Algebra For SBI PO : Set – 20

D.1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship between x and y cannot be established

1. I. 11x + 5y = 117

II. 7x + 13y = 153

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship between x and y cannot be established

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(c)

eqn (I) × 7

eqn (II) × 11

77x + 35y = 819

– 77x ± 143y = 1683

– 108y = – 864

y = 8, x = 7 ie x < y

2. I. 6x2 + 51x + 105 = 0

II. 2y2 + 25y + 78 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship between x and y cannot be established

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(a) I. 6x2 + 21x + 30x + 105 = 0

or, 3x(2x + 7) + 15(2x + 7) = 0

or, (3x + 15) (2x + 7) = 0

x = -5,-7/2

II. 2y2 + 12y + 13y + 78 = 0

or, 2y(y + 6) + 13(y + 6) = 0

or, (2y + 13) (y + 6) = 0

y=-13/2,-6

3. I. 6x + 7y = 52

II. 14x + 4y = 35

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship between x and y cannot be established

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(c) eqn (I) × 4

eqn (II) × 7

24x + 28y = 208

– 98x ± 28y = 245

– 74x = – 37

x=1/2,7

4. I. x2 + 11x + 30 = 0

II. y2 + 12y + 36 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship between x and y cannot be established

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(b) I. x2 + 5x + 6x + 30 = 0

or, x(x + 5) + 6(x + 5) = 0

or, (x + 5) (x + 6) = 0

x = – 5, – 6

II. y2 + 12y + 36 = 0

or, (y + 6)2 = 0

or, y + 6 = 0

y = – 6

5. I. 2x2 + x – 1=0

II. 2y2 – 3y + l = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship between x and y cannot be established

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(d) I. 2x2+ 2x – x – 1 = 0

or, 2x(x + 1) – 1(x + 1) = 0

or, (2x – 1) (x + 1) = 0

x=1/2,-1

II. 2y2 – 2y – y + 1 = 0

or, 2y(y – 1) – 1(y – 1) = 0

or, (2y – 1)(y – 1) = 0

y=1/2,1

D.6-10) In the following questions three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately, or by any other method and give answer If

(a) x < y = z

(b) x < y < z

(c)x < y > z

(d) x = y > z

(e) x = y = z or if none of the above relationship is established

6. I. 7x + 6y + 4z = 122

II. 4x + 5y + 3z = 88

III. 9x + 2y + z = 78

(a) x < y = z

(b) x < y < z

(c)x < y > z

(d) x = y > z

(e) x = y = z or if none of the above relationship is established

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(a)

7x + 6y + 4z = 122 … (i)

4x + 5y + 3z = 88 … (ii)

9x + 2y + z = 78 … (iii)

From (i) and (ii)

5x – 2y = 14… (iv)

From (ii) and (iii)

23x + y = 146 … (v)

From (iv) and (v),

x = 6, y = 8

Putting the value of x and y in eqn (i), we get

z = 8

:. x < y = z

7. I. 7x + 6y =110

II. 4x + 3y = 59

III. x + z = 15

(a) x < y = z

(b) x < y < z

(c)x < y > z

(d) x = y > z

(e) x = y = z or if none of the above relationship is established

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7. (c)

7x + 6y = 110 … (i)

4x + 3y = 59 … (ii)

x + z = 15 … (iii)

From eqn (i) and (ii), x = 8, y = 9

Put the value of x in eqn (iii).

Then, z = 7

x < y > z

8. I. x =

II. 2y + 3z = 33

III. 6y + 5z = 71

(a) x < y = z

(b) x < y < z

(c)x < y > z

(d) x = y > z

(e) x = y = z or if none of the above relationship is established

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(b)

2y + 3z = 33 … (ii)

6y + 5z = 71 … (iii)

From eqn (ii) and (iii),

y = 6 and z = 7

x = y , z

9. I. 8x + 7y= 135

II. 5x + 6y = 99

III. 9y + 8z = 121

(a) x < y = z

(b) x < y < z

(c)x < y > z

(d) x = y > z

(e) x = y = z or if none of the above relationship is established

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(d)

8x + 7y = 135 … (i)

5x + 6y = 99 … (ii)

9y + 8z = 121 … (iii)

From eqn (i) and (ii),

x = 9, and y = 9

Putting the value of y in eqn (iii),

z = 5

:. x = y > z

10. I.(x + y)3= 1331

II. x – y + z = 0

III. Xy = 28

(a) x < y = z

(b) x < y < z

(c)x < y > z

(d) x = y > z

(e) x = y = z or if none of the above relationship is established

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(e)

(x + y)3 = 1331

or, x + y = 11 … (i)

(x + y)2 = 121

(x – y)2 + 4xy = 121

x – y = 3… (ii)

[value of xy from eqn (iii)]

From eqn (i) and (ii), x = 7, y = 4

Put the value x and y in the eqn

x – y + z = 0

7 – y + z = 0

3 + z = 0

z = -3