Algebra Chapter 7 Formation Of Equation And Solutions
Question 1. Choose the correct answer:
1. If ax-bx = a-b then the value of ‘x’ is
- 0
- 1
- ab
- a-b
Solution:
ax-bx = a-b
x(a-b) = a-b
x = \(\frac{a-b}{a-b}\)
x = 1
∴ The option (2) 1 is the correct answer.
Read and Learn More Class 7 Maths Solutions
2. If \(\frac{x}{a}-\frac{x}{b}=b-a\) then the value of x is.
- 0
- 1
- ab
- b-a
Solution:
⇒ \(\frac{x}{a}-\frac{x}{b}=b-a\)
⇒ \(\frac{b x-a x}{a b}=b-a\)
⇒ \(\frac{x(b-a)}{a b}=b-a\)
x = \((b-a) \times \frac{a b}{(b-a)}\)
x = ab
∴ The option (3) ab is the correct answer.
3. If the difference of 1/3 rd and 1/4 th of a number is 12 then the number is 12 then the number is
- 144
- 12
- 1
- None of these.
Solution:
Let the number be ‘x’
⇒ \(\frac{x}{3}-\frac{x}{4}=12\)
⇒ \(\frac{4 x-3 x}{12}=12\)
⇒ \(\frac{x(4-3)}{12}=12\)
x = 12×12
x = 144
∴ The option (1) 144 is the correct answer.
Wbbse Class 7 Algebra Chapter 7
4. The Sum of the present ages of the father and his son is 62 years. If the age of the father is 2 years more than three times of his son’s age then the Son’s age will be.
- 20 years
- 30 years
- 15 years
- 17 year.
Solution:
Let’s define the son’s age as – ‘x’
According to the given information.
Father’s age is 2 Years more than three times of son’s age,
So the father’s age would be 3x+2
Now, the sum of their ages is 62
x+(3x+2)=62
2+3x+2 = 62
4x+2=62
4x =62-2
x = \(\frac{60}{4}\)
x = 15
∴ The Son’s age is 15 years
Option (3) 15 years is the correct answer.
5. The root of the equation 3x-2(x+5)=7x-16 is
- 2
- 1
- 3
- 4
Solution:
3x-2(x+5)=7x-16
32-2x-10 = 7x-16
x-10 = 7x-16
16-10=7x-x
6 = 6x
x = \(\frac{6}{6}\)
x = 1
∴ The option (2) 1 is the correct answer.
Class 7 Maths Algebra Solutions WBBSE
Question 2. Write true or False:
1. The Solution of the equation \(\frac{x}{3}\)-1\(\frac{1}{2}\)= \(\frac{x}{6}\) + 4 is x =-3
Solution:
⇒ \(\frac{x}{3}-1 \frac{1}{2} x=\frac{x}{6}+4\)
⇒ \(\frac{x}{3}+\frac{3 x}{2}=\frac{x}{6}+\frac{4}{1}\)
⇒ \(\frac{2 x-9 x}{6}=\frac{x+24}{6}\)
Denominators are equal Numaninators are equalized
2x-9x = x+24
-7x = x+24
-7x-x = 24
-8x = 24
x = \(\frac{-24}{8}\)
∴ \(\frac{x}{3}-1 \frac{1}{2} x=\frac{x}{6}+4\) is -3
True
2. If \(\frac{x-1}{5}+\frac{x-2}{5}+\frac{x-3}{5}=1\) then the Value of x is 10.
Solution:
⇒ \(\frac{x-1}{5}+\frac{x-2}{5}+\frac{x-3}{5}=1\)
⇒ \(\frac{x-1+x-2+x-3}{5}=1\)
3x-6 = 5
3x = 5+6
3x = 11
x = \(\frac{11}{3}\)
∴ The value of x is 10
False.
3. If \(\frac{a x-b}{c}\) and \(\frac{b x-a}{d}\) are equal then the value of ‘x’ is 1.
Solution:
⇒ \(\frac{a x-b}{c}\) = \(\frac{b x-a}{d}\)
d(ax-b) = c(bx-a)
ada-bd = bcx-ac
Now let’s isolate the terms with ‘x’ on one side.
adx – bcx = bd – ac
x(ad-bc) = bd-ac
х = \(\frac{b d-a c}{a d-b c}\)
∴ The value of ‘x’ is 1 i.e., False
WBBSE Class 7 Maths Chapter 7 Answers
Question 3. Fill in the blanks.
1. The Specific value of an unknown number for which the two sides of the ‘equal to sign are equal is called the ___________ of the equation.
