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 1000 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.

 cbse ncert 10th mathematics chapter 3 exercise 3.5

pair of linear equations in two variables

 

4

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 Rs.1000 as hostel charges whereas a student B, who takes
food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the
cost of food per day.

 

let x = fised charges for a month

and y = charge of food per day 

 

student A takes food for 20 days has to pay Rs.1000 means

x + 20y =1000 ----------------------(1)

 

student B, who takes  food for 26 days, pays Rs.1180 means

x +26 y =1180 ---------------------(2)

x + 20y =1000 ----------------------(1)

x +26 y =1180 ---------------------(2)

-----------------------------------------------------subtracting

(-6)y =-180

y = 30

substitute y =30 in x + 20y =1000

x+20(30)  =1000

x =1000 - 20*30

x=400

fixed charge =Rs.400 /- per month

charge for food = Rs.30 /- per day

(ii) A fraction changes to (1/3) when 1 is subtracted from the numerator and it changes to (1/4) when 8 is added to its denominator. Find the fraction.

let fraction =(x/y)

 

fraction changes to (1/3) when 1 is subtracted from the numerator means

[(x-1)/y] =(1/3)

3(x-1) = 1y

3x-3 = y

3x-y =3----------------------(1)

fraction changes to (1/4) when 8 is added to its denominator means

[x /(y+8)] =(1/4)

4x =1(y+8)

4x=1y+8

4x-1y =8-----------------(2)

3x-y =3----------------------(1)     eliminating y

4x-y =8-------------------(2)

----------------------------------------subtracting

(-1)x =(-5)

x = 5

substitute x=5 in 3x-y=3

3(5)-y=3

(-y)= 3-15

(-y)= =(-12)

y=12

fraction = [5/12]

 

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

(iv)

(a – b)x + (a + b) y = (a^2) – 2ab – (b^2)

(a + b)(x + y) = (a^2) + (b^2 )

solution

 

(v)

152x – 378y = – 74

–378x + 152y = – 604

solution   

 ABCD is a cyclic quadrilateral  Find the angles of the cyclic quadrilateral,

if angles are A =(4y+20) , B =(3y-5) , C=(-4x), D=(-7x+5)

solution

 

exercise 3.6

2

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. 

solution

 

 (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to
finish the work, and also that taken by 1 man alone.

solution

 

(iii) 

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

 

solution

 

exercise 3.5

4

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 Rs.1000 as hostel charges whereas a student B, who takes
food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the
cost of food per day.

solution

 

(ii) A fraction changes to (1/3) when 1 is subtracted from the numerator and it changes to (1/4) when 8 is added to its denominator. Find the fraction. 

 solution

 
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

 ncert cbse 10th mathematics chapter 3 exercise3.6 pair of linear equations in two variables

2

 (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to
finish the work, and also that taken by 1 man alone.

let x = number of days for a woman alone to finish the work

and y = number of days for a man alone to finish the work

 

This is a problem on time and work so remember to take reciprocals of x , y and time

 

2 women and 5 men can together finish an embroidery work in 4 days

so

[2/x] +[5/y]=[1/4]

3 women and 6 men can finish it in 3 days

so

[3/x]+[6/y] =[1/3]

equations are

[2/x] +[5/y]=[1/4]

[3/x]+[6/y] =[1/3]

to get rid of (1/x) and (1/y)

use a=(1/x) and b=(1/y)

 

equations change to 

2a+5b= [1/4]      *3

3a+6b= [1/3]      *2

-------------------------------------eliminate a

6a +15b=[3/4]

6a+12b=[2/3]

------------------------------------subtracting

     3b =[3/4] -[2/3]

     3b =[1/12]    

      

 b=[1/36]

 

substitute in 3a+6b= [1/3]

3a +6*(1/36) =[1/3]

3a = [1/3] - [1/6]

3a = [1/6]

 

a=[1/18]

a=[1/18] ,b=[1/36]

take reciprocal

x =18 days  , y = 36 days

 

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. 

