H.C.F and L.C.M Problems Question & Answers



madhu9669

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

A) 1677 B) 1683
C) 2523 D) 3363
Answer & Explanation Answer: B) 1683

Explanation:

 L.C.M of 5, 6, 7, 8 = 840


\inline \fn_jvn \therefore Required Number is of the form 840k+3.


Least value of k for which (840k+3) is divisible by 9 is k = 2 


\inline \fn_jvn \therefore  Required  Number = (840 x 2+3)=1683

Report Error

view answer Workspace Report Error Discuss

Submitted By: madhu9669
Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

madhu9669

 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

Report Error

view answer Workspace Report Error Discuss

Submitted By: madhu9669
Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

madhu9669

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

Report Error

view answer Workspace Report Error Discuss

Submitted By: madhu9669
Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

madhu9669

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:

Largest size of the tile  = H.C.F of 378 cm and 525 cm = 21 cm

Report Error

view answer Workspace Report Error Discuss

Submitted By: madhu9669
Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

madhu9669

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

Report Error

view answer Workspace Report Error Discuss

Submitted By: madhu9669
Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability