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71. If the product of two numbers is 324 and their HCF is 3, then their LCM will be = ?

A. 972

B. 327

C. 321

D. 108

Discuss

Solution:

LCM  = \frac{324}{3} = 108

72. What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder ?

A. 312

B. 242

C. 1562

D. 1586

Discuss

Solution:

\begin{aligned} LCM of (3, 5, 6, 8, 10, 12) \\ = 3 \times 5 \times 2 \times 4 \\ = 120 \\ Required number is \\ \frac{120 K + 2}{22} = \frac{10 K + 2}{22} \\ at K =  2, \frac{10 K + 2}{22} \\ \Rightarrow Remainder  =  0 \\ The given condition satisfied \\ = 120 K + 2 \\ = 240 + 2 \\ = 242 \end{aligned}

73. The LCM of two numbers is 20 times their HCF. The sum of HCF ans LCM is 2520. If one of the number 480, the other number is = ?

A. 400

B. 480

C. 520

D. 600

Discuss

Solution:

\begin{aligned} Let HCF  = x \\ LCM  =  20 x \\ sum of HCF + LCM  =  2520 \\ \Rightarrow x + 20 x = 2520 \\ \Rightarrow 21 x = 2520 \\ \Rightarrow x = 120 \\ HCF  =  120 \\ LCM  =  120 \times 20 = 2400 \\ One number  =  480 \\ Let another number  = y \\ y \times 480 = 120 \times 2400 \\ y = \frac{120 \times 2400}{480} = 600 \end{aligned}

74. The largest 4-digit number exactly divisible by each of 12, 15, 18 and 27 is?

A. 9690

B. 9720

C. 9930

D. 9960

Discuss

Solution:

LCM of (12, 15, 18, 27)
⇒ 4 × 3 × 5 × 3 × 3 = 540
largest 4-digits number = 9999 on dividing by 540 to number =

\frac{9999}{540}

remainder = 279
Required number = 9999 - 279 = 9720

75. If HCF of p and q is x and q = xy, then the LCM of p and q is = ?

A. pq

B. qy

C. xy

D. py

Discuss

Solution:

\begin{aligned} Product of numbers \\ =  HCF \times LCM \\ \Rightarrow p q = x \times LCM \\ \Rightarrow LCM = \frac{p q}{x} \\ = \frac{p (x y)}{x} = p y \end{aligned}

76. The HCF and LCM of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is = ?

A. 12

B. 48

C. 84

D. 108

Discuss

Solution:

\begin{aligned} Let the numbers be x and 4 x \\ Then , x \times 4 x = 84 \times 21 \\ \Leftrightarrow x^{2} = (\frac{84 \times 21}{4}) \\ \Leftrightarrow x = 21 \\ Hence larger number \\ =  4 x \\ = 84 \end{aligned}

77. Three numbers are in the ratio 2 : 3 : 4, If their LCM is 240 the smaller of the three numbers is = ?

A. 20

B. 60

C. 30

D. 80

Discuss

Solution:

\begin{aligned} Let number are \\ = 2 x, 3 x, 4 x \\ given, \\ LCM of (2 \times 3 \times 4) x = 24 x \\ 24 x = 240 \\ x = 10 \\ ∴ numbers are 2 \times 10 =  20 \\ 3 \times 10 = 30 \\ 4 \times 10 = 40 \\ ∴ Smaller is 20 \end{aligned}

78. The HCF of two numbers , each having three digits, is 17 and their LCM is 714. The sum of the numbers will be ?

A. 289

B. 391

C. 221

D. 731

Discuss

Solution:

HCF = 17
Let numbers are = 17x, 17y
LCM = 17xy = 714 (given)
xy = 42
Possible pairs are (1, 42), (2, 21), (3, 14), (6, 7)
Possible numbers are (17, 714), (34, 357), (51, 238), (102, 119)
but given that both numbers are of three digits
∴ numbers are = (102, 119)
∴ sum of numbers = 102 + 119 = 221

79. The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case , but when divided by 7 leaves no remainder , is = ?

A. 189

B. 182

C. 175

D. 91

Discuss

Solution:

LCM of (5, 10, 12, 15) = 5 × 2 × 6 = 60
smallest number divided by (5, 10, 12, 15)
leaves remainder 2 and when divided by 7 leaves no remainder is

\begin{aligned} = \frac{60 K + 2}{7} = \frac{4 K + 2}{7} \\ At K  =  3, \frac{4 K + 2}{7} \\ \Rightarrow remainder  =  0 \\ ∴ number  =  60K + 2 \\ =  60 \times 3 + 2 \\ = 182 \end{aligned}

80. If the sum of two numbers is 36 and their HCF and LCM are 3 and 105 respectively, the sum of the reciprocals of the numbers is = ?

A. $\frac{2}{35}$

B. $\frac{3}{35}$

C. $\frac{4}{35}$

D. None of these

Discuss

Solution:

Let the number be a and b

\begin{aligned} Then, \\ a + b = 36 and \\ a b = 3 \times 105 = 315 \\ ∴ Required sum \\ = \frac{1}{a} + \frac{1}{b} \\ = \frac{a + b}{a b} \\ = \frac{36}{315} \\ = \frac{4}{35} \end{aligned}

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