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Algebra Level 3 Test 4
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Question 1 
The total number of integers pairs (x, y) satisfying the equation x + y = xy is
0  
1  
2  
None of the above. 
Question 1 Explanation:
Given equation is x + y = xy
â‡’ xy â€“ x â€“ y + 1 = 1
â‡’ (x â€“ 1)(y â€“ 1) = 1
x â€“ 1= 1
andÂ y âˆ’1= 1or
x âˆ’1= â€“1
y â€“ 1= â€“1
Clearly (0, 0) and (2, 2) are the only pairs that will satisfy the equation
â‡’ xy â€“ x â€“ y + 1 = 1
â‡’ (x â€“ 1)(y â€“ 1) = 1
x â€“ 1= 1
andÂ y âˆ’1= 1or
x âˆ’1= â€“1
y â€“ 1= â€“1
Clearly (0, 0) and (2, 2) are the only pairs that will satisfy the equation
Question 2 
If  b â‰¥ 1 and x = âˆ’  a  b , then which one of the following is necessarily true?
a â€“ xb < 0  
a â€“ xb â‰¥ 0  
a â€“ xb > 0  
a â€“ xb â‰¤ 0 
Question 2 Explanation:
x = â€“a b
Now a â€“ xb = a â€“ (â€“ a b) b
= a + ab^{2}
Therefore Â a â€“ xb = a + ab^{2} â€¦a â‰¥ 0
OR
a â€“ xb
= a â€“ ab^{2} â€¦a < 0
= a(1 + b^{2}) = a(1 â€“ b^{2})
Consider first case:
As a â‰¥ 0 and b â‰¥ 1, therefore (1 + b^{2}) is positive.
Therefore a (1 + b^{2}) â‰¥ 0
Therefore a â€“ xb â‰¥ 0
Consider second case.
As a < 0 and b â‰¥ 1, therefore (1 â€“ b^{2}) â‰¤ 0
Therefore a (1 â€“ b^{2}) â‰¥ 0 (Since â€“ve Ã— ve = +ve and 1 â€“ b^{2} can be zero also), i.e. a â€“ xb â‰¥ 0
Therefore, in both cases a â€“ xb â‰¥ 0.
Now a â€“ xb = a â€“ (â€“ a b) b
= a + ab^{2}
Therefore Â a â€“ xb = a + ab^{2} â€¦a â‰¥ 0
OR
a â€“ xb
= a â€“ ab^{2} â€¦a < 0
= a(1 + b^{2}) = a(1 â€“ b^{2})
Consider first case:
As a â‰¥ 0 and b â‰¥ 1, therefore (1 + b^{2}) is positive.
Therefore a (1 + b^{2}) â‰¥ 0
Therefore a â€“ xb â‰¥ 0
Consider second case.
As a < 0 and b â‰¥ 1, therefore (1 â€“ b^{2}) â‰¤ 0
Therefore a (1 â€“ b^{2}) â‰¥ 0 (Since â€“ve Ã— ve = +ve and 1 â€“ b^{2} can be zero also), i.e. a â€“ xb â‰¥ 0
Therefore, in both cases a â€“ xb â‰¥ 0.
Question 3 
If 13x + 1 < 2z and z + 3 = 5y^{2}, then
x is necessarily less than y  
x is necessarily greater than y  
x is necessarily equal to y  
None of the above is necessarily true 
Question 3 Explanation:
13x + 1 < 2z and z + 3 = 5y^{2}
13x + 1 < 2 (5y^{2} âˆ’ 3)
13x + 1< 10y^{2} âˆ’ 6
13x + 7 < 10y^{2} put x = 1
20 < 10y^{2} and y^{2} > 2
Therefore y^{2} > 2 = Â (y^{2} âˆ’ 2) > 0
so option d is the right answer
13x + 1 < 2 (5y^{2} âˆ’ 3)
13x + 1< 10y^{2} âˆ’ 6
13x + 7 < 10y^{2} put x = 1
20 < 10y^{2} and y^{2} > 2
Therefore y^{2} > 2 = Â (y^{2} âˆ’ 2) > 0
so option d is the right answer
Question 4 
If n is such that 36 â‰¤ n â‰¤ 72 , Then a = {(n^{2} + 2âˆšn(n+ 4) +16)} / (n+ 4âˆšn +4) satisfies
20 < x < 54  
23 < x < 58  
25 < x < 64  
28 < x < 60 
Question 4 Explanation:
36 â‰¤ n â‰¤ 72
We are given by
a = {(n^{2} + 2âˆšn(n+ 4) +16)} / (n+ 4âˆšn +4)
Put a = 36.
And get the value 28 which is the least value
We are given by
a = {(n^{2} + 2âˆšn(n+ 4) +16)} / (n+ 4âˆšn +4)
Put a = 36.
And get the value 28 which is the least value
Question 5 
Consider the sets Tn = {n, n +1, n + 2, n + 3, n + 4} , where n = 1, 2, 3,â€¦, 96. How many of these sets contain 6 or any integral multiple thereof (i.e. any one of the numbers 6, 12, 18, â€¦)?
80  
81  
82  
83 
Question 5 Explanation:
From the question we can observe that 6 will appear in 5
sets T2, T3, T4, T5 and T6.
Similarly, 12 will also appear in 5 sets
But the multiple of 6 appears in the next 5^{th} set
that is in T12 Thus, each multiple of 6 will appear in 5 distinct sets.
Therefore we can say that there will be 16 multiplies of 6 till 96 .
Therefore 16 multiples of 6 will appear in 16 Ã— 5 = 80 sets.
sets T2, T3, T4, T5 and T6.
Similarly, 12 will also appear in 5 sets
But the multiple of 6 appears in the next 5^{th} set
that is in T12 Thus, each multiple of 6 will appear in 5 distinct sets.
Therefore we can say that there will be 16 multiplies of 6 till 96 .
Therefore 16 multiples of 6 will appear in 16 Ã— 5 = 80 sets.
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