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**Excercise 1**

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

The Correct answer is :40

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

The Correct answer is :127

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

The Correct answer is :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

The Correct answer is :12

$36= 2^{2}\times3^{2}$

$84= 2^{2}\times3\times7$

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

The Correct answer is :85

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

The Correct answer is :1015

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

The Correct answer is :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

The Correct answer is :630

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

The Correct answer is :308

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

The Correct answer is :1683

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

The Correct answer is :23

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

The Correct answer is :360

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

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