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

171.

The number of pair of positive integers whose sum is 99 and HCF is 9 is =

• (A) 5
• (B) 4
• (C) 3
• (D) 2

Show Answer

 Number of pair of positive integers whose sum is 99 and HCF is 9 is (9, 90) (18, 81) (27, 72) (36, 63) (45, 54)

172.

If r is the remainder when each of 7654, 8506, and 9997 is divided by the greatest number d (d > 1) then d - r is equal to = ?

• (A) 14
• (B) 18
• (C) 24
• (D) 28

Show Answer

 d = HCF of (8506 - 7654), (9997 - 8506), (9997 - 7654) = HCF of 852, 1491, 2343 = 213 $$\eqalign{ & {\text{213}}\overline {){\text{ 7654 (}}} 35 \cr & \,\,\,\,\,\,\,\,\,\,\frac{{639}}{{{\text{ }}1264}} \cr & \,\,\,\,\,\,\,\,\,\,\frac{{{\text{ }}1065}}{{{\text{ }}199}} \cr & {\text{Clearly r = 199}} \cr & \therefore d - r = 213 - 199 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$

173.

Find the greatest number of five digits which when divided by 3, 5, 8, 12 leaves 2 as remainder ?

• (A) 99999
• (B) 99948
• (C) 99962
• (D) 99722

Show Answer

 LCM of (3, 5, 8, 12) ⇒ 3 × 5 × 4 × 2 = 120 ⇒ Now greatest five digits number is 99999 On dividing 99999 by = 120 (LCM) We get remainder = $$\frac{{99999}}{{120}},$$  remainder = 39 ⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12) ∴ 99999 - 39 = 99960 ⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case. ⇒ add 2 in the 99960 ⇒ 99960 + 2 ⇒ 99962

174.

The H.C.F. of $$\frac{a}{b}\text{,}\frac{c}{d}\text{,}\frac{e}{f}$$  is equal to = ?

• (A) $$\frac{\text{LCM of }a,c,e}{\text{HCF of }b,d.f}$$  
• (B) $$\frac{\text{HCF of }a,c,e}{\text{LCM of }b,d,f}$$  
• (C) $$\frac{\text{HCF of }a,c,e}{\text{HCF of }b,d.f}$$  
• (D) $$\frac{ace}{bdf}$$

Show Answer

 $$\eqalign{ & {\text{HCF of fractions}} \cr & {\text{ = }}\frac{{{\text{HCF of Numerators}}}}{{{\text{LCM of Denominators}}}} \cr & {\text{HCF of }}\frac{a}{b}{\text{,}}\frac{c}{d}{\text{,}}\frac{e}{f} \cr & = \frac{{{\text{HCF of }}a,c,e}}{{{\text{LCM of }}b,d,f}} \cr} $$

175.

The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is

• (A) 2520
• (B) 842
• (C) 2522
• (D) 840

Show Answer

 LCM of (4, 5, 6, 7, 8) = 4 × 5 × 6 × 7 = 840 ⇒ Required number = 840k + 2, which is divisible by 13 $$\eqalign{ & {\text{For }}\frac{{840k + 2}}{{13}},\,\,\left( {{\text{remainder}} = 0} \right) \cr & {\text{Remainder}} = \frac{{8k + 2}}{{13}} \cr} $$ Put k = 3 Then, remainder = 0 For least multiple value of k is minimum ⇒ at k = 3 we get 840k + 2 = 840 × 3 + 2 = 2520 + 2 = 2522

176.

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

177.

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 $$\eqalign{ & {\text{ = }}\frac{{{\text{Area of ceiling}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{1517 \times 902}}{{41 \times 41}}} \right) \cr & = 814 \cr} $$

178.

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

• (A) 28
• (B) 32
• (C) 40
• (D) 64

Show Answer

 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

179.

The HCF and LCM of two numbers are 12 and 924 respectively. Then the number of such pair is = ?

• (A) 0
• (B) 1
• (C) 2
• (D) 3

Show Answer

 HCF = 12 Let numbers are 12x and 12y respectively LCM ⇒ 12xy = 924(given) ⇒ xy = 77 ⇒ Possible pairs are = (1 × 77) (7 × 11) ∴ Only two pairs are possible

180.

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 $$\eqalign{ & {\text{ = }}\left( {\frac{{78 + 104 + 117 + 169}}{{13}}} \right) \cr & = \frac{{468}}{{13}} \cr & = 36 \cr} $$