100+ TNPSC Aptitude HCF and LCM Questions and Answers - 1
Question: 1
The number of number-pairs lying between 40 and 100 with their H.C.F. as 15 is
(A) 3
(B) 4
(C) 5
(D) 6
Ans: B
Numbers with H.C.F. 15 must contain 15 as a factor.
Now, multiple of 15 between 40 and 100 are 45, 60, 75 and 90.
∴ Number pairs with H.C.F. 15 are (45, 60), (45, 75), (60, 75) and (75, 90)
[∴ H.C.F. of (60, 90) is 30 and that of (45, 90) is 45]
Clearly, there are 4 such pairs.
Question: 2
A person has to completely put each of three liquids 403 litres of petrol, 465 litres of diesel and 496 litres of Mobil oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required?
(A) 24
(B) 34
(C) 44
(D) 54
Ans: C
For the least number of bottles, the capacity of each bottle must be maximum.
∴ Capacity of each bottle = H.C.F. of 403 litres, 465 litres and 496 litres = 31 litres.
Hence, required number of bottles = $$({403 + 465 + 496} / {31})$$ = $${1364} / {31}$$ = 44.
Question: 3
A milkman has 3 jars containing 57 litres, 129 litres and 177 litres of pure milk respectively. A measuring can, after a different number of exact measurements of milk in each jar, leaves the same amount of milk unmeasured in each jar. What is the volume of the largest such can?
(A) 12 litres
(B) 16 litres
(C) 24 litres
(D) 28 litres
Ans: C
Required volume = [H.C.F. of (129 - 57), (177 - 129) and (177 - 57)] litres
= (H.C.F. of 72, 48 and 120) litres = 24 litres.
Question: 4
A number less than 500, when divided by 4, 5, 6, 7 leaves remainder 1 in each case. The number is
(A) 211
(B) 311
(C) 411
(D) 421
Ans: D
L.C.M. of 4, ,5 6, 7 = 420.
∴ Required number = 420 + 1 = 421.
Question: 5
Find the greatest number that will divide 964, 1238 and 1400 leaving remainders 41, 31 and 51 respectively.
(A) 71
(B) 81
(C) 61
(D) 73
Ans: A
964 - 31 = 923
1238 – 31 = 1207
1400 – 51 = 1349
H.C.F. of 923 and 1207 is 71.
H.C.F. of 71 and 1349 is 71.
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