Solution: Root or solution
2. The method of finding the value of the unknown number is called ________ of the equation.
Solution: Solving
3. The root of \(\frac{2 x}{5}\) = ________
Solution: \(\frac{5 x}{2}-\frac{7}{15}\)
Question 4. If \(\frac{x}{6}\) + 3(x+1) = \(\frac{x}{2}\) – 1 then find the value of ‘x’
Solution:
⇒ \(\frac{x}{6}\) + 3(x+1) = \(\frac{x}{2}\) – 1
⇒ \(\frac{x+(3 x+3)}{6}=\frac{x-2}{2}\)
⇒ \(\frac{x+18 x+18}{6}=\frac{x-2}{2}\)
⇒ \(\frac{19 x+18}{6}=\frac{x-2}{2}\)
2(19x+18)=6(x-2)
38x+36 = 6x-12
38x-6x = -36-12
32x = -48
x = \(\frac{48}{32}\) = \(\frac{3}{2}\)
x = 1 \(\frac{1}{2}\)
∴ The value of the x = 1 \(\frac{1}{2}\)
Question 5. If \(\frac{2 x}{7}+\frac{x-1}{5}=\frac{x}{3}+\frac{x}{7}\) then find the value of ‘x’.
Solution:
⇒ \(\frac{2 x}{7}+\frac{x-1}{5}=\frac{x}{3}+\frac{x}{7}\)
⇒ \(\frac{10 x+7 x-7}{35}=\frac{7 x+3 x}{21}\)
⇒ \(\frac{17 x-7}{35}=\frac{10 x}{21}\)
21(17x-7)=35(10x)
357x – 147 = 350x
357x-350x = 147
7x = 147
х = \(\frac{147}{7}\)
∴ x = 21
WBBSE Class 7 Algebra Exercise Solutions
Question 6. Solve the equations.
1. \(\frac{1}{2}\)(x-2)+ \(\frac{1}{3}\)(x-3) = \(\frac{1}{4}\)(x-4) + \(\frac{1}{5}\)(x-5) + 23
Solution:
Given, \(\frac{1}{2}\)(x-2)+ \(\frac{1}{3}\)(x-3) = \(\frac{1}{4}\)(x-4) + \(\frac{1}{5}\)(x-5) + 23
⇒ \(\frac{1}{2} x-1+\frac{1}{3} x-1=\frac{1}{4} x-1+\frac{1}{5} x-1+23\)
⇒ \(\frac{1}{2} x+\frac{1}{3} x-2=\frac{1}{4} x+\frac{1}{5} x+23-2\)
⇒ \(\left(\frac{1}{2}+\frac{1}{3}\right) x-2=\left(\frac{1}{4}+\frac{1}{5}\right) x+21\)
⇒ \(\left(\frac{3}{6}+\frac{2}{6}\right) x-2=\left(\frac{5}{20}+\frac{4}{20}\right) x+21\)
⇒ \(\frac{5}{6} x-2 \quad=\frac{9}{20} x+21\)
multiplying every term by the least common denominator
120\(\left(\frac{5}{6} x-2\right)=120\left(\frac{9}{20} x+21\right)\)
120 \(\times \frac{5}{6} x-120 \times 2=120 \times \frac{9}{20} x+120 \times 21\)
100x – 240 = 54x + 2520 (isolate x)
100x – 54x = 2520 + 240.
46x = 2760
x = \(\frac{2760}{46}\)
x = 60
∴ The Solution to the equation is x = 60
2. 0.5x – 0.75x + 0·85x = 1·2
⇒ 0.5x – 0.75x + 0·85x = 1.2
⇒ 1.35x – 0.75x = 1.2
⇒ 0.6x = 1.2
⇒ х = \(\frac{1.2}{0.6}\)
∴ The Solution to the equation is x = 2
3. \(\frac{b}{a x}-\frac{a}{b x}=\frac{a}{b}-\frac{b}{a}\)
Solution:
⇒ \(\frac{b}{a x}-\frac{a}{b x}=\frac{a}{b}-\frac{b}{a}\)
⇒ \(\frac{b^2-a^2}{a b x}=\frac{a^2-b^2}{a b}\)
The right side of the equation by multiplying both the Numerator and the denominator by ‘x’
⇒ \(\frac{b^2-a^2}{a b x}=\frac{\left(a^2-b^2\right) x}{a b x}\)
⇒ \(b^2-a^2=a^2 x-b^2 x\)
⇒ \(b^2-a^2=\left(a^2-b^2\right) x\)
⇒ \(x =\frac{b^2-a^2}{a^2-b^2}\)
⇒ \(x =\frac{(b+a)(b-a)}{(a+b)(a-b)}\)
x = \(-\left\{\frac{b+a}{a+b}\right\}\)
x =-1
∴ The Solution of the equation is x =-1
4. \(\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+3}{4}=0\)
