Question 12 Marks
Solve the riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
AnswerLet us assume the number is $x$
So, according to the riddle given in the question:
$(7x + 50) + 40 = 300$
$7x + 90 = 300$
$7x = 300 – 90$
$7x = 210$
$\frac{7 x}{7}=\frac{210}{7}$ [Divide both the sides by $7]$
$7x = 210$
$x = 30$
Hence, the required number is $30.$
View full question & answer→Question 22 Marks
Irfan says that he has $7$ marbles more than five times the marbles Permit has. Irfan has $37$ marbles. How many marbles does Permit have?
AnswerLet the number of marbles of Parmit be $x.$
Then according to the question, we have
$7 + 5x = 37$
$\therefore 5x = 37 – 7$
$\therefore 5x = 30$
$\therefore x = \frac{30}{2} = 6$
Therefore, Parmit has $6$ marbles.
View full question & answer→Question 32 Marks
Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
AnswerLet runs scored by Rahul $= x$
Then, runs scored by Sachin $= 2x$
According to problem, $x + 2x = 200 – 2$
$3x = 198$
$x = \frac{{198}}{3} = 66$
Hence, runs scored by Rahul $= 66$ and
Runs scored by Sachin $= 2(66) = 132$
View full question & answer→Question 42 Marks
In an isosceles triangle, the base angles are equal. The vertex angle is $40°$. what are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180°$)
AnswerLet ' $x$ ' be each base angle of the isosceles triangle
According to the question, we get
$2 x^{\circ}+40^{\circ}=180^{\circ}$
$\therefore 2 x^{\circ}=180^{\circ}-40^{\circ}$
$\therefore 2 x^{\circ}=140^{\circ}$
$\therefore x^{\circ}=\frac{140^{\circ}}{2}=70^{\circ}$
Therefore, the base angles of the triangle are $70^{\circ}$ each.
View full question & answer→Question 52 Marks
The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus $7$. The highest score is $87$. What is the lowest score?
AnswerLet the lowest marks $= x$
Highest marks $= 2x + 7$
According to the problem, $2x + 7 = 87$
$2x = 87 – 7 = 80$
$x = \frac{{80}}{2} = 40$
Hence, the lowest score $= 40.$
View full question & answer→Question 62 Marks
Solve the equation: $2q = 6$
AnswerThe given equation is
$2q = 6$
Divide both sides by $2,$
$\frac{2 q}{2}=\frac{6}{2}$
$\therefore q = 3$
It is the required solution
View full question & answer→Question 72 Marks
Solve the equation: $3s = 0$
AnswerThe given equation is
$3s = 0$
Divide both sides by $3$,
$\frac{35}{3}=-\frac{0}{3}$
$\therefore s = 0$
It is the required solution
View full question & answer→Question 82 Marks
Solve the equation: $3s + 12 = 0$
AnswerThe given equation is
$3s + 12 = 0$
Subtract $–12$ from both sides
$3s + 12 – 12 = 0 – 12$
$\therefore 3s = – 12$
Divide both sides by $3,$
$\frac{35}{3}=-\frac{12}{3}$
$\therefore s = –4$
It is the required solution
View full question & answer→Question 92 Marks
Solve the equation: $3s = –9$
AnswerThe given equation is
$3s = –9$
Divide both sides by $3$,
$\frac{35}{3}=-\frac{9}{3}$
$\therefore s = –3$
It is the required solution
View full question & answer→Question 102 Marks
Solve the equation: $\frac{3 p}{4} = 6$
AnswerThe given equation is
$\frac{3 p}{4} = 6$
Multiply both sides by $4,$
$\frac{3 p}{4} \times 4 = 6 \times 4$
$\therefore 3p = 24$
Divide both sides by $3,$
$\frac{3 p}{3}=\frac{24}{3}$
$\therefore p = 8$
It is the required solution
View full question & answer→Question 112 Marks
Solve the equation:$\frac{-p}{3}=5$
AnswerThe given equation is
$\frac{-p}{3}=5$
Multiply both sides by $(–3)$
$\left(\frac{-p}{3}\right) \times (-3) = 5 \times (-3)$
$\therefore p = –15$
It is the required solution
View full question & answer→Question 122 Marks
Solve the equation: $\frac{p}{4}=5$
AnswerThe given equation is
$\frac{p}{4} = 5$
Multiply both sides by $4,$
$\frac{p}{4} \times 4 = 5 \times 4$
$\therefore p = 20$
It is the required solution
View full question & answer→Question 132 Marks
Solve the equation: $10p + 10 = 100$
AnswerThe given equation is
$10p + 10 = 100$
Subtract 10 from both sides
$10p + 10 – 10 = 100 – 10$
$\therefore 10p = 90$
Divide both sides by $10,$
$\frac{10 p}{10}=\frac{90}{10}$
$\therefore p = 9$
It is the required solution.
