Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 35

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Question 1
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Solve: $$\displaystyle \frac{20}{x\, +\, y}\, +\, \displaystyle \frac{3}{x\, -\,y}\, =\, 7$$ and $$\displaystyle \frac{8}{x\, -\, y}\, -\, \displaystyle \frac{15}{x\, +\, y}\, =\, 5$$

A
$$x\, =\, 3$$ and $$y\, =\, 2$$

B
$$x\, =\, 2$$ and $$y\, =\, 7$$

C
$$x\, =\, 9$$ and $$y\, =\, 4$$

D
$$x\, =\, 1$$ and $$y\, =\, 9$$

Solution

$$ \dfrac {20}{x+y} + \dfrac {3}{x-y} = 7 $$ ...(1)
$$ \dfrac {8}{x-y} - \dfrac {15}{x+y} = 5 $$ ...(2)

Multiplying eq(1) with $$3 $$ we get,
$$ \dfrac {60}{x+y} + \dfrac {9}{x-y} = 21 $$ ...(3)
Multiplying eq(2) with $$ 4 $$ we get,
$$\dfrac {32}{x-y} - \dfrac {60}{x+y} = 20 $$ ...(4)
Adding eq(3) and eq(4), we get
$$\dfrac {41}{x-y} = 41 => x - y = 41 $$ ...(5)
Substituting $$ x-y = 41 $$ in the eq(1), we get
$$ \dfrac {20}{x+y} + \dfrac {3}{1} = 7 => \dfrac {20}{x+y} = 4 => x +y = 5 $$ ... (6)
Adding eq(5) and eq(6), we get
$$ 2x = 6 => x = 3 $$
Substituting $$ x = 3 $$ in the eq(5), we get
$$ 3-y = 1 => y = 2 $$
Question 2
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Solve the following pair of equations:
$$\displaystyle \frac{x}{a}\, -\, \displaystyle \frac{y}{b}\, =\,0$$, $$ax\, +\, by\, =\, a^{2}\, +\, b^{2}$$

A
$$x\, =\, b-a$$ and $$y\, =\, b$$

B
$$x\, =\, 0$$ and $$y\, =\, ba$$

C
$$x\, =\, ba$$ and $$y\, =\, ba$$

D
$$x\, =\, a$$ and $$y\, =\, b$$

Solution

The equation $$\frac { x }{ a }-\frac { y }{ b }=0$$ can be rewritten as:

$$bx-ay=0...........(1)$$

The other equation given is:

$$ax+by=a^2+b^2..........(2)$$

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $$bx-ay=0$$ and write it as

$$x=\frac { ay }{ b }$$

Substitute the value of $$x$$ in equation $$ax+by=a^2+b^2$$. We get

$$a\left( \frac { ay }{ b } \right) +by=a^{ 2 }+b^{ 2 }\\ \Rightarrow \frac { a^{ 2 }y }{ b } +by=a^{ 2 }+b^{ 2 }\\ \Rightarrow \frac { a^{ 2 }y+b^{ 2 }y }{ b } =a^{ 2 }+b^{ 2 }\\ \Rightarrow \left( a^{ 2 }+b^{ 2 } \right) y=b\left( a^{ 2 }+b^{ 2 } \right) \\ \Rightarrow y=b$$

Therefore, $$y = b$$

Substituting this value of $$y$$ in the equation $$x=\frac { ay }{ b }$$, we get

$$x=\frac { ab }{ b } =a$$

Hence, the solution is $$x = a, y = b$$.
Question 3
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A and B both have some pencils. If A gives $$10$$ pencils to B, then B will have twice as many as A. And if B gives $$10$$ pencils to A, then they will have the same number of pencils. How many pencils does each has?

A
A = $$20$$ and B = $$55$$

B
A = $$20$$ and B = $$60$$

C
A = $$30$$ and B = $$ 50$$

D
A = $$50$$ and B = $$ 70$$

Solution

Let A has $$ x $$ pencils and B has $$ y $$.
As per the statement, "If A gives 10 pencils to B, then B will have twice as many as A":
$$ \implies 2(x - 10) = y + 10$$
$$ 2x - y = 30 $$ --- (1)
Also, as per the statement, "if B gives 10 pencils to A, then they will have the same number of pencils"
$$\implies x + 10 = y - 10 $$
$$x - y = -20 $$ --- (2)
Subtracting equation $$ (2) $$ from $$ (1) $$, we get:
$$2x-y-x+y=30+20$$
$$ x = 50$$
Substituting $$ x = 50 $$ in equation $$ (2) $$, we get:
$$ 50 -y = -20$$
$$\implies y = 70 $$
So, A has $$50$$ pencils and B has $$70$$ pencils.
Question 4
1 / -0

Divide 80 into two numbers, such that 5 times one number is equal to 3 times the other number.

