LCM and HCF Questions and Answers
L.C.M. and H.C.F Basic Concepts, Formula, Short tricks
- The L.C.M of two or more numbers is greater than or equal to the greatest number of given numbers.
- L.C.M. ≥ Greatest Number
- The smallest number which is exactly divisible by a, b and c is L.C.M of a, b, c.
L.C.M. ≥ Greatest Number
Q.1: The LCM and HCF of the three numbers 48, 144 and p are 720 and 24 respectively, find the least value of p.
A) 192
B) 180
C) 120
D) 360
Explanation: 1.LCM of numbers is divisible by these numbers (720 should be divisible by 48,144,p),720 is not divisible by 192 so it cannot be the value of p. HCF is highest divisor of all numbers, 180 is not divisible by 24,Rest of the option divide 720 completely and divisible by 24 also but the minimum number is 120 so the answer is 120.
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Q.2:Two numbers having their LCM 480 are in a ratio 3 : 4. What will be the smaller number of this pair?
A) 120
B) 140
C) 180
D) 260
Explanation: 2. Let two number be 3x and 4x , LCM = 480
3*4*x = 480 , x = 40 the smaller number 3x = 3*40 = 120 .
Q.3: A magician wants to hide his magical rod inside a cubical box whose total surface area is 3042 CM2. What can be the maximum length of the rod?
A) 37
B) 39
C) 42
D) 33
Explanation: Total surface area = 3042 cm².
Total surface area of the cube = 6a² , a is the side of the cube 6a² = 3042 ⇒ a² = 507 ⇒ a =13√3
The maximum length of rod which can be placed inside the cubical box is equal to diagonal of cube = √3 a = √3 * 13 √3 = 13*3 = 39 cm
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Q.4: H.C.F of two Number is 27 and their sum is 216 then how many number pairs are possible to satisfy these condition.
A) 2
B) 4
C) 5
D) 1
Explanation: Lets two numbers be 27a and 27b (numbers can be written as multiple of HCF) 27a+27b=216 ⇒ a+b=8
now check how many values of a and b satisfy this equation ( values should not be repeat and co factors) (1,7) (3,5) only two pairs are possible because (2,6) co factors(4,4)(5,3) repeated values, numbers are (27,189)(81,135)
now check how many values of a and b satisfy this equation ( values should not be repeat and co factors) (1,7) (3,5) only two pairs are possible because (2,6) co factors(4,4)(5,3) repeated values, numbers are (27,189)(81,135)
Q.5: length and width of a room are 5.44m and 3.74m respectively. How many least number of square tiles are required to cover this room?
A) 34
B) 170
C) 172
D) 176
Explanation: Number of tiles = Area of room/ area of tile. now find area of tile
Area of square tile = (side)2.
to calculate side of title find HCF of length and breadth of room
L= 544cm ,B=374cm HCF of 544 and 374 is 34
Number of tiles = 544*374/ (34*34) = 176 tiles.
Area of square tile = (side)2.
to calculate side of title find HCF of length and breadth of room
L= 544cm ,B=374cm HCF of 544 and 374 is 34
Number of tiles = 544*374/ (34*34) = 176 tiles.
Q.6: How many pairs of two numbers are possible Whose H.C.F is 13 and LCM is 156.
A) 2
B) 3
C) 4
D) 1
Explanation: Lets two numbers be 13a and 13b (numbers can be written as multiple of HCF) 13ab= 156 ⇒ ab = 12 Possible pairs (1,12) (3,4), numbers are (13,156) (39,52) so answer is 2
Q.7: 6 balls falling together at an interval of 2,4,6,8,10,12 sec., how many times they will fall together in an interval of 30 min.
A) 15
B) 16
C) 20
D) 22
Explanation: LCM of ((2,4,6,8,12) is 120 sec.,120 sec = 2 min, in interval of 30 min they will meet (30/2)= 15 times, at starting they fall together so include one more, answer is 15+1 = 16
Q.8: P is the smallest number which leaves remainder 1 when it is divided by 9, 15, 21, 27 but it leaves no remainder when divided by 11, find the value of P..
A) 944
B) 945
C) 940
D) 946
Explanation: LCM of numbers is the smallest number which leaves no remainder when divided by these numbers.
Value of P = LCM of (9,15,21,27)K+ 1 ⇒ P = 945K + 1 ⇒ if K=1 ⇒ P = 946 which is divisible by 11 also therefore required value of P= 946
Value of P = LCM of (9,15,21,27)K+ 1 ⇒ P = 945K + 1 ⇒ if K=1 ⇒ P = 946 which is divisible by 11 also therefore required value of P= 946
Q.9: LCM of two numbers is 45 times of their HCF, if one number is 125 and sum of LCM and HCF is 1150, find the value of second number.
A) 220
B) 225
C) 235
D) 25
Explanation: Let HCF = x then LCM = 45x
sum of LCM and HCF = 1150 ⇒ x + 45x = 1150 ⇒ 46x= 1150 ⇒ x = 25
Product of Numbers = HCF × LCM
one number is 125 let second number = N
N×125 = x(45x)
⇒ N = 25(25*45)/125 ⇒ N = 225
sum of LCM and HCF = 1150 ⇒ x + 45x = 1150 ⇒ 46x= 1150 ⇒ x = 25
Product of Numbers = HCF × LCM
one number is 125 let second number = N
N×125 = x(45x)
⇒ N = 25(25*45)/125 ⇒ N = 225
Q.10: Which least number must be subtracted from 1856 so that when it is divided by 7, 12, 16 it leaves remainder 4 .
A) 336
B) 330
C) 172
D) 180
Explanation: LCM of (7, 12, 16) = 336
now find number nearest to 1856 which is multiple of 336
336×2 =672….336×3….336xx5 = 1680 required number which leaves rem. 4 = 1680 + 4 = 1684
difference 1856−1684 = 172
now find number nearest to 1856 which is multiple of 336
336×2 =672….336×3….336xx5 = 1680 required number which leaves rem. 4 = 1680 + 4 = 1684
difference 1856−1684 = 172
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