Let x = speed of her rowing in still water in km/hr

and y = speed of the current in km/hr

speed downstream = (x+y)

speed upstream = (x-y)

time = distance / speed

 

downstream 20 km in 2 hours

means 

2 =20/(x+y) 

or 

x+y = 20/2

x+y =10-------------------(1)

 

upstream 4 km in 2 hours

means

2 =4/(x-y) 

or

x-y =4/2

x-y =2----------------------(2)

x+y =10-------------------(1)

x-y =2----------------------(2)

------------------------------------------adding

2x = 12

 

x= 6km/hr

 

substitute in x-y =2

to get

y=4km/hr

 

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

(iv)

(a – b)x + (a + b) y = (a^2) – 2ab – (b^2)

(a + b)(x + y) = (a^2) + (b^2 )

solution

 

(v)

152x – 378y = – 74

–378x + 152y = – 604

solution   

 ABCD is a cyclic quadrilateral  Find the angles of the cyclic quadrilateral,

if angles are A =(4y+20) , B =(3y-5) , C=(-4x), D=(-7x+5)

solution

 

exercise 3.6

2

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. 

solution

 

 (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to
finish the work, and also that taken by 1 man alone.

solution

 

(iii) 

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

 

solution

 

 

 
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

 

ABCD is a cyclic quadrilateral Find the angles of the cyclic quadrilateral.

10th ncert cbse mathematics  chapter 3 miscellaneous / optional exercise

pair of linear equations in two variables

 

 ABCD is a cyclic quadrilateral  Find the angles of the cyclic quadrilateral,

if angles are A =(4y+20) , B =(3y-5) , C=(-4x), D=(-7x+5)

 

We know that the opposite angles of a cyclic quadrilateral are supplementary, 

which means they add up to 180 degrees.

 

A+C=180 degrees

and 

B+D = 180 degrees

 

A+C=180 

gives 

(4y+20) +(-4x)=180

or

 -4x +4y =180-20

or

-4x +4y =160 

dividing by 4 makes it

-x+y=40--------------(1)

 

 B+D = 180

gives

 (3y-5)+(-7x+5) =180

or

-7x+3y=180-------------(2)

eliminating y

-x+y=40--------------(1)  *3

-7x+3y=180-------------(2)

 

-3x+3y =120

-7x+3y =180 

--------------------------subtracting

4x     =(-60)

x = (-60)/4

 

x=(-15)

 

substitute in

-x+y=40

-(-15) +y =40

15+y-40

y=40-15

 

y=25

 

use  x=(-15) and y=25

A =(4y+20) =4*25+20=120 degrees

 B =(3y-5)=3*25-5=75-5=70 degrees

 C=(-4x) =(-4)*(-15) = 60 degrees

 D=(-7x+5) =(-7)*(-15) +5 =105+5=110 degrees

 

exercise 3.6

2

(iii) 

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

 

let x = speed of train in km/hr

and y = speed of bus in km/hr

 

plan 1

60 km by train and remaining 300-60 = 240 km by bus

time = distance / speed

t1=(60/x) in the train 

and 

t2=(240/y) hours in the bus

total of 4 hours means

t1 +t2 =4 hours

(60/x) +(240/y) =4--------------------(1)

 

plan 2

100 km by train and remaining 300-100 = 200 km by bus

time = distance / speed

t3=(100/x) in the train 

and 

t4=(200/y) hours in the bus

10 minutes longer means 4hours 10minutes

This has to be changed in term of hours 

4hours 10minutes =4+(10/60)=4+(1/6) = (25/6)hours

so 

t3 +t4 = (25/6)hours

(100/x)+(200/y) =(25/6)----------------(2)

 

equations are

 

(60/x) +(240/y) =4--------------------(1)

(100/x)+(200/y) =(25/6)----------------(2)

 

we have to use substitution (1/x)=u , (1/y)=v

 

equations change to

 

60u   + 240v =4   ----------------------(3)  *5

100u + 200v =(25/6)-------------------(4)  *3 

--------------------------------------------------------------------eliminate u

300u + 1200v =20

300u +600v =(25/2)

