### Wipro Sample Algebraic Problems

Below are three model problems dealing with the formulae (x+y)

Find X when X - Y = 3 and X

a)10 b)-5 c)-10 d)-3

Solution :

We know that (x - y)

Sub. the given values,

3

9 - 89 = -2XY

80 = 2XY

XY = 40

Since X and Y are integers and XY = 40,the possibilities of X and Y are as follows:

(1,40), (2,20), (4,10), (5,8), (-1,-40), (-2,-20), (-4,-10) and (-5,-8)

X - Y is a positive integer(3), so X will be greater than Y.

By checking the given condition X - Y = 3, we have X = 8, Y = 5 or X = -5, Y = -8.

i.e., X is either 8 or -5.

From the given options we can conclude that X = -5.

If the summation and multiplication of two integer is 24 and 143 respectively then the difference of them is:

a)2 b)1 c)12 d)4

Solution :

Let A and B be two integers.

Then A + B = 24 and AB = 143.

We know that (x+y)

Here, 24

576 - 286 = A

A

i.e., Summation of the square of two integers is 290.

i.e., A

Since A and B are integers, the possibilities are {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289}.

From the above set of square numbers, the values may be

A

A

The condition A + B = 24 and AB = 143, is satisfied by A = 11, B = 13 or A = 13, B = 11

We have to find the difference of A and B.

Hence the difference is 2.

If X - Y = 9, X

a)21,12 b)15,6 c)none of these d)cannot be determined

Solution :

Here we use the formula, (x-y)

9

XY = (257-81)/2 = 176/2 = 88

XY = 88

Since X and Y are integers and XY = 88 then the possibilities of X and Y are (1,88), (2,44), (4,22), (8,11), (-1,-88), (-2,-44), (-4,-22) and (-8,-11).

By observing, none of the above possibility of X and Y satisfies X - Y = 9.

Hence, the values of X and Y cannot be determined by given conditions.

^{2}= x^{2}+ y^{2}+ 2xy and (x-y)^{2}= x^{2}+ y^{2}- 2xy**Question 1**Find X when X - Y = 3 and X

^{2}+ Y^{2}= 89 where X and Y are integers.a)10 b)-5 c)-10 d)-3

**Answer :**b)-5.Solution :

We know that (x - y)

^{2}= x^{2}+ y^{2}- 2xySub. the given values,

3

^{2}= 89 - 2XY9 - 89 = -2XY

80 = 2XY

XY = 40

Since X and Y are integers and XY = 40,the possibilities of X and Y are as follows:

(1,40), (2,20), (4,10), (5,8), (-1,-40), (-2,-20), (-4,-10) and (-5,-8)

X - Y is a positive integer(3), so X will be greater than Y.

By checking the given condition X - Y = 3, we have X = 8, Y = 5 or X = -5, Y = -8.

i.e., X is either 8 or -5.

From the given options we can conclude that X = -5.

**Question 2**If the summation and multiplication of two integer is 24 and 143 respectively then the difference of them is:

a)2 b)1 c)12 d)4

**Answer :**a)2Solution :

Let A and B be two integers.

Then A + B = 24 and AB = 143.

We know that (x+y)

^{2}= x^{2}+ y^{2}+ 2xyHere, 24

^{2}= A^{2}+ B^{2}+ 2(143)576 - 286 = A

^{2}+ B^{2}A

^{2}+ B^{2}= 290.i.e., Summation of the square of two integers is 290.

i.e., A

^{2}or B^{2}is < or = 290.Since A and B are integers, the possibilities are {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289}.

From the above set of square numbers, the values may be

A

^{2}+ B^{2}= 1 + 289 or 121 + 169 = 290A

^{2}+ B^{2}= 1^{2}+ 17^{2}or 11^{2}+ 13^{2}= 290The condition A + B = 24 and AB = 143, is satisfied by A = 11, B = 13 or A = 13, B = 11

We have to find the difference of A and B.

Hence the difference is 2.

**Question 3**If X - Y = 9, X

^{2}+ Y^{2}= 257 and X,Y are integers then what will be the value of X and Y ?a)21,12 b)15,6 c)none of these d)cannot be determined

**Answer :**d)cannot be determinedSolution :

Here we use the formula, (x-y)

^{2}= x^{2}+ y^{2}- 2xy9

^{2}= 257 - 2XYXY = (257-81)/2 = 176/2 = 88

XY = 88

Since X and Y are integers and XY = 88 then the possibilities of X and Y are (1,88), (2,44), (4,22), (8,11), (-1,-88), (-2,-44), (-4,-22) and (-8,-11).

By observing, none of the above possibility of X and Y satisfies X - Y = 9.

Hence, the values of X and Y cannot be determined by given conditions.

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