Formation Of An Equation And Its Solution
Class 8th Mathematics West Bengal Board Solution
- Sima has written a number. If 2 is added to twice the number, it becomes 5 less than…
- Let’s write down three consecutive integers such that 5 less than the sum of the number…
- Let’s find a number such that one-fourth of the number is 1 less than one-third of the…
- Let’s find a fraction whose denominator is 2 more than the numerator and when 3 is…
- Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again she…
- Raju wrote a number having 2 digit where the digit in the ten’s position is thrice the…
- If the sum of two numbers is 89 and their difference is 15, let’s find the two numbers.…
- Divide 830 into two parts in such a way, that 30% if one part will be 4 more than 40%…
- Divide 56 into parts, so that thrice of the first part becomes 48 more than one third…
- {1}/{5} part of a pole lies in mud, {3}/{5} part lies in water and the…
- At present, my father’s age in 7 times of my age. After 10 years my father’s age is 3…
- My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee…
- Salem uncle of our village used {1}/{2} of this savings to buy a house after his…
- There was a food provision for twenty days in a refugee camp of the village Goplapur.…
- {3}/{x+3} = frac {5}/{x+2} Let’s find the roots of the equations (i.e. let’s solve…
- {3}/{3x+4} = frac {5}/{ 5 (x-3) } Let’s find the roots of the equations (i.e. let’s…
- 14 (x-2) + 3 (x+5) = 3(x+8) + 5 Let’s find the roots of the equations (i.e. let’s…
- {x}/{2} + 5 = frac {x}/{3} + 7 Let’s find the roots of the equations (i.e. let’s…
- {x+1}/{8} + frac {x-2}/{5} = frac {x+3}/{10} + frac {3x-1}/{20} Let’s find the roots…
- {x+1}/{4} + 3 = frac {2x+4}/{5} + 2 Let’s find the roots of the equations (i.e. let’s…
- {x+1}/{7} + x = frac {3x-4}/{14} + 6 Let’s find the roots of the equations (i.e.…
- {3}/{5} (x-4) - frac {1}/{3} (2x-9) = frac {1}/{4} (x-1) - 2 Let’s find the roots of…
- {x+5}/{3} + frac {2x-1}/{7} = 4 Let’s find the roots of the equations (i.e. let’s…
- 25 + 3(4x - 5) + 8(x + 2) = x + 3 Let’s find the roots of the equations (i.e. let’s…
- {x-8}/{3} + frac {2x+2}/{12} + frac {2x-1}/{18} = 3 Let’s find the roots of the…
- {t+12}/{6} - t = 6 frac {1}/{2} - frac {1}/{12} Let’s find the roots of the equations…
- {x+1}/{2} - frac {5x+9}/{28} = frac {x+6}/{21} + 5 - frac {x-12}/{3} Let’s find the…
- {9x+5}/{14} + frac {8x-7}/{7} = frac {18x+11}/{28} + frac {5}/{4} Let’s find the…
- {3y+1}/{16} + frac {2y-3}/{7} = frac {y+3}/{8} + frac {3y-1}/{14} Let’s find the…
- 5x - (4x-7) (3x-5) = 6-3 (4x-9) (x-1) Let’s find the roots of the equations (i.e.…
- 3(x-4)2+ 5(x-3)2 = (2x-5) (4x-1)-40 Let’s find the roots of the equations (i.e. let’s…
- 3(y-5)2 + 5y = (2y-3)2 +1 Let’s find the roots of the equations (i.e. let’s solve the…
- x = 5 → Let’s write mathematical stories and from equations.
- y=-11 → Let’s write mathematical stories and from equations.
- = {7}/{8} arrow Let’s write mathematical stories and from equations.…
- x=24 → Let’s write mathematical stories and from equations.
- x= → Let’s write mathematical stories and from equations.
Lets Do
Question 1.Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than 7/3 part of marbles of mine then calculate the possible number of marbles of Murad.
If Murad got 40 marbles, then let's find the number of marbles Shibani has given to me…
Answer:
Let the no. of marbles of mine = x
According to the question,
Now, it is given that Murad got 40 marbles
⇒ 7x = 42 × 3
⇒ x = 18
Hence, the no. of marbles of mine = 18
Lets Work Out 19
Question 1.Sima has written a number. If 2 is added to twice the number, it becomes 5 less than thrice the number. Let’s write the numbers that Sima has written.
Answer:
Let the number be x
2 is added to twice the number is 2x + 2.
5 less than thrice the number is 3x – 5
So,
2x + 2 = 3x – 5
Solve it,
2x – 3x = - 5 – 2
- X = -7
X = 7
Question 2.
Let’s write down three consecutive integers such that 5 less than the sum of the number is equal to 11 more than twice the second number. Let’s find the three consecutive integers.