Solution:
⇒ \(\frac{6 x+6+4 x+8+3 x+9}{12}=0\)
6x+4x+3x+6+8+9 = 0
13x+23 = 0
13x = -23
x = \(-\frac{23}{13}\)
x = \(-1 \frac{10}{13}\)
∴ The Solution of the equation is x = -1 \(\frac{10}{13}\)
5. \(\frac{2}{15}\left(\frac{x}{3}-1\right)-\frac{5}{4}\left(2-\frac{x}{10}\right)=\frac{3}{5}\left(\frac{x}{4}-4\right)\)
Solution:
⇒ \(\left(\frac{2 x}{45}-\frac{2}{15}\right)-\left(\frac{10}{4}-\frac{5 x}{40}\right)=\frac{3 x}{20}-\frac{12}{5}\)
⇒ \(\left(\frac{2 x-6}{45}\right)-\left(\frac{100-5 x}{40}\right)=\left(\frac{3 x-48}{20}\right)\)
⇒ \(\frac{8(2 x-6)-9(100-5 x)}{360}=\frac{3 x-48}{20}\)
⇒ \(\frac{16 x-48-900+45 x}{360}=\frac{3 x-48}{20}\)
⇒ \(\frac{61 x-948}{360}=\frac{3 x-48}{20}\)
⇒ (61x-948)20 = (3x-48)360
⇒ 1220x – 18960 = 1080x-17280
⇒ 1220x-1080x = 18960-17280
⇒ 140x = 1680
x = \(\frac{1680}{40}\)
x = 12
∴ The Solution of the equation is x = 12
Class 7 Algebra Problems With Solutions
6. \(\frac{3 x+1}{5}+\frac{2 x-3}{8}=\frac{13 x+15}{16}-\frac{4 x+3}{15}\)
Soluton:
⇒ \(\frac{8(3 x+1)+5(2 x-3)}{40}=\frac{15(13 x+15)-16(4 x+3)}{240}\)
⇒ \(\frac{24 x+8+10 x-15}{40}=\frac{195 x+225-64 x-48}{240}\)
⇒ \(\frac{34 x-7}{11}=\frac{131 x+177}{240}\)
⇒ 240(34x-7) = 40(131x+177)
⇒ 8160x -1680 = 5240x + 7080
⇒ 8160x-5240x = 1680+7080
⇒ 2920x = 8760
⇒ x = \(\frac{8760}{2920}\)
x = 3
∴ The Solution of the equation is x = 3
7. 54-8(5+x) = (5-3x)-13(5x+27)
Solution:
54-8(5+x) = (5-3x)-13(5x+27)
⇒ 54-40-8x = 5-3x-65x-351
⇒ 14-8x = -68x-346
⇒ 14+346 = −68x+8x
⇒ 360 = -60x
⇒ x = \(\frac{360}{60}\)
x = -6
∴ The Solution of the equation is x = -6
8. \(\frac{2 x}{3}-\frac{x}{4}=\frac{x}{5}-\frac{x}{6}+23\)
Solution:
⇒ \(\frac{8 x-3 x}{12}=\frac{6 x-5 x+23 \times 30}{30}\)
⇒ \(\frac{5 x}{12}=\frac{x+690}{30}\)
150x = 12x + 690×12
150x – 12x = 8280
138x = 8280
x = \(\frac{8280}{138}\)
x = 60
∴ The Solution of the equation is x = 60
Question 7. The measurements of angles of a quadrilateral are (x-5)°, (2x-3)°; (3x+10)° and (42+8)°. Find the measurement of the greatest angle.
Solution:
Sum of all angles in a quadrilateral which is 360° (x-5)° + (2x-3)° + (3x+10)° + (4x+8)° = 360
x-5+2x-3+3x+10+4x+8 = 360
10x+10 = 360
10x = 350
x = \(\frac{350}{10}\)
x = 35
First Angle: (x-5)°= (35-5)= 30°
Second Angle: (2x-3)° (2×35-3)
⇒ 70-3
⇒ 67°
Third Angle; (3x+10)° = (3×35+10) ⇒ 105+10=115°
Fourth Angle; 4x+8= (4×35+8)
⇒ 140 +8 = 148°
∴ The Greatest Angle = 148°
Class 7 Maths Chapter 7 Solved Exercises
Question 8. If the sum of three consecutive even numbers is 60 then find the numbers
Solution:
The sum of three consecutive even numbers is 60
∴ n + (n+2) + (n+4)=60
n+n+2+n+4 = 60
30+6 = 60
3n = 60-6
3n = 54
n = \(\frac{54}{3}\)
n = 18
∴ n+2 = 18+2
= 20
∴ n+4 = 18+4
= 22
∴ The three consecutive even numbers are 18, 20, 22.