View full question & answer→Question 142 Marks
Solve the equation: $2q + 6 = 12$
AnswerHere,
We have to find the value of $q$
Thus,
$2q + 6 - 6 = 12 - 6$ [Subtracting both sides by $6]$
$2q = 6$
Now,
Dividing both sides by $2$, we get,
$\frac{2 q}{2}=\frac{6}{2}$
Therefore, $q = 3$
View full question & answer→Question 152 Marks
Solve the equation: $2q + 6 = 0$
AnswerThe given equation is
$2q + 6 = 0$
Subtract 6 from both sides
$2q + 6 – 6 = 0 – 6$
$\therefore 2q = – 6$
Divide both sides by $2,$
$\frac{2 q}{2}=-\frac{6}{2}$
$\therefore q = – 3$
It is the required solution
View full question & answer→Question 162 Marks
Solve the equation: $2q – 6 = 0$
AnswerThe given equation is
$2q – 6 = 0$
Add 6 to both sides
$2q – 6 + 6 = 0 + 6$
$2q = 6$
Divide both sides by $2,$
$\frac{2 q}{2}=\frac{6}{2}$
$\therefore q = 3$
It is the required solution
View full question & answer→Question 172 Marks
Solve the equation: $10p = 100$
AnswerThe given equation is
$10p = 100$
Divide both sides by $10,$
$\frac{10 p}{10}=\frac{100}{10}$
$\therefore p = 10$
It is the required solution
View full question & answer→Question 182 Marks
Give the step you will use to separate the variable and then solve the equation: $\frac{3 p}{10} = 6$
AnswerThe given equation is
$\frac{3 p}{10} = 6$
Multiply both sides by $10$,
$\frac{3 p}{10} \times0 = 6 \times 10$
$\therefore 3p = 60$
Divide both sides by $3,$
$\frac{3 p}{3}$ = $\frac{60}{3}$
$\therefore p = 20$
It is the required solution
View full question & answer→Question 192 Marks
Give the step you will use to separate the variable and then solve the equation: $\frac{20 p}{3}= 40$
AnswerThe given equation is
$\frac{20 p}{3} = 40$
Multiply both sides by $3,$
$ \frac{20 p}{3} \times 3 = 40 \times 3$
$ \therefore 20p = 120$
Divide both sides by $20,$
$ \frac{20 p}{20}=\frac{120}{20} $
$ \therefore p = 6$
It is the required solution
View full question & answer→Question 202 Marks
Give the step you will use to separate the variable and then solve the equation:
$5m + 7 = 17$
AnswerThe given equation is
$5m + 7 = 17$
Subtract 7 from both sides,
$5m + 7 – 7 = 17 – 7$
$5m = 10$
Divide both sides by $5,$
$\frac{5 m}{5}=\frac{10}{5}$
$\therefore m = 2$
It is the required solution.
