Question 13 Marks
Solve the following equations by trial and error method: $4x = 28$
Answer$4x = 28$ Here, $L.H.S. = 4x$ and $R.H.S. = 28$
|
x
|
L.H.S.
|
R.H.S.
|
Is L.H.S. = R.H.S.
|
|
$1$
|
$4 \times 1 = 4$
|
$28$
|
No
|
|
$2$
|
$4 \times 2 = 8$
|
$28$
|
No
|
|
$3$
|
$4 \times 3 = 12$
|
$28$
|
No
|
|
$4$
|
$4 \times 4 = 16$
|
$28$
|
No
|
|
$5$
|
$4 \times 5 = 20$
|
$28$
|
No
|
|
$6$
|
$4 \times 6 = 24$
|
$28$
|
No
|
|
$7$
|
$4 \times 7 = 28$
|
$28$
|
Yes
|
Therefore, if $x = 7, L.H.S. = R.H.S.$
Hence, $x = 7$ is the solution to this equation. View full question & answer→Question 23 Marks
$\frac{\text{x}}{4}=\frac78$
Answer$\frac{\text{x}}{4}=\frac78$
Multiplying both sides by $4$,
we get $\frac{\text{x}}{4}\times4=\frac{7}{8}\times4$
$\text{x}=\frac72$
Verification: Substituting $\text{x}=\frac72$ in $L.H.S$.,
we get $\text{L.H.S.}=\frac{\frac72}{4}=\frac78,$ and $\text{R.H.S.}=\frac78$
$\text{L.H.S.}=\text{R.H.S.}$ Hence, verified.
View full question & answer→Question 33 Marks
One day, during their vacation at a beach resort, Shella found twice as many sea shells as Anita and Anita found $5$ shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?
AnswerLet the number of sea shells found by Sandy $= 'x'$.
So, the number of sea shells found by Anita $= (x + 5)$.
The number of sea shells found by Shella $= 2 (x + 5)$.
According to the question:
$\Rightarrow x + 2 (x + 5) = 16$
$\Rightarrow x + 2x + 10 = 16$
$ \Rightarrow 3x + 10 = 16$
Subtracting $10$ from both sides, we get
$\Rightarrow 3x + 10 - 10 = 16 - 10 $
$\Rightarrow 3x = 6$
Dividing both sides by $3$, we get
$\Rightarrow\frac{\text{3x}}{3}=\frac{6}{3}$
$\Rightarrow\text{x}=2$
Thus, the number of sea shells found by Sandy $= x = 2,$
the number of sea shells found by Anita $= x + 5 = 2 + 5 = 7,$
and the number of sea shells found by Shella $= 2(x + 5) = 2(2 + 5) = 2(7) = 14.$
View full question & answer→Question 43 Marks
The length of a rectangular field is twice its breadth. If the perimeter of the field is $228$ metres, find the dimensions of the field.
AnswerLet the breadth of the rectangle $= ‘x’$ metres.
According to the question: Length of the rectangle $= ‘2x’$ metres
Perimeter of a rectangle $= 2$ (length + breadth)
So, $2(2x + x) = 228 \Rightarrow 2(3x) = 228 \Rightarrow 6x = 228$
Dividing both sides by $6$,
we get $\Rightarrow\frac{6\text{x}}{6}=\frac{228}{6}$
$\Rightarrow\text{x}=38$
So, the breadth of the rectangle $= x = 38$ metres, and the length of the rectangle $= 2x = 2(38) = 76$ metres.
View full question & answer→Question 53 Marks
$\frac{3}{\text{x}}=0$
Answer$\frac{3}{\text{x}}=0$ Dividing both sides by $3$,
we get $\frac{3\text{x}}{3}=\frac{0}{3}$
$\text{x}=0$
Verification: Substituting $x = 0$ in $L.H.S. = 3x$,
we get $L.H.S. = 3 \times 0 = 0$ and $R.H.S. = 0$ $L.H.S. = R.H.S$. Hence, verified.
