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Question 1 of 5
1. Question
The digit in the unit’s place of the resulting number of the expression (234)100 + (234)101, is
Correct
Ans: d
We have, (234)100+ (234)101 (234)00(1+ 234) = 235(234)100
We know that square of any number having 4 at unit place is a number in which 6 at unit place. Any exponent of a number 6 at unit place is always 6 is at unit place.
Because (235)(…..6)=…30
Resulting number have 0 at unit place.
Incorrect
Ans: d
We have, (234)100+ (234)101 (234)00(1+ 234) = 235(234)100
We know that square of any number having 4 at unit place is a number in which 6 at unit place. Any exponent of a number 6 at unit place is always 6 is at unit place.
Because (235)(…..6)=…30
Resulting number have 0 at unit place.
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Question 2 of 5
2. Question
The value of [1/(1 ´4)] +[1/(4´ 7)]+[1/(7´ 10)]+……+[1/(16 ´ 19)], is
Correct
Ans: b
We have, [1/(1 x4)] +[1/(4x 7)]+[1/(7x 10)]+……+[1/(16 x 19)]
Then nth term of 1.4.7,……is (3n-2) and of 4,7,10…….is(3n+1)
Hence, [1/(1×4)] +[1/(4x 7)]+[1/(7x 10)]+……+[1/(16 x 19)]
= =
=1/3[(1/1-1/4)+(1/4-1/7)+….+(1/16-1/9)]
=1/3(1-1/9)=1/3 x 18/19 = 6/19
Incorrect
Ans: b
We have, [1/(1 x4)] +[1/(4x 7)]+[1/(7x 10)]+……+[1/(16 x 19)]
Then nth term of 1.4.7,……is (3n-2) and of 4,7,10…….is(3n+1)
Hence, [1/(1×4)] +[1/(4x 7)]+[1/(7x 10)]+……+[1/(16 x 19)]
= =
=1/3[(1/1-1/4)+(1/4-1/7)+….+(1/16-1/9)]
=1/3(1-1/9)=1/3 x 18/19 = 6/19
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Question 3 of 5
3. Question
A number when divided by 7 leaves a remainder 3 and the resulting quotient, when divided by 11 leaves a remainder 6, If the same number when divided by 11 leaves a remainder m and the resulting quotient when divided by 7 leaves a remainder n.
What are the values of m and n, respectively
Correct
Ans: a
This is an example of successive division. Let the number be N.
The number and successive quotients, the successive divisors and the corresponding remainders are tabulated below.
Quotients N q1 q2 Divisors 7 11 Remainders 3 6 One value of N is 6 x7 +3=45
∴ N = 77k + 45 = 11(7k+4)+1
Þ m = 1
And q1 = 7k +4,q2 = k and n = 4
∴(m,n) = (1,4)
Incorrect
Ans: a
This is an example of successive division. Let the number be N.
The number and successive quotients, the successive divisors and the corresponding remainders are tabulated below.
Quotients N q1 q2 Divisors 7 11 Remainders 3 6 One value of N is 6 x7 +3=45
∴ N = 77k + 45 = 11(7k+4)+1
Þ m = 1
And q1 = 7k +4,q2 = k and n = 4
∴(m,n) = (1,4)
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Question 4 of 5
4. Question
The seven digit number 876p37q is divisible by 225. The values of p and q can be respectively
Correct
Ans: d
Seven digits number 876p37q is divisible by 225, then this number is also divisible by 9 and 25.
If this number is divisible by 9.
∴Sum of its digits is divisible by 9
Now, sum of digits= 8+ 7 + 6+ P+ 3+ 7+ q = 31+ p+q
p+q= 5 or p + q=14,q must be 5, if q= 5, p = 0 and 9
Incorrect
Ans: d
Seven digits number 876p37q is divisible by 225, then this number is also divisible by 9 and 25.
If this number is divisible by 9.
∴Sum of its digits is divisible by 9
Now, sum of digits= 8+ 7 + 6+ P+ 3+ 7+ q = 31+ p+q
p+q= 5 or p + q=14,q must be 5, if q= 5, p = 0 and 9
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Question 5 of 5
5. Question
Let x and y be positive integers such that x˃y. The expression 3x + 2y and 2x +3y, when divided by 5 leave remainders 2 and 3, respectively. What is the remainder when (x-y), is divided by 5?
Correct
Ans: a
We have, 3x+2y is divided by 5 remainder is 2.
∴3x+2y=5q+2 ……………………..(1)
And 2x +3y is divided by 5 remainder is 3.
∴ 2x + 3y = 5m+3 ……………………..(2)
Subtract Eq. (2 ) from Eq.(1) , we get
x-y = 5(q-m)-1
x-y=5(q-m)-5+4
x-y=5(q-m-1)+4
∴x-y is divided by 5 remainder is 4.
Incorrect
Ans: a
We have, 3x+2y is divided by 5 remainder is 2.
∴3x+2y=5q+2 ……………………..(1)
And 2x +3y is divided by 5 remainder is 3.
∴ 2x + 3y = 5m+3 ……………………..(2)
Subtract Eq. (2 ) from Eq.(1) , we get
x-y = 5(q-m)-1
x-y=5(q-m)-5+4
x-y=5(q-m-1)+4
∴x-y is divided by 5 remainder is 4.
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