# GSEB Solutions for ધોરણ ૧૦ English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

How many solutions does the pair of equations y = 0 and y = -5 have? (2013)

જવાબ : y = 0 and y = -5 are Parallel lines, hence no solution.

In the figure, ABCDE is a pentagon with BE 11 CD and BC 11 DE. BC is perpendicular to CD. If the perimeter of ABCDE is 21 cm, find the value of x and y.

જવાબ : If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). (2013)

જવાબ : Calculate the area bounded by the line x + y = 10 and both the co-ordinate axes. (2012)

જવાબ : Area of triangle
= 12 × base × corresponding altitude
= 12 × 10 × 10 = 50 cm2

Find whether the following pair of linear equations is consistent or inconsistent: (2015)
3x + 2y = 8 6x – 4y = 9

જવાબ : Therefore, given pair of linear equations is con-sistent.

Solve by elimination: (2014)
3x = y + 5
5x – y = 11

જવાબ : We have, 3x = y + 5, and 5x – y = 11 Putting the value of x in (i), we get
3x – y = 5 3(3) – y = 5
9 – 5 = y y = 4
x = 3, y = 4

Solve by elimination: 2015
3x – y – 7
2x + 5y + 1 = 0

જવાબ : 3 x – y = 7 …(i)
2x + 5y = -1 -00
Multiplying equation (i) by 5 & (ii) by 1, x = 2
Putting the value of x in (i), we have
3(2)-y = 7 6 – 7 = y
y = -1 x = 2, y = -1

For what value of k, the pair of equations 4x – 3y = 9, 2x + ky = 11 has no solution? (2017D)

જવાબ : We have, 4x – 3y = 9 and 2x + ky = 11 A man earns ₹600 per month more than his wife. One-tenth of the man’s salary and l/6th of the wife’s salary amount to ₹1,500, which is saved every month. Find their incomes. (2014)

જવાબ : Let wife’s monthly income = ₹x
Then man’s monthly income = ₹(x + 600)
According to the question,
110 (x + 600) + 16 (x) = ₹1,500
3(x+600)+5x30 = ₹1,500
3x + 1,800 + 5x = ₹45,000
8x = ₹45,000 – ₹1,800
x = ₹343,2008 = ₹5,400
Wife’s income = ₹x = ₹5,400
Man’s income = ₹(x + 600) = ₹6,000

Solve for x and y: xa=yb;
ax + by = a2+ b2 (2017D)

જવાબ : Putting the value of x in (i), we get
b(a) – ay = 0 ba = ay
baa = y b = y
x = a, y = b

Solve for x and y: (2017OD)
27x + 31y = 85;
31x + 2 7y = 89

જવાબ : Putting the value of ‘x’ in (i), we get
2 + y = 3
y = 3 – 2 = 1
x = 2, y = 1

Find the value of a and p for which the following pair of linear equations has infinite number of solutions:
2x + 3y = 7;
αx + (α + β)y = 28 (2013)

જવાબ : We have, 2x + 3y = 7 and αx + (α + β)y = 28 Solve the following pair of linear equations:
y -4x= 1
6x- 5y= 9

જવાબ : Solve for x and y
x + 2y – 3= 0
3x – 2y + 7 = 0

જવાબ : A part of monthly Hostel charge is fixed and the remaining depends on the number of days one has taken food in the mess. When Swati takes food for 20 days, she has to pay 3000 as hostel charges whereas, Mansi who takes food for 25 days pays Rs 3500 as hostel charges. Find the fixed charges and the cost of food per day.

જવાબ : Solve using cross multiplication method:
x+y=1
2x – 3y = 11

જવાબ : Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: 2x – 3y + 6 = 0,4x – 5y + 2 = 0

જવાબ : Given a linear equation 3x-5y = 11. Form another linear equation in these variables such that the geometric representation of the pair so formed is:
(i) intersecting lines
(ii) coincident lines
(iii) parallel lines

જવાબ : Find those integral values of m for which the c-coordinate of the point of intersection of lines represented by y = mx + 1 and 3x + 4y = 9 is an integer.

