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## Algebra Level 3 Test 5

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*Algebra Level 3 Test 5*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

The number of solutions of the equation 2p + q = 40 where both p and q are positive integers and p≤ q is:

A | 7 |

B | 13 |

C | 14 |

D | 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 a

_{1}= 1and a_{n+1 }– 3a_{n}+ 2 = 4n for every positive integer n, then a_{100}equals?A | 3 ^{99} – 200 |

B | 3 ^{99} + 200. |

C | 3 ^{100} – 200 |

D | 3 ^{100 }+ 200 |

Question 2 Explanation:

a

a

when n = 2 then a

when n = 3 then a

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) 3

3

(b) 3

(c) 3

(d) 3

_{1}= 1, a_{n+1 }– 3a_{n}+ 2 = 4na

_{n+1 }= 3a_{n }+ 4n – 2when n = 2 then a

_{2}= 3 + 4 – 2 = 5when n = 3 then a

_{3}= 3 × 5 + 4 × 2 – 2 = 21From the options, we get an idea that an can be expressed in a combination of some power of

3 & some multiple of 100.

(a) 3

^{99}– 200; tells us that a_{n}could be:3

^{n–1 }– 2 × n; but it does not fit a_{1}or a_{2}or a_{3}(b) 3

^{99}+ 200; tells us that a_{n}could be: 3^{n–1 }+ 2 × n; again, not valid for a_{1}, a_{2}etc.(c) 3

^{100}– 200; tells 3^{n}– 2n: valid for all a_{1}, a_{2}, a_{3.}(d) 3

^{100}+ 200; tells 3^{n}+ 2n: again not valid. so, (C) is the correct answerQuestion 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?

A | 5 |

B | 3 |

C | 2 |

D | 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?

A | 15 |

B | 14 |

C | 12 |

D | 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?

A | 3 |

B | 4 |

C | 6 |

D | 7 |

Question 5 Explanation:

Let number of elements in progression be n, then

1000 = 1+ (n −1)d

⇒ (n −1)d = 999 = 3

Possible values of d = 3, 37, 9, 111, 27, 333, 999 Hence 7 progressions.

1000 = 1+ (n −1)d

⇒ (n −1)d = 999 = 3

^{3}× 37Possible values of d = 3, 37, 9, 111, 27, 333, 999 Hence 7 progressions.

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