View full question & answer→Question 63 Marks
$x - 3 = 5$
Answer$x - 3 = 5$
Adding $3$ to both sides, we get $x - 3 + 3 = 5 + 3 x = 8$
Verification: Substituting $x = 8$ in $L.H.S.$,
we get $L.H.S. = x - 3$ and $R.H.S. = 5$ $L.H.S. = 8 - 3 = 5$ and $R.H.S. = 5 L.H.S. = R.H.S$.
Hence, verified.
View full question & answer→Question 73 Marks
In a hostel mess, $50\ kg$ rice are consumed everyday. If each student gets $400\ gm$ of rice per day, find the number of students who take meals in the hostel mess.
AnswerLet the number of students in the hostel be $‘x’$.
Quantity of rice consumed by each student $= 400\ gm$.
So, daily rice consumption in the hostel mess $= 400(x)$.
But, daily rice consumption $= 50\ kg = 50 \times 1000 = 50000\ gm$ [since $1\ kg = 1000\ gm$].
According to the question: $400x = 50000$ Dividing both sides by $400$,
we get $\Rightarrow\frac{400\text{x}}{400}=\frac{50000}{400}$
$\Rightarrow\text{x}=125$
Thus, $125$ students have their meals in the hostel mess.
View full question & answer→Question 83 Marks
Solve the following equations by trial and error method: $\frac{\text{x}}{4}= 12$
Answer$\frac{\text{x}}{4}=12$ Here, $\text{L.H.S.}=\frac{\text{x}}{4}$ and $R.H.S. = 12$ Since $R.H.S.$ is a natural number, $\frac{\text{x}}{4}$ must also be a natural number, so we must substitute values of $x$ that are multiples of $4$.
| X |
L.H.S. |
R.H.S. |
Is L.H.S. = R.H.S. |
| $16$ |
$\frac{16}{4}=4$ |
$12$ |
No |
| $20$ |
$\frac{20}{4}=5$ |
$12$ |
No |
| $24$ |
$\frac{24}{4}=6$ |
$12$ |
No |
| $28$ |
$\frac{28}{4}=7$ |
$12$ |
No |
| $32$ |
$\frac{32}{4}=8$ |
$12$ |
No |
| $36$ |
$\frac{36}{4}=9$ |
$12$ |
No |
| $40$ |
$\frac{40}{4}=10$ |
$12$ |
No |
| $44$ |
$\frac{44}{4}=11$ |
$12$ |
No |
| $48$ |
$\frac{48}{4}=12$ |
$12$ |
Yes |
Therefore, if $x = 48, L.H.S. = R.H.S.$ Hence, $x = 48$ is the solution to this equation. View full question & answer→Question 93 Marks
A man is $4$ times as old as his son. After $16$ years, he will be only twice as old as his son. Find the their present ages.
AnswerLet the present age of the son $= ‘x’$ years.Therefore, the present age of his father $= ‘4x’$ years.
So, after $16$ years,
Son’s age $= (x + 16)$ and father’s age $= (4x + 16)$ years
According to question:
$\Rightarrow 4x + 16 = 2(x + 16)$
$\Rightarrow 4x + 16 = 2x + 32$
Transposing $2x$ to $L.H.S.$ and $16$ to $R.H.S.$, we get
$⇒ 4x - 2x = 32 - 16$
$⇒ 2x = 16$
Dividing both sides by $2$, we get
$\frac{2\text{x}}{2}=\frac{16}{2}$
$\Rightarrow x = 8$
So, the present age of the son $= x = 8$ years, and the present age of the father $= 4x = 4(8) = 32$ years.
View full question & answer→Question 103 Marks
Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. If Andy has $75$ marbles more than Sandy. How many does each of them have?
AnswerLet the number of marbles with Pandy = $‘x’$.
So, the number of marbles with Andy $= ‘2x’$.
Thus, the number of marbles with Sandy $=\frac{\text{x}}{2}+\frac{2\text{x}}{2}=\frac{3\text{x}}{2}$
According to the question:
$2\text{x}+75=\frac{3\text{x}}{2}$
$2\text{x}-\frac{3\text{x}}{2}=-75$
$\frac{4\text{x}-3\text{x}}{2}=-75$
$\frac{\text{x}}{2}=-75$
$\text{x}=-150$
Since, no. of marbles can not be negative.
