LCM and HCF Questions Answers - Quantitative Aptitude

Q1 :

Which one of the following numbers will divide 129 and 281 leaving 3 and 5 remainders respectively?

A

23

B

8

C

7

D

6

Check Answer

Q2 :

Find the LCM of $frac{1}{2}, frac{3}{5}, frac{4}{7} text{ and } frac{5}{21}$

A

$frac{2}{27}$

B

21

C

$frac{1}{60}$

D

60

Check Answer

Right Answer: D 60

Description:

LCM of the Fractions

$= frac{text{LCM of } 1,3,4,5}{text{HCF of } 2,7,21} = frac{60}{1} = 60$

Q3 :

The product of two numbers is $396 times 576$ and the L.C.M. is 6336. Their G.C.M. is

A

36

B

26

C

59

D

76

Check Answer

Right Answer: A

36

Description:

$text{G.C.M } =frac{396times576}{6336} = 36$

Q4 :

The greatest number that will divide 2930 and 3250, and will leave as remainder 7 and 11 respectively is:

A

37

B

53

C

59

D

79

Check Answer

Q5 :

HCF of two quadratic expressions is $(x + 2)$ and their LCM is $x^3 + 2x^2 - x - 2$ . The two expressions are:

A

$x^2 - 3x + 2, x^2 - x -2$

B

$x^2 + 3x + 2, x^2 + x -2$

C

$x^2 + 3x + 2, x^2 - x -2$

D

$x^2 - 3x + 2, x^2 + x -2$

Check Answer

Right Answer: B

$x^2 + 3x + 2, x^2 + x -2$

Description:

$begin{aligned} text{Given HCF } &= (x + 2) text{LCM } &= x^3 + 2x^2 - x - 2 &= (x + 2)(x - 1)(x + 1) text{ Now } p(x) q(x) &= text{ HCF and LCM} therefore p(x) &= (x + 2)(x - 1) = x^2 + x - 2 therefore q(x ) &= (x + 2) (x + 1) = x^2 + 3x + 2 end{aligned}$

Q6 :

Find the greatest number which will divide 400, 435 and 541 leaving 9,10 and 14 as remainders respectively.

A

19

B

17

C

13

D

9

Check Answer

Q7 :

For any integers 'a' and 'b' with HCF (a, b) = 1, what is HCF (a + b, a − b) equal to?

A

It is always 1

B

It is always 2

C

Either 1 or 2

D

None of these

Check Answer

Right Answer: C

Either 1 or 2

Description:

Put arbitrary values of a and b.
Illustration 1 Let a = 9 and b = 8.
$therefore$ HCF (8 + 9, 9 − 8) $Rightarrow$ HCF (17,1) = 1

Illustration 2 Let a = 23 and b = 17.
$therefore$ HCF (17 + 23, 23 − 17) $Rightarrow$ HCF (40,6) = 2
Hence, HCF (a + b, a − b) can either be 1 or 2.

Q8 :

What is the greatest possible size of a measuring vessel that can be used to measure the petrol in any of the three drums (1653 liters, 2261 liters, and 2527 liters), ensuring that the vessel is completely filled each time ?

A

31

B

27

C

19

D

41

Check Answer

Q9 :

If $x = 2^3 times 3^2 times 5^4$ and $y = 2^2 times 3^2 times 5 times 7$ , then HCF of $x$ and $y$ is

A

180

B

360

C

540

D

35

Check Answer

Right Answer: A

180

Description:

Given $x = 2^3 times 3^2 times 5^4$

and $y = 2^2 times 3^2 times 5 times 7$

$therefore$ HCF $= 2^2 times 3^2 times 5 = 4 times 9 times 5 = 180$

Q10 :

LCM of $2^3 times 3 times 5$ and $2^4 times 5 times 7$ is

A

$2^{12} times 3 times 5^2 times 7$

B

$2^4 times 5 times 7 times 9$

C

$2^4 times 3 times 5 times 7$

D

$2^3 times 3 times 5 times 7$

Check Answer

Right Answer: C

$2^4 times 3 times 5 times 7$

Description:

Here, say $a = 2^3 times 3 times 5$ and
$b = 2^4 times 5 times 7$ , then
LCM $= 2^4 times 3 times 5 times 7$

Q11 :

Two numbers have a ratio of 13:15, and their least common multiple (LCM) is 39,780. What are the two numbers?

