First Degree Equation in One Variable — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step worked answers to every textbook question of ASSEB (Assam State School Education Board) Class 8 General Mathematics Chapter 2, First Degree Equation in One Variable — the “Try Yourself” box, Exercise 2.1, Exercise 2.2 and the Multiple Choice Questions.
Summary
The highest power of the variable in an algebraic expression is called its degree. When an expression contains only one variable whose highest power is 1, it is a first degree expression in one variable, e.g. $x$, $y+3$, $2m-\frac{11}{6}$, $ax+b$. An algebraic representation of a problem is an equation, and $2x=12$, $3x=x+5$, $2y-7=8y$ are examples of first degree equations in one variable.
The value of the variable for which the Left Hand Side (LHS) and the Right Hand Side (RHS) of an equation become equal is the solution or root of the equation. To solve an equation, the terms containing the variable are kept on the LHS and the remaining terms on the RHS, by adding, subtracting, multiplying, or dividing both sides by the same (non-zero) number.
When a term is moved from one side of the ‘=’ sign to the other, its sign changes — ‘+’ becomes ‘−’ and ‘−’ becomes ‘+’. This is transposition. An equation of the form $\frac{A}{B}=\frac{C}{D}$ can be solved directly by writing $A\times D = B\times C$; this is the cross-multiplication method. Many everyday problems can be solved by forming first degree equations in one variable.
Summary: This ASSEB Class 8 General Mathematics Chapter 2 solution covers First Degree Equation in One Variable — identifying linear equations in one variable, finding the solution or root, solving by transposition and by the cross-multiplication method, verifying results, and forming equations to solve practical word problems. Every question of the Try Yourself box, Exercise 2.1, Exercise 2.2 and the Multiple Choice Questions is solved with full step-by-step working.
Textbook Questions and Answers
Try Yourself
Select the first degree equations in one variable from the following ($a, b, c, p, q$ are constants): (i) $5x+3y-7=9$ (ii) $5m-8=0$ (iii) $5=3l$ (iv) $x^2-9y+11=9$ (v) $ax^2+bx+c=0$ (vi) $px+q=10$ (vii) $a^2x+b=0$ (viii) $ax^2+b=0$ (ix) $y=0$ (x) $z=p^3$ (xi) $3y+8=3y-2$ (xii) $5z=-z+6$
Answer: A first degree equation in one variable must have exactly one variable whose highest power is 1.
The first degree equations in one variable are — (ii) $5m-8=0$, (iii) $5=3l$, (vi) $px+q=10$, (vii) $a^2x+b=0$, (ix) $y=0$, (x) $z=p^3$, (xi) $3y+8=3y-2$, (xii) $5z=-z+6$.
Not selected — (i) and (iv) contain two variables; (v) and (viii) have a variable of degree 2. Note that (xi) $3y+8=3y-2$ simplifies to $8=-2$, so although it is of first degree in one variable, it has no solution.
Exercise 2.1
1. Solve the following equations:
(i) $4x+5=21$
Answer: $4x=21-5=16$, so $x=\frac{16}{4}=4$.
(ii) $17y-3=48$
Answer: $17y=48+3=51$, so $y=\frac{51}{17}=3$.
(iii) $-8+2x=-4$
Answer: $2x=-4+8=4$, so $x=\frac{4}{2}=2$.
(iv) $\frac{6x}{7}=42$
Answer: $6x=42\times 7=294$, so $x=\frac{294}{6}=49$.
(v) $\frac{6y}{11}=\frac{54}{99}$
Answer: $\frac{54}{99}=\frac{6}{11}$, so $\frac{6y}{11}=\frac{6}{11}\Rightarrow 6y=6\Rightarrow y=1$.
(vi) $3x=180+6x$
Answer: $3x-6x=180\Rightarrow -3x=180$, so $x=\frac{180}{-3}=-60$.
(vii) $2x+3=x+4$
Answer: $2x-x=4-3\Rightarrow x=1$.
(viii) $2-5x=3x-9$
Answer: $-5x-3x=-9-2\Rightarrow -8x=-11$, so $x=\frac{11}{8}$.
(ix) $5(p-3)=3(p+2)$
Answer: $5p-15=3p+6\Rightarrow 2p=21$, so $p=\frac{21}{2}$.