Therefore, $x = 150$
So, Pandy has $150$ marbles, Andy has $2x = 2(150) = 300$ marbles, and Sandy has $\frac{3\text{x}}{2}=225$ marbles.
View full question & answer→Question 113 Marks
$x + 9 = 13$
Answer$x + 9 = 13$
Subtracting $9$ from both sides, we get
$\Rightarrow x + 9 - 9 = 13 – 9$
$\Rightarrow x = 4$
Verification:
Substituting $x = 4$ on $L.H.S.$, we get
$L.H.S. = 4 + 9 = 13 = R.H.S.$
$L.H.S. = R.H.S.$
Hence, verified.
View full question & answer→Question 123 Marks
Solve the following equations by trial and error method: $x + 3 = 12$
Answer$x + 3 = 12$ Here, $L.H.S. = x + 3$ and $R.H.S. = 12$
|
x
|
L.H.S.
|
R.H.S.
|
Is L.H.S. = R.H.S.
|
|
$1$
|
$1 + 3 = 4$
|
$12$
|
No
|
|
$2$
|
$2 + 3 = 5$
|
$12$
|
No
|
|
$3$
|
$3 + 3 = 6$
|
$12$
|
No
|
|
$4$
|
$4 + 3 = 7$
|
$12$
|
No
|
|
$5$
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$5 + 3 = 8$
|
$12$
|
No
|
|
$6$
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$6 + 3 = 9$
|
$12$
|
No
|
|
$7$
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$7 + 3 = 10$
|
$12$
|
No
|
|
$8$
|
$8 + 3 = 11$
|
$12$
|
No
|
|
$9$
|
$9 + 3 = 12$
|
$12$
|
Yes
|
Therefore, if $x = 9, L.H.S. = R.H.S$. Hence, $x = 9$ is the solution to this equation. View full question & answer→Question 133 Marks
There are only $25$ paise coins in a purse. The value of money in the purse is $Rs. 17.50$. Find the number of coins in the purse.
AnswerLet the number of $25$-paise coins in the purse be $‘x’$.
So, the value of money in the purse $= 0.25x$. But $0.25x = 17.5$.
Dividing both sides by $0.25$,
we get $\Rightarrow\frac{0.25\text{x}}{0.25}=\frac{17.5}{0.25}$
$\Rightarrow\text{x}=70$
Thus, the number of $25$-paise coins in the purse $= 70$.
View full question & answer→Question 143 Marks
$\text{x}-\frac35=\frac75$
Answer$\text{x}-\frac35=\frac75$
Adding $\frac35$ to both sides,
we get $\Rightarrow\text{x}-\frac{3}{5}+\frac35=\frac{7}{5}+\frac35$
$\Rightarrow\text{x}=\frac{7}{5}+\frac{3}{5}$
$\Rightarrow\text{x}=\frac{10}{5}$
$\Rightarrow\text{x}=2$
Verification: Substituting $x = 2$ in $L.H.S.$,
we get $\text{L.H.S.}=2-\frac{3}{5}=10-\frac35=\frac75$ and $\text{R.H.S.}=\frac75$ $L.H.S. = R.H.S.$ Hence, verified.
View full question & answer→Question 153 Marks
A man says, "I am thinking of a number. When I divide it by $3$ and then add $5$, my answer is twice the number I thought of". Find the number.
AnswerLet the number thought of by the man be $‘x’$.
So, According to question: $\frac{\text{x}}{3}+5=2\text{x}$
Transposing $\frac{\text{x}}{3}$ to $R.H.S.$, we get $5=\text{2x}-\frac{\text{x}}{3}$
$5=6-\frac{\text{x}}{3}$ $5=\frac{5\text{x}}{3}$
Multiplying both sides by $3$, we get $5\times3=\frac{5\text{x}}{3}\times3$
$15=5\text{x}$
Dividing both sides by $5$,
we get $\frac{15}{5}=\frac{\text{5x}}{5}$
$\text{x}=3$
Thus, the number thought of by the man is $3$.