A
$$30 \ and \ 50$$

B
$$35 \ and \ 52$$

C
$$40 \ and \ 60$$

D
$$22 \ and \ 49$$

Solution

Let the two numbers be $$x$$ and $$y$$
Given,
  $$x+y=80$$ (i) and
  $$5x=3y$$
$$\implies \displaystyle x=\frac{3y}{5}$$ (ii)
Putting the value of $$x$$ in equation (i) we get,
$$x+y=80$$
$$\implies \displaystyle \frac{3y}{5}+y=80$$
$$\implies \displaystyle \frac{3y+5y}{5}=80$$
$$\implies \displaystyle 8y=400$$
$$\implies \displaystyle y=50$$
Putting the value of $$y$$ in equation (i) we get,
  $$x+y=80$$
$$\implies \displaystyle x+50=80$$
$$\implies \displaystyle x=80-50$$
$$\implies \displaystyle x=30$$
$$\therefore x=30, y=50$$
Question 5
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$$A$$'s age is twice as $$B$$'s age. $$4$$ years ago, $$A$$ was three times as old as $$B$$. Find their present ages.

A
$$30$$ years and $$15$$ years

B
$$20$$ years and $$10$$ years

C
$$10$$ years and $$5$$ years

D
$$16$$ years and $$ 8$$ years

Solution

$$\\ Let\quad the\quad numbers\quad be\quad x\quad and\quad y.\\ x=2y;\\ x-4=3(y-4);\\ Substituting\quad the\quad value\quad in\quad the\quad 2nd\quad equation:\\ 2y-4=3(y-4);\\ Therefore:\\ y=8\\ x=16\\ $$
Question 6
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Divide $$32$$ into two parts such that if the larger is divided by the smaller, the quotient is $$2$$ and the remainder is $$5$$.

A
$$13$$ and $$5$$

B
$$23$$ and $$9$$

C
$$17$$ and $$39$$

D
$$28$$ and $$45$$

Solution

Let the larger part be $$a$$ and smaller part be $$b$$

Given,

$$a+b=32$$ ....(i)

and $$\dfrac{a}{b}=2+\dfrac{5}{b}$$

Multiplying both sides by $$ b $$, we get

$$\Rightarrow a=2b+5$$ .....(ii)

Substituting the value of $$a$$ in equation (i), we get

$$a+b=32$$

$$\Rightarrow 2b+5+b=32$$

$$\Rightarrow 3b=32-5$$

$$\Rightarrow b=\dfrac{27}{3}$$

$$\Rightarrow b=9$$

Putting the value of $$b$$ in equation (i), we get

$$a+b=32$$

$$\Rightarrow a+9=32$$

$$\Rightarrow a=32-9$$

$$\Rightarrow a=23$$
Question 7
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A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you". B replies, "if you give me 10. I will have thrice as many as left with you." How many mangoes does each have ?

A
A has 34 Mangoes and B has 62 Mangoes

B
A has 14 Mangoes and B has 35 Mangoes

C
A has 70 Mangoes and B has 148 Mangoes

D
A has 96 Mangoes and B has 182 Mangoes

Solution

$$\\ Let\quad the\quad numbers\quad be\quad x\quad and\quad y.\\ x+30=2(y-30);\\ 3(x-10)=(y+10);\\ x=2y-90;\\ Substituting\quad the\quad value\quad in\quad the\quad 1st\quad equation:\\ 3(2y-90-10)=(y+10);\\ Therefore:\\ y=62\\ x=34\\ $$
Question 8
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Divide 184 into two parts such that one- third of one part may exceed one-seventh of the other part by 4.

A
61.6 and 120.4

B
62.6 and 120.4

C
63.6 and 120.4

D
64.6 and 120.4

Solution

Let the two parts be $$x $$ and $$ y $$

=> $$ x + y = 184 $$ --- (1)

Given, one- third of one part may exceed one-seventh of the other part by $$ 4 $$
$$ => \frac {x}{3} - \frac {y}{7} = 4 $$
$$ => 7x -3y = 84 $$ --- (2)

Multiplying equation $$

(1) $$ with $$ 3 $$ we get, $$ 3x + 3y = 552 $$ ----- equation $$

(3) $$
Adding equations $$2 $$ and $$ 3 $$, we get $$ 10x = 636 => x = 63.6 $$
Substituting

$$ x = 63.6 $$ in the equation $$ (2) $$, we get $$ 63.6 + y = 184 =>y = 120.4

$$
Thus , the parts are $$ 63.6 ; 120.4 $$
Question 9
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Two articles $$A$$ and $$B$$ are sold for Rs. $$1,167$$ making $$5\%$$ profit on $$A$$ and $$7\%$$ profit on B. If the two articles are sold for Rs. $$1,165$$, a profit of $$7\%$$ is made on $$A$$ and a profit of $$5\%$$ is made on $$B$$. Find the cost price of each article.