------------------------------------------subtracting

600v =  20 -(25/2)

600v = (15/2)

v = [ 15 / (2*600) ]

v = [1/80]

resubstitute v = [1/80] in 60u   + 240v =4  

60u+{240/80} =4

60u+3=4

60u=4-3

60u=1

u=(1/60)

 

u=(1/60)

v = [1/80]

 

take reciprocal

x=60 km/hr 

y=80 km/hr 

speed of train =60km/hr

speed of bus =80km/hr

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

(iv)

(a – b)x + (a + b) y = (a^2) – 2ab – (b^2)

(a + b)(x + y) = (a^2) + (b^2 )

solution

 

(v)

152x – 378y = – 74

–378x + 152y = – 604

solution   

 ABCD is a cyclic quadrilateral  Find the angles of the cyclic quadrilateral,

if angles are A =(4y+20) , B =(3y-5) , C=(-4x), D=(-7x+5)

solution

 

exercise 3.6

2

(iii) 

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

 

solution

 

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).   
solution
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other mean

 

 

 

 

(a – b)x + (a + b) y = a 2 – 2ab – b 2 (a + b)(x + y) = a 2 + b 2

 10th cbse mathematics  chapter 3 miscellaneous / optional exercise 

solve

(iv)

(a – b)x + (a + b) y = (a^2) – 2ab – (b^2)

(a + b)(x + y) = (a^2) + (b^2 )

 

rearrange the second equation 

making the RHS zero for both equations

(a – b)x + (a + b) y -[ (a^2) – 2ab – (b^2)] = 0

(a + b)x +(a + b)y -[ (a^2) + (b^2 )] =0

 

using method of cross multiplication

 

                 x                                           y                     1

(a+b)            {-[ (a^2)–2ab–(b^2)]}        (a-b)                 (a+b)

(a+b)             { -[ (a^2) + (b^2 )]}         (a+b)               (a+b)

 

 

x /{(a+b){-[ (a^2) + (b^2 )]} -(a+b){-[ (a^2)–2ab–(b^2)]}}

=

y/ {{-[ (a^2)–2ab–(b^2)]}(a+b) - { -[ (a^2) + (b^2 )]}(a-b)} 

=

1 /{(a-b)(a+b) - (a+b)(a+b)}

simplifying

x / {(a+b)[-2(b^2 ) -2ab]} taking (a+b) out in the Dr.

=

y/ {(a^2) [-a-b+a-b] +2ab(a+b) +(b^2)[a+b+a-b]} grouping (a^2) and (b^2)

= 

1/ {(a+b)[a-b-a-b]} taking (a+b) out

 

OR

x /{(a+b)(-2b)(b+a)}

=

y / {[(a^2)(-2b) +2ab(a+b) +(b^2)(2a)]}

=

1 / {(a+b)(-2b)}

 

OR

x /{(a+b)(-2b)(b+a)}

=

y / {[(a^2)(-2b) + 2(a^2)b + 2a(b^2) + (b^2)(2a)] } opening middle term

=

1 / {(a+b)(-2b)}

 

OR

x /{(a+b)(-2b)(b+a)}

=

y / {[2a(b^2) + (b^2)(2a)] } cancelling off

=

1 / {(a+b)(-2b)}

 

x /{(a+b)(-2b)(b+a)}=y / {4a(b^2) }=1 / {(a+b)(-2b)}

 

using the first and third

x /{(a+b)(-2b)(b+a)} = 1 / {(a+b)(-2b)}

 

x = {(a+b)(-2b)(b+a)} /{(a+b)(-2b)}

x= (a+b) 

using the second and the third

y /{4a(b^2) }=1 / {(a+b)(-2b)

 

y =  {{4a(b^2) } / {(a+b)(-2b)}

y =(-2ab) / (a+b) 

(v)

152x – 378y = – 74

–378x + 152y = – 604

here note that the coefficients of x and y are interchanged

namely 152 and (-378)

Therefore we add and subtract the two equations to get another two equations which will usually be easier to solve than the two given equations