View full question & answer→Question 163 Marks
Solve the following equations by trial and error method:
$\frac{\text{x}}{2}+7=11$
Answer$\frac{\text{x}}{2}+7=11$
Here, $\text{L.H.S.} = \frac{\text{x}}{2}+7$ and $R.H.S. = 11$.
Since $R.H.S.$ is a natural number, $\frac{\text{x}}{2}$ must also be a natural number, so we must substitute values of $x$ that are multiples of $2$.
| x |
L.H.S. |
R.H.S. |
Is L.H.S. = R.H.S. |
| $2$ |
$\frac{2}{2}+7=8$ |
$11$ |
No |
| $4$ |
$\frac{4}{2}+7=9$ |
$11$ |
No |
| $6$ |
$\frac{6}{2}+7=10$ |
$11$ |
No |
| $8$ |
$\frac{8}{2}+7=11$ |
$11$ |
Yes |
Therefore, if $x = 8, L.H.S. = R.H.S$.
Hence, $x = 8$ is the solution to this equation. View full question & answer→Question 173 Marks
If a number is tripled and the result is increased by $5$, we get $50$. Find the number.
AnswerLet the required number be $‘x’$.
So, According to question:
$\Rightarrow 3x + 5 = 50$
Subtracting $5$ from both sides,
we get $\Rightarrow 3x + 5 - 5 = 50 - 5 \Rightarrow 3x = 45$
Dividing both sides by $3$,
we get $\frac{3\text{x}}3 = \frac{45}3$ $x = 15$
Thus, the required number is $15$.
View full question & answer→Question 183 Marks
The difference between two numbers is $7$. Six times the smaller plus the larger is $77$. Find the numbers.
AnswerLet the smaller number be $‘x’$.So, the larger number $= x + 7$.
According to question:
$6x + (x + 7) = 77$
$6x + x + 7 = 77$
$7x + 7 = 77$
Subtracting $7$ from both sides, we get
$7x + 7 - 7 = 77 - 7$
$7x = 70$
Dividing both sides by $7$, we get
$\frac{7\text{x}}{7}=\frac{70}{7}$
$\text{x}=10$
Thus, the smaller number $= x = 10,$ and the larger number $= x + 7 = 10 + 7 = 17.$
The two required numbers are $10$ and $17$.
View full question & answer→Question 193 Marks
Find three consecutive natural numbers such that the sum of the first and second is $15$ more than the third.
AnswerLet the first number be $‘x’$.
Hence, the second number $= x + 1$ and the third number $= x + 2$.
$\Rightarrow $ Sum of first and second numbers $= (x) + (x + 1)$.
According to question:
$(x) + (x + 1) = 15 + (x + 2)$
$\Rightarrow 2x + 1 = 17 + x$
Transposing $x$ to $L.H.S.$ and $1$ to $R.H.S.$, we get
$\Rightarrow 2x - x = 17 - 1$
$\Rightarrow x = 16$
So, first number $= x = 16$, second number $= x + 1 = 16 + 1 = 17$ and third number $= x + 2 = 16 + 2 = 18$
Thus, the required consecutive natural numbers are $16, 17$ and $18$.
View full question & answer→Question 203 Marks
A bag contains $25$ paise and $50$ paise coins whose total value is $Rs. 30.$ If the number of $25$ paise coins is four times that of $50$ paise coins, find the number of each type of coins.
AnswerLet the number of $50$ paise coins $= ‘x’.$
So, the money value contribution of $50$ paise coins $= 0.5x.$
The number of $25$ paise coins $= ‘4x’.$
The money value contribution of $25$ paise coins $= 0.25(4x) = x.$
According to the question: $0.5x + x = 30$
$ \Rightarrow 1.5x = 30$
Dividing both sides by $1.5,$
we get $\Rightarrow\frac{1.5\text{x}}{1.5}=\frac{30}{1.5}$
$\Rightarrow\text{x}=20$
Thus, the number of $50$ paise coins $= ‘x’ = 20$, and the number of $25$ paise coins $= ‘4x’ = 4 (20) = 80.$
View full question & answer→Question 213 Marks
If 5 is subtracted from three times a number, the result is $16$. Find the number.