A
$$A =$$ Rs. $$200$$ and $$B =$$ Rs. $$300$$

B
$$A =$$ Rs. $$600$$ and $$B =$$ Rs.$$ 700$$

C
$$A =$$ Rs.$$ 100$$ and $$B =$$ Rs.$$ 200$$

D
$$A =$$ Rs. $$500$$ and $$B =$$ Rs. $$600$$

Solution

Let the cost price of $$A =$$ Rs. $$ x $$ and that of $$B =$$ Rs. $$ y $$
Given,
They are sold for Rs. $$ 1,167 $$ making $$ 5\% $$ profit on $$A$$ and $$ 7\% $$ profit on $$B$$
$$\Rightarrow 1.05x + 1.07y = 1167 $$ ------ $$(1)$$

Also, when the two articles are sold for Rs. $$ 1,165 $$ a profit of $$ 7\% $$ is made on $$A$$ and a profit of $$ 5\% $$ is made on $$B$$.
$$\Rightarrow 1.07x + 1.05y = 1165 $$ ------ $$(2)$$

Multiplying equation $$ (1) $$ with $$ 1.07 $$ we get, $$ 1.1235x + 1.1449y = 1248.69 $$ ----- equation $$(3) $$
Multiplying equation $$ (2) $$ with $$ 1.05 $$ we get, $$ 1.1235x + 1.1025y = 1223.25 $$ ----- equation $$ (4) $$
Subtracting equation $$(4) $$ from $$ (3) $$, we get $$ 0.0424y = 25.44 \Rightarrow y = 600 $$

Substituting $$ y = 600 $$ in the equation $$ (2) $$, we get $$ 1.07x + 1.05(600) = 1165 \Rightarrow x = 500 $$

Hence, cost price of $$A$$ is Rs. $$ 500 $$ and of $$B$$ is Rs. $$ 600 $$
Question 10
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The class XI students of a school wanted to give a farewell party to the out going students of class XII. They decided to purchase two kinds of sweets, one costing Rs. 250 per kg and the other costing Rs. 350 per kg. They estimated that 40 kg of sweets were needed. if the total budget for the sweets was Rs. 11,800; find how much sweets of each kind were bought ?

A
20 kg and 18 kg

B
22 kg and 18 kg

C
28 kg and 18 kg

D
26 kg and 18 kg

Solution

Let $$ x $$ kg of one sweet be bought and $$ y $$ kg of other.

Given,
$$ x + y = 40 $$ --- (1)

And $$250x + 350y = 11800 $$
=> $$ 5x + 7y = 236 $$ --- (2)

Multiplying equation $$ (1) $$ with $$ 5 $$ we get, $$ 5x + 5y = 200 $$ ----- equation $$ (3) $$

Subtracting equation $$ (3) $$from $$ (2) $$, we get $$ 2y = 36 => y = 18 $$

Substituting $$ y = 18 $$ in the equation $$ (1) $$, we get $$ x + 18 = 40 => x = 22 $$
Hence $$ 22 kg $$ and $$ 18 kg $$ of two types of sweets were bought.

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Pair of Linear Equations in Two Variables Test - 35

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Pair of Linear Equations in Two Variables Test - 35

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Question 1

$$x\, =\, 3$$ and $$y\, =\, 2$$

$$x\, =\, 2$$ and $$y\, =\, 7$$

$$x\, =\, 9$$ and $$y\, =\, 4$$

$$x\, =\, 1$$ and $$y\, =\, 9$$

Solution

Question 2

$$x\, =\, b-a$$ and $$y\, =\, b$$

$$x\, =\, 0$$ and $$y\, =\, ba$$

$$x\, =\, ba$$ and $$y\, =\, ba$$

$$x\, =\, a$$ and $$y\, =\, b$$

Solution

Question 3

A = $$20$$ and B = $$55$$

A = $$20$$ and B = $$60$$

A = $$30$$ and B = $$ 50$$

A = $$50$$ and B = $$ 70$$

Solution

Question 4

$$30 \ and \ 50$$

$$35 \ and \ 52$$

$$40 \ and \ 60$$

$$22 \ and \ 49$$

Solution

Question 5

$$30$$ years and $$15$$ years

$$20$$ years and $$10$$ years

$$10$$ years and $$5$$ years

$$16$$ years and $$ 8$$ years

Solution

Question 6

$$13$$ and $$5$$

$$23$$ and $$9$$

$$17$$ and $$39$$

$$28$$ and $$45$$

Solution

Question 7

A has 34 Mangoes and B has 62 Mangoes

A has 14 Mangoes and B has 35 Mangoes

A has 70 Mangoes and B has 148 Mangoes

A has 96 Mangoes and B has 182 Mangoes

Solution

Question 8

61.6 and 120.4

62.6 and 120.4

63.6 and 120.4

64.6 and 120.4

Solution

Question 9

$$A =$$ Rs. $$200$$ and $$B =$$ Rs. $$300$$

$$A =$$ Rs. $$600$$ and $$B =$$ Rs.$$ 700$$

$$A =$$ Rs.$$ 100$$ and $$B =$$ Rs.$$ 200$$

$$A =$$ Rs. $$500$$ and $$B =$$ Rs. $$600$$

Solution

Question 10

20 kg and 18 kg

22 kg and 18 kg

28 kg and 18 kg

26 kg and 18 kg

Solution

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