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## Number System: Remainders Test-3

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Question 1 |

A number is divided by 713 gives the remainder 115 . If the same number is divided by 31 , then what will be the remainder

22 | |

34 | |

23 | |

none |

Question 1 Explanation:

Number will be in the form of (713n + 115)

where n is the quotient (713n + 115) = (31 x 23 x n ) + (31 x 3) + 22

From this we can extract 31 as common the left value would be = 31 x (23n +3) + 22

Therefore the left term is 22 hence 22 will be the remainder

where n is the quotient (713n + 115) = (31 x 23 x n ) + (31 x 3) + 22

From this we can extract 31 as common the left value would be = 31 x (23n +3) + 22

Therefore the left term is 22 hence 22 will be the remainder

Question 2 |

What will be the remainder when 3257

^{683}is divided by 32563255 | |

1 | |

3254 | |

none |

Question 2 Explanation:

As we do earlier (a

(3257

So whenÂ 3257

as the remainder

^{n}-1) is completely divisible by (a-1)(3257

^{683}-1) is exactly divisible by 3256 ,So whenÂ 3257

^{683}is divided by 3256 then there would be 1as the remainder

Question 3 |

On dividing a number by 4, we get 2 as the remainder. On dividing the quotient so obtained by 5, we get 3 as the remainder. On dividing the quotient so obtained by 6, we get 5 as the remainder. If the last quotient is 7 then find the number

954 | |

955 | |

987 | |

none |

Question 3 Explanation:

On dividing by 4 number N would be N= 4Q + 2 where q is the quotient â€¦â€¦â€¦â€¦â€¦â€¦â€¦1

On dividing by 5 number N would be N= 5R + 3 where r is the quotient â€¦â€¦â€¦â€¦â€¦â€¦â€¦2

By 1 and 2

N = 20R +14â€¦â€¦â€¦....................3

Now quotient R is divided by 6 we get 5 as the remainder and 7 as the quotient

so the value of R becomes 47, put R = 47 in 3

We get N = 954

So the right option is A

On dividing by 5 number N would be N= 5R + 3 where r is the quotient â€¦â€¦â€¦â€¦â€¦â€¦â€¦2

By 1 and 2

N = 20R +14â€¦â€¦â€¦....................3

Now quotient R is divided by 6 we get 5 as the remainder and 7 as the quotient

so the value of R becomes 47, put R = 47 in 3

We get N = 954

So the right option is A

Question 4 |

There are two numbers. When 3 times the larger number is divided by the smaller number we get 4 as the quotient and 3 as the remainder, Also 7 times smaller number is divided by the larger number we get 5 as the quotient and 1 as the remainder. Find the numbers.

34 , 24 | |

25 , 18 | |

34 , 40 | |

none |

Question 4 Explanation:

There are two numbers one is larger and second is smaller

Let us assume the larger number is A and the smaller one is B

We know that Dividend = (divisor x quotient + remainder )

So 3A = 4B + 3

7B = 5A + 1

On solving both the equations we bet B = 18

On substituting the B = 18

We get the number A= 25 so the right option is B

Let us assume the larger number is A and the smaller one is B

We know that Dividend = (divisor x quotient + remainder )

So 3A = 4B + 3

7B = 5A + 1

On solving both the equations we bet B = 18

On substituting the B = 18

We get the number A= 25 so the right option is B

Question 5 |

If there are two numbers â€˜aâ€™ and â€˜bâ€™ are separately divided by number c, then respective remainder are 4375 and 2986. If the sum of these numbers namely (a+b) is divided by d then 2361 is obtained as the remainder. Find d

5000 | |

6000 | |

7000 | |

none |

Question 5 Explanation:

Let the numbers are A and B

So as we said in the previous question

Dividend = (divisor x quotient + remainder)

Therefore

A = Quotient (q1) x divisor (d) + 4375

B = Quotient (q2) x divisor (d) + 2986

By adding we will get

(A+B) = (q1 +q2) x d + 7361

d = 5000, option A

So as we said in the previous question

Dividend = (divisor x quotient + remainder)

Therefore

A = Quotient (q1) x divisor (d) + 4375

B = Quotient (q2) x divisor (d) + 2986

By adding we will get

(A+B) = (q1 +q2) x d + 7361

d = 5000, option A

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