Practice | Hashtable

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

A. 120

B. 240

C. 360

D. 480

Solution:

\begin{aligned} 2 | 24 - 36 - 40 \\ - - - - - - - - - - - - - \\ 2 | 12 - 18 - 20 \\ - - - - - - - - - - - - - \\ 2 | 6 - 9 - 10 \\ - - - - - - - - - - - - - \\ 3 | 3 - 9 - 5 \\ - - - - - - - - - - - - - \\ | 1 - 3 - 5 \\ L . C . M = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360 \end{aligned}

12. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A. 3

B. 13

C. 23

D. 33

Discuss

Solution:

L.C.M. of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37
∴ Number to be added = (60 - 37) = 23

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

A. $\frac{3}{4}$

B. $\frac{5}{13}$

C. $\frac{7}{13}$

D. $\frac{9}{13}$

Discuss

Solution:

\begin{aligned} 128352 \bar{) 238368 (} 1 \\ 128352 \\ \bar{110016)128352(} 1 \\ 110016 \\ \bar{183336)110016(} 6 \\ \frac{110016}{\times} \\ So, H .C .F of 128352 and 238368 \\ =  18336 \\ ∴ \frac{128352}{238368} = \frac{128352 \div 18336}{238368 \div 18336} = \frac{7}{13} \end{aligned}

14. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A. 1677

B. 1683

C. 2523

D. 3363

Discuss

Solution:

L.C.M. of 5, 6, 7, 8 = 840.
∴ Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2
∴ Required number = (840 x 2 + 3) = 1683

15. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A. 26 minutes and 18 seconds

B. 42 minutes and 36 seconds

C. 45 minutes

D. 46 minutes and 12 seconds

Discuss

Solution:

L.C.M. of 252, 308 and 198 = 2772
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

16. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A. 279

B. 283

C. 308

D. 318

Discuss

Solution:

\begin{aligned} Other number \\ = \frac{11 \times 7700}{275} \\ = 308 \end{aligned}

17. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

A. 196

B. 630

C. 1260

D. 2520

Discuss

Solution:

12 = 2 × 2 × 3
18 = 2 × 3 × 3
21 = 3 × 7
30 = 2 × 3 × 5

L.C.M. of 12, 18, 21 30= 2 × 3 × 2 × 3 × 7 × 5 = 1260
Required number =

\frac{1260}{2}

= 630

18. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

A. 12

B. 16

C. 24

D. 48

Discuss

Solution:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4
So, the numbers 12 and 16
L.C.M. of 12 and 16 = 48

19. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

A. 1008

B. 1015

C. 1022

D. 1032

Discuss

Solution:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015

20. 252 can be expressed as a product of primes as:

A. 2 × 2 × 3 × 3 × 7

B. 2 × 2 × 2 × 3 × 7

C. 3 × 3 × 3 × 3 × 7

D. 2 × 3 × 3 × 3 × 7

Discuss

Solution:

Clearly, 252 = 2 × 2 × 3 × 3 × 7