(x) $\frac{3}{4y}=-9$
Answer: $3=-9\times 4y=-36y$, so $y=\frac{3}{-36}=-\frac{1}{12}$.
(xi) $\frac{4x}{5}+1=\frac{7}{15}$
Answer: $\frac{4x}{5}=\frac{7}{15}-1=\frac{7-15}{15}=-\frac{8}{15}$, so $4x=-\frac{8}{15}\times 5=-\frac{8}{3}$ and $x=-\frac{8}{3\times 4}=-\frac{2}{3}$.
(xii) $\frac{17x}{3}-\frac{16}{9}=2$
Answer: $\frac{17x}{3}=2+\frac{16}{9}=\frac{34}{9}$, so $17x=\frac{34}{9}\times 3=\frac{34}{3}$ and $x=\frac{34}{3\times 17}=\frac{2}{3}$.
2. In each equation below, some values of the variable are given. Determine which value is the solution of that equation.
(i) $2x-4=0$; $x=1, 2, -2$
Answer: $2x=4\Rightarrow x=2$. So $x=2$ is the solution.
(ii) $11y+5=-6$; $y=0, 1, -1$
Answer: $11y=-11\Rightarrow y=-1$. So $y=-1$ is the solution.
(iii) $\frac{3y}{5}=3$; $y=3, -3, 5$
Answer: $3y=15\Rightarrow y=5$. So $y=5$ is the solution.
(iv) $x+5=7-x$; $x=1, -1, 2$
Answer: $2x=2\Rightarrow x=1$. So $x=1$ is the solution.
(v) $2x+\frac{1}{3}=1$; $x=-\frac{1}{2}, \frac{1}{2}, \frac{1}{3}$
Answer: $2x=1-\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{3}$. So $x=\frac{1}{3}$ is the solution.
(vi) $10p-4=4(2p+1)$; $p=2, 4, -4$
Answer: $10p-4=8p+4\Rightarrow 2p=8\Rightarrow p=4$. So $p=4$ is the solution.
3. Solve the following equations and verify the result:
(i) $\frac{x}{3}-\frac{x-1}{2}=1$
Answer: Multiplying both sides by 6 (LCM of 3 and 2): $2x-3(x-1)=6\Rightarrow 2x-3x+3=6\Rightarrow -x=3\Rightarrow x=-3$. Verify: LHS $=\frac{-3}{3}-\frac{-3-1}{2}=-1+2=1=$ RHS. ✓
(ii) $\frac{n}{6}-\frac{2}{3}=\frac{n}{3}+\frac{5}{6}$
Answer: Multiplying both sides by 6: $n-4=2n+5\Rightarrow -n=9\Rightarrow n=-9$. Verify: LHS $=\frac{-9}{6}-\frac{2}{3}=-\frac{13}{6}$; RHS $=\frac{-9}{3}+\frac{5}{6}=-\frac{13}{6}$. ✓
(iii) $2x+7-\frac{6x}{5}=10-\frac{5x}{2}$
Answer: Multiplying both sides by 10: $20x+70-12x=100-25x\Rightarrow 8x+70=100-25x\Rightarrow 33x=30\Rightarrow x=\frac{10}{11}$. Verify: both sides equal $\frac{85}{11}$. ✓
(iv) $\frac{2y}{5}-\frac{3}{2}=\frac{y}{2}+1$
Answer: Multiplying both sides by 10: $4y-15=5y+10\Rightarrow -y=25\Rightarrow y=-25$. Verify: both sides equal $-\frac{23}{2}$. ✓
(v) $\frac{x}{7}+\frac{x-4}{3}=2$
Answer: Multiplying both sides by 21: $3x+7(x-4)=42\Rightarrow 10x-28=42\Rightarrow 10x=70\Rightarrow x=7$. Verify: $\frac{7}{7}+\frac{7-4}{3}=1+1=2$. ✓
(vi) $\frac{2x+(3x+1)+(4x+2)}{3}=13$
Answer: $2x+3x+1+4x+2=39\Rightarrow 9x+3=39\Rightarrow 9x=36\Rightarrow x=4$. Verify: $\frac{8+13+18}{3}=\frac{39}{3}=13$. ✓