  152x – 378y = – 74       --------------------(1)

–378x + 152y = – 604   ---------------------(2)

---------------------------------------adding

-226x -226y = (-678) 

divide each term with (-226)

we get

x + y = 3 -------------------------(3)

  152x – 378y = – 74       --------------------(1)

–378x + 152y = – 604   ---------------------(2)

---------------------------------------subtracting

530x -530y = 530

dividing each term with 530

x - y = 1 ----------------------------(4)

now we concentrate on the new equations (3) and (4)

x + y = 3 -------------------------(3)

x -  y = 1 ----------------------------(4)

-------------------adding

2x   = 4

x =2

 

substitute x=2 in  x + y = 3

2+y = 3 

y = 3-2

y =1

 

 

 

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

(iv)

(a – b)x + (a + b) y = (a^2) – 2ab – (b^2)

(a + b)(x + y) = (a^2) + (b^2 )

solution

 

(v)

152x – 378y = – 74

–378x + 152y = – 604

solution   

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).   
solution
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other mean

 

 

(ii) ax + by = c bx + ay = 1 + c

 10th cbse mathematics chapter 3 miscellaneous / optional exercise 

solve 

(ii) ax + by = c
     bx + ay = 1 + c

eliminating x

ax + by = c             ----------------------(1)  *b

bx + ay = 1 + c       ---------------------(2)  *a

abx +(b^2)y = bc

abx +(a^2)y =a +ac

-------------------------------------subtracting

[(b^2)- (a^2) ]y = bc-ac -a

[(b^2)- (a^2) ]y = c(b-a) - a

y = {c(b-a) - a} / [(b^2)- (a^2) ]

introducing (-1)  in Nr.  and Dr.

y ={a -c(b-a)} / [(a^2)- (b^2) ]

or , changing the second term in the Nr.

 

y = {a+ c(a-b)} / [(a^2)- (b^2) ]  

 

eliminating y

ax + by = c             ----------------------(1)  *a

bx + ay = 1 + c       ---------------------(2)  *b

 

(a^2) x + aby = ac

(b^2)x +aby =b+bc

------------------------------------------subtracting

[(a^2)- (b^2) ]x =ac-bc-b

x ={c(a-b) -b }/ [(a^2)- (b^2) ]

 

OR 

using cross multiplication method

Remember to make the RHS zero

ax + by - c =0

 bx + ay -( 1 + c )=0

            x             y           1

 b               -c            a            b

a             - (1+c)       b            a

{ x /[-b(1+c) - (-ac)] } = {y / [-cb - (-a(1+c))]  } = {1 / [ (a^2)- (b^2) ]}

using the first and third expression

{ x /[-b(1+c) + ac] } = {1 / [ (a^2)- (b^2) ]}

x =   [[-b(1+c) +ac] / [ (a^2)- (b^2) ]

or

x = [c(a-b)  - b] / [ (a^2)- (b^2) ]

 

using the first and third expression

{y / [-cb - (-a(1+c))]  } = {1 / [ (a^2)- (b^2) ]}

y =[-bc+a+ac]  / [ (a^2)- (b^2) ]

y = [c(a-b)+a]  / [ (a^2)- (b^2) ]

 

(iii) 

solve

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

multiplying the first equation with ab

and rearrange the second equation to get the RHS to zero before starting

method of  crossmultiplication

 

bx-ay   +0                        = 0 

ax +by -[ (a^2)  + (b^2)] =0

 

 

                x                               y                        1

(-a)                              0                        b                          (-a)  

b                  { -[ (a^2)  + (b^2)] }       a                          b

 

x/{ 0 - (-a){-[(a^2) + (b^2)]}}=y/{0-b{ -[ (a^2)  + (b^2)] }}=1/{(b^2)+(a^2)}

using the first and third expressions

x / {a[(b^2)+(a^2)]}=1/{(b^2)+(a^2)}

 

x = a 

using the second and third expressions

y/{0-b{ -[ (a^2)  + (b^2)] }}=1/{(b^2)+(a^2)}

 

y / [b [{(b^2)+(a^2)} ] ] = 1/{(b^2)+(a^2)}

y =b

 

