Download LCM and HCF Questions for XAT PDF \u2013 XAT LCM and HCF questions pdf by Cracku. Practice XAT solved LCM and HCF Questions paper tests and These sample questions are to analyse the XAT exam pattern. Top 15 very Important LCM and HCF Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers, and solutions.<\/p>\n
Download LCM and HCF Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>0ne-eighth of a number is 17.25. What will 73% of the number be ?<\/p>\n a)\u00a0100.74<\/p>\n b)\u00a0138.00<\/p>\n c)\u00a096.42<\/p>\n d)\u00a082.66<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number be $8x$<\/p>\n Acc to ques,<\/p>\n => $\\frac{1}{8} * 8x = 17.25$<\/p>\n => $x = 17.25$<\/p>\n $\\therefore$ 73 % of the number = $\\frac{73}{100} * 8x$<\/p>\n = 0.73 * 8 * 17.25 = 100.74<\/p>\n Question 2:\u00a0<\/b>\u2018 A\u2019 ,\u2019B\u2019 and \u2018C\u2019 are three consecutive even integers such that four times \u2018A\u2019 is equal to three times \u2018C\u2019. What is the value of B\u2019?<\/p>\n a)\u00a012<\/p>\n b)\u00a010<\/p>\n c)\u00a016<\/p>\n d)\u00a014<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n let A be 2x ,B be 2x +2 and C be 2x +4<\/p>\n Given that 4A = 3C<\/p>\n 4(2x) = 3 (2x +4)<\/p>\n 8x = 6x +12<\/p>\n 2x = 12<\/p>\n x = 6<\/p>\n B = 2*6 + 2 = 14<\/p>\n Question 3:\u00a0<\/b>What is the LCM of the following fractions? 3\/11, 2\/5, 1\/9<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n e)\u00a06<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The LCM of the fractions = LCM of the numerators\/ HCF of the denominators Question 4:\u00a0<\/b>What is the LCM of the following fractions? \u2154, 4\/7, 1\/4<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a07<\/p>\n e)\u00a08<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n LCM of fractions = LCM of the numerators\/ HCF of the denominators. Question 5:\u00a0<\/b>The LCM of two numbers is 360, and their HCF is 15. If one of the numbers is 45, what is the value of the remainder when the other number is divided by 7?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n e)\u00a00<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the second number be x. Question 6:\u00a0<\/b>What is the greatest number which divides 1070 and 1265 and leaves remainders 3 and 4 respectively?<\/p>\n a)\u00a091<\/p>\n b)\u00a093<\/p>\n c)\u00a095<\/p>\n d)\u00a097<\/p>\n e)\u00a0101<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The required number is the HCF of (1070 – 3, 1265 – 4) = HCF of (1067, 1261) Take XAT 2023 Mock Tests<\/a><\/p>\n Question 7:\u00a0<\/b>The sum of two numbers is 65 and the HCF of the two numbers is 5. If the LCM of the two numbers is 180, what is the sum of the reciprocals of the two numbers?<\/p>\n a)\u00a011\/13<\/p>\n b)\u00a011\/625<\/p>\n c)\u00a013\/625<\/p>\n d)\u00a011\/180<\/p>\n e)\u00a013\/180<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the numbers be ha and hb. HCF of the numbers = h and LCM of the numbers = hab. Question 8:\u00a0<\/b>The ratio of two numbers is 5:6. The product of the HCF and LCM of the two numbers is 120. What is the HCF of the two numbers?<\/p>\n a)\u00a02<\/p>\n b)\u00a04<\/p>\n c)\u00a06<\/p>\n d)\u00a08<\/p>\n e)\u00a01<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the two numbers be 5h and 6h where h is the HCF of the numbers. Product of the numbers = Product of LCM and HCF. Question 9:\u00a0<\/b>What is the smallest number which when multiplied by 10 is exactly divisible by 12, 18, 24 and 32?<\/p>\n a)\u00a0288<\/p>\n b)\u00a0144<\/p>\n c)\u00a01440<\/p>\n d)\u00a0552<\/p>\n e)\u00a0None of the above<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The LCM of the numbers is 288. So, multiples of 288 are divisible by all the numbers. LCM of 288 and 10 = 1440. So, the required number = 144<\/p>\n Question 10:\u00a0<\/b>In a church, the first bell rings at intervals of 2 seconds, the second bell rings at intervals of 4 seconds and so on till the fifth bell, which rings at intervals of 10 seconds. How many times do all the bells ring together in a half-an-hour period?<\/p>\n a)\u00a060<\/p>\n b)\u00a045<\/p>\n c)\u00a0120<\/p>\n d)\u00a015<\/p>\n e)\u00a030<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The LCM of 2, 4, 6, 8 and 10 is 120. So, the bells rings together once every 2 minutes. So, they ring together 15 times in a half-an-hour period.<\/p>\n Question 11:\u00a0<\/b>The largest number that divides 43, 91 and 183 and leaves the same remainder in each case is<\/p>\n a)\u00a016<\/p>\n b)\u00a08<\/p>\n c)\u00a09<\/p>\n d)\u00a013<\/p>\n e)\u00a04<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The required number is the HCF of (91 – 43), (183 – 91) and (183 – 43) = HCF of (48, 92, 140) = 4<\/p>\n Question 12:\u00a0<\/b>The LCM of two numbers x and y is 20. Find the number of unordered pairs (x,y)?<\/p>\n a)\u00a08<\/p>\n b)\u00a09<\/p>\n c)\u00a07<\/p>\n d)\u00a010<\/p>\n e)\u00a0None of the above<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pairs of numbers are (1,20), (2,20), (4,20) , (5,20) , (10,20) , (20,20) , (4,5) , (4,10). So, the number of unordered pairs is 8.<\/p>\n
\nLCM of the numerators = LCM of 3, 2 and 1 = 6
\nHCF of the denominators = HCF of 11, 5 and 9 = 1
\nSo, the required LCM = 6\/1 = 6<\/p>\n
\nLCM of 2, 4 and 1 is 4
\nHCF of 3, 7 and 4 is 1
\nSo, the LCM of the given fractions is 4\/1 = 4<\/p>\n
\nWe know that LCM*HCF = Product of the two numbers
\nSo 360*15 = x *45
\n=> x = $\\frac{360*15}{45}$
\n=> x = 120
\nNow 120 on being divided by 7, leaves a remainder of 1.<\/p>\n
\n1067 = 11*97 and 1261 = 13*97
\nSo, HCF = 97<\/p>\n
\nSum of reciprocals = (a+b)\/hab
\nh(a+b) = 65
\nh = 5
\n(a+b) = 13
\nhab = LCM = 180
\nSo, sum of reciprocals = 13\/180<\/p>\n
\n=> $30 h^2$ = 120 => h = 2
\nSo, the HCF of the two numbers is 2<\/p>\n