HCF and LCM CAT Questions consists of important LCM and HCF questions in CAT. This 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 Least common multiple and Highest Common factor is a very basic topic but the applications are very important.<\/p>\n You can download the HCF and LCM CAT Problems or you can go through the details below.<\/p>\n Download HCF & LCM PDF<\/a><\/p>\n Enroll to CAT 2018 Courses<\/a><\/p>\n HCF and LCM CAT Questions:<\/strong><\/p>\n Question 1:<\/p>\n Number of students who have opted for subjects A, B and C are 60, 84 and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions?<\/p>\n A. 28<\/p>\n B. 60<\/p>\n C. 12<\/p>\n D. 21<\/p>\n Question 2: A. 30<\/p>\n B. 24<\/p>\n C. 20<\/p>\n D. 60<\/p>\n Question 3: A. 0<\/p>\n B. 1<\/p>\n C. 2<\/p>\n D. None of these<\/p>\n Question 4:<\/strong><\/p>\n In Sivakasi, each boy’s quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?<\/p>\n A. 200<\/p>\n B. 150<\/p>\n C. 125<\/p>\n D. 175<\/p>\n Question 5:<\/strong><\/p>\n A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is<\/p>\n A. 12 \u00d7 81<\/p>\n B. 16 \u00d7 192<\/p>\n C. 20 \u00d7 125<\/p>\n D. 24 \u00d7 216<\/p>\n Download CAT maths Tricks PDF<\/a> and also read HCF and LCM tricks for CAT<\/a>.<\/p>\n Download CAT Quantitative Aptitude Questions PDF<\/a><\/p>\n Answer and Solutions for HCF and LCM CAT Questions:<\/strong><\/p>\n Solutions: Answer 2:<\/strong><\/p>\n A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1\/3 min and green light flashes every 2\/5 min. LCM of both the fractions is 2 min . Answer 3:<\/strong><\/p>\n Let the number ‘n’ belong to the set A. Answer 4:<\/strong><\/p>\n Let the number of sticks assigned to each boy be N. Answer 5:<\/strong><\/p>\n Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K Hope you will find this HCF and LCM CAT questions useful, you can also download all our CAT Verbal Ability PDF’s<\/a>\u00a0and CAT Quantitative Aptitude questions PDF<\/a>.<\/p>\n
\n<\/strong>
\nA red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?<\/p>\n
\n<\/strong>
\nA is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?<\/p>\n
\n<\/strong>
\nAnswer 1:
\n<\/strong>
\nAs we can see here that total number of students are = 60+84+108 = 252
\nNow given condition is that in one room only the students of the same subject can be there and the number of rooms should be minimum that means the number of students in a particular room will be maximum.
\nThis Maximum number of students will be HCF (Highest common factor) of 60, 84 and 108 and that will be 12
\nHence, number of rooms will be = 252\/12 = 21<\/p>\n
\nHence they flash together after every 2 min. So in an hour they flash together 30 times .<\/p>\n
\nHence, the remainder when n is divided by 2 is 1
\nThe remainder when n is divided by 3 is 2
\nThe remainder when n is divided by 4 is 3
\nThe remainder when n is divided by 5 is 4 and
\nThe remainder when n is divided by 6 is 5
\nSo, when (n+1) is divisible by 2,3,4,5 and 6.
\nHence, (n+1) is of the form 60k for some natural number k.
\nAnd n is of the form 60k-1
\nBetween numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1<\/p>\n
\nLet the number of boxes be M.
\nSo, number of sticks per box = N\/M
\nNow, if he reduces the number of sticks in each box, the equation becomes N\/(M+3) = N\/M – 25
\nSo, 25 = N\/M – N\/(M+3)
\nFrom the options, if N = 150, then, we get 25 = 150 [ 1\/M – 1\/(M+3) ]
\n=> 1\/6 = 1\/M – 1\/(M+3) => M = 3
\nSo, the number of sticks assigned to each boy = 150<\/p>\n
\nLet k = 1; the number becomes 53
\nIf it is divided by 84, the remainder is 53.
\nOption d) is the correct answer.<\/p>\n