Answer:
Let the three consecutive numbers be x, (x+1), (x+2).
Now according to the given criteria,
5 less than the sum of the number = x+(x+1)+(x+2)-5…..(i)
11 more than twice the second number = 2(x+1)+11…….(ii)
So from question equation (i) and equation (ii) are equal,
Hence
x+(x+1)+(x+2)-5 = 2(x+1)+11
⇒ 3x+3-5=2x+2+11
⇒ 3x-2=2x+13
⇒ 3x-2x=13+2
⇒ x=15
⇒ x+1=15+1=16
⇒ x+2=15+2=17
Hence the three consecutive numbers are
15, 16 and 17
Question 3.
Let’s find a number such that one-fourth of the number is 1 less than one-third of the number.
Answer:
Let a number be x
Now according to the given criteria,
one-fourth of the number = 
…..(i)
1 less than one-third of the number = 
…….(ii)
So from question equation (i) and equation (ii) are equal,
Hence
By cross multiplying we get
⇒ 3x=4(x-3)
⇒ 3x=4x-12
⇒ 12=4x-3x
⇒ x=12
Hence the number is 12
Question 4.
Let’s find a fraction whose denominator is 2 more than the numerator and when 3 is added to the numerator and 3 subtracted from the denominator, the fraction becomes equal to 
.
Answer:
Let the numerator of a fraction be x
Then given the denominator of the fraction is 2 more than the numerator, so the denominator = x+2
So the fraction is 
…..(i)
Now according to the given criteria,
when 3 is added to the numerator and 3 subtracted from the denominator = 
…..(ii)
So from question equation (ii) is equal to 
,
Hence
By cross multiplying we get
⇒ 3(x+3)=7(x-1)
⇒ 3x+9=7x-7
⇒ 9+7=7x-3x
⇒ 4x=16
From equation (i), the fraction is
Hence the fraction is 
Question 5.
Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 with the denominator she will get two fraction whose product is 
. Let’s write the fraction written by Sucheta.
Answer:
Let the numerator of a fraction written by Sucheta be x
Then given the denominator of the fraction is 3 more than the numerator, so the denominator = x+3
So the fraction is 
…..(i)
Now according to the given criteria,
she added 2 to the numerator and subtracted 1 from the denominator = 
…..(ii)
she subtracted 1 from the numerator and added 2 with the denominator = 
……….(iii)
So Sucheta got equation (ii) and equation (iii) as two fractions.
So from question product of equation (ii) and equation (iii) is equal to 
,
Hence
By cross multiplying we get
⇒ 5(x-1)=2(x+5)
⇒ 5x-5=2x+10
⇒ 5x-2x=10+5
⇒ 3x=15
From equation (i), the fraction is
Hence the fraction written by Sucheta is 
Question 6.
Raju wrote a number having 2 digit where the digit in the ten’s position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number. Let’s find the number writer by Raju.
Answer:
Let the digit in the unit’s place be x
Then given the digit in the ten’s place is thrice the digit in the unit position = 3x
So the 2 digit number is 10(3x)+x………(i)
By reversing the position of the digits we get the new number as
10(x)+(3x)……..(ii)
Now according to the given criteria,
36 less than the initial number = 10(3x)+x-36=31x-36…..(iii)
So from question equation (ii) and equation (iii) are equal,
Hence
10(x)+(3x)= 31x-36
⇒ 13x=31x-36
⇒ 36=31x-13x
⇒ 18x=36
From equation (i), the 2 digit number is
10(3x)+x=10(3×2)+2=10(6)+2=62
Hence the 2 digit number written by Raju is 62
Question 7.
If the sum of two numbers is 89 and their difference is 15, let’s find the two numbers.
Answer:
Let two numbers be x and y,
Then according to the given criteria
x+y=89
Hence the y=89-x is another number
Given the difference of both the numbers is 15
Hence
x-(89-x)=15
⇒ x-89+x=15
⇒ 2x=89+15
⇒ 2x=104
And the other number is y=89-x=89-52=37
Hence the two numbers are 52 and 37
Question 8.
Divide 830 into two parts in such a way, that 30% if one part will be 4 more than 40% of the other.
Answer:
Let the two parts be x and y such that
x+y=830…..(i)
So y=830-x
So the two parts will be x and (830-x)
Now according to the given criteria,
30% of x=40% of (830-x) +4
⇒ 3x=4(830-x)+40
⇒ 3x=3320-4x+40
⇒ 3x+4x=3360
⇒ 7x=3360
So the other part is 830-x=830-480=350
Hence the two parts are 480 and 350.
Question 9.
Divide 56 into parts, so that thrice of the first part becomes 48 more than one third of the second part.