જવાબ : In a two digit number, the digit in the unit place is twice of the digit in the tenth place. If the digits are reversed, the new number is 27 more than the given number. Find the number.

જવાબ : The owner of a taxi company decides to run all the taxi on CNG fuels instead of petrol/ diesel. The taxi charges in city comprises of fixed charges together with the charge for the distance covered.
For a journey of 13 km, the charge paid is? 129 and for journey of 22 km, the charge paid is ^ 210.
(i) What will a person have to pay for travelling a distance of 32 km?
(ii) Why did he decide to use CNG for his taxi as a fuel?

જવાબ : The area of a rectangle reduces by 160 m if its length is increased by 5 m and breadth is reduced by 4 m. However, if length is decreased by 10 m and breadth is increased by 2 m, then its area is decreased by 100 m2. Find the dimensions of the rectangle.

જવાબ : At a certain time in a zoo, the number of heads and the number of legs of tiger and peacocks were counted and it was found that there were 47 heads and 152 legs. Find the number of tigers and peacocks in the zoo.
Why it is necessary to conserve these animals?

જવાબ : If the system of equations
6x + 2y = 3 and kx + y = 2 has a unique solution, find the value of k.

જવાબ : Determine the value of m and n so that the following pair of linear equations have infinite number of solutions.

જવાબ : For what values of p and q will the following pair of linear equations has infinitely many solutions?
4x + 5y = 2;
(2p + 7q)x + (p + 8q)y = 2q-p + 1

જવાબ : Solve the following pair of equations for x and y જવાબ : 8 men and 12 boys can finish a piece of work in 10 days, while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.

જવાબ : A two digit number is equal to 7 times the sum of its digits. The number formed by reversing its digits is less than the original number by 18. Find the original number.

જવાબ : For what value of k will the pair of equations have no solution?
3x + y = 1
(2k-l)x+ (k-l)y = 2k+l

જવાબ : Solve for x and y: જવાબ : The sum of the numerator and denominator of a fraction is 12. If 1 is added to both the numerator and the denominator the fraction becomes 3/4. Find the fraction.

જવાબ : 4 men and 6 boys can finish a piece of work in 5 days while 3 men and 4 boys can finish it in 7 days. Find the time taken by 1 man alone or that by 1 boy alone.

જવાબ : A man travels 600 km partly by train and partly by car. It takes 8 hours and 40 minutes if he travels 320 km by train and the rest by car. It would take 30 minutes more if he travels 200 km by train and the rest by car. Find the speed of the train and the car separately.

જવાબ : Find the value of m for which the pair of linear equations 2x + 3y – 7 = 0 and (m – l)x + (m + l)y = (3m – 1) has infinitely many solutions.

જવાબ : For what value of k will the following pair of linear equations have no solution?
2x + 3y = 9;
6x + (k – 2)y = (3k – 2).

જવાબ : For what value of p will the following pair of linear equations have infinitely many solutions?
(p – 3)x + 3y = p;
px + py = 12

જવાબ : Find the values of a and b for which the following pair of linear equations has infinitely many solutions: 2x + 3y = 7; 3/4(a + b)x + (2a – b)y = 21

જવાબ : Solve the following pair of equations: જવાબ : The sum of the numerator and the denominator of a fraction is 4 more than twice the numerator. If 3 is added to each of the numerator and denominator, their ratio becomes 2 : 3. Find the fraction. [All India]

જવાબ : A number consists of two digits. When the number is divided by the sum of its digits, the quotient is 7. If 27 is subtracted from the number, the digits interchange their places, find the number.