View full question & answer→Question 212 Marks
Give the step you will use to separate the variable and then solve the equation:
$3n – 2 = 46$
AnswerThe given equation is
$3n – 2 = 46$
Add $2$ to both sides
$3n – 2 + 2 = 46 + 2$
$3n = 48$
Divide both sides by
$\frac{3 n}{3}=\frac{48}{3}$
$\therefore n = 16$
It is the required solution
View full question & answer→Question 222 Marks
Give the first step you will use to separate the variable and then solve the equation: $20t = -10$
AnswerThe given equation is
$20t = -10$
Divide both sides by $20,$
$\frac{20 t}{20}=-\frac{10}{20}$
$\therefore$ t = - $\frac{1}{2}$
It is the required solution
View full question & answer→Question 232 Marks
Give the first step you will use to separate the variable and then solve the equation: $\frac{a}{5}=\frac{7}{15}$
AnswerThe given equation is
$\frac{a}{5}=\frac{7}{15}$
Multiply both sides by $5,$
$\frac{a}{5} \times 5=\frac{7}{15} \times$5
$\therefore$ a = $\frac{7}{3}$
It is the required solution
View full question & answer→Question 242 Marks
Give the first step you will use to separate the variable and then solve the equation: $\frac{z}{3}=\frac{5}{4}$
AnswerThe given equation is
$\frac{z}{3}=\frac{5}{4}$
Multiply both sides by $3,$
$3 \times\left(\frac{2}{3}\right)=3 \times\left(\frac{5}{4}\right)$
$z = \frac{15}{4}$
It is the required solution
View full question & answer→Question 252 Marks
Give the first step you will use to separate the variable and then solve the equation:
$8y = 36$
AnswerThe given equation is
$8y = 36$
Divide both sides by $8, $
$ \frac{8 y}{8}=\frac{36}{8} $
$ \therefore y = \frac{36}{8} $
$ \therefore y = 36 \div 4/8 \div 4$
$ \therefore y = \frac{9}{2}$
It is the required solution.
View full question & answer→Question 262 Marks
Give first the step you will use to separate the variable and then solve the equation $4x = 25$
AnswerHere,
We have to separate the variables.
Here,
$4x = 25$
Dividing both sides by $4$, we get,
$\frac{4 x}{4}=\frac{25}{4}$
Therefore, $x = \frac{25}{4}$
View full question & answer→Question 272 Marks
Give the first step you will use to separate the variable and then solve the equation:$\frac{p}{7}= 4$
AnswerThe given equation is
$\frac{p}{7} = 4$
Multiply both sides by $7,$
$(p \div 7) \times 7 = 4 \times 7$
$ \therefore p = 28$
It is the required solution
View full question & answer→Question 282 Marks
Give the first step you will use to separate the variable and then solve the equation: $\frac{b}{2}$ = 6
AnswerThe given equation is
$\frac{b}{2} = 6$
Multiplyboth sides by
$(b \div 2) \times 2 =6 \times 2$
$ \therefore b = 12$
It is the required solution
View full question & answer→Question 292 Marks
Give the first step you will use to separate the variable and then solve the equation:
3l = 42
AnswerThe given equation is
3l = 42
Divide both sides by 3,
$\frac{31}{3}=\frac{42}{3}$
l = 14
It is the required solution.
View full question & answer→Question 302 Marks
Give the first step you will use to separate the variable and then solve the equation : $y+ 4 = – 4$
AnswerThe given equation is
$y+ 4 = –4$
Subtracting 4 from both sides,
$y+ 4 – 4 = – 4 – 4$
$\therefore y = – 8$
It is the required solution
View full question & answer→Question 312 Marks
Give the first step you will use to separate the variable and then solve the equation :$y+ 4 = 4$
AnswerThe given equation is
$y+ 4 = 4$
Subtracting 4 from both sides,
$y + 4 – 4 = 4 – 4$
$\therefore y = 0$
It is the required solution
View full question & answer→Question 322 Marks
Give the first step that you will use to separate the variable and then solve the equation $x + 6 = 2$
AnswerHere, we have to separate the variables
Equation is, $x + 6 = 2$
Subtracting 6 on both sides of the equation, we get
$x + 6 - 6 = 2 - 6$
$x = -4$
View full question & answer→Question 332 Marks
Give the first step you will use to separate the variable and then solve the equation :$x + 1 = 0$
AnswerThe given equation is
$x + 1 = 0$
Subtracting $1$ from both sides.
$x + 1 – 1 = 0 – 1$
$\therefore x = – 1$
It is the required solution
View full question & answer→Question 342 Marks
Give the first step that you will use to separate the variable and then solve the equation $x - 1 = 0$
AnswerWe have to separate the variables
Given equation is, $x - 1 = 0$
Now, Adding $1$ on both side of the equation, we get
$x - 1 + 1 = 0 + 1$
$x = 1$
View full question & answer→Question 352 Marks
Set up an equation in the case: In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is $180$ degrees).
AnswerLet the base angle $= b$
Now, as per the question, we have,
Vertex angle $= 2 \times$ (base angle) $= 2b$
Now,
Sum of interior angles of a triangle $= 180^\circ $
We can write this as,
$b + b + 2b = 180^\circ $
$4b = 180^\circ , b = 45^\circ $
View full question & answer→Question 362 Marks
Set up an equation in the case: The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus $7$. The highest score is $87.$ (Take the lowest score to be l.)
AnswerLet 'x' be the lowest score
According to the question, we get
$2x + 7 = 87$
$ \therefore 2x = 87 -7$
$ \therefore 2x = 80$
$ \therefore x = \frac{80}{2} = 40$
Therefore, the lowest score is $40$ marks
View full question & answer→Question 372 Marks
Set up an equation in the case: Laxmi’s father is $49$ years old. He is $4$ years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
AnswerLet Laxmi's age be $'x'$ years.
Then according to the question, we have
$4 + 3x = 49$
$ \therefore 3x = 49 - 4$
$ \therefore 3x = 45$
$ \therefore x= \frac{45}{3} = 15$
Therefore, Laxmi's age is $15$ years.
View full question & answer→Question 382 Marks
Set up an equation in the case: Irfan says that he has $7$ marbles more than five times the marbles Parmit has. Irfan has $37$ marbles. (Take m to be the number of Parmit’s marbles.)
AnswerLet Parmit has m marbles
As per the question,
$5 \times$ (Number of marbles Parmit has) $+ 7 = $ number of marbles Irfan has
We can write this as,
$5m + 7 = 37$ [Number of marbles irfan has $= 37]$
$5m - 30 = 0$
Which is the required equation.
View full question & answer→Question 392 Marks
Write the equation for the statement:
If you add $3$ to one-third of $z$, you get $30.$
AnswerIf you add $3$ to one-third of $z$, you get $30.$
$3 +$ One third of $z = 30$
i,e, $3 + \frac{1}{3} z = 30$
Therefore, the required equation of the statement is, $\frac{1}{3} z+3=30$
View full question & answer→Question 402 Marks
Write the equation for the statement:
if you take away $6$ from $6$ times $y$, you get $60.$
AnswerClearly, $6$ times $y$ can be represented as: $6y$
Taking away $6$ from $6y$ means subtracting $6$ from $6y$ and we get: $6y - 6$
As per the question, the above quantity equals $60$, so we have
$6y - 6 = 60$
Thus, the required equation of the statement is, $6y - 6 = 60$
View full question & answer→Question 412 Marks
Write the equation for the statement: $2$ subtracted from y is $8.$
AnswerClearly, $2$ subtracted from y can be written as $y - 2.$
Now this difference equals $8$, therefore
$y - 2 = 8$
Thus, the required equation is $y - 2 = 8.$
View full question & answer→Question 422 Marks
Solve the equation by trial and error method: $5p + 2 = 17$
Answer$5p + 2 = 17$
| $L.H.S.$ |
Value ofP |
Value of $L.H.S.$ |
$R.H.S.$ |
| $5p + 2$ |
$0$ |
$2$ |
$17$ |
| $5p + 2$ |
$1$ |
$7$ |
$17$ |
| $5p + 2$ |
$2$ |
$12$ |
$17$ |
| $5p + 2$ |
$3$ |
$17$ |
$17$ |
So, $p = 3$ is the solution of the given equation $5p + 2 = 17$ View full question & answer→Question 432 Marks
Check whether the value given in the bracket is a solution to the given equation or not.
