HCF and LCM<\/strong>\u00a0for CAT<\/strong> are one of the key topics in the CAT Quant section. You can check out these<\/span> HCF and LCM CAT Previous year questions<\/strong>.<\/span><\/a>\u00a0 In this article, we will look into some important HCF and LCM Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT HCF and LCM Questions PDF, which is completely Free.<\/p>\n HCF and LCM<\/strong> CAT Questions form an important part of CAT<\/strong> Quant Number Systems<\/strong><\/a>. Applications of LCM and HCF questions in CAT were asked in the past years. These CAT questions will be very useful for quantitative aptitude for CAT.<\/p>\n Take a free CAT mock test <\/a>and also try to solve CAT previous questions<\/a> to get good understanding of the LCM and HCF CAT questions. This topic comes under the number system for CAT. Though the Least Common Multiple and Highest Common Factor is a very basic topic, their applications are very important.<\/p>\n You can download the HCF and LCM CAT Problems or you can go through the questions below.<\/p>\n Download HCF and LCM Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Abhi is a very superstitious person. He went to an astrologer who suggested him that the next lottery winning ticket will be a 2-digit number which is less than 100, not a multiple of 2,3 or 5 and is not a perfect square or a perfect cube. Abhi wants to buy all the tickets which have a possibility of becoming the winning ticket. How many tickets should he buy ?<\/p>\n a)\u00a022<\/p>\n b)\u00a021<\/p>\n c)\u00a024<\/p>\n d)\u00a023<\/p>\n e)\u00a026<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n To find all the numbers that suits Abhi, we just subtract the ones that are not possible from the total number of possibilities.<\/p>\n Since it is a 2-digit number, so the number of possibilities = 99-9 = 90<\/p>\n Thus, within all the 2-digit numbers :<\/p>\n Number of multiples of 2 = 49 – 4 = 45<\/p>\n Number of multiples of 3 = 33 – 3=30<\/p>\n Number of multiples of 5 = 19 – 1 = 18<\/p>\n Number of perfect squares = 1 (only 49 matches the requirements)<\/p>\n Number of perfect cubes = 0 (all the cubes are already multiples of the prohibited numbers)<\/p>\n Here, we need to account for using a few numbers from the set more than once.<\/p>\n Thus, multiples of 6,10,15,30 (LCM of (2,3),(2,5),(3,5),(2,3,5) respectively) have been counted more than once.<\/p>\n Multiples of 6 = 16 – 1 = 15<\/p>\n Multiples of 10 = 9<\/p>\n Multiples of 15 = 6<\/p>\n Multiples of 30 = 3<\/p>\n Thus, we need to add the multiples of 6,10,15 to the sum and deduct the multiples of 30 to get the actual number of numbers.<\/p>\n Thus, number of numbers = 90 – (45+30+18+1) + (15+9+6) – 3 = 23 numbers<\/p>\n Alternatively, Question 2:\u00a0<\/b>Three friends P, Q and R started running on a circular track of length 128 m at the same time and from the same spot. P and R were running in the clockwise direction with speeds 4 m\/sec and 8 m\/sec respectively. Q was running in counter clockwise direction with speed 6 m\/sec. What is the time taken by all of them to meet together for the first time?<\/p>\n a)\u00a032 seconds<\/p>\n b)\u00a096 seconds<\/p>\n c)\u00a064 seconds<\/p>\n d)\u00a0192 seconds<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Ratio of speeds of P and R = 4 : 8 = 1 : 2<\/p>\n Both P and R are running in same direction. Hence, number of meeting points = |2-1| = 1<\/p>\n They meet at one point only, i.e they meet at starting point<\/p>\n Therefore, three of them will meet at starting point for the first time<\/p>\n Time taken by P to complete one round $\\frac{128}{4}$ = 32 seconds<\/p>\n Time taken by R to complete one round\u00a0= $\\frac{128}{8}$ = 16 seconds<\/p>\n Time taken by Q to complete one round =\u00a0$\\frac{128}{6}$ =\u00a0$\\frac{64}{3}$ seconds<\/p>\n Time taken by all of them to meet together for the first time = LCM(32, 16,\u00a0$\\frac{64}{3}$) =\u00a0$\\ \\frac{\\ LCM\\left(32,\\ 