AnswerLet the required number be $‘x’$. Then, $5$ subtracted from $3$ times $x = 3x - 5.$
$\Rightarrow 3x - 5 = 16$
Adding 5 to both sides, we get $= 3x - 5 + 5 = 16 + 5 \Rightarrow 3x = 21$
Dividing both sides by $3$, we get $\frac{3\text{x}}{3}=\frac{21}{3}$
$\text{x}=7$
Thus, the required number is $7$.
View full question & answer→Question 223 Marks
$3(x + 2) = 15$
Answer$3 (x + 2) = 15$ Dividing both sides by $3$,
we get $\frac{3(\text{x}+2)}{3}=\frac{15}{3}$ $(x + 2) = 5$
Subtracting $2$ from both sides, we get $x + 2 - 2 = 5 - 2 x = 3$
Verification: Substituting $x = 3$ in $L.H.S.,$
we get $L.H.S. = 3 (x + 2) = 3(3 + 2) = 3(5) = 15,$ and $R.H.S. = 15$ $L.H.S. = R.H.S.$ Hence, verified.
View full question & answer→Question 233 Marks
Shikha is $3$ years younger to her brother Ravish. If the sum of their ages is $37$ years, what are their present ages?
AnswerLet the present age of Shikha $= ‘x’$ years.So, the present age of Shikha’s brother Ravish $= (x + 3)$ years.
So, sum of their ages $= x + (x + 3)$
$\Rightarrow x + (x + 3) = 37$
$\Rightarrow 2x + 3 = 37$
Subtracting $3$ from both sides, we get
$\Rightarrow 2x + 3 - 3 = 37 - 3$
$\Rightarrow 2x = 34$
Dividing both sides by $2,$ we get
$\Rightarrow\frac{\text{2x}}{2}=\frac{34}{2}$
$\Rightarrow\text{x}=17$
So, the present age of Shikha $= 17$ years, and the present age of Ravish $= x + 3 = 17 + 3 = 20$ years.
View full question & answer→Question 243 Marks
Mrs. Jain is $27$ years older than her daughter Nilu. After $8$ years she will be twice as old as Nilu. Find their present ages.
AnswerLet the present age of Nilu = $‘x’$ years
Therefore, the present age of Nilu’s mother, Mrs. Jain $= (x + 27)$ years
So, after $8$ years, Nilu’s age $= (x + 8)$, and
Mrs. Jain’s age $= (x + 27 + 8) = (x + 35)$ years
$\Rightarrow x + 35 = 2(x + 8)$ Expanding the brackets,
we get $\Rightarrow x + 35 = 2x + 16$ Transposing $x$ to $R.H.S.$ and $16$ to $L.H.S.$,
we get $\Rightarrow 35 - 16 = 2x - x \Rightarrow x = 19$
So, the present age of Nilu $= x = 19$ years, and the present age of Nilu’s mother $= x + 27 = 19 + 27 = 46$ years.
View full question & answer→Question 253 Marks
$\frac{\text{x}}{2}=0$
Answer$\frac{\text{x}}{2}=0$ Multiplying both sides by $2$,
we get $\Rightarrow\frac{\text{x}}{2}\times2=0\times2$
$\Rightarrow\text{x}=0$
Verification: Substituting $x = 0$ in $L.H.S.$,
we get $L.H.S. =$ $\frac02$ $= $0 and $R.H.S. = 0 L.H.S. = 0$ and $R.H.S. = 0 L.H.S. = R.H.S$.
View full question & answer→Question 263 Marks
Solve the following equations by trial and error method: $x - 7 = 10$
Answer$x - 7 = 10$ Here, $L.H.S. = x - 7$ and $R.H.S. = 10.$
|
x
|
L.H.S.
|
R.H.S.
|
Is L.H.S. = R.H.S.
