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

11.

A milk vendor has 21 litres of cow milk, 42 litres of toned milk and 63 litres of double toned milk. If he wants to pack them in cans, so that each can contains same litres of milk and does not want to mix any two kinds of milk in a can, then the least number of cans required is = ?

• (A) 3
• (B) 6
• (C) 9
• (D) 12

Show Answer

 For least or minimum number of cans we should have maximum capacity cans for required quantity ⇒ For this we take HCF of given quantities. HCF (21, 42, 63) = 21 ∴ Maximum capacity of a can = 21 litres ∴ Number of cans of cow milk $${\text{ = }}\frac{{21}}{{21}}{\text{ = 1}}$$ ∴ Number of cans of toned milk $${\text{ = }}\frac{{42}}{{21}} = 2$$ ∴ Number of cans of double toned milk $${\text{ = }}\frac{{63}}{{21}} = 3$$ ∴ Total number of cans = 1 + 2 + 3 = 6

12.

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

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

13.

The LCM of two numbers is 520 and their HCF is 4. If one of the numbers is 52, then the other number is = ?

• (A) 40
• (B) 42
• (C) 50
• (D) 52

Show Answer

 LCM = 520 HCF = 4 one number = 52 Let other number is y 52y = 4 × 520 ⇒ y = 40

14.

Find the smallest 5-digit number which when divided by 15, 20 and 25 leaves remainders 7, 12 and 17, respectively.

• (A) 10195
• (B) 10192
• (C) 10193
• (D) 10194

Show Answer

 Five-digit smallest number = 10000 Five-digit system number which is exactly divisible by 20, 25, 15 and leaves remainder 12, 17, 7 - means gap = 20 - 12 = 25 - 17 = 15 - 7 = 8 ⇒ 10000 + 200 - 8 = 10192 Answer

15.

The greatest possible length which can be used to measure exactly the lengths 7m, 3m 85cm, 12m 95cm is = ?

• (A) 15cm
• (B) 25cm
• (C) 35cm
• (D) 42cm

Show Answer

 Required length = HCF of 700cm, 385cm and 1295cm = 35cm

16.

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

Show Answer

 $$\eqalign{ & {\text{Let number are}} \cr & {\text{ = }}2x,3x,4x \cr & {\text{given,}} \cr & {\text{LCM of }}\left( {2 \times 3 \times 4} \right)x = 24x \cr & 24x = 240 \cr & x = 10 \cr & \therefore {\text{numbers are 2}} \times {\text{10 = 20}} \cr & {\text{3}} \times 10 = 30 \cr & {\text{4}} \times 10 = 40 \cr & \therefore {\text{Smaller is 20}} \cr} $$

17.

The L.C.M. of 22, 54, 108, 135 and 198 is = ?

• (A) 330
• (B) 1980
• (C) 5940
• (D) 11880

Show Answer

  2   22   -   54   -   108   -   135   -   198   3   11   -   27   -   54   -   135   -   99   3   11   -   9   -   18   -   45   -   33   3   11   -   3   -   6   -   15   -   11   11   11   -   1   -   2   -   5   -   11     1   -   1   -   2   -   5   -   1  L.C.M. = 2 × 3 × 3 × 3 × 11 × 2 × 5             = 5940

18.

An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 am. The next time, when they would beep together at the earliest is = ?

• (A) 10.30 am
• (B) 10.31 am
• (C) 10.59 am
• (D) 11 am

Show Answer

 Interval of change = (LCM of 60 and 62) sec = 1860 sec = 31 min So, the devices would beep together 31 min after 10 am i,e; at 10.31 am

19.

Let x be the last number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves remainder 0, the sum of digits of x is = ?

• (A) 24
• (B) 21
• (C) 22
• (D) 18

Show Answer

 $$\eqalign{ & {\text{LCM of 5, 6, 7, 8 = 840}} \cr & \frac{{840n + 3}}{9} \cr & \Rightarrow \frac{{3n + 3}}{9} \cr & \Rightarrow {\text{Take }}n{\text{ = 2}} \cr & \Rightarrow 3\left( 2 \right) + 3 \cr & \Rightarrow \frac{9}{9} = {\text{Remainder = 0}} \cr & \therefore {\text{Number is }}840n + 3 \cr & \Rightarrow 840\left( 2 \right) + 3\left[ {\because n = 2} \right] \cr & \Rightarrow 1683 \cr & {\text{Sum of digits}} \cr & {\text{ = }}\left( {1 + 6 + 8 + 3} \right){\text{ = 18}} \cr} $$

20.

The sum of and difference between the LCM and HCF of two numbers are 512 and 496, respectively. If one numbers is 72, then the other number is:

• (A) 56
• (B) 64
• (C) 80
• (D) 40

Show Answer

 LCM + HCF = 512 LCM - HCF = 496 $$\overline {\,2{\text{LCM}}\,\,\,\,\,\,\, = 1008\,} $$ LCM = 504 HCF = 8 I = 72 II = ? I × II = LCM × HCF 72 × II = 504 × 8 II = 56