16,\\ 64\\right)}{HCF\\left(1,1,3\\right)}$ =\u00a0$\\ \\frac{64}{1}$ = 64 seconds<\/p>\n Alternate Explanation:<\/strong><\/p>\n Length of the circular track = 128m<\/p>\n Speed of P, Q and R is 4 m\/sec, 6 m\/sec and 8 m\/sec<\/p>\n Relative speed of P and R = 8 – 4 = 4 m\/sec<\/p>\n Time taken by P and R to meet, t1 = $\\frac{128}{4} =32$seconds<\/p>\n Relative speed of R and Q = 8 + 6 = 14 m\/sec<\/p>\n Time taken by Q and R to meet, t2 = $\\frac{128}{14} $ seconds<\/p>\n Time taken by all the three friends to meet together for the first time = $ LCM(t1, t2) $= $LCM(\\frac{128}{4}, \\frac{128}{14}) $= $\\frac{LCM(128, 128)}{HCF(4, 14)} $=$ \\frac{128}{2}$ = 64 seconds<\/p>\n Hence, option C is the right choice.<\/p>\n Question 3:\u00a0<\/b>If LCM of two numbers 28!*32! and 30!*31! is $\\frac{a!*b!}{c!}$, where a, b (a > b) are two-digit numbers and c is one digit number. Find out the value of a -(b+c)?<\/p>\n 3)\u00a0Answer:\u00a00<\/b><\/p>\n Solution:<\/b><\/p>\n Let us first calculate HCF of 28!*32! and 30!*31!<\/p>\n $\\Rightarrow A = 28! *(32 *31!)$<\/p>\n $\\Rightarrow A = 32*28! *31!$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026 (1)<\/p>\n $\\Rightarrow B = 30!*31!$<\/p>\n $\\Rightarrow B = 30*29*28!*31!$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026 (2)<\/p>\n From equation 1 and 2 it is clear that HCF = 2*28!*31!<\/p>\n We know that: Product of 2 number = LCM * HCF<\/p>\n $\\Rightarrow LCM = \\frac{Product of A and B}{HCF of A and B}$<\/p>\n $\\Rightarrow LCM = \\frac{(28!*32!)*(30!31!)}{2*28!*31!}$<\/p>\n $\\Rightarrow LCM = \\frac{32!*30!}{2} = \\frac{32!*30!}{2!}$<\/p>\n Hence we can say that a = 30, b = 30 and c = 2<\/p>\n Hence a-(b+c) = 32 -(30 +2) = 0 (Answer :0)<\/p>\n Question 4:\u00a0<\/b>Ram has a certain number of chocolates with him. If he divided these chocolates among 12 children, he would be left with 9 chocolates. If he divides the chocolates among 17 children then he would be left with 14 chocolates. Similarly, if he decides to divide the chocolates among 7 people, he will be left with 4 chocolates. What is the minimum possible number of chocolates with Ram?<\/p>\n a)\u00a0116<\/p>\n b)\u00a01425<\/p>\n c)\u00a01723<\/p>\n d)\u00a0201<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total number of chocolates with Ram be \u2018n\u2019. Question 5:\u00a0<\/b>Three wheels can complete 60, 36 and 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?<\/p>\n a)\u00a0$\\frac{5}{2}$ s<\/p>\n b)\u00a0$\\frac{5}{3}$ s<\/p>\n c)\u00a0$5$ s<\/p>\n d)\u00a0$7.5$ s<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The first wheel completes a revolution in $\\frac{60}{60}=1$ second The three wheels touch the ground simultaneously at time which are multiples of the above three times. So, the correct option is option (c)<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>What is the least number that must be subtracted from 2015 so that the resultant number when divided by 9, 12 and 16 leaves the same remainder 7?<\/p>\n a)\u00a0143<\/p>\n b)\u00a0136<\/p>\n c)\u00a0164<\/p>\n d)\u00a0150<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The LCM of 9,12 and 16 is 144. Question 7:\u00a0<\/b>Car A starts from city P towards city Q. At the same time,\u00a0another car B starts from the city Q towards city P. They will meet at point R if the ratio of the speed of car A and car B is 4:5. They meet at point S if the ratio of the speed of car A and car B is 3:4. The ratio of the distance between R and S and the distance between P and Q is m:n where m and n are co-primes. Find the value of m+n.<\/p>\n a)\u00a062<\/p>\n b)\u00a063<\/p>\n c)\u00a064<\/p>\n d)\u00a065<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Assuming the distance between P and\u00a0Q is 63 units (LCM of 4+5=9 and 3+4=7) Then, the distance of S from P = Distance travelled by A from P to S = Speed of A\u00a0$\\times$ Time taken to meet =\u00a0$\\dfrac{63}{3x+4x}\\times\\ 3x$ = 27 units<\/p>\n Hence, the distance between R and S = 1 unit Question 8:\u00a0<\/b>Krishna and Shyam can finish\u00a0work together in 18 days. After 6 days, Krishna left and Shyam finished the work alone. If the total time taken to finish the work in this way is 24 days, what is the ratio of work done by Krishna and Shyam?