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## Algebra Level 3 Test 4

Congratulations - you have completed Algebra Level 3 Test 4.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
 Question 1
The total number of integers pairs (x, y) satisfying the equation x + y = xy is
 A 0 B 1 C 2 D 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
 Question 2
If | b |â‰¥ 1 and x = âˆ’ | a | b , then which one of the following is necessarily true?
 A a â€“ xb < 0 B a â€“ xb â‰¥ 0 C a â€“ xb > 0 D a â€“ xb â‰¤ 0
Question 2 Explanation:
x = â€“|a| b
Now a â€“ xb = a â€“ (â€“ |a| b) b
= a + |a|b2
Therefore Â a â€“ xb = a + ab2 â€¦a â‰¥ 0
OR
a â€“ xb
= a â€“ ab2 â€¦a < 0
= a(1 + b2) = a(1 â€“ b2)
Consider first case:
As a â‰¥ 0 and |b| â‰¥ 1, therefore (1 + b2) is positive.
Therefore a (1 + b2) â‰¥ 0
Therefore a â€“ xb â‰¥ 0
Consider second case.
As a < 0 and |b| â‰¥ 1, therefore (1 â€“ b2) â‰¤ 0
Therefore a (1 â€“ b2) â‰¥ 0 (Since â€“ve Ã— -ve = +ve and 1 â€“ b2 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 = 5y2, then
 A x is necessarily less than y B x is necessarily greater than y C x is necessarily equal to y D None of the above is necessarily true
Question 3 Explanation:
13x + 1 < 2z and z + 3 = 5y2
13x + 1 < 2 (5y2 âˆ’ 3)
13x + 1< 10y2 âˆ’ 6
13x + 7 < 10y2 put x = 1
20 < 10y2 and y2 > 2
Therefore y2 > 2 = Â (y2 âˆ’ 2) > 0
so option d is the right answer
 Question 4
If n is such that 36 â‰¤ n â‰¤ 72 , Then a = {(n2 + 2âˆšn(n+ 4) +16)} / (n+ 4âˆšn +4) satisfies
 A 20 < x < 54 B 23 < x < 58 C 25 < x < 64 D 28 < x < 60
Question 4 Explanation:
36 â‰¤ n â‰¤ 72
We are given by
a = {(n2 + 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€¦)?
 A 80 B 81 C 82 D 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 5th 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.
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
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