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

171.

The least number of square tiles required to pave the ceiling of a room 15m 17cm long and 9m 2cm broad is = ?

• (A) 656
• (B) 738
• (C) 814
• (D) 902

Show Answer

 For the least number of tiles, the size of the tiles must be the maximum. Maximum size of the tile = HCF of 1517 cm and 902 cm = 41 cm. Hence, required number of tiles $$\eqalign{ & {\text{ = }}\frac{{{\text{Area of ceiling}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{1517 \times 902}}{{41 \times 41}}} \right) \cr & = 814 \cr} $$

172.

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:

• (A) $$\frac{{55}}{{601}}$$
• (B) $$\frac{{601}}{{55}}$$
• (C) $$\frac{{11}}{{120}}$$
• (D) $$\frac{{120}}{{11}}$$

Show Answer

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{numbers}}\,{\text{be}}\,a\,{\text{and}}\,b \cr & {\text{Then}},\,a + b = 55\,{\text{and}} \cr & ab = 5 \times 120 = 600 \cr & \therefore {\text{The}}\,{\text{required}}\,{\text{sum}} \cr & = \frac{1}{a} + \frac{1}{b} = \frac{{a + b}}{{ab}} = \frac{{55}}{{600}} = \frac{{11}}{{120}} \cr} $$

173.

A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number ?

• (A) 1 litre
• (B) 5 litres
• (C) 15 litres
• (D) 25 litres

Show Answer

 75 litres, 45 litres For maximum capacity take HCF (75, 45) = 15

174.

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

175.

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

• (A) $$\frac{5}{{14}}$$
• (B) $$\frac{1}{{84}}$$
• (C) $$\frac{1}{{63}}$$
• (D) $$\frac{1}{{168}}$$

Show Answer

 For HCF of fractions take HCF of numerators and LCM denominators HCF of 3, 5, 6 = 1 LCM of 4, 6, 7 = 84 Hence, HCF of fractions = $$\frac{1}{{84}}$$

176.

Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut ?

• (A) 27
• (B) 36
• (C) 43
• (D) 480

Show Answer

 Maximum length of each part = HCF of 78 cm, 104 cm, 117 cm, 169 cm = 13 cm ∴ Number of pieces $$\eqalign{ & {\text{ = }}\left( {\frac{{78 + 104 + 117 + 169}}{{13}}} \right) \cr & = \frac{{468}}{{13}} \cr & = 36 \cr} $$

177.

Reduce $$\frac{{128352}}{{238368}}$$  to its lowest terms.

• (A) $$\frac{{3}}{{4}}$$
• (B) $$\frac{{5}}{{13}}$$
• (C) $$\frac{{7}}{{13}}$$
• (D) $$\frac{{9}}{{13}}$$

Show Answer

 $$\eqalign{ & {\text{128352}}\overline {)238368{\text{(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{128352}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {110016{\text{)128352(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,110016 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {183336{\text{)110016(}}} {\text{6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{110016}}{ \times } \cr & {\text{So, H}}{\text{.C}}{\text{.F of 128352 and 238368}} \cr & {\text{ = 18336}} \cr & \therefore \frac{{128352}}{{238368}} = \frac{{128352 \div 18336}}{{238368 \div 18336}} = \frac{7}{{13}} \cr} $$

178.

The LCM of three different numbers is 120. Which of the following cannot be their HCF?

• (A) 8
• (B) 12
• (C) 24
• (D) 35

Show Answer

 LCM = 120 (given) LCM is the product of one common factor and other different factors of the given numbers. ∴ Factorize the given LCM = 120 $$\frac{{\,\,\,{\text{2}} \times {\text{2}} \times {\text{3}} \times {\text{5}} \times {\text{2}}\,\,\,}}{{4\left( {3 \times 5 \times 2} \right)}}$$ = Here 4 is common factor (common factor is the HCF of the given number) ∴ HCF = 4 So, for the given numbers the HCF should be multiple of 4 ⇒ Hence go through options which is not a multiple of 4 is 35 Hence answer is 35

179.

The G.C.D. of 1.08, 0.36 and 0.9 is:

• (A) 0.03
• (B) 0.9
• (C) 0.18
• (D) 0.108

Show Answer

 Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18, ∴ H.C.F. of given numbers = 0.18

180.

A and B are two prime numbers such that A > B and their LCM is 209. The value of A² - B is:

• (A) 350
• (B) 372
• (C) 361
• (D) 339

Show Answer

 A and B is prime number LCM of A and B = 209 = 11 × 19 ∵ A > B ∴ A = 19 B = 11 A2 - B = (19)2 - 11 = 361 - 11 = 350