<\/p>\n a)\u00a03:8<\/p>\n b)\u00a04:9<\/p>\n c)\u00a01:8<\/p>\n d)\u00a02:9<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n It is given that Krishna and Shyam can finish work in 18 days. So after 6 days, one-third of work is finished.<\/p>\n Shyam finishes 2\/3 of the work in 24-6=18 days.<\/p>\n Hence, Shyam can finish total work in 18*3\/2 = 27 days<\/p>\n Assuming the total work to be 54 units.\u00a0 (LCM(27,18)=54)<\/p>\n Work done by both Krishna and Shyam in 1 day = 54\/18 = 3 units<\/p>\n Work done by Shyam in 1 day = 54\/27 = 2 units<\/p>\n Hence, work done by Krishna = 3-2 = 1 units in 1 day<\/p>\n Then work done by Krishna in 6 days = 6 units<\/p>\n Work done by Shyam in 24 days = 24*2=48 units<\/p>\n The ratio = 6:48= 1:8<\/p>\n Question 9:\u00a0<\/b>Find the sum of the terms which are common to the following series: a)\u00a02187<\/p>\n b)\u00a04374<\/p>\n c)\u00a02836<\/p>\n d)\u00a03186<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The given series are:<\/p>\n S1: 4, 11, 18 \u2026\u2026 704 The common difference of the first series is 7 and the common difference of the second series is 11.<\/p>\n The terms which are common to them will form an AP with a common difference of LCM(7, 11) = 77<\/p>\n The first common term of these two series is 46.<\/p>\n The last term of the common series cannot be greater than 704.<\/p>\n Let\u2019s find the number of common terms.<\/p>\n 704 = 46 + ( n – 1) 77<\/p>\n Taking only the integral value, n = 9<\/p>\n The sum of the series = $\\frac{9}{2}*[2*46 + (8) * 77] = 3186$<\/p>\n Question 10:\u00a0<\/b>Find the largest number which leaves the same remainder when it divides 722, 1097 and 1347?<\/p>\n a)\u00a05<\/p>\n b)\u00a045<\/p>\n c)\u00a0125<\/p>\n d)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the greatest number which leaves the same remainder when it divides 722, 1097 and 1347 be N and let the remainder be R.<\/p>\n So, N divides 722-R, 1097-R and 1347-R. The remainder when 125 divides 722, 1097 and 1347 is 97.<\/p>\n Question 11:\u00a0<\/b>3 friends A, B and C run around a circular track of length 120 m continuously. All of them start at the same point. A and B run in the clockwise direction while C runs in the anti-clockwise direction. The speeds of A, B and C are in the ratio 13:5:7. At how many distinct points will all three of them meet?<\/p>\n a)\u00a06<\/p>\n b)\u00a04<\/p>\n c)\u00a03<\/p>\n d)\u00a05<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Length of the track is 120 m. B and C run in the opposite directions. Their speeds are in the ratio 5:7. A and C run in the opposite directions. Their speeds are in the ratio 13:7. All 3 of them will meet at points that are multiples of the LCM of 15, 10 and 6. Alternatively, Number of distinct points = HCF(8, 12, 20)<\/p>\n Therefore, in a sufficiently large number of rounds, all 3 of them will meet at points at a distance of 0 m, 30 m, 60 m and 90 m from the starting point. There are 4 points in total. Therefore, option D is the right answer.<\/p>\n Question 12:\u00a0<\/b>Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same\u00a0point at the same time, and going in the clockwise direction. If they run at speeds of 15\u00a0km\/hr, 10 km\/hr, and 8 km\/hr, respectively, how much distance in km will Ravi have run\u00a0when Anil and Sunil meet again for the first time at the starting point?<\/p>\n a)\u00a04.8<\/p>\n b)\u00a04.6<\/p>\n c)\u00a05.2<\/p>\n d)\u00a04.2<\/p>\n
\nThere are 25 primes less than 100, 21 of which are 2 digit primes
\nThese primes are relatively co-prime to 2, 3 and 5
\nWe know the prime numbers are 2,3,5,7,11,13,17….
