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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 3257683 is divided by 3256
3255 | |
1 | |
3254 | |
none |
Question 2 Explanation:
As we do earlier (an-1) is completely divisible by (a-1)
(3257683 -1) is exactly divisible by 3256 ,
So when 3257683 is divided by 3256 then there would be 1
as the remainder
(3257683 -1) is exactly divisible by 3256 ,
So when 3257683 is divided by 3256 then there would be 1
as 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 …………………1
On dividing by 5 number N would be N= 5R + 3 where r is the quotient …………………2
By 1 and 2
N = 20R +14………....................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 …………………2
By 1 and 2
N = 20R +14………....................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 ‘a’ and ‘b’ 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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