Answer:
Let the two parts be x and y such that
x+y=56…..(i)
So y=56-x
So the two parts will be x and (56-x)
Now according to the given criteria,
3x=
of (56-x) +48
By cross multiplying we get
⇒ 9x=(56-x)+48×3
⇒ 9x=56-x+144
⇒ 9x+x=144+56
⇒ 10x=200
So the other part is 56-x=56-20=36
Hence the two parts are 20 and 36.
Question 10.
part of a pole lies in mud,
part lies in water and the remaining 5 meters lies above the water. Let’s find the length of the pole and write it.
Answer:
Let the length of pole be x,
part of a pole is
.
part of pole is 
Now remaining part is,
According to question,
Solve it,
x = 25
Question 11.
At present, my father’s age in 7 times of my age. After 10 years my father’s age is 3 times of my age. Let’s calculate and write the present age of my father and me.
Answer:
Let my present age be x.
Then my father’s present age is 7x.
After 10 years
My age will be (x+10)
And my father’s age will be (7x+10)
According to the given criteria,
7x+10=3(x+10)
⇒ 7x+10=3x+30
⇒ 7x-3x=30-10
⇒ 4x=20
So my present age is 5 years old
And my father’s present age is 7x=7×5=35 years old
Question 12.
My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes. If he received 137 notes in total then find the number of 5 rupee notes that he got.
Answer:
Let the number of 5 rupee notes got by uncle be x.
As it is given he gets total of 137 notes, so number of 10 rupee notes he get will be
137-x
It is given adding these notes the total amount should be 1000.
Hence the equation becomes,
5(x)+10(137-x)=1000
⇒ 5x+1370-10x=1000
⇒ 1370-5x=1000
⇒ 1370-1000=5x
⇒ 5x=370
So uncle got 74,5-rupee notes,
And 137-x=137-74=63,10-rupee notes.
Question 13.
Salem uncle of our village used 
of this savings to buy a house after his retirement from a government job. One day after falling in trouble, he sold his house and got 5% more than his cost price. If he would have taken Rs. 3450 rupees more then he would get 8% more than his cost price. Let’s find the amount of money Salem uncle used to buy his house, and his entire savings.
Answer:
Let salem uncle savings is x.
� part of savings is 
which he used to buy the house.
After selling it he got 5% more than the cost price i.e 
8% more than cost price is e 
According to the question,
3x = 3450 × 200
3x = 690000
X = Rs 230000
So, the cost of house is 
= Rs 115000
Question 14.
There was a food provision for twenty days in a refugee camp of the village Goplapur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days. Let’s write the number of refugees that were present initially.
Answer:
Let number of refugees be x.
For 20 days food for x refugees will be 20 x.
When 7 days, food available is (20-7)x.
Now 100 refugees were added so the refugees will be x + 100.
Now total refugees consume food in 11 days i.e total food consumption will be 11(x+100).
As the quantity of food will be same,
So,
(20-7) x = 11(x+100)
13 x = 11 x + 1100
13 x – 11 x = 1100
2x = 1100
x =550
Hence there were 550 regugees.
Question 15.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
3(x+2)=5(x+3)
⇒ 3x+6=5x+15
⇒ 3x-5x=15-6
⇒ -2x=9
This is the required root of the given equation.
Question 16.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
3(5x-15) = 5(3x+4)
15x – 45 = 15x + 20
15x – 15x = 20+45
0 = 65
Which is not possible.
Hence there are no roots.
Question 17.
Let’s find the roots of the equations (i.e. let’s solve the equations).
14 (x-2) + 3 (x+5) = 3(x+8) + 5
Answer:
14 (x-2) + 3 (x+5) = 3(x+8) + 5
⇒ 14x-28+3x+15=3x+24+5
⇒ 17x-13=3x+29
⇒ 17x-3x=29+13
⇒ 14x=42
This is the required root of the given equation
Question 18.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
3(x+10)=2(x+21)
⇒ 3x+30=2x+42
⇒ 3x-2x=42-30
⇒ x=12
This is the required root of the given equation.
Question 19.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
4(13x-11)=40(x+1)
⇒ 52x-44=40x+40
⇒ 52x-40x=40+44
⇒ 12x=84
This is the required root of the given equation.
Question 20.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
5(x+13)=4(2x+14)
⇒ 5x+65=8x+56
⇒ 65-56=8x-5x
⇒ 3x=9
This is the required root of the given equation.
Question 21.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
2(8x+1)=3x+80
⇒ 16x+2=3x+80
⇒ 16x-3x=80-2
⇒ 13x=78
This is the required root of the given equation.
Question 22.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
4(9-x)=15(x-9)
⇒ 36-4x=15x-135
⇒ 135+36=15x+4x
⇒ 19x=171
This is the required root of the given equation.