\nThe lowest possible number which is not a prime,
\nnot divisible by 2,3, and 5 and not a perfect square is 7*11=77
\nThe only other possible number is 7*13=91
\nHence, total such numbers = 21+2 = 23<\/p>\n
\nThen \u2018n\u2019 can be expressed as
\nn = 12k + 9 = 17m + 14 = 7p + 4
\nWe can see that in each of the above cases, the remainder is -3.
\nHence, the minimum number of chocolates with Ram will be LCM (12, 7, 17) – 3
\nThus, the required number will be 12*7*17 – 3 = 1428 – 3 = 1425.<\/p>\n
\nThe second wheel completes a revolution in $\\frac{60}{36}=1\\frac{2}{3}$ second
\nThe third wheel completes a revolution in $\\frac{60}{24}=2\\frac{1}{2}$ second<\/p>\n
\nHence, the required number is $LCM(1,\\frac{5}{3},\\frac{5}{2}) = 5$ seconds.<\/p>\n
\nThe closest multiple of 144 which is less than 2015 is 144*13 = 1872.
\nTo get a remainder of 7, we need to add 7 to the multiple of the LCM, in this case 1872+7=1879.
\nHence, the required number to be subtracted equals 2015-1879=136<\/p>\n
\nThe fraction of the distance travelled by each car will be proportional to their speeds because the time taken is the same in each case.
\nAssume the speed of car A in the first case to be 4x and 5x
\nHence, the distance of R from P = Distance travelled by A from P to R = Speed of A\u00a0$\\times$ Time taken to meet=\u00a0$\\dfrac{63}{4x+5x}\\times\\ 4x$ = 28 units
\nSimilarly, let the speed of car A in the second case be 3x and 4x<\/p>\n
\nThe ratio\u00a0$\\ \\dfrac{\\ m}{n}=\\ \\dfrac{\\ 1}{63}$
\nHence, the sum = 1+63 = 64<\/p>\n
\nS1: 4, 11, 18 \u2026\u2026 704
\nS2: 2, 13, 24\u2026\u2026 827<\/p>\n
\nS2: 2, 13, 24\u2026\u2026 827<\/p>\n
\nIt also divides (1097-R) – (722-R) and (1347-R) – (722-R).
\nSo, N divides 1097-722 and 1347-722.
\nSo, N divides 375 and 625.
\nThe HCF of 375 and 625 is 125 and hence, the greatest possible value of N is 125.<\/p>\n
\nSpeeds of A and B are in the ratio 13:5. They are running in the same direction.
\nTherefore, they will meet at 13-5 = 8 distinct points.
\nLet us use the distance from the starting point to label the meeting points.
\nA and B will meet at points at a distance of 0 m, 15 m, 30 m, 45 m, 60 m, 75 m, 90 m, and 105 m from the starting point.<\/p>\n
\nTherefore, they will meet at 5+7 = 12 distinct points.
\nThey will meet at points at a distance of 0 m, 10 m , 20 m, 30 m, 40 m, 50 m, 60 m, 70 m ,80 m, 90 m, 100 m and 110 m.<\/p>\n
\nTherefore, they will meet at 13+7 = 20 distinct points.
\nThey will meet at points at a distance of 6 m, 12 m, 18 m, \u2026 114 m.<\/p>\n
\nLCM (15,6,10) = 30 m.<\/p>\n