જવાબ : Find the value of a so that the point (3, a), lies on the line represented by 2x-3y = 5

જવાબ : Find the number of solutions of the following pair of linear equations .
x + 2y – 8 = 0
2x + 4y = 16

જવાબ : Write whether the following pair of linear equations is consistent or not
x +y = 14,
x-y = 4

જવાબ : Find the value of k for which the pair of linear equations
kx + 3y = k – 2 and 12x + ky = k has no solution.

જવાબ : Solve for x and y: જવાબ : Without drawing the graph, find out the lines representing the following pair of linear equations intersect at a point, are parallel or coincident. જવાબ : Solve the following system of equations for x and y જવાબ : Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

જવાબ : Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 ……….(1)
Given, the length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36
2x + 4 = 36
2x = 32
x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

On comparing the ratios a1/a2, b1/b2, and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

જવાબ : (i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0
and 2x – 3y = 7 or 2x – 3y – 7 = 0

Comparing the above equations with a1x + b1y + c1=0
And a2x + b2y + c2 = 0
We get,
a= 3, b= 2, c= -5
a= 2, b= -3, c= -7

a1/a= 3/2, b1/b= 2/-3, c1/c= -5/-7 = 5/7
Since, a1/a2≠b1/bthe lines intersect each other at a point and have only one possible solution.

Hence, the equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9
Therefore,
a= 2, b= -3, c= -8
a= 4, b= -6, c= -9

a1/a= 2/4 = 1/2, b1/b= -3/-6 = 1/2, c1/c= -8/-9 = 8/9

Since, a1/a2=b1/b2≠c1/c2

Therefore, the lines are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

જવાબ : 2x + 3y = 11…………………………..(i)
2x – 4y = -24………………………… (ii)
From equation (ii), we get;
x = (11 – 3y)/2 ……….…………………………..(iii)

Putting the value of x in equation (ii), we get

2[(11 – 3y)/2] – 4y = −24
11 – 3y – 4y = -24
-7y = -35
y = 5……………………………………..(iv)
Putting the value of y in equation (iii), we get;

x = (11 – 15)/2 = -4/2 = −2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore, the value of m is -1.

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14
x – y = 4

(ii) 3x – y = 3
9x – 3y = 9

જવાબ : (i) Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get,
x = 14 – y
Now, put the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the value of x;
x = 14 – y
x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.

(ii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3 + y)/3
Now, substitute the value of x in the given second equation to get,
9[(3 + y)/3] – 3y = 9

3(3+y) – 3y = 9
9 + 3y – 3y = 9
9 = 9
Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

જવાબ : Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (i)
3x + 5y = 1750 ………………. (ii)
From (i), we get;
y = (3800 – 7x)/6 …………………… (iii)
Substituting (iii) in (ii). we get,
3x + 5[(3800 – 7x)/6] = 1750
3x + (9500/3) – (35x/6) = 1750
3x – (35x/6) = 1750 – (9500/3)

(18x – 35x)/6 = (5250 – 9500)/3
-17x/6 = -4250/3
-17x = -8500
x = 500
Putting the value of x in (iii), we get;

y = (3800 – 7 × 500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

જવાબ : Let the fraction be x/y.

According to the question,
(x + 2)/(y + 2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)
(x + 3)/(y + 3) = 5/6

6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2), we get
6[(-4 + 9y)/11] – 5y = -3

-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y in (3), we get
x = (-4 + 81)/11 = 77/11 = 7

Hence, the fraction is 7/9.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

જવાબ : (i)

Let us assume, the present age of Nuri is x.
And the present age of Sonu is y.
According to the given condition, we can write as;
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2, to get,
y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,
x – 3(20) = -10
x – 60 = -10
x = 50
Therefore,
Age of Nuri is 50 years
Age of Sonu is 20 years.

(ii)

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i) we get,
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i) we get,
A + 12 = 27
A = 15
Hence, the fixed charge is Rs. 15.
And the Additional charge per day is Rs. 3.

Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4

જવાબ : 8x + 5y = 9 …………………..(1)
3x + 2y = 4 ……………….….(2)
From equation (2) we get;

x = (4 – 2y) / 3 ……………………. (3)
Substituting this value in equation 1, we get
8[(4 – 2y)/3] + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5 ……………………………….(4)
Substituting this value in equation (2), we get
3x + 10 = 4

3x = -6
x = -2
Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:
8x +5y – 9 = 0
3x + 2y – 4 = 0

x/(-20 + 18) = y/(-27 + 32 ) = 1/(16 – 15)
-x/2 = y/5 = 1/1

x = -2 and y =5.

Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

જવાબ : (i) Let us consider,
Speed of boat is still water = x km/hr
Speed of current = y km/hr
Now, speed of Ritu, during,
Downstream = x + y km/hr
Upstream = x – y km/hr
As per the question given,
2(x + y) = 20
Or x + y = 10……………………….(1)
And, 2(x – y) = 4
Or x – y = 2………………………(2)
Adding both the eq.1 and 2, we get,
2x = 12
x = 6
Putting the value of x in eq.1, we get,
y = 4
Therefore,

Speed of Ritu is still water = 6 km/hr
Speed of current = 4 km/hr

On reversing the digits of a two digit number, number obtained is 9 less than three times the original number. If difference of these two numbers is 45, find the original number. (2014)

જવાબ : Let unit’s place digit be x and ten’s place digit bey.
Original number = x + 10y Reversed number = 10x + y
According to the Question,
10x + y = 3(x + 10y) – 9

10x + y = 3x + 30y – 9
10x + y – 3x – 30y = -9
7x – 29y = -9 …(i)
10x + y – (x + 10y) = 45
9x – 9y = 45
x – y = 5 …[Dividing both sides by 9
x – 5 + y …(ii)
Solving (i),
7x – 29y = -9
7(5 + y) – 29y = -9 …[From (ii)
35+ 7y – 29y = -9
-22y = -9 – 35
-22y = -44
y = 4422 = 2
Putting the value of y in (ii),
x = 5 + 2 = 7

Original number = x + 10y
= 7 + 10(2) = 27

One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

જવાબ : Let the money with the first person and second person be Rs x and Rs y respectively.
According to the question,
x + 100 = 2(y – 100) x + 100 = 2y – 200 x – 2y = 300 … (1)
Again,   6(x – 10) = (y + 10) 6x – 60 = y + 10 6x – y = 70 … (2)
Multiplying equation (2) by 2, we obtain:
12x – 2y = 140 … (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300 11x = 440 x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = –300 40 + 300 = 2y 2y = 340 y = 170
Thus, the two friends had Rs 40 and Rs 170 with them.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

જવાબ : Let the number of rows be x and number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row = xy
According to the question,
Total number of students = (x – 1) (y + 3) xy = (x – 1) (y + 3) xy= xy – y + 3x – 3 3x – y – 3 = 0 3x – y = 3 … (1)
Total number of students = (x + 2) (y – 3) xy = xy + 2y – 3x – 6 3x – 2y = –6 … (2)
Subtracting equation (2) from (1), we obtain:
y = 9
Substituting the value of y in equation (1), we obtain:
3x – 9 = 3 3x = 9 + 3 = 12 x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Hence, Total number of students in a class = xy = 4 x 9 = 36

Solve: ax + by = a – b and bx – ay = a + b

જવાબ : The given system of equations may be written as
ax + by – (a – b) = 0
bx – ay – (a + b) = 0
By cross-multiplication, we have Hence, the solution of the given system of equations is x = 1, y = -1

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

જવાબ : Let the length and breadth of a rectangle be x and y respectively.
Then area of the rectangle = xy
According to question, we have
(x – 5) 6 + 3) = xy – 9
xy + 3x – 5y – 15 = xy – 9
3x – 5y = 15 – 9 = 6 3x – 5y = 6 …(i)
Again, we have
(x + 3) 6 + 2) = xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y = 67 – 6 = 61 2x + 3y = 61…. (ii)
Now, from equation (i), we express the value of x in terms of y as Putting the value of y in equation (i), we have
3x – 5 x 9 = 6 3x = 6 + 45 = 51
x = 513 = 17
Hence, the length of rectangle = 17 units and breadth of rectangle = 9 units.  A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay 31000 as hostel charges whereas a student B, who takes food for 26 days, pays 1180 as hostel charges. Find the fixed charges and the cost of food per day.

