Answer - B
Explanation
Prime numbers which are common to all the given numbers are 2,5 ,7.
∴ H.C.F = (22*5*73)= (4*5*343) = 6860
Answer - A
Explanation
108 = (22*33) , 360 = (23*32*5) and 600 = (23*52*3)
∴ H.C.F = (22* 3) = (4* 3)=12
Answer - A
Explanation
Remainder of 185/148 = 37
Remainder of 148/37 = 0
∴ H.C.F. = 37
Answer - B
Explanation
Remainder of 1190/204 = 170
Remainder of 204/170 = 34
Remainder of 170/34 = 0
∴ H.C.F. of 204, 1190 = 34
Remainder of 1145/34 = 17
Remainder of 34/17 = 0
∴ H.C.F. of 204, 1190 and 1145 = 17
Answer - C
Explanation
First we find the H.C.F of 391 and 667.
Remainder of 667/391 = 276
Remainder of 391/276 =115
Remainder of 276/115 = 46
Remainder of 115/46 = 23
Remainder of 46/23= 0
∴ H.C.F. of 391, 667 = 23
∴ 391/667 =( 391/23)/ (667/23)= 17/29
Answer - A
Explanation
We have L.C.M = product of terms containing highest powers of (2,3,5,7,11)
= (2³* 32* 52*72*11) = (8*9*25*11*49)= 970200
Answer - D
Explanation
15 = 3 * 5
18 = 2* 3 * 3 = 2 * 3 2
24 = 2 *2 * 2 * 3 = 2 3 * 3
27 = 3 * 3 * 3 = 3 3
56 = 2 * 2 * 2 * 7= 2 3 * 7
L.C.M = product of terms containing highest powers of (2,3,5,7) =2 3 * 3 3 * 5 * 7 = 7560
Answer - D
Explanation
H.C.F of 2,8,10,16 = 2
L.C.M of 3,9,27,81 = 81
H.C.f = H.C.F of 2,8,10,16/L.C.M of 3,9,27,81 = 2/81
L.C.M = L.C.M of 2,8,10,16/H.C.F of 3,9, 27,81 = 80/3
Answer - B
Explanation
Let the numbers be 8x and 11 x. then, their H.C.F = x
So, the numbers are (8*6), (11*6) i.e 48 and 66.
Q 10 - Given the H.C. F of two numbers as 7 and their L.C.M as 210. If one of the numbers is 35, find the other.
A - 32
B - 42
C - 52
D - 62
Answer - B
Explanation
Let the Other number be X. then,
Product of numbers = product of their H.C .F and L.C.M
35*x= 7* 210 ⇒ x= 7*210/35 = 42
Hence, the other number is 42.
Answer - A
Explanation
Required measure = H.C.F of 36 L, 45 L, and 72 L
= (32) liters = 9 liters
[As 36 = 22*32, 45 = 32*5 and 72 = 24* 34]
Q 12 - Four electronic devices make a beep after duration of 30 minutes, 1 hour, 3/2 hours and 1 hour 45 min. respectively. If all the devices beeped together at 12 noon at what time will they beep together again?
A - 9 am
B - 10 am
C - 11 am
D - 11:30 am
Answer - A
Explanation
Intervals of beeping 30 min, 60 min, 90 min, 105 min.
Interval of beeping together= L.C.M of 30 min. 60 min. 90 min. 105 min
= (3*5*2*2*3*7) min. = 1260 min = 21 hrs.
So, they will beep together again next morning at 9 am.
Answer - B
Explanation
Remainder of 783/513 = 270
Remainder of 513/270 =243
Remainder of 270/243 = 27
Remainder of 243/27 = 0
Remainder of 46/23= 0
∴ H.C.F. of 513, 783 = 23
Remainder of 1107/23 = 0
∴ H.C.F. of 513, 783 and 1107= 23
Q 14 - Find the smallest number which is exactly divisible by each one of the numbers 12, 15, 20 and 27.
A - 540
B -530
C - 520
D - 510
Answer - A
Explanation
Required no. = L.C.M of 12,15, 20 and 27
= (3*2*2*5*9) = 540
Answer - A
Explanation
Required number = (L.C.M of 6,7,8,9,12)+2 = (2*3*2*7*2*3)+ 2 = (504+2)= 506.
Q 16 - Find the largest natural number which can divide the product of any 4 consecutive natural numbers.
A - 23
B - 24
C - 25
D - 26
Answer - B
Explanation
(1*2*3*4) = 24
∴ Required number = 24
Answer - A
Explanation
Here (35-18) = 17 , (45-28)= 17 and (55- 38) = 17
Required number = (L.C.M of 35,45, 55)- 17 = (3465 -17) = 3448
Answer - A
Explanation
H.C.F = H.C.F of 1,2,3,4/ L.C.M of 2,3,4,5 = 1/120
Answer - C
Explanation
H.C.F = H.C.F of 2, 8,10, 32/ L.C.M of 3,9, 27, 81 = 2/81
Answer - B
Explanation
H.C.F of 18 and 25 is 1.
∴ 18 and 25 are co-primes.
