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Problems On H.C.F And L.C.M - Aptitude MCQ Questions and Solutions with Explanations

Problems On H.C.F And L.C.M MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Problems On H.C.F And L.C.M with easy and logical explanations. Problems On H.C.F And L.C.M MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-3 section-1

Each section contains 100 MCQs on Problems on H.C.F and L.C.M .
21
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm
                          = 35 cm
22
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29
$$\eqalign{ & {\text{First}}\,{\text{number}} \cr & = {\frac{{551}}{{29}}} \cr & = 19 \cr & {\text{Third}}\,{\text{number}} \cr & = {\frac{{1073}}{{29}}} \cr & = 37 \cr & \therefore {\text{Required}}\,{\text{sum}} \cr & = {19 + 29 + 37} \cr & = 85 \cr} $$
23
Find the highest common factor of 36 and 84
Discuss
Answer & Solution
Answer: Option C
Solution:
36 = 22 × 32
84 = 22 × 3 x 7
∴ H.C.F. = 22 × 3 = 12
24
Which of the following fraction is the largest ?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\,{\text{of}}\,8,\,16,\,40\,{\text{and}}\,80 = 80 \cr & \frac{7}{8} = \frac{{70}}{{80}};\,\frac{{13}}{{16}} = \frac{{65}}{{80}};\,\frac{{31}}{{40}} = \frac{{62}}{{80}} \cr & {\text{Since}},\,\frac{{70}}{{80}} > \frac{{65}}{{80}} > \frac{{63}}{{80}} > \frac{{62}}{{80}}, \cr & {\text{So}},\,\frac{7}{8} > \frac{{13}}{{16}} > \frac{{63}}{{80}} > \frac{{31}}{{40}} \cr & {\text{So}},\,\frac{7}{8}\,{\text{is}}\,{\text{the}}\,{\text{largest}}{\text{.}} \cr} $$
25
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Required number
= (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548
26
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Required number
= H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127
27
Which of the following has the most number of divisors?
Discuss
Answer & Solution
Answer: Option C
Solution:
99 = 1 × 3 × 3 × 11
101 = 1 × 101
176 = 1 × 2 × 2 × 2 × 2 × 11
182 = 1 × 2 × 7 × 13
So, divisors of 99 are 1, 3, 9, 11, 33, 99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182
Hence, 176 has the most number of divisors.
28
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the numbers be 2x and 3x
Then, their L.C.M. = 6x
So, 6x = 48 or x = 8
∴ The numbers are 16 and 24
Hence, required sum = (16 + 24) = 40
29
The H.C.F. of $$\frac{{9}}{{10}},$$ $$\frac{{12}}{{25}},$$ $$\frac{{18}}{{35}}$$ and $$\frac{{21}}{{40}}$$ is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Required}}\,{\text{H}}{\text{.C}}{\text{.F}}{\text{.}} \cr & = \frac{{{\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\,{\text{of}}\,9,\,12,\,18,\,21}}{{{\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\,{\text{of}}\,10,\,25,\,35,\,40}} \cr & = \frac{3}{{1400}} \cr} $$
30
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{numbers}}\,{\text{be}}\,a\,{\text{and}}\,b \cr & {\text{Then}},\,a + b = 55\,{\text{and}} \cr & ab = 5 \times 120 = 600 \cr & \therefore {\text{The}}\,{\text{required}}\,{\text{sum}} \cr & = \frac{1}{a} + \frac{1}{b} = \frac{{a + b}}{{ab}} = \frac{{55}}{{600}} = \frac{{11}}{{120}} \cr} $$