જવાબ : Let the fixed charge be *x and the cost of food per day be by.
Therefore, according to question,
x + 20y = 1000 …(i)
x + 26y = 1180 …(ii)
Now, subtracting equation (ii) from (i), we have Putting the value of y in equation (i), we have
x + 20 x 30 = 1000
x + 600 = 1000 x = 1000 – 600 = 400
Hence, fixed charge is ₹400 and cost of f
ood per day is ₹30.

If x = a, y = b is the solution of the pair of equation x – y = 2 and x + y = 4, then find the value of a and b.

જવાબ : x – y = 2 … (i)
x + y = 4 … (ii)
On adding (i) and (ii), we get 2x = 6 or x = 3
From (i), 3 – y 2 = y = 1
a = 3, b = 1.
On comparing the ratios a1a2,b1b2 , and, c1c2 find out whether the following pair of linear equations consistent or inconsistent. is consistent or inconsistent. (Q. 5 to 6)

32 x + 53 y = 7
9x – 10y = 14

જવાબ : Jamila sold a table and a chair for 1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair she would have got 1065. Find cost price of each.

જવાબ : Let cost price of table be ₹x and the cost price of the chair be ₹y. From equation (i) and (ii) we get
110x + 125y = 105000
and 125x + 110y = 106500
On adding and subtracting these equations, we get
235x + 235y = 211500
and 15x – 15y = 1500
i.e., x + y = 900 …(iii)
x – y = 100 …… (iv)
Solving equation (iii) and (iv) we get
x = 500, y = 400
So, the cost price of the table is ₹500 and the cost price of the chair is ₹400.

The sum of a two digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the first number.

જવાબ : Let the digits at unit and tens places be x and y respectively.
Then, number = 10y + x …(i)
Number formed by interchanging the digits = 10x + y
According to the given condition, we have
(10y + x) + (10x + y) = 110

11x + 11y = 110
x + y – 10 = 0
Again, according to question, we have
(10y + x) – 10 = 5 (x + y) + 4

10y + x – 10 = 5x + 5y + 4
10y + x – 5x – 5y = 4 + 10
5y – 4x = 14 or 4x – 5y + 14 = 0
By using cross-multiplication, we have . Putting the values of x and y in equation (i), we get
Number 10 × 6 + 4 = 64.

Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) Roobi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by bus and the remaining by train. If she travels 100 km by bus and the remaining by train, she takes 10 minutes longer. Find the speed of the train and the bus separately.

જવાબ : (i) Let her speed of rowing in still water be x km/h and the speed of the current be y km/h.
Case I: When Ritu rows downstream
Her speed (downstream) = (x + y) km/h Putting the value of x in equation (i), we have
6 + y = 10 y = 10 – 6 = 4
Hence, speed of Ritu in still water = 6 km/h.
and speed of current = 4 km/h.

(ii) Let the speed of the bus be x km/h and speed of the train be y km/h.
According to question, we have The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

જવાબ : Let the length and breadth of a rectangle be x and y respectively.
Then area of the rectangle = xy
According to question, we have
(x – 5) 6 + 3) = xy – 9
xy + 3x – 5y – 15 = xy – 9
3x – 5y = 15 – 9 = 6 3x – 5y = 6 …(i)
Again, we have
(x + 3) 6 + 2) = xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y = 67 – 6 = 61 2x + 3y = 61…. (ii)
Now, from equation (i), we express the value of x in terms of y as Putting the value of y in equation (i), we have
3x – 5 x 9 = 6
3x = 6 + 45 = 51
x = 513 = 17
Hence, the length of rectangle = 17 units and breadth of rectangle = 9 units.

Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deduced for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

જવાબ : Let x be the number of questions of right answer and y be the number of questions of wrong answer.
According to question,
3x – y = 40 … (i)
and 4x – 2y = 50
or 2x – y = 25 …(ii)
Subtracting (ii) from (i), we have Putting the value of x in equation (i), we have
3 x 15 – y = 40
45 – y = 40
y = 45 – 40 = 5
Hence, total number of questions is x + yi.e., 5 + 15 = 20.

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay 31000 as hostel charges whereas a student B, who takes food for 26 days, pays 1180 as hostel charges. Find the fixed charges and the cost of food per day.

જવાબ : Let the fixed charge be *x and the cost of food per day be by.
Therefore, according to question,
x + 20y = 1000 …(i)
x + 26y = 1180 …(ii)
Now, subtracting equation (ii) from (i), we have  Putting the value of y in equation (i), we have
x + 20 x 30 = 1000
x + 600 = 1000 x = 1000 – 600 = 400
Hence, fixed charge is ₹400 and cost of food per day is ₹30.

Form the pair of linear equations in this problem, and find its solution graphically: 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

જવાબ : Let x be the number of girls and y be the number of boys.
According to question, we have
x = y + 4
x – y = 4 ……(i)
Again, total number of students = 10
Therefore, x + y = 10 …(ii)
Hence, we have following system of equations
x – y = 4
and x + y = 10

From equation (i), we have the following table: Here, the two lines intersect at point (7,3) i.e., x = 7, y = 3.
So, the number of girls = 7
and number of boys = 3.

Solve the following pairs of equations by reducing them to a pair of linear equations: જવાબ :   3 x 1 + y = 4
y = 4 – 3 = 1
Hence, the solution of given system of equations is x = 1, y = 1
.

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) for which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1

જવાબ : (i) We have, 2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2 … (ii)
Here, a1 = 2, b1 = 3, c1 = 7 and
a2 = a – b, b2 = a + b, c2 = 3a + b – 2
For infinite number of solutions, we have 9a – 7a + 3b – 75 -6 = 0 2a – 45 – 6 = 0 => 2a – 4b = 6
a – 2b = 3 …(iv)
Putting a = 5b in equation (iv), we get
56 – 2b = 3 or 3b = 3 i.e., b = 33 =1
Putting the value of b in equation (ii), we get a = 5(1) = 5
Hence, the given system of equations will have an infinite number of solutions for a = 5 and b = 1.

(ii) We have, 3x + y = 1, 3x + y − 1 = 0 …(i)
(2k – 1) x + (k – 1) y = 2k + 1
(2k – 1) x + (k – 1) y – (2k + 1) = 0 ……(ii)
Here, a1 = 3, b1 = 1, C1 = -1
a2 = 2k – 1, b2 = k – 1, C2 = -(2k + 1)
For no solution, we must have 3k – 2k = 3 – 1 k = 2
Hence, the given system of equations will have no solutions for k = 2.

Find whether the following pair of linear equations has a unique solution. If yes, find the
7x – 4y = 49 and 5x – y = 57

જવાબ : We have, 7x – 4y = 49 ……..(i)
and 5x – 6y = 57 ……..(ii) Put y = -7 in equation (ii)
5x – 6(-7)57 5x = 57 – 42 x = 3
hence, x = 3 and y = -7.

Solve for x and y જવાબ : Putting the value of y in (ii), we get
x + ab = 2ab x = 2ab – ab x = ab
x = ab, y = ab

Solve the following linear equations:
152x – 378y = -74 and -378x + 152y = -604

જવાબ : We have, 152x – 378y = -74 …(i)
-378x + 152y = -604 ……(ii) Putting the value of x in (iii), we get
2 + y = 3
y = 1
Hence, the solution of given system of equations is x
= 2, y = 1.