HCF And LCM(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

HCF AND LCM MCQs

Problems On Hcf And Lcm, H.C.F And L.C.M. Of Numbers, Lcm & Hcf, Hcf And Lcm

Total Questions : 1401 | Page 10 of 141 pages

Question 91.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

Discuss Question

Answer: Option C. -> 40

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

Therefore The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

Question 92.

The H.C.F. of \(\frac{9}{10}, \frac{12}{25}, \frac{18}{35}and\frac{21}{40} is:\)

\(\frac{3}{5}\)
\(\frac{252}{5}\)
\(\frac{3}{1400}\)
\(\frac{63}{700}\)

Discuss Question

Answer: Option C. -> \(\frac{3}{1400}\)

\( Required H.F.C = \frac{H.F.C of 9,12,18,21}{L.C.M.of 10,25,35,40}= \frac{3}{1400}\)

Question 93.

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:

\(\frac{55}{601}\)
\(\frac{601}{55}\)
\(\frac{11}{120}\)
\(\frac{120}{11}\)

Discuss Question

Answer: Option C. -> \(\frac{11}{120}\)

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

Threrefore The required sum = \(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{55}{600}=\frac{11}{120}\)

Question 94.

The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

15 cm
25 cm
35 cm
42 cm

Discuss Question

Answer: Option C. -> 35 cm

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

Question 95.

Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is sum of the numbers?

Discuss Question

Answer: Option B. -> 40

Let the numbers be 2x2x and 3x3x
LCM of 2x2x and 3x3x =6x =6x (âˆµ LCM of 2 and 3 is 6. Hence LCM of 2x2x and 3x3x is 6x6x)
Given that LCM of 2x2x and 3x3x is 48
=> 6x=486x=48
=> x=486=8x=486=8
Sum of the numbers
=2x+3x=5x=2x+3x=5x
= 5 × 8 = 40

Question 96.

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

123
127
235
305
None of these

Discuss Question

Answer: Option B. -> 127
- Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

Question 97.

Which of the following has the most number of divisors?

99
101
176
182
None of these

Discuss Question

Answer: Option C. -> 176
- 99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 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.

Question 98.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

28
32
40
64
None of these

Discuss Question

Answer: Option C. -> 40
- 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.

Question 99.

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:

55/601
601/55
11/120
120/11
None of these

Discuss Question

Answer: Option C. -> 11/120
- Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1 + 1 = a + b = 55 = 11 a b ab 600 120

Question 100.

The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is: