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1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is sum of the numbers? | |

A. 64 | B. 42 |

C. 28 | D. 40 |

Discuss |

answer with explanation

Answer: Option D

Explanation:

Let the numbers be $2x$ and $3x$

LCM of $2x$ and $3x$ $=6x~~$ (∵ LCM of 2 and 3 is 6. Hence LCM of $2x$ and $3x$ is $6x$)

Given that LCM of $2x$ and $3x$ is 48

=> $6x=48$

=> $x=\dfrac{48}{6}=8$

Sum of the numbers

$=2x+3x\\=5x$

= 5 × 8 = 40

2. What is the greatest number of four digits which is divisible by 15, 25, 40 and 75 ? | |

A. 9200 | B. 9600 |

C. 9800 | D. 9400 |

Discuss |

answer with explanation

Answer: Option B

Explanation:

Greatest number of four digits = 9999

LCM of 15, 25, 40 and 75 = 600

9999 ÷ 600 = 16, remainder = 399

Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75

= 9999 - 399 = 9600

3. Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is: | |

A. 40 | B. 20 |

C. 10 | D. 30 |

Discuss |

answer with explanation

Answer: Option B

Explanation:

Let the numbers be $2x$, $3x$ and $4x$

LCM of $2x$, $3x$ and $4x$ = $12x$

$12x=240\\~\\\Rightarrow x=\dfrac{240}{12}=20$

H.C.F of $2x$, $3x$ and $4x$ $=x=20$

4. What is the lowest common multiple of 12, 36 and 20? | |

A. 120 | B. 160 |

C. 220 | D. 180 |

Discuss |

answer with explanation

Answer: Option D

Explanation:

LCM = 2 × 2 × 3 × 1 × 3 × 5 = 180

5. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder? | |

A. 1683 | B. 1108 |

C. 2007 | D. 3363 |

Discuss |

answer with explanation

Answer: Option A

Explanation:

**Solution 1**

LCM of 5, 6, 7 and 8 = 840

Hence the number can be written in the form (840k + 3) which is divisible by 9.

If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.

If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.

Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

**Solution 2 - Hit and Trial Method**

Just see which of the given choices satisfy the given condtions.

Take 3363. This is not even divisible by 9. Hence this is not the answer.

Take 1108. This is not even divisible by 9. Hence this is not the answer.

Take 2007. This is divisible by 9.

2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer

Take 1683. This is divisible by 9.

1683 ÷ 5 = 336, remainder = 3

1683 ÷ 6 = 280, remainder = 3

1683 ÷ 7 = 240, remainder = 3

1683 ÷ 8 = 210, remainder = 3

Hence 1683 is the answer

Ashok Bhukhar

2015-04-05 16:37:55

How to find out sum of numbers if their lcm given..?

plz . tell me

plz . tell me

surajsingh

2015-03-08 13:45:32

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is?

Jyoti Kumari

2015-09-09 06:27:52

Since h.c.f. is 13, these numbers can be written as 13a and 13b.

Now, according to the question, we have product of these numbers is 2028.

Then, 13a × 13b = 2028.

=> a × b=12

Prime factors of 12 are 2,3.

12 = 2×2×3.

Number of factors of 12 =(2+1)(1+1)=6

Since 12 is not a perfect Square, so we can express 12 in $\frac{6}{2}$ = 3 ways as a product of two numbers.

Hence, Required no. of pairs = 3 Ans.

Now, according to the question, we have product of these numbers is 2028.

Then, 13a × 13b = 2028.

=> a × b=12

Prime factors of 12 are 2,3.

12 = 2×2×3.

Number of factors of 12 =(2+1)(1+1)=6

Since 12 is not a perfect Square, so we can express 12 in $\frac{6}{2}$ = 3 ways as a product of two numbers.

Hence, Required no. of pairs = 3 Ans.

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Javed Khan

2015-09-09 10:47:43

Going by your argument, the 3 ways of expressing 12 as product of two number are

12 = 1×12

12 = 2×6

12 = 3×4

In this, you can not take 2,6 because they are not co-prime. Suppose we take (2,6) then

numbers are 26, 78

But HCF(26,78) is not 13.

Generally, you can only take such pairs which are co-prime

(1,12) and (3,4) are fine and as explained by Jay, answer is 2

12 = 1×12

12 = 2×6

12 = 3×4

In this, you can not take 2,6 because they are not co-prime. Suppose we take (2,6) then

numbers are 26, 78

But HCF(26,78) is not 13.

Generally, you can only take such pairs which are co-prime

(1,12) and (3,4) are fine and as explained by Jay, answer is 2

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Jay

2015-03-11 15:35:07

Take numbers as 13a and 13b (because 13 is the HCF)

13a * 13b = 2028

ab = 12

12 can be written as a product of co-prime numbers in the following ways

a. 1*12

b. 3*4

(i.e., two ways)

So required number of ways = 2

The pairs are (13*1, 13*12) and (13*3, 13*4)

13a * 13b = 2028

ab = 12

12 can be written as a product of co-prime numbers in the following ways

a. 1*12

b. 3*4

(i.e., two ways)

So required number of ways = 2

The pairs are (13*1, 13*12) and (13*3, 13*4)

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Jay

2015-03-01 17:54:14

Many values are possible.

15 = 3*5

20 = 2*2*5

LCM of 15 and 20 = 3*2*2*5=60

(3*2*2*5) * 3 = 180

ie, an additional 3 should be there to make the LCM 180

number can be 3*3=9, 3*5*3=45, 2*2*5*3*3 = 180, etc

15 = 3*5

20 = 2*2*5

LCM of 15 and 20 = 3*2*2*5=60

(3*2*2*5) * 3 = 180

ie, an additional 3 should be there to make the LCM 180

number can be 3*3=9, 3*5*3=45, 2*2*5*3*3 = 180, etc

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priya

2015-02-13 17:44:46

the HCF of 2 number is 98 and their LCM is 2352.the sum of the number may be

a.1372

b.1398

c.1426

d.1484

gaukaran

2015-05-20 11:56:00

X*y= 98*2352

X*y=98*98*12*2

X*y= (98*2)*(98*12)

As compair to x and y then

X= 98*2

X= 196

Y = 98*12

Y= 1176

Then

X+y

196+1176= 1372

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Jay

2015-02-13 20:55:10

In the question, 'none of these' is not given as an option and 1372 is the only number divisible by 98. So one can directly write the answer as 1372

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