|
|
$9$
|
$9 - 7 = 2$
|
$10$
|
No
|
|
$10$
|
$10 - 7 = 3$
|
$10$
|
No
|
|
$11$
|
$11 - 7 = 4$
|
$10$
|
No
|
|
$12$
|
$12 - 7 = 5$
|
$10$
|
No
|
|
$13$
|
$13 - 7 = 6$
|
$10$
|
No
|
|
$14$
|
$14 - 7 = 7$
|
$10$
|
No
|
|
$15$
|
$15 - 7 = 8$
|
$10$
|
No
|
|
$16$
|
$16 - 7 = 9$
|
$10$
|
No
|
|
$17$
|
$17 - 7 = 10$
|
$10$
|
Yes
|
Therefore, if $x = 17, L.H.S. = R.H.S$. Hence, $x = 17$ is the solution to this equation. View full question & answer→Question 273 Marks
The difference in age between a girl and her younger sister is $4$ years. The younger sister in turn is $4$ years older than her brother. The sum of the ages of the younger sister and her brother is $16$. How old are the three children?
AnswerLet the age of the girl $= ‘x’$ years.
So, the age of her younger sister $= (x - 4)$ years.
Thus, the age of the brother $= (x - 4 - 4)$ years $= (x - 8)$ years.
According to question:
$(x - 4) + (x - 8) = 16$
$x + x - 4 - 8 = 16$
$2x - 12 = 16$
Adding $12$ to both sides, we get
$2x - 12 + 12 = 16 + 12$
$2x = 28$
Dividing both sides by $2$, we get
$\frac{2\text{x}}{2}=\frac{28}{2}$
$x = 14$
Thus, the age of the girl $= x = 14$ years,
the age of the younger sister $= x - 4 = 14 - 4 = 10$ years, and
the age of the younger brother $= x - 8 = 14 - 8 = 6$ years.
View full question & answer→Question 283 Marks
Solve the following equations by trial and error method: $\frac{15}{\text{x}}=3$
Answer$\frac{15}{\text{x}}=3$ Here, $\text{L.H.S.}=\frac{15}{\text{x}}$ and $R.H.S. = 3$.
Since $R.H.S.$ is a natural number, $\frac{15}{\text{x}}$ must also be a natural number,
so we must substitute values of $x$ that are factors of $15$.
| x |
L.H.S. |
R.H.S. |
Is L.H.S. = R.H.S. |
| $1$ |
$\frac{15}{1}=15$ |
$3$ |
No |
| $3$ |
$\frac{15}{3}=5$ |
$3$ |
No |
| $5$ |
$\frac{15}{5}=3$ |
$3$ |
Yes |
Therefore, if $x = 5, L.H.S. = R.H.S.$ Hence, $x = 5$ is the solution to this equation. View full question & answer→Question 293 Marks
Solve the following equations by trial and error method: $\frac{\text{x}}{18}=20$
Answer$\frac{\text{x}}{18}=20$ Here, $\text{L.H.S.}=\frac{\text{x}}{18}$ and $R.H.S. = 20$.
Since $R.H.S.$ is a natural number, $\frac{\text{x}}{18}$ must also be a natural number,
so we must substitute values of $x$ that are multiples of $18$.
| X |
L.H.S. |
R.H.S. |
Is L.H.S. = R.H.S. |
| $324$ |
$\frac{324}{18}=18$ |
$20$ |
No |
| $342$ |
$\frac{342}{18}=19$ |
$20$ |
No |
| $360$ |
$\frac{360}{18}=20$ |
$20$ |
Yes |
Therefore, if $x = 360, L.H.S. = R.H.S$. Hence, $x = 360$ is the solution to this equation. View full question & answer→Question 303 Marks
Solve the following equations by trial and error method: $2x + 4 = 3x$
Answer$2x + 4 = 3x$ Here, $L.H.S. = 2x + 4$ and $R.H.S. = 3x$
|
x
|
L.H.S.
|
R.H.S.
|
Is L.H.S. = R.H.S.
|
|
$1$
|
$2(1) + 4 = 6$
|
$3(1) = 3$
|
No
|
|
$2$
|
$2(2) + 4 = 8$
|
$3(2) = 6$
|
No
|
|
$3$
|
$2(3) + 4 =10$
|
$3(3) = 9$
|
No
|
|
$4$
|
$2(4) + 5 = 12$
|
$3(4) = 12$
|
Yes
|
Therefore, if $x = 4, L.H.S. = R.H.S$. Hence, $x = 4$ is the solution to this equation. View full question & answer→Question 313 Marks
$3(x + 6) = 24$
Answer$3(x + 6) = 24$
Dividing both sides by $3$, we get $\frac{3(\text{x}+6)}{3}=\frac{24}{3}$ $(x + 6) = 8$
Subtracting $6$ from both sides, we get $x + 6 - 6 = 8 - 6 x = 2$
Verification: Substituting $x = 2$ in $L.H.S.$,
we get $L.H.S. = 3(x + 6) = 3(2 + 6) = 24$, and $R.H.S. = 24$ $L.H.S. = R.H.S$. Hence, verified.