A

273, 315

B

2652, 3060

C

516, 685

D

None of these

Check Answer

Right Answer: B

2652, 3060

Description:

The numbers are $13x$ and $15x$ .
So, $x$ is the HCF. Now,
HCF $times$ LCM = Product of numbers

$begin{aligned} & x times 39780 = 13x times 15x & Rightarrow x times 39780 = 13x times 15 times x^2 & Rightarrow x = dfrac{39780}{13 times 15} = 204 end{aligned}$

$therefore$ Numbers are $13 × 204 = 2652$ and $15 × 204 = 3060$

Q12 :

If the HCF of three numbers 144, $x$ and 192 is 12, then the number $x$ cannot be

A

180

B

84

C

60

D

48

Check Answer

Right Answer: D

48

Description:

Here, we know that
144 = 12 × 2 × 2 × 3
and 192 = 12 × 2 × 2 × 2 × 2

By taking option (d), 48 = 12 × 2 × 2

Hence, the value of $x$ will not be 48 otherwise the HCF of given numbers becomes 48.

Q13 :

Find the sum of all numbers between 300 and 400 that leave a remainder of 4 when divided by 6, 9, and 12.

A

692

B

764

C

1080

D

1092

Check Answer

Right Answer: A

692

Description:

LCM of 6, 9 and 12 = 36
So, number is the form of 36 $p$ + 4.
Since, the required numbers are between 300 and 400.
$therefore p$ = 9 and 10
$therefore$ Required sum = 328 + 364 = 692

Q14 :

Find the smallest positive integer that, when divided by 4, 5, 8, and 9, leaves remainders of 3, 4, 7, and 8, respectively.

A

119

B

319

C

359

D

719

Check Answer

Right Answer: C

359

Description:

Here,
$4 - 3 = 5 - 4 = 8 - 7 = 9 - 8 = 1$

Now, $4 = 2 times 2, 5 = 5$
$8 = 2 times 2 times 2, 9 = 3 times 3$

$therefore$ LCM $= 5 times 2 times 2 times 2 times 3 times 3 = 360$
Required number = 360 − 1 = 359

Q15 :

What is the HCF of $a^2b^4 + 2a^2b^2$ and $(ab)^7 - 4a^2b^9$ ?

A

ab

B

$a^2b^2$

C

$a^2b^3$

D

$a^3b^2$

Check Answer

Right Answer: B

$a^2b^2$

Description:

$a^2 b^4 + 2a^2b^2 = a^2b^2 (b^2 + 2)$ ...(i)

and $(ab)^7 - 4a^2b^9 = a^7b^7 - 4a^2b^9$

$= a^2b^2 (a^5b^5 - 4b^7)$ ...(ii)

From Eqs. (i) and (ii), we get HCF $= a^2b^2$

Q16 :

What is the sum of the digits of the smallest number that leaves a remainder of 33 when divided by 52, a remainder of 59 when divided by 78, and a remainder of 98 when divided by 117 ?

A

17

B

18

C

19

D

21

Check Answer

Right Answer: A

17

Description:

$begin{aligned} & text{Here, } 52 - 33 = 78 - 59 &= 117 - 98 = 19 &text{Now, } 52 = 13 times 2 times 2 &Rightarrow 78 = 13 times 2 times 3 &Rightarrow 117 = 13 times 3 times 3 &therefore text{LCM } = 13 times 2 times 2 times 3 times 3 = 468 end{aligned}$

$therefore$ Required number $= 468 - 19 = 449$

Hence, the sum of the digits is 17.

Q17 :

For any integer $n$ , what is the HCF of integers $m = 2n + 1$ and $k = 9n + 4$ ?

A

3

B

1

C

2

D

4

Check Answer

Right Answer: B

1

Description:

Since, $m = 2n + 1$ is an odd integer, so its factors may be 1 or 3 and $k = 9 n + 4$ its factors may be 1, 2 and 4.
Hence, HCF of ( $m, k$ ) is 1.