## Practice 3 Algebra Problems To Prepare For IBM Placement Tests

Dear Reader,

Below are four problems based on numbers with basic algebraic calculations.

**Question 1**

If pq=28 where p and q are whole numbers then which of the following will be (p^{2} - q^{2})?

a) 32 b) 83 c) 783 d) cannot be determined.

**Answer : **c) 783.

Solution :

Given that, pq=28 and p, q are integers.

The factors of 28 are 1, 2, 4, 7, 14 and 28.

Then the possibilities of pq = 28 are (p,q) = (1, 28), (2, 14), (4, 7), (7,4), (14, 2) and (28,1)

We would have p > q since the given options are positive.

(That is, if p < q then (p^{2} - q^{2}) will be negative).

Therefore, the possibilities of (p, q) reduced to (7,4), (14, 2) and (28,1).

If (p,q) = (28,1) then (p^{2} - q^{2}) = 28^{2} - 1^{2} = 784 - 1 = 783.

If (p,q) = (14, 2) then (p^{2} - q^{2}) = 14^{2} - 2^{2} = 196 - 4 = 192

If (p,q) = (7,4) = (p^{2} - q^{2}) = 7^{2} - 4^{2} = 49 - 16 = 33

From the given options, required answer is option c.

**Question 2**

If pq= 289 and p, q are integers then find the integer value of p/q.

a) 289 b) 1 c) a&b d) none of these

**Answer : **c) a&b

Solution :

Given that, pq= 289 and p, q are integers.

The factors of 289 are 1, 17 and 289.

Then the possibilities of pq = 289 are (p,q) = (1,289), (17,17) and (289,1).

We have to find the integer value of p/q.

Now, we would have p/q is either 289/1 or 17/17 since p/q is an integer.

(That is, if (p,q) = (1,289) then p/q = 1/289 which is not an integer).

Therefore, we have p/q = 289 or p/q = 17

Hence the answer is option c.

**Question 3**

If pq = 143 where p,q are integers then (p^{2} - q^{2})^{2} = ?

a) 20736 b) 43 c) cannot be determined d) none of these

**Answer : **d) none of these

Solution :

Given that, pq= 143 and p, q are integers.

The factors of 143 are 1, 11, 13 and 143.

Then the possibilities of pq = 143 are (p,q) = (1,143), (11,13), (13,11) and (143,1).

We have to find the value of (p^{2} - q^{2})^{2}.

Now, (1^{2} - 143^{2})^{2} = (143^{2} - 1^{2})^{2} and (11^{2} - 13^{2})^{2} = (13^{2} - 11^{2})^{2} since square of any negative number is always a positive.

Remember that, “if last digit of a number is 3 then its square will end with 9 and if last digit of a number is 8 then its square must end with 4"

Now, If (p^{2} - q^{2})^{2} = (13^{2} - 11^{2})^{2} = (169-121)^{2} = 48^{2} = a number end with 4

If (p^{2} - q^{2})^{2} = (143^{2} - 1^{2})^{2} = ( a number end with 9 - 1)^{2}

= (a number end with 8)^{2} = a number end with 4.

Therefore, required answer must end with 4.

But none of the options end with 4.

Hence the answer is option d.

**Question 4**

If pq = 361, p, q are integers then the value of p+(q-1)^{2} is:

a) 343 b) 111 c) 109 d) none of these

**Answer : **a) 325

Solution :

Given that, pq= 361 and p, q are integers.

The factors of 361 are 1, 19 and 361.

Then the possibilities of pq = 361 are (p,q) = (1, 361), (19,19) and (361,1).

We have to find the value of p+(q-1)^{2}

If (p,q) = (1,361) then p+(q-1)^{2} = 1 + (361-1)^{2} = 1+360^{2} = 129601

If (p,q) = (361,1) then p+(q-1)^{2} = 361 + (1-1)^{2} = 361

If (p,q) = (19,19) then p+(q-1)^{2} = 19 + (19-1)^{2} = 19+18^{2} = 343

From the given options, required answer is option a.

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