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Algebra Level 3 Test 5
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Question 1 |
The number of solutions of the equation 2p + q = 40 where both p and q are positive integers and p≤ q is:
7 | |
13 | |
14 | |
18 |
Question 1 Explanation:
We are given by the equations
2p + q = 40
p ≤ q
⇒ q = 40 – 2p
Values of p and q that satisfy the equation
Let’s solve the problem using the hit and trial method
The first minimum value that satisfies the equation will be
If p= 1 then q= 38
If p = 2 then q = 36
If p= 3 then q = 34
And so on up to p=13 then q = 14
Therefore there are 13 values of (p, q) which satisfy the equation such that x ≤ y
2p + q = 40
p ≤ q
⇒ q = 40 – 2p
Values of p and q that satisfy the equation
Let’s solve the problem using the hit and trial method
The first minimum value that satisfies the equation will be
If p= 1 then q= 38
If p = 2 then q = 36
If p= 3 then q = 34
And so on up to p=13 then q = 14
Therefore there are 13 values of (p, q) which satisfy the equation such that x ≤ y
Question 2 |
If a1 = 1and an+1 – 3an + 2 = 4n for every positive integer n, then a100 equals?
399 – 200 | |
399 + 200. | |
3100 – 200 | |
3100 + 200 |
Question 2 Explanation:
a1 = 1, an+1 – 3an + 2 = 4n
an+1 = 3an + 4n – 2
when n = 2 then a2 = 3 + 4 – 2 = 5
when n = 3 then a3 = 3 × 5 + 4 × 2 – 2 = 21
From the options, we get an idea that an can be expressed in a combination of some power of
3 & some multiple of 100.
(a) 399 – 200; tells us that an could be:
3n–1 – 2 × n; but it does not fit a1 or a2 or a3
(b) 399 + 200; tells us that an could be: 3n–1 + 2 × n; again, not valid for a1, a2 etc.
(c) 3100 – 200; tells 3n – 2n: valid for all a1, a2, a3.
(d) 3100 + 200; tells 3n + 2n: again not valid. so, (C) is the correct answer
an+1 = 3an + 4n – 2
when n = 2 then a2 = 3 + 4 – 2 = 5
when n = 3 then a3 = 3 × 5 + 4 × 2 – 2 = 21
From the options, we get an idea that an can be expressed in a combination of some power of
3 & some multiple of 100.
(a) 399 – 200; tells us that an could be:
3n–1 – 2 × n; but it does not fit a1 or a2 or a3
(b) 399 + 200; tells us that an could be: 3n–1 + 2 × n; again, not valid for a1, a2 etc.
(c) 3100 – 200; tells 3n – 2n: valid for all a1, a2, a3.
(d) 3100 + 200; tells 3n + 2n: again not valid. so, (C) is the correct answer
Question 3 |
Let g(x) be a function such that g(x + 1) + g(x – 1) = g(x) for every real x. Then for what value of p is the relation g(x+p) = g(x) necessarily true for every real x?
5 | |
3 | |
2 | |
6 |
Question 3 Explanation:
g(x + 1) + g(x –1) = g(x)
g(x+ 2) + g(x) = g(x + 1)
Adding these two equations we get
g(x + 2) + g(x – 1) = 0
⇒ g(x + 3) + g(x) = 0
⇒ g(x + 4) + g(x + 1) = 0
⇒ g(x + 5) + g(x + 2) = 0
⇒ g(x + 6) + g(x + 3) = 0 ⇒ g(x + 6) – g(x) = 0
g(x+ 2) + g(x) = g(x + 1)
Adding these two equations we get
g(x + 2) + g(x – 1) = 0
⇒ g(x + 3) + g(x) = 0
⇒ g(x + 4) + g(x + 1) = 0
⇒ g(x + 5) + g(x + 2) = 0
⇒ g(x + 6) + g(x + 3) = 0 ⇒ g(x + 6) – g(x) = 0
Question 4 |
A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?
15 | |
14 | |
12 | |
10 |
Question 4 Explanation:
There are two equations to be formed 40 m + 50 f = 1000
250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
850 m + 8000 f = A
m and f are the number of males and females A is amount paid by the employer.
Then the possible values of f = 8, 9, 10, 11, 12
If f = 8
M = 15
If f = 9, 10, 11 then m will not be an integer while f = 12 then m will be 10.
By putting f = 8 and m = 15, A = 18800. When f = 12 and
m = 10 then A = 18100
Therefore the number of males will be 10.
250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
850 m + 8000 f = A
m and f are the number of males and females A is amount paid by the employer.
Then the possible values of f = 8, 9, 10, 11, 12
If f = 8
M = 15
If f = 9, 10, 11 then m will not be an integer while f = 12 then m will be 10.
By putting f = 8 and m = 15, A = 18800. When f = 12 and
m = 10 then A = 18100
Therefore the number of males will be 10.
Question 5 |
Consider the set S = {1, 2, 3, . . ., 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?
3 | |
4 | |
6 | |
7 |
Question 5 Explanation:
Let number of elements in progression be n, then
1000 = 1+ (n −1)d
⇒ (n −1)d = 999 = 33 × 37
Possible values of d = 3, 37, 9, 111, 27, 333, 999 Hence 7 progressions.
1000 = 1+ (n −1)d
⇒ (n −1)d = 999 = 33 × 37
Possible values of d = 3, 37, 9, 111, 27, 333, 999 Hence 7 progressions.
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