Question 9. If a train maintains an average speed of 48km/hr it arrives at its destination Punctually, if however, the average speed is 36 km/hr it arrives 10 minutes late. Find the length of the Journey.
Solution:
Speed (S1) = 48km/hr, t1 = T
Speed (S2) = 36km/hr, t2 = T+\(\frac{10}{6}\)
we know that Speed x time
Distance = Speed x time
when the distances are the same
48 x T = \(36 \times\left(T+\frac{10}{60}\right)\)
48T = \(36 \times\left(T+\frac{1}{6}\right)\)
48T = \(36 T+\frac{366}{66}\)
48T-36T = 6
12T = 6
T = \(\frac{6}{12}\)
T = \(\frac{1}{2}\)
Substitute in the above value
48 X T = Distance.
48x – \(\frac{1}{2}\)
24 = Distance
D = 24km/hr
Algebra Formulas For Class 7 WBBSE
Question 10. The sum of the present ages of a father and his Son is 80 years, 10 years ago the ratio of their ages was 5:1 Find their present ages.
Solution:
Let’s denote:
F is the present age of the father.
S is the present age of the son.
F+S = 80
10years ago the father’s age was F-10 and the son’s age was S-10.
The ratio of their ages 10 years ago was 5:1
⇒ \(\frac{F-10}{S-10}=\frac{5}{1}\)
(F-10) = 5(S-10)
F-10 = 5S -50
F = 80-S
(80-5)-10 = 5s-50
70-s = 5s-50
70+50 = 5s+s
120 = 6s
120 = 6s
s = \(\frac{120}{6}\)
s = 20
Now, F = 80-S
F = 80-20
F = 60
The father and son’s present ages are 60 and 20.
Algebra Formulas For Class 7 WBBSE
Question 11. If the difference between the five-times and three-times of number is 40. then Find the numbers.
Solution:
Given Condition:
The difference between the five-times and three-times of number is 40.
∴ Let the number be ‘x’
5x-3x = 40
2x = 40
x = \(\frac{40}{2}\)
x = 20
Substitute the value of ‘x’ in the above equation.
5x-3x = 40
5(20)-3(20) = 40
100 – 60 = 40
∴ The numbers are 100, 60.
Question 12. The denominator of a fraction exceeds its numerator by 7. If 3 is added to both its numerator and denominator the fraction becomes \(\frac{8}{15}\). Find the original fraction.
Solution:
Let’s denote the original numerator of the Fraction as’n! and the original denominator is ‘d’.
the denominator exceeds its numerator by 7 so we have:
d = n+7 → 1
n = d-7
when 3 is added to both the numerator and denominator the Fraction becomes \(\frac{8}{15}\), so we have
⇒ \(\frac{n+3}{d+3}=\frac{8}{15}\)
⇒ \(\frac{d-7+3}{d-3}=\frac{8}{15}\)
⇒ \(\frac{d-4}{d-3}=\frac{8}{15}\)
15d – 60 = 8d+24
15d-8d = 60+24
7d = 84
d = \(\frac{84}{7}\)
d = 12
Now, Finding the value of n
n = d-7 = 12-7
n = 5
∴ The original Fraction is \(\frac{5}{12}\).
Question 13. The length of a rectangle is 1 \(\frac{1}{2}\) times its breadth. If the perimeter of the rectangle is 200 meters, then find its area.
Solution:
Length of a rectangle(l) = 1 \(\frac{1}{2}\) x b
⇒ \(\frac{3}{2}\) b
⇒ 2l = 3b
perimeter of rectangle 200m
P = 2(l+b)
= 2l+2b
= 3b+2b
200 = 5b
b = \(\frac{200}{5}\)
b = 40
l = \(\frac{3}{2}\) x b
= \(\frac{3}{2}\) x 40
l = 60
∴ Area of the rectangle (A) = l x b
= 60×40
= 2400 Sq.m
Class 7 Maths Algebra Solutions WBBSE
Question 14. Gopal is at present 12 years older than his younger brother 7 years hence his age will be double that of his brother Find the present age of Gopal.
Solution:
Let’s denote:
G as the present age of Gopal.
B is the present age of his younger brother.
According to the problem.
G = B+12….(1)
G+7 = 2(B+7) …..(2)
Substitute 1 in 2
B+12+7 = 2(B+7)
B+19 = 2B+14
19-14 = 2B-B
B = 5
Substitute the ‘B’ value in 1
G = 5+12
G = 17
The present age of the Gopal is 17