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If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?


A0

B2

C3

D5

Answer: Option C

Explanation:

Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60
If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a

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NA
SHSTTON
20
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76
Solv. In. Corr.
96
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Three wheels make 18, 24, 54 revolutions per minute. Each has a black mark on it. It is aligned at the start. When does it align again for the first time?


A2 Seconds

B10 Seconds

C12 seconds

DNone of these

Answer: Option B

Explanation:

Let the time be x minutes.
Therefore 18x 24x 54x
Hcf = 6
Therefore time = 1/6 minute ie 10 secs

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NA
SHSTTON
112
Solv. Corr.
97
Solv. In. Corr.
209
Attempted
2 M:2 S
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3 / 239

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find the number between 100 to 400 which is divisible by either 2,3,5,7..


A120

B100

C205

D210

 View Answer |  Submit Your Solution | Important Formulas | Topic: H.C.F and L.C.M | Asked In AccentureTCS NQTHireMee |

Answer: Option D

Explanation:

LCM of 2,3,5,7=210

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NA
SHSTTON
79
Solv. Corr.
63
Solv. In. Corr.
142
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0 M:47 S
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What is the least amount that a person can have, such that when he distributes it into groups of Rs.16 or Rs. 18 or Rs. 20 or Rs. 25, he is always left with Rs. 4?


ARs. 1796

BRs. 1804

CRs. 2596

DRs. 3604

Answer: Option D

Explanation:

Find the lcm of 16 18 20 and 25 i.e 3600.
so the amount is (3600 + 4) = Rs. 3604

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NA
SHSTTON
135
Solv. Corr.
185
Solv. In. Corr.
320
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There are three numbers, these 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. What will be the sum of three numbers :


A80

B82

C85

D87

 View Answer |  Submit Your Solution | Important Formulas | Topic: H.C.F and L.C.M | Asked In NagarroHexaware TechnologiesHireMee |

Answer: Option C

Explanation:

Given that numbers are co primes,
and two products have the middle number in common.

=> Middle number = H.C.F. of 551 and 1073 = 29

so first number is = 551/29 = 19
=> Third number = 1073/29 = 37

Therefore, sum of these numbers is = (19 + 29 + 37) = 85

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NA
SHSTTON
8
Solv. Corr.
6
Solv. In. Corr.
14
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What is LCM of 36/225, 48/150 , 72/65 ?
A


A36/65

B144/5

C72/225

D288/5

Answer: Option B

Explanation:

LCM of fraction is LCM of Numerator upon LCM of Denominator.

Numerators = 36, 48 and 72

72 is largest number among them. 72 is not divisible by 36 or 48

start with table of 72.

72

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NA
SHSTTON
16
Solv. Corr.
31
Solv. In. Corr.
47
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The total number of prime factors of the product (8)20


A59

B98

C123

D4

E14

Answer: Option D

Explanation:

The prime numbers are 2,3,5,17 in the expression.

The expression can be written as (2

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NA
SHSTTON
6
Solv. Corr.
5
Solv. In. Corr.
11
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The least number which is divisible by 12, 15, 20 and is a perfect square, is -


A400

B900

C1600

D3600

E4200

Answer: Option B

Explanation:

12 = 2 x 2 x 3

15 = 3 x 5

20 = 2 x 2 x 5

So, LCM = 2 x 2 x 3 x 5 = 60

As 3 and 5 are not in pair in LCM’s factor so we need to multiply 60 by 5 and 3 to make it a perfect square.

Required smallest number = 60 × 3 × 5 = 900

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NA
SHSTTON
7
Solv. Corr.
3
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10
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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?


A7m

B14m

C42m

D63m

E78m

Answer: Option A

Explanation:

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

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SHSTTON
28
Solv. Corr.
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Solv. In. Corr.
51
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The HCF of two numbers is 12 and their difference is also 12. The numbers are -


A66,78

B94, 104

C70,82

D84,96

Answer: Option D

Explanation:

Let the numbers be 12a,12b where gcd(a,b)=1. (By the definition of HCF).
Also, 12b−12a=12.
⟹b−a=1
So, we are left with two equations,
gcd(a,b)=1 and
b−a=1

Which is true for all adjacent pairs of natural numbers. Therefore,

(a,b)=(1,2)(2,3),(3,4),(4,5),…


Which means,
(12a,12b)=(12,24),(24,36),(36,48),(48,60),(84,96)…

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