(vii) $\frac{x-3}{2}-\frac{x-1}{5}=\frac{2x-3}{5}$
Answer: Multiplying both sides by 10: $5(x-3)-2(x-1)=2(2x-3)\Rightarrow 3x-13=4x-6\Rightarrow -x=7\Rightarrow x=-7$. Verify: both sides equal $-\frac{17}{5}$. ✓
(viii) $0.25(5x-4)=0.05(10x-5)$
Answer: $1.25x-1=0.5x-0.25\Rightarrow 0.75x=0.75\Rightarrow x=1$. Verify: LHS $=0.25\times 1=0.25$; RHS $=0.05\times 5=0.25$. ✓
(ix) $0.5y+\frac{5y}{6}=21+0.75y$
Answer: $\frac{y}{2}+\frac{5y}{6}-\frac{3y}{4}=21$; multiplying by 12: $6y+10y-9y=252\Rightarrow 7y=252\Rightarrow y=36$. Verify: LHS $=18+30=48$; RHS $=21+27=48$. ✓
(x) $\frac{10x+7}{4x}=2$
Answer: $10x+7=8x\Rightarrow 2x=-7\Rightarrow x=-\frac{7}{2}$. Verify: $\frac{10(-\frac{7}{2})+7}{4(-\frac{7}{2})}=\frac{-28}{-14}=2$. ✓
(xi) $\frac{x-9}{x-4}=\frac{2}{3}$
Answer: Cross-multiplying: $3(x-9)=2(x-4)\Rightarrow 3x-27=2x-8\Rightarrow x=19$. Verify: $\frac{19-9}{19-4}=\frac{10}{15}=\frac{2}{3}$. ✓
(xii) $\frac{2y-3}{2y}=-\frac{1}{8}$
Answer: Cross-multiplying: $8(2y-3)=-2y\Rightarrow 16y-24=-2y\Rightarrow 18y=24\Rightarrow y=\frac{4}{3}$. Verify: $\frac{2(\frac{4}{3})-3}{2(\frac{4}{3})}=\frac{-\frac{1}{3}}{\frac{8}{3}}=-\frac{1}{8}$. ✓
(xiii) $\frac{p}{2p+6}=\frac{3}{8}$
Answer: Cross-multiplying: $8p=3(2p+6)\Rightarrow 8p=6p+18\Rightarrow 2p=18\Rightarrow p=9$. Verify: $\frac{9}{24}=\frac{3}{8}$. ✓
(xiv) $\frac{5x+2}{6x-2}=\frac{2}{3}$
Answer: Cross-multiplying: $3(5x+2)=2(6x-2)\Rightarrow 15x+6=12x-4\Rightarrow 3x=-10\Rightarrow x=-\frac{10}{3}$. Verify: both sides equal $\frac{2}{3}$. ✓
(xv) $\frac{3(2+x)-5(2x-3)}{5-3x}=9$
Answer: $3(2+x)-5(2x-3)=9(5-3x)\Rightarrow 21-7x=45-27x\Rightarrow 20x=24\Rightarrow x=\frac{6}{5}$. Verify: numerator $=\frac{63}{5}$, denominator $=\frac{7}{5}$, quotient $=9$. ✓
(xvi) $\frac{0.4b-2}{1.5b+15}=\frac{2}{3}$
Answer: Cross-multiplying: $3(0.4b-2)=2(1.5b+15)\Rightarrow 1.2b-6=3b+30\Rightarrow -1.8b=36\Rightarrow b=-20$. Verify: $\frac{0.4(-20)-2}{1.5(-20)+15}=\frac{-10}{-15}=\frac{2}{3}$. ✓
Exercise 2.2 (Word Problems)
1. Two numbers are in the ratio $5:7$. The smaller number is 12 less than the larger number. Find the numbers.
Answer: Let the numbers be $5x$ and $7x$. $5x=7x-12\Rightarrow -2x=-12\Rightarrow x=6$. So the numbers are $5\times 6=30$ and $7\times 6=42$.
2. The sum of three consecutive even numbers is 48. Find the numbers.
Answer: Let them be $x, x+2, x+4$. $x+(x+2)+(x+4)=48\Rightarrow 3x+6=48\Rightarrow x=14$. So the numbers are 14, 16, 18.
