Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
41.
The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?
(A) 24
(B) 48
(C) 56
(D) 60
Solution:
HCF = 8 ⇒ Now, LCM should have a factor 8 So, check also the option we have only 60 which does not have a factor 8. So, it will never be the LCM.
42.
Find the greatest number that will divide 964, 1238 and 1400 leaving remainders 41, 31, and 51 respectively = ?
(A) 61
(B) 71
(C) 73
(D) 81
Solution:
Required number = HCF of (964 - 41), (1238 - 31) and (1400 - 51) HCF of 923, 1207 and 1349 = 71
43.
What is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?
(A) 840
(B) 841
(C) 820
(D) 814
Solution:
Required number of tiles are $$ = \frac{{{\text{Area of floor}}}}{{{\text{Area of tiles}}}}$$ Sides of tiles is HCF (1517, 902) = 41 ∴ Area of tiles = 41 × 41 ∴ Number of tiles $$ = \frac{{1517 \times 902}}{{41 \times 41}} = 814$$
44.
If three number are 2a, 5a and 7a, what will be their LCM ?
(A) 70a
(B) 65a
(C) 75a
(D) 73a
Solution:
The given three numbers are 2a, 5a and 7a LCM of 2a, 5a and 7a = 2 × 5 × 7 × a = 70a
45.
The HCF of two numbers is 8. Which one of the following can never be their LCM ?
(A) 24
(B) 48
(C) 56
(D) 60
Solution:
HCF = 8 Now LCM should have a factors 8 So check also the option we have only 60 which does not have a factor 8. So it will never be the LCM.
46.
The sum of two positive numbers is 240 and their HCF is 15. Find the number of pairs of numbers satisfying the given condition.
The number nearest to 43582 divisible by each of 25, 50 and 75 is = ?
(A) 43500
(B) 43550
(C) 43600
(D) 43650
Solution:
LCM of 25, 50 and 75 = 150 On dividing 43582 by 150, remainder = 82 ∴ Required number = 43582 + (150 – 82) = 43650 Alternate LCM of 25, 50 and 75 = 5 × 5 × 2 × 3 = 150 On dividing 43582 by 150, the remainder is 82 and quotient is 290 So, required number = 150 × 291 = 43650
48.
HCF of 3240, 3600, and a third number is 36 and their LCM is 2 ÃÂÃÂÃÂÃÂÃÂÃÂ 3 ÃÂÃÂÃÂÃÂÃÂÃÂ 5 ÃÂÃÂÃÂÃÂÃÂÃÂ 7 . The third number is = ?4522
(A) 22 × 35 × 72
(B) 22 × 53 × 72
(C) 25 × 52 × 72
(D) 23 × 35 × 72
Solution:
3240 = 23 × 34 × 5 3600 = 24 × 32 × 52 HCF = 36 = 22 × 32 Since H.C.F. is the product of lowest powers of common factors, so the third number must have (22 × 32) as its factors. Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors. ∴ Third number = 22 × 35 × 72
49.
13, a, b, c are four distinct numbers and the HCF of each pair of numbers (13, a) ; (13, b) ; (13, c) is 13, where a, b, c are each less than 60 and a b c. What is the value of
(A) 3.5
(B) 2
(C) 5
(D) 4.5
Solution:
Four distinct numbers 13, a, b, c are given HCF of (13, a) : (13, b) : (13, c) = 13 and a b c It is clear that a = 26 b = 39 c = 52 $$\eqalign{ & {\text{Then, }}\frac{{{\text{a}} + {\text{c}}}}{{\text{b}}} \cr & = \frac{{26 + 52}}{{39}} \cr & = \frac{{78}}{{39}} \cr & = 2 \cr} $$
50.
The sides of a triangular field are 62 m, 186 m and 279 m. Find the greatest length of tape that would be able to exactly measure each of them without any fractions.
(A) 62 m
(B) 93 m
(C) 31 m
(D) 30 m
Solution:
Maximum length of tape = HCF of 62, 186, 279 = 31 × 2, 31 × 6, 31 × 9 = 31 Answer