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Test: HCF and LCM- 3 - Railways MCQ


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20 Questions MCQ Test - Test: HCF and LCM- 3

Test: HCF and LCM- 3 for Railways 2024 is part of Railways preparation. The Test: HCF and LCM- 3 questions and answers have been prepared according to the Railways exam syllabus.The Test: HCF and LCM- 3 MCQs are made for Railways 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: HCF and LCM- 3 below.
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Test: HCF and LCM- 3 - Question 1
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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

Detailed Solution for Test: HCF and LCM- 3 - Question 1

 L.C.M of 5, 6, 7, 8 = 840

Therefore, Required Number is of the form 840k+3.

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

Therefore, Required  Number = (840 x 2+3)=1683

Test: HCF and LCM- 3 - Question 2
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Find the 4-digit smallest number which when divided by 12, 15, 25, 30 leaves no remainder?

Detailed Solution for Test: HCF and LCM- 3 - Question 2

LCM of 12, 15, 25 and 30 is 300
least number of 4-digit divided by 300 is 1200

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Test: HCF and LCM- 3 - Question 3
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Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder?

Detailed Solution for Test: HCF and LCM- 3 - Question 3

Number = LCM (12, 17, 35) + 6 = 3780 + 6 = 3786

Test: HCF and LCM- 3 - Question 4
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The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one
Detailed Solution for Test: HCF and LCM- 3 - Question 4

LCM*HCF = a*b
840*84 = 168*b, b = 420

Test: HCF and LCM- 3 - Question 5
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Find the last number which when divided by 6, 8, 15 and 30 leaves remainder 2, 4, 11 and 26 respectively?
Detailed Solution for Test: HCF and LCM- 3 - Question 5

LCM (6, 8, 15, 30) – 4 = 120 – 4 = 116

Test: HCF and LCM- 3 - Question 6
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HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers
Detailed Solution for Test: HCF and LCM- 3 - Question 6

a*b = 5*275 and a+b = 80
(a+b)/(a*b) = 80/(5*275) = 16/275

Test: HCF and LCM- 3 - Question 7
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Three buckets contains balloons filled with water. First bucket contains 243 balloons. Second contains 304 balloons and last bucket contains 127 balloons. Find the largest number of balloons that can be given equally to the children such that 3, 4 and 7 balloons are left in first, second and third bucket respectively?
Detailed Solution for Test: HCF and LCM- 3 - Question 7

HCF (240, 300, 120) = 60

Test: HCF and LCM- 3 - Question 8
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Riya, Anil and Rishi start running around a circular stadium and complete one round in 15s, 12s and 21s respectively. In how much time will they meet again at the starting point?
Detailed Solution for Test: HCF and LCM- 3 - Question 8

LCM (15, 12, 21) = 420 second = 7 minutes

Test: HCF and LCM- 3 - Question 9
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In a college all the students are made to stand in four rows. 4 rows contains 12, 8, 22, 30 students respectively. Find the least number of students in the college?
Detailed Solution for Test: HCF and LCM- 3 - Question 9

LCM (12, 8, 22, 30) = 3960

Test: HCF and LCM- 3 - Question 10
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Find the greatest number that will divide 427 and 900 leaving the remainders 3 and 8 respectively?
Detailed Solution for Test: HCF and LCM- 3 - Question 10

HCF (427 – 3, 900 – 8) (424, 892) = 4

Test: HCF and LCM- 3 - Question 11
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What is the HCF of 3/5, 7/10, 2/15, 6/21?
Detailed Solution for Test: HCF and LCM- 3 - Question 11

HCF will be HCF of numerators/LCM of denominators
So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)
= 1/210

Test: HCF and LCM- 3 - Question 12
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The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.
Detailed Solution for Test: HCF and LCM- 3 - Question 12

Product of two numbers = HCF * LCM
So 2nd number = 84*840/168

Test: HCF and LCM- 3 - Question 13
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Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?
Detailed Solution for Test: HCF and LCM- 3 - Question 13

Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)

Test: HCF and LCM- 3 - Question 14
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Find the least number exactly divisible by 10, 15, 18 and 30.
Detailed Solution for Test: HCF and LCM- 3 - Question 14

It will be the LCM of these numbers.

Test: HCF and LCM- 3 - Question 15
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Find the least number which when divided by15, 21, 24 and 32 leaves the same remainder 2 in each case.
Detailed Solution for Test: HCF and LCM- 3 - Question 15

It will be the LCM of these numbers + 2
So 3360 + 2

Test: HCF and LCM- 3 - Question 16
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Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.
Detailed Solution for Test: HCF and LCM- 3 - Question 16

Since 10-6 = 4, 15-11 = 4, 18-14 = 4, 30-26 = 4
So answer will be the LCM of these numbers – 4
So 90 – 4

Test: HCF and LCM- 3 - Question 17
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Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.
Detailed Solution for Test: HCF and LCM- 3 - Question 17

Least 4 digit number = 1000
LCM of (12, 15, 21, 30) = 420
On dividing 420 by 1000, leaves remainder = 160
So answer = 1000 + (420 – 160)

Test: HCF and LCM- 3 - Question 18
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Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.
Detailed Solution for Test: HCF and LCM- 3 - Question 18

LCM(2, 5, 9, 12) = 180
So the number will be somewhat = 180x + 3
This number leaves no remainder when divided by 11, so x = 2 to get (180x+3) fully divided by 11
So number = 180*2 + 3

Test: HCF and LCM- 3 - Question 19
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HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.
Detailed Solution for Test: HCF and LCM- 3 - Question 19

x + y = 80, xy = 5*275 = 1375
sum of reciprocals = 1/x + 1/y = (x+y)/xy = 80/1375

Test: HCF and LCM- 3 - Question 20
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The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.
Detailed Solution for Test: HCF and LCM- 3 - Question 20

Produvt of two numbers = HCF * LCM
So 2nd number = 70*1050/210

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