Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 11

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Question 1
1 / -0

The solution of 217x + 131y = 913 and 131x + 217y = 827 is

A
x = 2 and y =2

B
x = 3 and y = 3

C
x = 3 and y = 2

D
x = 2 and y = 3

Solution

Fristly add up both eqn.

217x + 131y = 913,

131x + 217y = 827,

348x + 348y = 1740

Dividing both side by 348

We get x + y = 5-------(1)

Similarly Subtract given eqn 217x + 131y = 913 - (131x + 217y = 827 )

86x - 86y = 86

Dividing both side by 86

We get x - y =1------(2)equation
Now, solve equation 1&2

x + y = 5

x - y = 1
2x = 6 x = 3

Put x = 3 in equation 1 x + y = 5 3 + y = 5 y = 5 - 3

y = 2

Hence, x = 3 y = 2
Question 2
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If (x + 1) is a factor of 2x³ + ax² + 2bx + 1, then the values of ‘a’ and ‘b’, given that 2a – 3b = 4 are

A
a = – 5 and b = – 2

B
a = – 5 and b = 2

C
a = 5 and b = 2

D
a = 5 and b = – 2

Solution

2x³ + ax² + 2bx + 1
= 2x³ + 2x² + (a - 2)x² + (a - 2)x + (2b - a + 2)x + (2b - a + 2) - (2b - a + 2) + 1
= 2x²(x + 1) + (a - 2)x(x + 1) + (2b - a + 2)(x + 1) + a - 2b - 1
= (x + 1)[2x² + (a - 2)x + (2b - a + 2)] + (a - 2b - 1)
Since x+1 is a factor of 2x³+ax²+2bx+1, we conclude that:
a - 2b - 1 = 0
a = 2b + 1
b = (a - 1)/2
So any couple of reals of the form (a; (a-1)/2 ) will yield a polynomial that will have x+1 as a factor.
For instance:
If a=1, then b=0 and:
2x³ + ax² + 2bx + 1
= 2x³ + x² + 1
= 2x³ + 2x² - x² - x + x + 1
= 2x²(x + 1) - x(x + 1) + (x + 1)
= (x + 1)(2x² - x + 1)
and x+1 is indeed a factor.
If a=3, then b=1 and:
2x³ + ax² + 2bx + 1
= 2x³ + 3x² + 2x + 1
= 2x³ + 2x² + x² + x + x + 1
= 2x²(x + 1) + x(x + 1) + (x + 1)
= (x + 1)(2x² + x + 1)
and x+1 is indeed a factor.
So there is an infinity of (a; b) that would make x+1 a factor of 2x³+ax²+2bx+1
and they are all of the form (a; (a-1)/2 ) with 'a' being any real value.
Question 3
1 / -0

The angles of a triangle are x^∘, y^∘ and 40^∘ . The difference between the two angles ‘x’ and ‘y’ is 30^∘, then

A
x^∘ = 75^∘ and y^∘= 45^∘

B
None of these

C
x^∘ = 65^∘ and y^∘= 95^∘

D
x^∘ = 85^∘and y^∘ = 55^∘

Solution

According to the question, x^∘+ y^∘ + 40^∘ = 180^∘ ⇒ x^∘+ y^∘ + 140^∘ ..................(i) and x^∘ − y^∘ = 30^∘...................(ii)

On solving eq. (i) and eq. (ii),
x + y + x - y = 140 + 30

2x = 170

x = 85°

Putting the value of x in equation 1, we get

85° + y = 140°

y = 140° - 85°

y = 55°

we get x^∘ = 85^∘, y^∘ = 55^∘
Question 4
1 / -0

5 pencils and 7 pens together cost Rs.50 whereas 7 pencils and 5 pens together cost Rs.46. The cost of 1 pen is

A
Rs.5

B
Rs.4

C
Rs.3

D
Rs.6

Solution

Let, cost(in RS) of one pencil = x
and cost (in RS) of one pen = y
Therefore , according to question
5x + 7y = 50 ........ (1)
7x + 5y = 46 .........(2)
Multiply equation (1) by 7 and equation (2) by 5 we get
7(5x + 7y) = 7 × 50
35x + 49y = 350 .......(3)
and
5(7x + 5y) = 5 × 46
35x + 25y = 230 ....... (4)
Subtract equation (4) from equation 3 , we get
35x + 49y - 35x - 25y = 350 -230
49y - 25y = 120
24y = 120
y = 120/24
y = 5
Substitute y = 5 in equation 1 , we get
5x + 7 × 5 =50
5x + 35 = 50
5x = 50 - 35
5x = 15
x = 15/5
x = 3
Hence, Cost of One Pen = y =5