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Algebra Level 3 Test 6
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Question 1 |
For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of 7th and 6th terms of this sequence is 517, what is the 10th term of this sequence?
147 | |
76 | |
123 | |
Cannot be determined |
Question 1 Explanation:
Let the suppose 6th and the 7th terms be p and q
. Then 8th term = p + q
Also q2 – p2 = 517
⇒ (q + p)(q – p) = 517 = 47 × 11
So q + p = 47
q – p = 11
Taking q = 29 and p = 18, we have 8th term = 47,
9th term = 47 + 29 = 76 and 10th term = 76 + 47 = 123.
. Then 8th term = p + q
Also q2 – p2 = 517
⇒ (q + p)(q – p) = 517 = 47 × 11
So q + p = 47
q – p = 11
Taking q = 29 and p = 18, we have 8th term = 47,
9th term = 47 + 29 = 76 and 10th term = 76 + 47 = 123.
Question 2 |
If p, q and r are real numbers such that p + q + r = 5 and pq + qr + rp = 3, what is the largest value that p can have?
5/3 | |
√19 | |
13/3 | |
none |
Question 2 Explanation:
pq + qr + rp = 3
= pq + (q + p)r = 3
= pq + (q + r)(5 − p − q) = 3
= p2 + q2 + pq – 5p – 5q + 3 = 0
= q2 + (p − 5) q + p2 – 5p + 3 = 0
As it is given that q is a real number,
the D for above equation must be greater than or equal to zero.
Hence, (P − 5)2 − 4(p2 – 5p + 3) ≥ 0
⇒3p2 −10p −13 ≤ 0
⇒3p2 −13p + 3p −13 ≤ 0
Therefore p can have two values -1, 13/3
And the largest would be = 13/3
= pq + (q + p)r = 3
= pq + (q + r)(5 − p − q) = 3
= p2 + q2 + pq – 5p – 5q + 3 = 0
= q2 + (p − 5) q + p2 – 5p + 3 = 0
As it is given that q is a real number,
the D for above equation must be greater than or equal to zero.
Hence, (P − 5)2 − 4(p2 – 5p + 3) ≥ 0
⇒3p2 −10p −13 ≤ 0
⇒3p2 −13p + 3p −13 ≤ 0
Therefore p can have two values -1, 13/3
And the largest would be = 13/3
Question 3 |
Let x and y be two positive numbers such that p + q = 1, Then the minimum value of [(p + 1/p)2] + [(q + 1/q)2]
12 | |
20 | |
12.5 | |
13.3 |
Question 3 Explanation:
p + q = 1 and q > 0 p > 0
Taking p = y =1/2
Therefore value of [(p + 1/p)2] + [(q + 1/q)
2] = [(2 + 1/2)2] + [(2 + 1/2)
2] = 25/2
Taking p = y =1/2
Therefore value of [(p + 1/p)2] + [(q + 1/q)
2] = [(2 + 1/2)2] + [(2 + 1/2)
2] = 25/2
Question 4 |
Let q be a positive integer and p = q2 – q. If q < equal to 4 , then p2 – 2p is divisible by
15 | |
20 | |
24 | |
All of these |
Question 4 Explanation:
p = q2 – q, q ≥ 4
p2 – 2p = (q2 – q)2 – 2(q2 – q)
= (q2 – q)(q2 – q – 2)
Using different values to q ≥ 4 and we find that it is divisible by 15, 20, 24.
Hence. the right option for this question is all of these
p2 – 2p = (q2 – q)2 – 2(q2 – q)
= (q2 – q)(q2 – q – 2)
Using different values to q ≥ 4 and we find that it is divisible by 15, 20, 24.
Hence. the right option for this question is all of these
Question 5 |
Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?
(6, 1) | |
(–3, –4) | |
(4, 3) | |
(–4, –3) |
Question 5 Explanation:
Using the roots obtained by the first boy, the equation
formed will be:
x2 – 7q+ 12 = 0 ………………………………... (i)
In this equation, the constant term is incorrect..
Using the roots obtained by the second boy, the equation formed will be:
x2 – 5q + 6 = 0 ... (ii)
in this equation, the co-efficient of x is incorrect.
Combining the two, we have the correct equation:
x2 – 7x + 6 = 0
x2 – 6x – x + 6 = 0
x(x – 6) – 1(x – 6) = 0
(x – 6)(x – 1) = 0
Therefore x = 6, 1
Hence, the actual roots = (6, 1).
formed will be:
x2 – 7q+ 12 = 0 ………………………………... (i)
In this equation, the constant term is incorrect..
Using the roots obtained by the second boy, the equation formed will be:
x2 – 5q + 6 = 0 ... (ii)
in this equation, the co-efficient of x is incorrect.
Combining the two, we have the correct equation:
x2 – 7x + 6 = 0
x2 – 6x – x + 6 = 0
x(x – 6) – 1(x – 6) = 0
(x – 6)(x – 1) = 0
Therefore x = 6, 1
Hence, the actual roots = (6, 1).
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