**1. About the number of pairs which have 16 as their HCF and 136 as their LCM, the conclusion can be**

a. only one such pair exists

b. only two such pairs exist

c. many three pairs exist

d. many such pairs exist

e. no such pair exists

**2. The HCF of two numbers is 12 and their difference is also 12. The numbers are**

a. 66, 78

b. 94, 106

c. 70, 82

d. 84, 96

e. 50. 62

**3. The HCF of two numbers is 16 and their LCM is 160. If one of the numbers is 32, then the other number is**

a. 48

b. 80

c. 96

d. 112

e. 108

**4. HCF of three numbers is 12. If they are in the ratio 1:2:3, then the numbers are**

a. 12,24,36

b. 10,20,30

c. 5,10,15

d. 4,8,12

e. 15, 30, 45

**5. Six bells commence tolling together and toll at intervals of 2,4,6,8,10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?**

a. 4

b. 10

c. 15

d. 16

e. 18

**6. The largest natural number which exactly divides the product of any four consecutive natural numbers is :**

a. 6

b. 12

c. 24

d. 120

e. 150

**7. The traffic lights at three different road crossing change after every 48 sec; 72 sec; and 108 sec., respectively. If they all change simultaneously at 8:20:00 hrs, then they will again change simultaneously at**

a. 8:27:12 Hrs

b. 8:27:24 Hrs

c. 8:27:36 Hrs

d. 8:27:48 Hrs

e. 8: 27:53 Hrs

**8. The greatest number by which if 1657 and 2037 are divided the remainders will be 6 and 5 respectively is**

a. 127

b. 235

c. 260

d. 305

e. 310

**9.The total number of prime factors of the product (8)20×(15)24×(7)15 is**

a. 59

b. 98

c. 123

d. 4

e. 14

**10. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divisible by 3, then the first number is**

a. 264

b. 132

c. Both a and b

d. 33

e. 36

**11. What least number must be subtracted from 1294 so that the remainder when divided 9, 11, 13 will leave in each case the same remainder 6 ?**

a. 0

b. 1

c. 2

d. 3

e. 4

**12. The least number which is divisible by 12, 15, 20 and is a perfect square, is**

a. 400

b. 900

c. 1600

d. 3600

e. 4200

**13. The least perfect square number which is divisible by 3,4,5,6 and 8 is**

a. 900

b. 1200

c. 2500

d. 3600

e. 1500

**14. Three piece of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank?**

a. 7 m

b. 14 m

c. 42 m

d. 63 m

e. 78

**15. The greatest number which can divide 1354, 1866, 2762 leaving the same remainder 10 in each case is :**

a. 64

b. 124

c. 156

d. 260

e. 280

**Answers**

**1.e**

HCF is always a factor of LCM. ie., HCF always divides LCM perfectly.

**2.d**

The difference of required numbers must be 12 and every number must be divisible by 12. Therefore, they are 84, 96.

**3. b**

The number = HCF×LCMGiven number=16×16032=80

**4.A**

Let the numbers be a, 2a and 3a.

Then, their HCF = a so a=12

The numbers are 12,24,36

**5.d**

LCM of 2,4,6, 8,10 and 12 is 120. So, the bells will toll simultaneously after 120 seconds.

i.e.2 minutes. In 30 minutes, they (302+1) toll times ie.16 times.

**6. C**

The required number can be find out by following way.

1×2×3×4=24

**7.(A)**

The change of interval=(LCM of 48,72,108)sec.=432. So, for every 432 seconds i.e.7 min. 12 sec. the lights will change. So add 7 min.12 sec.to 8:20:00 Hrs.i.e.8:27:12 Hrs.

**8. A**

The needed number is HCF of (1657-6) and (2037-5)=HCF of 1651 & 2032=127.

**9.D**

The prime numbers are 2,3,5,17 in the expression. The expression can be written as (23)20×(3×5)24×(17)15⇒260×324×524×1715

So number of prime factors are 4. i.e., 2, 3, 5, 17

**10.C**

Let the numbers are ah, bh respectively. Here h is HCF of two numbers. (obviously a, b are coprimes i.e., HCF (a, b) = 1)

Given that HCF = h = 44 and LCM = abh = 264

Dividing LCM by HCF we get ab = 6.

ab can be written as 1 x 6, 2 x 3, 3 x 2, 6 x 1.

But given that the first number is divisible by 3. So only two options possible for A. 3 x 44, 6 x 44. So option C is correct

**11. B**

LCM of 9,11,13 is 1287. Dividing 1294 with 1287, the remainder will be 7, to get remainder 6, 1 is to be deducted from 1294 so that 1293 when divided by 9,11,13 leaves 6 as remainder.

**12. b**

LCM = 5 × 3 × 22 = 60

To make this number as a perfect square, we have to multiply this number by 5 × 3

The number is 60 × 15= 900

**13.D**

LCM = 2×2×2×3×5=23×3×5

But the least perfect square is = 23×3×5×(2×3×5)=24×32×52=3600 as the perfect squares have their powers even.

**14. A**

The maximum possible length = (HCF of 42, 49, 63) = 7

**15. A**

The needed number = HCF of 1344, 1856 and 2752=64

## No comments:

## Post a Comment