 

 

 

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

 

 

 

 

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).   
solution
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

 

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

 10th cbse mathematics chapter 3 pair of linear equations in two variables optional exercise 3.7

 

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

given C = 3B =2(A+B) --------------------(1)

using first and last expression

2(A+B) =C

or A+B = C/2---------------(2)

using angle sum property of triangle

A+B+C = 180 degrees

(A+B) +C = 180  

using (2)

(C/2) + C = 180

 (3C/2) = 180

C = 180*2/3

 

C = 120 degrees -----------------(3)

 

using (3) and the first two expressions of (1)

 

3B = C

3B = 120

B = 120/3

 

B=40 degrees----------------------(4)

 

using (3), (4) in (2)

A+B = C/2

A + 40 = 120/2 

A=60-40

 

A = 20 degrees

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

using elimination method 

eliminating x

 px + qy = p – q -------------------(1) *q
 qx – py = p + q -------------------(2) *p

pq x + (q^2) y =    pq   -  (q^2) 

pq x -  (p^2) y  = (p^2) + pq

-----------------------------------------------------------subtracting

[(q^2) +(p^2) ] y = {- [ (q^2)  + (p^2) ] }

y = {- [ (q^2)  + (p^2) ] } / [(q^2) +(p^2) ]

 

y = (-1) 

 

substitute  y = (-1)  in   qx – py = p + q

qx -p(-1)  = p+q

qx+p = p+q

qx = q

x =q/q

 

x=1

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

 

 

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).   
solution
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means  

 

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

 10th cbse ncert mathematics chapter 3 pair of linear equations in two variables optional exercise 3.7 for ncert cbse

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

x = normal speed of train in km/hr

y = normal time of travel in hours

distance covered = speed * time = xy

If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time

means 

speed = (x+10) 

and time taken = (y-2)

and distance covered = (x+10) * (y-2)

 

if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time

means

speed = (x-10) 

time =(y+3)

and distance =(x-10)*(y+3)

since distance covered is the same 

(x+10) * (y-2) = x*y

(x-10)*(y+3) = x*y

simplify

(x+10) * (y-2) = x*y

xy - 2x +10y -20 = xy  

OR 

-2x +10y =20 -------------(1)

 

 

simplify 

(x-10)*(y+3) = x*y 

xy+3x-10y-30 =xy

OR 

3x -10y = 30  -----------------(2)

using elimination method

 

-2x +10y =20 -------------(1)

3x -10y = 30  --------------(2)

--------------------------------------------adding

x           =  50

substitute in -2x +10y =20

-2(50)+10y =20

10y =20+100

10y = 120

y =12 

x =50 km/hr 

y =12 hours

distance travelled =xy = 50*12 = 600 km

4. 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.

x = normal number of rows

y = normal number of students in each row

total number of students = xy

 

If 3 students are extra in a row, there would be 1 row less

means

number of students in each row = (y+3)

number of rows =(x-1) 

total number of students =(x-1)(y+3)

 

If 3 students are less in a row, there would be 2 rows more

means

number of students in each row =(y-3) 

number of rows = (x+2)

total number of students =(x+2)(y-3)

since number of students does not change

(x-1)(y+3) = xy

(x+2)(y-3) = xy

 

simplify

(x-1)(y+3) = xy

xy +3x-1y-3 =xy

3x-y = 3 -------------------(1) 

simplify

(x+2)(y-3) = xy

xy-3x+2y-6=xy

-3x+2y=6 -------------------(2) 

using elimination method

 3x-y = 3 -------------------(1) 

-3x+2y=6 -------------------(2) 

--------------------------------------------------adding

        y =9

substitute in 3x-y = 3 

3x -9 = 3

3x = 12

x =4

 

x = 4 rows

y =9 students in each row

 

Total number of students in the class = xy =4*9 = 36

=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. 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? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. 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.

solution 

 

 

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).   
solution
 
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means