Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M

81.

The greatest 4-digit number exactly divisible by 10, 15, 20 is = ?

• (A) 9990
• (B) 9960
• (C) 9980
• (D) 9995

Show Answer

 LCM of (10, 15, 20) ⇒ 5 × 2 × 3 × 2 = 60 ⇒ Largest 4-digit number = 9999 divide 9999 by LCM of given number ⇒ We get remainder = 39 ⇒ So, to divide completely subtract it from (9999 - 39) = 9960 ∴ 9960 is the largest four digit number which is completely divided by the given numbers.

82.

The least number which when divided by 4, 6, 8, 12 and 16 leaves remainder of 2 in each case is = ?

• (A) 46
• (B) 48
• (C) 50
• (D) 56

Show Answer

 LCM of (4, 6, 8, 12, 16) ⇒ 16 × 3 = 48 ∴ The number when divided by (4, 6, 8, 12, 16) leaves remainder 2 is = 48 + 2 = 50

83.

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 $$\frac{{{\text{a}} + {\text{c}}}}{{\text{b}}}?$$

• (A) 3.5
• (B) 2
• (C) 5
• (D) 4.5

Show Answer

 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 $$\eqalign{ & {\text{Then, }}\frac{{{\text{a}} + {\text{c}}}}{{\text{b}}} \cr & = \frac{{26 + 52}}{{39}} \cr & = \frac{{78}}{{39}} \cr & = 2 \cr} $$

84.

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

Show Answer

 Let the number be a and b $$\eqalign{ & {\text{Then,}} \cr & a + b = 36\ \text{and} \cr & ab = 3 \times 105 = 315 \cr & \therefore {\text{Required sum}} \cr & {\text{ = }}\frac{1}{a} + \frac{1}{b} \cr & = \frac{{a + b}}{{ab}} \cr & = \frac{{36}}{{315}} \cr & = \frac{4}{{35}} \cr} $$

85.

The HCF and LCM of two numbers are 8 and 48 respectively. If one of the number is 24, then the other number is = ?

• (A) 48
• (B) 36
• (C) 24
• (D) 16

Show Answer

 HCF = 8 LCM = 48 One number = 24 Let other number be = y ∴ 24y = 48 × 8 ⇔ y = 16

86.

Find the lowest common multiple of 24, 36 and 40.

• (A) 120
• (B) 240
• (C) 360
• (D) 480

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 $$\eqalign{ & 2\,\,\,|\,\,\,24\,\,\,\,\, - \,\,\,\,\,36\,\,\,\,\, - \,\,\,\,\,40 \cr & - - - - - - - - - - - - - \cr & 2\,\,\,|\,\,\,12\,\,\,\,\, - \,\,\,\,\,18\,\,\,\,\, - \,\,\,\,\,20 \cr & - - - - - - - - - - - - - \cr & 2\,\,\,|\,\,\,\,\,\,6\,\,\,\,\, - \,\,\,\,\,\,\,\,9\,\,\,\,\, - \,\,\,\,\,10 \cr & - - - - - - - - - - - - - \cr & 3\,\,\,|\,\,\,\,\,\,\,3\,\,\,\,\, - \,\,\,\,\,\,\,\,9\,\,\,\,\, - \,\,\,\,\,\,\,\,5 \cr & - - - - - - - - - - - - - \cr & \,\,\,\,\,\,\,|\,\,\,\,\,1\,\,\,\, - \,\,\,\,\,\,\,\,\,\,3\,\,\,\,\, - \,\,\,\,\,\,\,\,5 \cr & L.C.M = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360 \cr} $$

87.

The H.C.F of 2 × 3 × 5 × 7, 2 × 3 × 5 × 7 and 3 × 5 × 7 × 11 is = ?4233322

• (A) 105
• (B) 1155
• (C) 2310
• (D) 27720

Show Answer

 H.C.F. = Product of lowest powers of common factors            = 3 × 5 × 7            = 105

88.

The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is = ?

• (A) 91
• (B) 910
• (C) 1001
• (D) 1911

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 Required number of students = HCF of 1001 and 910 = 91

89.

A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is = ?

• (A) 1539
• (B) 539
• (C) 359
• (D) 1359

Show Answer

 10 - 9 = 1 9 - 8 = 1 8 - 7 = 1 ∴ LCM of (10, 9, 8) = 5 × 2 × 9 × 4 = 360 ∴ Required result = 360 - 1 = 359

90.

HCF of $$\frac{2}{3}{\text{,}}\frac{4}{5}$$  and $$\frac{6}{7}$$ is = ?

• (A) $$\frac{{48}}{{105}}$$
• (B) $$\frac{2}{{105}}$$
• (C) $$\frac{1}{{105}}$$
• (D) $$\frac{{24}}{{105}}$$

Show Answer

 $$\eqalign{ & {\text{HCF of fractional number is}} \cr & = \left( {\frac{{{\text{HCF of numerator}}}}{{{\text{LCM of Denominator}}}}} \right) \cr & \therefore {\text{HCF }}\left( {\frac{2}{3},\frac{4}{5},\frac{6}{7}} \right) \cr & \Rightarrow \left( {\frac{{{\text{HCF of 2,4,6}}}}{{{\text{LCM of 3,5,7}}}}} \right) \cr & \Rightarrow \frac{2}{{3 \times 5 \times 7}} \cr & \Rightarrow \frac{2}{{105}} \cr} $$