Question 23.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
13x+32=4×21
⇒ 13x+32=84
⇒ 13x=84-32
⇒ 13x=52
This is the required root of the given equation.
Question 24.
Let’s find the roots of the equations (i.e. let’s solve the equations).
25 + 3(4x - 5) + 8(x + 2) = x + 3
Answer:
25+3(4x-5)+8(x+2)=x+3
⇒ 25+12x-15+8x+16=x+3
⇒ 26+20x=x+3
⇒ 20x-x=3-26
⇒ 19x=-23
This is the required root of the given equation.
Question 25.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
Take LCM of 3, 12 and 18, we get 36 as the LCM
By cross multiplying we get
22x-92=3×36
⇒ 22x-92=108
⇒ 22x=108+92
⇒ 22x=200
This is the required root of the given equation.
Question 26.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
2(12-5t)=77
⇒ 24-10t=77
⇒ 10t=24-77
⇒ 10t=-53
This is the required root of the given equation.
Question 27.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
By cross multiplying we get
3(9x+5)=4(195-6x)
⇒ 27x+15=780-24x
⇒ 27x+24x=780-15
⇒ 51x=765
This is the required root of the given equation.
Question 28.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
LCM of 14 and 7 is 14, and LCM of 28 and 4 is 28, so the above equation becomes
By cross multiplying we get
2(25x-9)=18x+46
⇒ 50x-18=18x+46
⇒ 50x-18x=46+18
⇒ 32x=64
This is the required root of the given equation.
Question 29.
Let’s find the roots of the equations (i.e. let’s solve the equations).
Answer:
⇒ 53y-41=38y+34
⇒ 53y-38y=41+34
⇒ 15y=75
=5
This is the required root of the given equation.
Question 30.
Let’s find the roots of the equations (i.e. let’s solve the equations).
5x - (4x-7) (3x-5) = 6-3 (4x-9) (x-1)
Answer:
5x - (4x-7) (3x-5) = 6-3 (4x-9) (x-1)
⇒ 5x-(4x(3x-5)-7(3x-5))=6-3(4x(x-1)-9(x-1))
⇒ 5x-(12x2-20x-21x+35)=6-3(4x2-4x-9x+9)
⇒ 5x-(12x2-41x+35)=6-3(4x2-13x+9)
⇒ 5x-12x2+41x-35=6-12x2+39x-27
⇒ 46x-39x=35+6-27
⇒ 7x=14
=2
This is the required root of the given equation.
Question 31.
Let’s find the roots of the equations (i.e. let’s solve the equations).
3(x-4)2+ 5(x-3)2 = (2x-5) (4x-1)-40
Answer:
3(x-4)2+ 5(x-3)2 = (2x-5) (4x-1)-40
⇒ 3(x2-8x+16)+5(x2-6x+9)=2x(4x-1)-5(4x-1)-40
⇒ 3x2-24x+48+5x2-30x+45=8x2-2x-20x+5-40
⇒ 8x2-54x+93=8x2-22x-35
⇒ 93+35=54x-22x
⇒ 32x=128
=4
This is the required root of the given equation.
Question 32.
Let’s find the roots of the equations (i.e. let’s solve the equations).
3(y-5)2 + 5y = (2y-3)2 +1
Answer:
3(y-5)2 + 5y = (2y-3)2 –(y+1)2+1
⇒ 3(y2 + 25 – 10y)+ 5y = (4y2 + 9 – 12y)- (y2+1+2y)+1
⇒3 y2 + 75 – 30 y + 5y= 4y2 + 9 – 12y - y2 – 1-2y+1
⇒3 y2 + 75 – 25 y = 3y2 + 9 – 14y
⇒3 y2 - 3y2 +75 – 9 = -14 y + 25y
⇒66 = 11y
Y = 6
Question 33.
Let’s write mathematical stories and from equations.
x = 5 →
Answer:
x = 5
When 3 is added to a number we get 8.
So,
Lets the number be x,
According to the condition:
x + 3= 8
x = 5
Question 34.
Let’s write mathematical stories and from equations.
y=-11 →
Answer:
y=-11
When a number is multiplied by 2 we get -22.
So,
Let a number be y
Then,
2y = -22
y = -11
Question 35.
Let’s write mathematical stories and from equations.
Answer:
When 2 is subtracted from a number we get 
So,
Let the number be t,
Then,
Question 36.
Let’s write mathematical stories and from equations.
x=24 →
Answer:
x=24
When 2 number is multiplied by a number we get 48.
Let a number be x,
So,
2x = 48
x = 24
Question 37.
Let’s write mathematical stories and from equations.
x= →
Answer:
Let ? be 16,
When a number is multiplied by 4 we get 64.
So,
4x= 64
x = 16