3. Divide Rs 17500 among three persons in the ratio $1:2:4$. Find the amount each person gets.
Answer: Let the shares be $x, 2x, 4x$. $x+2x+4x=17500\Rightarrow 7x=17500\Rightarrow x=2500$. So they receive Rs 2500, Rs 5000 and Rs 10000 respectively.
4. The perimeter of a rectangular playground is 280 m and its length is 2 m more than twice its breadth. Find the length and breadth.
Answer: Let breadth $=x$ m; then length $=(2x+2)$ m. Perimeter $=2(\text{length}+\text{breadth})$, so $2\{(2x+2)+x\}=280\Rightarrow 6x+4=280\Rightarrow x=46$. So breadth $=46$ m and length $=2\times 46+2=94$ m.
5. The unit’s place digit of a two-digit number is 5. The number is 5 times the sum of its digits. Find the number.
Answer: Let the ten’s digit be $x$; then the number $=10x+5$ and the sum of digits $=x+5$. $10x+5=5(x+5)\Rightarrow 10x+5=5x+25\Rightarrow 5x=20\Rightarrow x=4$. So the number is $10\times 4+5=45$.
6. In a scalene triangle, the first side is 2 cm more than the third side and the second side is 5 cm less than twice the third side. If the perimeter is 29 cm, find the lengths of the sides.
Answer: Let the third side $=x$; then first side $=(x+2)$ and second side $=(2x-5)$. $(x+2)+(2x-5)+x=29\Rightarrow 4x-3=29\Rightarrow x=8$. So the third side is 8 cm, the first side 10 cm and the second side $2\times 8-5=11$ cm.
7. Six times a number is the same as three times the number obtained by adding 12 to the number. Find the number.
Answer: Let the number $=x$. $6x=3(x+12)\Rightarrow 6x=3x+36\Rightarrow 3x=36\Rightarrow x=12$. So the number is 12.
8. The sum of three consecutive natural numbers is 45. Find the numbers.
Answer: Let them be $x, x+1, x+2$. $3x+3=45\Rightarrow x=14$. So the numbers are 14, 15, 16.
9. Three consecutive integers in ascending order, when multiplied by 2, 3 and 4 respectively, give a sum of 119. Find the numbers.
Answer: Let them be $x, x+1, x+2$. $2x+3(x+1)+4(x+2)=119\Rightarrow 9x+11=119\Rightarrow 9x=108\Rightarrow x=12$. So the numbers are 12, 13, 14.
10. After 20 years, Smita’s age will be 4 years less than 5 times her present age. What is her present age?
Answer: Let her present age $=x$ years. After 20 years, $x+20=5x-4\Rightarrow 24=4x\Rightarrow x=6$. So Smita’s present age is 6 years.
11. The present age of Raj is twice that of Rashmi. 10 years ago, Raj’s age was three times Rashmi’s age. Find their present ages.
Answer: Let Rashmi’s present age $=x$; then Raj $=2x$. Ten years ago: $2x-10=3(x-10)\Rightarrow 2x-10=3x-30\Rightarrow x=20$. So Rashmi is 20 years and Raj is 40 years.
12. Ranu took a Rs 500 note to a shop for change. The shopkeeper gave her a total of 19 notes of Rs 50 and Rs 20. How many notes of each did she get?
Answer: Let the number of Rs 50 notes $=x$; then Rs 20 notes $=(19-x)$. $50x+20(19-x)=500\Rightarrow 30x=120\Rightarrow x=4$. So there are 4 notes of Rs 50 and 15 notes of Rs 20.
13. Each drama ticket costs Rs 100 for children and Rs 250 for adults. Rs 8600 is collected from 50 persons. How many of them are children?
Answer: Let the number of children $=x$; then adults $=(50-x)$. $100x+250(50-x)=8600\Rightarrow -150x=-3900\Rightarrow x=26$. So there are 26 children.
14. $\frac{4}{5}$ of a number is 6 more than $\frac{2}{3}$ of that number. What is the number?
Answer: Let the number $=x$. $\frac{4}{5}x=\frac{2}{3}x+6\Rightarrow \frac{12x-10x}{15}=6\Rightarrow \frac{2x}{15}=6\Rightarrow x=45$. So the number is 45.
15. Find a rational number which, when multiplied by $\frac{4}{5}$ and then $\frac{2}{3}$ subtracted from the product, gives $-\frac{8}{15}$.
Answer: Let the number $=x$. $\frac{4}{5}x-\frac{2}{3}=-\frac{8}{15}\Rightarrow \frac{4}{5}x=-\frac{8}{15}+\frac{2}{3}=\frac{2}{15}\Rightarrow x=\frac{2}{15}\times\frac{5}{4}=\frac{1}{6}$. So the number is $\frac{1}{6}$.
16. Two buses 575 km apart start at the same time towards each other. One travels at 60 km/h and the other at 55 km/h. After how much time will they meet?
Answer: Let them meet after $t$ hours. Together they cover $60t+55t=575$ km. $115t=575\Rightarrow t=5$. So the buses meet after 5 hours.
17. A man spends $\frac{1}{4}$ of his total money on vegetables, $\frac{3}{5}$ on fruits and $\frac{1}{8}$ on sweets. He spends the remaining Rs 8 on bus fare. How much money did he take for shopping?
Answer: Let the total $=x$. Spent $=\frac{x}{4}+\frac{3x}{5}+\frac{x}{8}=\frac{10x+24x+5x}{40}=\frac{39x}{40}$. Remaining $x-\frac{39x}{40}=\frac{x}{40}=8\Rightarrow x=320$. So he took Rs 320.
18. A fraction has a denominator 4 more than its numerator. If 6 is added to the numerator and 6 is subtracted from the denominator, the fraction becomes $\frac{11}{3}$. Find the fraction.
Answer: Let numerator $=x$; then denominator $=(x+4)$. $\frac{x+6}{(x+4)-6}=\frac{x+6}{x-2}=\frac{11}{3}\Rightarrow 3(x+6)=11(x-2)\Rightarrow -8x=-40\Rightarrow x=5$. So numerator $=5$, denominator $=9$; the fraction is $\frac{5}{9}$.
19. The denominator of a rational number is 5 more than its numerator. If the numerator is decreased by 1 and the denominator by 3, the new number becomes $\frac{1}{4}$. Find the rational number.
Answer: Let numerator $=x$; then denominator $=(x+5)$. $\frac{x-1}{(x+5)-3}=\frac{x-1}{x+2}=\frac{1}{4}\Rightarrow 4(x-1)=x+2\Rightarrow 3x=6\Rightarrow x=2$. So numerator $=2$, denominator $=7$; the rational number is $\frac{2}{7}$.
20. Mother is 25 years older than Rohan. After 8 years, the ratio of Rohan’s age to Mother’s age will be $4:9$. Find their present ages.
Answer: Let Rohan’s present age $=x$; then Mother $=(x+25)$. After 8 years, $\frac{x+8}{x+33}=\frac{4}{9}\Rightarrow 9(x+8)=4(x+33)\Rightarrow 5x=60\Rightarrow x=12$. So Rohan is 12 years and Mother is 37 years.
21. Mondeep sells his car to Raktim at an 8% profit. Raktim spends Rs 5400 on repairs and then sells it to Nripen for Rs 113400 without any profit or loss. At what price did Mondeep buy the car?
Answer: Let Mondeep’s cost price $=x$. Selling price at 8% profit $=x+\frac{8x}{100}=\frac{108x}{100}$. Raktim’s total cost $=\frac{108x}{100}+5400=113400\Rightarrow \frac{108x}{100}=108000\Rightarrow x=108000\times\frac{100}{108}=100000$. So Mondeep bought the car for Rs 100000.
22. In a school week, $\frac{1}{5}$ of the students take part in the 100 m race and $\frac{1}{3}$ in the 200 m race. Twice the difference between the students in the 200 m and 100 m races take part in the $4\times100$ m race. The remaining 15 students only watch. How many students are there in the playground?
Answer: Let the total $=x$. 100 m: $\frac{x}{5}$; 200 m: $\frac{x}{3}$; difference $=\frac{x}{3}-\frac{x}{5}=\frac{2x}{15}$, twice of which $=\frac{4x}{15}$ (in the $4\times100$ m race). So $\frac{x}{5}+\frac{x}{3}+\frac{4x}{15}+15=x\Rightarrow \frac{4x}{5}+15=x\Rightarrow 15=\frac{x}{5}\Rightarrow x=75$. So there are 75 students in the playground.
