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Q:

The smallest number which when diminished by 7,  is divisible by  12, 16, 18, 21 and 28 is

A) 1008 B) 1015
C) 1022 D) 1032
 
Answer & Explanation Answer: B) 1015

Explanation:

Required Number = (L.C.M  of 12, 16, 18,21,28)+7

= 1008 + 7

= 1015

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Q:

Find the Greatest Number that will devide 43, 91  and 183  so as to leave the same remainder in each case

A) 4 B) 7
C) 9 D) 13
 
Answer & Explanation Answer: A) 4

Explanation:

Required Number = H.C.F  of  (91- 43), (183- 91) and (183-43)

                          = H.C.F of 48, 92, and 140 = 4

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Q:

A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square  tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?

A) 14 cms B) 21 cms
C) 42 cms D) None of these
 
Answer & Explanation Answer: B) 21 cms

Explanation:

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

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Q:

The maximum number of students  among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

A) 91 B) 910
C) 1001 D) 1911
 
Answer & Explanation Answer: A) 91

Explanation:

Required number of students = H.C.F  of 1001 and 910  = 91

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Q:

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A) -2 B) -1
C) 1 D) 2
 
Answer & Explanation Answer: A) -2

Explanation:

 H. C. F of two prime numbers is 1.  Product of numbers = 1 x 161 = 161.

 Let the numbers be a and b . Then , ab= 161.

 Now, co-primes with  product 161 are (1, 161) and (7, 23).

 Since x and y are prime numbers and x >y , we have x=23 and y=7.

 Therefore,  3y-x = (3 x 7)-23 = -2

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Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77 B) 88
C) 99 D) 110
 
Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

 

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 35

 

Now, co-primes with product  35 are (1,35) and (5,7)

 

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

 

Since one number lies 75 and 125, the suitable pair is  (55,77)

 

Hence , required number = 77

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Q:

Product of two co-prime numbers is 117. Their L.C.M  should be

A) 1 B) 117
C) Equal to their H.C.F D) cannot be calculated
 
Answer & Explanation Answer: B) 117

Explanation:

H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117

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Q:

If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to

A) 55/601 B) 601/55
C) 11/120 D) 120/11
 
Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.

 

Therefore, Required sum = 1a+1b=a+bab=55600=11120

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