$4p – 3 = 13 (p = –4)$
Answer$4p – 3 = 13 (p = –4)$
$L.H.S. = 4p – 3$
$= 4 (–4) –3 . . . . $[When $p = –4]$
$= –16 – 3$
$= –19$
$R.H.S. = 13$
$\because L.H.S. \neq R.H.S.$
$\therefore P = – 4$ is not a solution to the given equation $4p – 3 = 13$
View full question & answer→Question 442 Marks
Check whether the value given in the bracket is a solution to the given equation or not.
$4p – 3 = 13 (p = 1)$
Answer$4p – 3 = 13 (p = 1)$
$L.H.S. = 4p – 3$
$= 4 (1) – 3 . . . .$ [When $p = 1]$
$= 4 – 3$
$= 1$
$R.H.S. = 13$
$\because L.H.S. \neq R.H.S.$
$\therefore p = 1$ is not a solution to the given equation. $4p – 3 = 13$
View full question & answer→Question 452 Marks
Check whether the value given in the bracket is a solution to the given equation or not.
$n + 5 = 19 (n = 1)$
Answer$n + 5 = 19 (n = 1)$
$L.H.S. = n + 5 = 1 + 5 = 6 .. . .$ [ When $n = 1]$
$R.H.S. = 19$
$\because L.H.S. \neq R.H.S.$
$\therefore n = 1$ is not a solution to the given equation $n + 5 = 19.$
View full question & answer→Question 462 Marks
The sum of three times a number and $11$ is $32$. Find the number.
AnswerLet the unknown number be $x$, then three times the number is $3x$ and the sum of $3x$ and $11$ is $32.$
i,e, $3x + 11 = 32.$
$3x = 32 – 11$ [Subtracting $11$ from both sides]
or $3x = 21$
Now, divide both sides by $3$, we get,
$x=\frac{21}{3}=7$
Therefore, the required number is $7.$
View full question & answer→Question 472 Marks
Solve: $– 2(x + 3) = 8$
AnswerGiven:$ -2(x + 3) = 8$
We divide both sides by $(-2),$ we get,
$x+3=-\frac{8}{2}$
or $x + 3 = – 4$
i.e., $x = – 4 – 3$ (Subtarcting $3$ from both sides)
or $x = –7$
Verification:
$LHS = - 2 (-7+3) = -2 (-4)$
$= 8 = RHS$
View full question & answer→Question 482 Marks
Solve: $4(m + 3) = 18$
AnswerGiven: $4(m + 3) = 18$
Divide both the sides by $4$, we get,
$m+3=\frac{18}{4}$
or $m+3=\frac{9}{2}$
or $m=\frac{9}{2}-3$ (Subtracting $3$ on both sides)
or $m=\frac{3}{2}$
Verification:
Check $LHS = 4\left[\frac{3}{2}+3\right]=4 \times \frac{3}{2}+4 \times 3=2 \times 3+4 \times 3$
$= 6 + 12 = 18 = RHS$
View full question & answer→Question 492 Marks
Solve: $2p – 1 = 23$
AnswerGiven: $2p – 1 = 23$
or, $2p – 1 + 1 = 23 + 1$ [Adding $1$ on both sides]
or $2p = 24$
Now divide both sides by $2$, we get,
$\frac{2 p}{2}=\frac{24}{2}$
or $p = 12$, which is the solution.
View full question & answer→Question 502 Marks
Solve: $3n + 7 = 25$
AnswerHere,
$3n + 7 = 25$
$3n + 7 – 7 = 25 – 7$ [Subtracting both sides by $7]$
or $3n = 18$
Now divide both sides by $3,$
$\frac{3 n}{3}=\frac{18}{3}$
or $n = 6$, which is the solution.
View full question & answer→