- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Number System: LCM and HCF Test-4

Congratulations - you have completed

*Number System: LCM and HCF Test-4*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

81 owls, 108 bats, andÂ 162 crows haveÂ to be put in different cages and then arranged in a row in a such way that each row has equal number of cages and each row cage has a particular type of bird only .The least number of rows needed are:

23 | |

43 | |

27 | |

none |

Question 1 Explanation:

Since each row should have same number of cages and

thus the number of cage in a row has to be a common factor of 81, 108, 162 i.e 27

So the number of rows required for this are

[81/27 + 108/27 + 162/27]= 3 +4+6=13 rows required

thus the number of cage in a row has to be a common factor of 81, 108, 162 i.e 27

So the number of rows required for this are

[81/27 + 108/27 + 162/27]= 3 +4+6=13 rows required

Question 2 |

How many 4 digit numbers are there which are completely divisible by all even numbers (0 and 9)?

375 | |

567 | |

324 | |

none |

Question 2 Explanation:

Since we are asked to find all 4 digit numbers which are completely divisible by all even digits. So the numbers obtained here will be divisible by 2, 4, 6 and 8. On the first step, we find the LCM of 2,4,6 and 8: 24

Numbers between 1 to 999 divisible by 24= 41 numbers are there

Numbers between 1000 to 9999 divisible by 24= 9999/ 24 = 416

Subtract first from second, we get the numbers between 1000 to 9999 = 375.

Numbers between 1 to 999 divisible by 24= 41 numbers are there

Numbers between 1000 to 9999 divisible by 24= 9999/ 24 = 416

Subtract first from second, we get the numbers between 1000 to 9999 = 375.

Question 3 |

How many unique pairs of numbers are there if the HCF of two numbers is 12 and the product of numbers is 432?

2 | |

3 | |

4 | |

none |

Question 3 Explanation:

Since the HCF of two numbers is 12, so we can say that the

numbers can be in the form ofÂ 12a and 12b

and we have theÂ product of numbers as 432,

therefore: So 144(a x b) = 432

a x b = 3

So the factors of 36 are = 1,3

The pairs of numbers is (12, 36). Thus only one pair is possible.

Hence, correct answer is option D.

numbers can be in the form ofÂ 12a and 12b

and we have theÂ product of numbers as 432,

therefore: So 144(a x b) = 432

a x b = 3

So the factors of 36 are = 1,3

The pairs of numbers is (12, 36). Thus only one pair is possible.

Hence, correct answer is option D.

Question 4 |

Find the smallest 3 digit number which when divided by 7 and 9 leaves the remainder 1 and 3 respectively?

321 | |

120 | |

119 | |

none |

Question 4 Explanation:

When we divide the number by 9 the number leaves the remainder 3

so the number must be in the form of (9a+3) such numbers are 12, 21, 30, 39 48,57,66..

The smallest number which is divisible by 7 and 9 and leaves remainder 1 and 3 is 57.

As we know from the properties, we can obtain all numbers in the series with the help of the formula:

N = F (L.C.M. of 7, 9)+ 57

Smallest value can be obtained for F= 0, that is 57 itself.

Since we need the smallest 3 digit number, we take F=1.

Hence, the number is 120.

so the number must be in the form of (9a+3) such numbers are 12, 21, 30, 39 48,57,66..

The smallest number which is divisible by 7 and 9 and leaves remainder 1 and 3 is 57.

As we know from the properties, we can obtain all numbers in the series with the help of the formula:

N = F (L.C.M. of 7, 9)+ 57

Smallest value can be obtained for F= 0, that is 57 itself.

Since we need the smallest 3 digit number, we take F=1.

Hence, the number is 120.

Question 5 |

A red light flashes 7 times per minute and the green light flashes 13 times in 2 minutes at regular intervals. If both the lights start flashing at the same time, how many times do they flash together in an hour?

30 | |

35 | |

31 | |

45 |

Question 5 Explanation:

First light flashes after 60/7 sec

Second light flashes after 120/13 sec they blink together after

( LCM of 60/7, 120/13) = 120sec = 2 min,

So in one hour they blink together = 60/2=30 times.

Second light flashes after 120/13 sec they blink together after

( LCM of 60/7, 120/13) = 120sec = 2 min,

So in one hour they blink together = 60/2=30 times.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |