LCM and HCF: Solved 50 LCM and HCF Questions and answers section with explanation for various online exam preparation, various interviews, Logical Reasoning Category online test. Category Questions section with detailed description, explanation will help you to master the topic.
LCM and HCF Questions
1. Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder?
Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder:
4. Four Iron metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?
Since each iron rod must be cut into parts of equal length and each part must be as long as possible, so HCF should be taken. HCF of 78, 104, 117, 169 = 13 Number of pieces from 78 cm rod = 78/13 = 6 Number of pieces from 104 cm rod = 104/13 = 8 Number of pieces from 117 cm rod = 117/13 = 9 Number of pieces from 78 cm rod = 169/13 = 13 Total number of pieces = 6 + 8 + 9 + 13 = 36.
6. The product of LCM and HCF of two numbers is 48. The difference of two numbers is 8. Find the numbers?
Correct Ans:4, 12
Let the numbers be 'a' and 'b'. WKT, Product of two numbers = LCM Ã— HCF Product of two numbers = 48 ab = 48 Difference of two numbers = 8 a - b = 8 Factors of ab are (1,48) , (2,24), (3,16),(4,12),(6,8). Here, the only pair which satisfies the given condition is (4, 12).
7. Four different electronic devices make a beep after every 30 minutes, 1 hour, 1 hour 30 minutes and 1 hour 45 minutes respectively. All the device beeped together at 12 noon. They will again beep together at
Correct Ans:9:00 AM
Interval after the devices will beep together = (L.C.M. of 30, 60, 90, 105) min = 1260 min = 1260/60 = 21 hrs So, the devices will again beep together after 21 hrs i.e., at 9 AM(after 12 noon).
9. On dividing a number by 4, the remainder is 2. The quotient so obtained when divided by 5, leaves the remainder 3. Now the quotient so obtained when divided by 6, leaves the remainder 5. The last quotient is 7. The number was
WKT, Divisor x Quotient + Remainder = Number 6 x 7 + 5 = 47 5 x 47 + 3 = 238 4 x 238 + 2 = 954 Therefore, required number is 954.
15. The sum of two number is 528 and their HCF is 33. How many pairs of such numbers can be?
Here, the HCF = 33 Then, let the two numbers be 33a and 33b, Respectively, where a and b are coprime. Then, 33a + 33b = 528 a + b = 16 Hence, values of (a, b) which are coprime to each other can be (1, 15), (3, 13), (5, 11) and (7, 9). Hence, required number of pairs is 4.
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