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## Number System: Level 1 Test - 3

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Question 1 |

1. The LCM and HCF of two numbers are 4641 and 21 respectively. If one of the numbers lies between 200 and 300, find the numbers which satisfy the condition

273, 357 | |

273, 361 | |

273, 359 | |

273, 363 |

Question 1 Explanation:

Product of the numbers = HCF x LCM

= 21 x 4641 = 21 x3x7 x13x17

= 3x7 x3x7 x13x17 The numbers can be 3x7 x13 and 3x7 x17,

= 21 x 4641 = 21 x3x7 x13x17

= 3x7 x3x7 x13x17 The numbers can be 3x7 x13 and 3x7 x17,

*ie,*273 and 357.Question 2 |

Members of a club have their ages are in AP, the common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250 yr, Find the number of members in the club is

15 | |

20 | |

25 | |

30 |

Question 2 Explanation:

Let

Then 250 =n/2[2x7+(n-1)3/12]

250= n/2[14+1/4n-1/4]

250= 7n +n^2/8-n/8

n= 25

*n*be the number of members in the club.Then 250 =n/2[2x7+(n-1)3/12]

250= n/2[14+1/4n-1/4]

250= 7n +n^2/8-n/8

n= 25

Question 3 |

6 identical machines can produce a total of 270 bottles per min by running at constant speed. At this rate, how many bottles could 10 such machines produce in 4 min?

1800 | |

648 | |

2700 | |

10800 |

Question 3 Explanation:

Answer a) Number of machines =270x10x4/6=1800

Question 4 |

There are two examination rooms A and B. If 10 candidates from A are sent to

*B,*the number ofÂ students in each room is the same. If 20 candidates from B are sent to A, the number of students in A is double the number of students in B. Find the number of students in each room.
(a) 100 in A and 80 in B | |

(b) 80 in A and 100 in B | |

(e) 120 in A and 100 in B | |

(d) 100 in A and 120 in B |

Question 4 Explanation:

A -10 = B + 10

A - B = 20 Â Â Â Â Â Â Â Â Â Â Â ... 1)

and A + 20 = 2 (B - 20)

A - 2B = - 60 Â Â Â Â Â Â ... 2

From Eqs. (i) And (ii), A = 100, B = 80

A - B = 20 Â Â Â Â Â Â Â Â Â Â Â ... 1)

and A + 20 = 2 (B - 20)

A - 2B = - 60 Â Â Â Â Â Â ... 2

From Eqs. (i) And (ii), A = 100, B = 80

Question 5 |

The LCM of two numbers is 4800 and their HCF is 160. One of the numbers is 480, Find the other number is

16 | |

16000 | |

160 | |

1600 |

Question 5 Explanation:

Product of two numbers=LCM*HCF

4800*160=480*x

x=1600

4800*160=480*x

x=1600

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