# HCF and LCM - Excercise 1

## Questions

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

• 28
• 32
• 40
• 64

Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x
So, 6x = 48 or x = 8.
Therefore The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.

2.   The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

• 123
• 127
• 235
• 305

Required number = H.C.F. of (1657-6) and (2037-5 )
=> H.C.F. of 1651 and 2032 = 127.

3.   The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

• 504
• 536
• 544
• 548

Required number = (L.C.M. of 12, 15, 20, 54) + 8
=> (540+8) =548

4.   Find the highest common factor of 36 and 84.

• 4
• 6
• 12
• 18

$36= 2^{2}\times3^{2}$
$84= 2^{2}\times3\times7$
Therefore H.C.F = $2^{2}\times3=12$

5.   Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

• 75
• 81
• 85
• 89

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29
First number = ${\left(\begin{array}{c}\frac{551}{29}\end{array}\right)}$ =19
Third number = ${\left(\begin{array}{c}\frac{1073}{29}\end{array}\right)}$ =37
Therefore Required sum = (19 + 29 + 37) = 85.

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

• 15 cm
• 25 cm
• 35 cm
• 42 cm
The Correct answer is :35 cm

Required length =( H.C.F. of 700 cm, 385 cm and 1295) cm = 35 cm.

7.   252 can be expressed as a product of primes as:

• 2 x 2 x 3 x 3 x 7
• 2 x 2 x 2 x 3 x 7
• 3 x 3 x 3 x 3 x 7
• 2 x 3 x 3 x 3 x 7
The Correct answer is :2 x 2 x 3 x 3 x 7

Clearly, 252 = 2 x 2 x 3 x 3 x 7

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

• 1008
• 1015
• 1022
• 1032

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

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

• 12
• 16
• 24
• 48

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.

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

• 196
• 630
• 1260
• 2520

L.C.M. of 12, 18, 21 30
${ 2\times3\times2\times3\times7\times5}$
Required number = ${1260\div2}$ =630

11.   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:

• 279
• 283
• 308
• 318

Other number = ${\left(\begin{array}{c}\frac{11\times7700}{275}\end{array}\right)}$=308

12.   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 ?

• 26 minutes and 18 seconds
• 42 minutes and 36 seconds
• 45 minutes
• 46 minutes and 12 seconds
The Correct answer is :46 minutes and 12 seconds

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

13.   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:

• 1677
• 1683
• 2523
• 3363

L.C.M. of 5, 6, 7, 8 = 840.
therefore Required number is of the form 840 k+3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Therefore Required number = ${(840\times2+3\frac{}{})}$ =1683

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

• 3
• 13
• 23
• 33

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

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

• 120
• 240
• 360
• 480
L.C.M of => ${2\times2\times2\times3\times3\times5\frac{}{}}$ =360