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
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
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