Multiple Choice Questions (MCQ)
1. The first degree equation in one variable among the following is — (a) $\frac{2}{x}=\frac{x}{2}$ (b) $\frac{1}{x}+\frac{1}{x+1}=1$ (c) $\frac{x}{3}+\frac{x}{5}=\frac{1}{4}$ (d) $x^2+2x-5=c$
Answer: (c) $\frac{x}{3}+\frac{x}{5}=\frac{1}{4}$ — only this is a first degree equation in the single variable $x$.
2. ‘If 15 is added to a number, it becomes 40.’ The equation of this statement is — (a) $15x=40$ (b) $x-15=40$ (c) $x+15=40$ (d) $\frac{x}{15}=40$
Answer: (c) $x+15=40$.
3. ‘8 subtracted from a number gives $-15$.’ The equation of this statement is — (a) $x+8=-15$ (b) $x-8=15$ (c) $x+8=15$ (d) $x-8=-15$
Answer: (d) $x-8=-15$.
4. The root of $x\div 4=8$ is — (a) 12 (b) 32 (c) 4 (d) $-12$
Answer: (b) 32, since $x=8\times 4=32$.
5. The root of $8x-\frac{20}{7}=4x$ is — (a) $-\frac{5}{7}$ (b) $\frac{5}{7}$ (c) $\frac{10}{7}$ (d) $\frac{20}{21}$
Answer: (b) $\frac{5}{7}$. $4x=\frac{20}{7}\Rightarrow x=\frac{5}{7}$.
6. The root of $x=0$ is — (a) 0 (b) 4 (c) 2 (d) no root
Answer: (a) 0.
7. $y$ is an odd number. The immediate preceding odd number of $y$ is — (a) $y-1$ (b) $y-2$ (c) $y-3$ (d) $y-4$
Answer: (b) $y-2$.
8. For a two-digit number, the unit’s place digit is 4 and the ten’s place digit is $y$. The number is — (a) $10y-4$ (b) $10-40y$ (c) $10+40y$ (d) $10y+4$
Answer: (d) $10y+4$.
9. The root of $8x-15=9-4x$ is — (a) 1 (b) 2 (c) 3 (d) 4
Answer: (b) 2. $12x=24\Rightarrow x=2$.
10. The value of $x$ when $\frac{5x}{3}=30$ is — (a) 15 (b) 9 (c) 18 (d) 12
Answer: (c) 18. $5x=90\Rightarrow x=18$.
11. ‘$\frac{2}{3}$ of a number is 5 less than $\frac{3}{4}$ of that number.’ The equation of this statement is — (a) $\frac{2}{3}x-\frac{3}{4}x=5$ (b) $\frac{2}{3}x=\frac{3}{4}x-5$ (c) $\frac{3}{4}x-5=\frac{2}{3}x$ (d) $\frac{3}{4}x-5=-\frac{2}{3}x$
Answer: (b) $\frac{2}{3}x=\frac{3}{4}x-5$.
12. Of two complementary angles, one is $20^\circ$ greater than the other. The measure of the smaller angle is — (a) $90^\circ$ (b) $45^\circ$ (c) $55^\circ$ (d) $35^\circ$
Answer: (d) $35^\circ$. Complementary angles add to $90^\circ$; with smaller $=x$, $x+(x+20)=90\Rightarrow x=35$.
13. Of two supplementary angles, the larger is twice the smaller. The larger angle is — (a) $180^\circ$ (b) $120^\circ$ (c) $90^\circ$ (d) $60^\circ$
Answer: (b) $120^\circ$. Supplementary angles add to $180^\circ$; with smaller $=x$, $x+2x=180\Rightarrow x=60$, larger $=120^\circ$.
14. The value of $x$ when $bx=0$ is — (a) 0 (b) $b$ (c) $-b$ (d) $\frac{1}{b}$
Answer: (a) 0 (for $b\neq 0$).
15. The value of $m$ when $\frac{m}{2}=-7$ is — (a) 9 (b) $-9$ (c) $-14$ (d) 14
Answer: (c) $-14$. $m=-7\times 2=-14$.
