X\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = (11*693)\/77 = 99.<\/p>\n

      4. <\/strong>Find the least number which is exactly divisible by 32,36,45,60 and 80\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(RRB 2006)<\/strong><\/p>\n

      Answer:<\/strong> Required number = LCM of 32,36,45,60,80<\/p>\n

      = 2x2x2x3x5x3x2x2<\/p>\n

      = 1440.<\/p>\n

      \"lcm<\/p>\n

       <\/p>\n

      5.<\/strong>Find the smallest number which when increased by 10 is completely divisible by 12, 15, 18, 20 and 24. (PCS 2006)<\/strong><\/p>\n

      Answer:<\/strong> Required number = LCM of (12, 15, 18, 20, 24)-10<\/p>\n

      = 360-10<\/p>\n

      = 350.<\/p>\n

      Check this article to learn more about how to improve calculation speed to crack the bank exams\u00a0<\/a><\/strong><\/p>\n

      6. <\/strong>Find the greatest number of five digits which is divisible by 15, 21 and 36.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(PCS 2006)<\/strong><\/p>\n

      Answer:<\/strong> Greatest number of five digits = 99999<\/p>\n

      Required number must be divisible by LCM of 15, 21, 36 = 1260<\/p>\n

      On dividing 99999 by 1260, we get 459 as remainder.<\/p>\n

      Required number = 99999-459 = 99540.<\/p>\n

      7. <\/strong>Find the largest number of five digits which when divided by 16, 24, 30 or 36, leaves the same remainder 10 in each case.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(CPO 2007)<\/strong><\/p>\n

      Answer:<\/strong> Largest five-digit number = 99999<\/p>\n

      LCM of 16, 24, 30, 36 = 720<\/p>\n

      On dividing 99999 by 720, remainder obtained = 639<\/p>\n

      Thus, largest number of five-digits divisible by 16, 24, 30, 36 = 99999-639 = 99360<\/p>\n

      Hence, required number = 99360 + 10 = 99370.<\/p>\n

      8. <\/strong>Seema, Meena and Reema begin to jog around a circular stadium and they complete their revolutions in 54 seconds, 42 seconds and 63 seconds respectively. After how much time will they come together at the starting point?\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Bank P.O 2010)<\/strong><\/p>\n

      Answer:<\/strong>\u00a0 LCM of 54, 42 and 63 = 378<\/p>\n

      So, the three girls will come together at the starting point in 378 seconds i.e., 6 minutes 18 seconds.<\/p>\n

      9. <\/strong>The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8:20:00 hours, then at what time will they again change simultaneously?\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (RRB 2006)<\/strong><\/p>\n

      Answer:<\/strong> Interval of change = LCM of 48,72,108 = 432 seconds.<\/p>\n

      So, the lights will again change simultaneously after every 432 seconds, i.e., 7 minute 12 seconds.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Hence, next simultaneous change will take place at 8:27:12 hours.<\/p>\n

      10.<\/strong>When Seeta made necklaces of either 16 beads, 20 beads or 36 beads, not a single bead was left over. What could be the least number of beads Seeta had?\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Bank recruitment 2008)<\/strong><\/p>\n

      Answer<\/strong>: Required number of beads = LCM of 16, 20, 36 = 720.<\/p>\n

      Prepare for all exams using Entri app<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

      HCF and LCM is one of the most frequently asked topics in most of the exams. HCF stands for Highest Common Factor. It\u00a0is the greatest number that divides each of the two or more numbers. HCF\u00a0is also called the Greatest Common Measure (GCM) and Greatest Common Divisor (GCD). LCM stands for Least Common Multiple. It […]<\/p>\n","protected":false},"author":56,"featured_media":7432490,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[802,246],"tags":[],"class_list":["post-7310853","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-articles","category-banking"],"acf":[],"yoast_head":"\nLCM And HCF Questions for Bank, SSC, RRB Exam 2020 - Entri Blog<\/title>\n<meta name=\"description\" content=\"LCM and HCF is one of the most commonly asked questions in any exam. 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