Solve for Fun (Dandiram Dutta’s Puzzle)
“Where are you going, 100 brothers?” “We are not 100. The number of persons who have come, the same number will come next, then half of them, then half of the previous number, and together with you we will be 100.” How many have come?
Answer: Let the number who have come $=x$. $x+x+\frac{x}{2}+\frac{x}{4}+1=100\Rightarrow 2x+\frac{3x}{4}=99\Rightarrow \frac{11x}{4}=99\Rightarrow x=36$. Check: $36+36+18+9+1=100$. So 36 persons have come.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. In a first degree equation in one variable, the highest power of the variable is — (a) 0 (b) 1 (c) 2 (d) 3
Answer: (b) 1.
2. The solution of $5x=35$ is — (a) 5 (b) 7 (c) 30 (d) 40
Answer: (b) 7.
3. The root of $x+9=4$ is — (a) 5 (b) $-5$ (c) 13 (d) $-13$
Answer: (b) $-5$.
4. If $\frac{x}{4}=3$ then $x=$ — (a) 12 (b) $\frac{3}{4}$ (c) 7 (d) $\frac{4}{3}$
Answer: (a) 12.
5. Two numbers are in the ratio $2:3$ and their sum is 20; the smaller number is — (a) 6 (b) 8 (c) 12 (d) 10
Answer: (b) 8. $2x+3x=20\Rightarrow x=4$; smaller $=2\times 4=8$.
6. The solution of $2(x-1)=x+3$ is — (a) 3 (b) 5 (c) 4 (d) 2
Answer: (b) 5. $2x-2=x+3\Rightarrow x=5$.
7. By cross-multiplication, $\frac{a}{b}=\frac{c}{d}$ gives — (a) $ac=bd$ (b) $ad=bc$ (c) $ab=cd$ (d) $a+d=b+c$
Answer: (b) $ad=bc$.
8. If the sum of three consecutive natural numbers is 30, the middle number is — (a) 9 (b) 10 (c) 11 (d) 12
Answer: (b) 10. $3x+3=30\Rightarrow x=9$; middle $=x+1=10$.
Fill in the Blanks
1. The value of the variable for which the LHS and RHS of an equation become equal is called the ______ of the equation.
Answer: root (solution).
2. If the highest power of the variable is 1, the equation is called a ______ degree equation.
Answer: first.
3. The solution of $3x=21$ is $x=$ ______.
Answer: 7.
4. Moving a term from one side of the ‘=’ sign to the other by changing its sign is called ______.
Answer: transposition.
5. The method used to solve an equation of the form $\frac{A}{B}=\frac{C}{D}$ directly is called ______.
Answer: cross-multiplication.
True or False
1. $x^2+3=0$ is a first degree equation in one variable.
Answer: False (it is a second degree / quadratic equation).
2. Adding the same number to both sides of an equation keeps the equality unchanged.
Answer: True.
3. $5=3l$ is a first degree equation in one variable.
Answer: True.
4. During transposition, a ‘+’ sign remains a ‘+’ sign.
Answer: False (it becomes a ‘−’ sign).
5. A first degree equation in one variable usually has exactly one root.
Answer: True.
Short Answer Questions
1. What is an equation?
Answer: An algebraic representation of a problem is called an equation; it has expressions on both sides of the ‘=’ sign, which become equal for certain values of the variable.
2. Solve $4x-3=9$.
Answer: $4x=12\Rightarrow x=3$.
3. What is transposition?
Answer: Moving a term or expression from one side of the ‘=’ sign to the other, changing its sign, is called transposition.
4. Solve $\frac{2x}{3}=8$.
Answer: $2x=24\Rightarrow x=12$.
Key Terms
| Term | Meaning |
|---|---|
| Equation | An algebraic representation of a problem |
| Variable | A symbol whose value can change |
| Constant | A quantity with a fixed value |
| Degree | The highest power of the variable |
| First degree equation | Highest power of the variable is 1 |
| Solution / Root | Value of the variable making both sides equal |
| Transposition | Moving a term across ‘=’ with a sign change |
| Cross-multiplication | From $\frac{A}{B}=\frac{C}{D}$ we get $AD=BC$ |
| LHS / RHS | Left / Right Hand Side of the ‘=’ sign |
| Verification | Checking whether the solution is correct |