View full question & answer→Question 323 Marks
Find the number which when multiplied by $7$ is increased by $78$.
AnswerLet the required number be $‘x’.$
Thus, when multiplied by $7$, it gives $7x$, and $x$ increases by $78$.
$7x = x + 78$
Transposing $x$ to $L.H.S.$, we get
$7x - x = 78$
$6x = 78$
Dividing both sides by $6$, we get
$\frac{6\text{x}}{6}=\frac{78}{6}$
$\text{x}=13$
Thus, the required number is $13$.
View full question & answer→Question 333 Marks
Find the number which when multiplied by 7 is increased by 78.
Question 343 Marks
If 5 is subtracted from three times a number, the result is 16 . Find the number.
Question 353 Marks
Solve the following equation. Also, verify the result.
$\frac{6 x-2}{9}+\frac{3 x+5}{18}=\frac{1}{3}$
View full question & answer→Question 363 Marks
Solve the following equation. Also, verify the result.
$x-\frac{x}{4}-\frac{1}{2}=3+\frac{x}{4}$
View full question & answer→Question 373 Marks
Solve the following equation. Also, verify the result.
$3 x-2(2 x-5)=2(x+3)-8$
View full question & answer→Question 383 Marks
Solve the following equation. Also, verify the result.
$3(x-3)=5(2 x+1)$
View full question & answer→Question 393 Marks
Solve the following equation. Also, verify the result.
$\frac{3}{4}(x-1)=x-3$
View full question & answer→Question 403 Marks
Solve the following equation. Also, verify the result.
$\frac{x}{2}+\frac{3}{2}=\frac{2 x}{5}-1$
View full question & answer→Question 413 Marks
Solve the following equation. Also, verify the result.
$\frac{x}{2}=\frac{x}{3}+1$
View full question & answer→Question 423 Marks
Solve the following equation. Also, verify the result.
$2(5 x-3)-3(2 x-1)=9$
View full question & answer→Question 433 Marks
Solve the following equation. Also, verify the result.
$0.5 x+\frac{x}{3}=0.25 x+7$
View full question & answer→Question 443 Marks
Solve the following equation. Also, verify the result.
$0.6 x+\frac{4}{5}=0.28 x+1.16$
View full question & answer→Question 453 Marks
Solve the following equation. Also, verify the result.
$\frac{(5 x-1)}{3}-\frac{(2 x-2)}{3}=1$
View full question & answer→Question 463 Marks
Solve the following equation. Also, verify the result.
$m-\frac{m-1}{2}=1-\frac{m-2}{3}$
View full question & answer→Question 473 Marks
Solve the following equation. Also, verify the result.
$6 x+5=2 x+17$
View full question & answer→Question 483 Marks
Solve the following equation and check your answer:
$7+4 y=-5$
View full question & answer→Question 493 Marks
Solve the following equation and check your answer:
$10-y=6$
View full question & answer→Question 503 Marks
Solve each of the following equation and check your answer:
$x+\frac{1}{2}=\frac{7}{2}$
View full question & answer→Question 513 Marks
Solve the following equation and check your answer:
$x-\frac{1}{3}=\frac{2}{3}$
View full question & answer→Question 523 Marks
Solve the following equation and check your answer:
$\frac{x}{2}=0$
View full question & answer→Question 533 Marks
Solve the following equation and check your answer:
$3 x=0$
View full question & answer→Question 543 Marks
Solve the following equation and check your answer:
$x-\frac{3}{5}=\frac{7}{5}$
View full question & answer→Question 553 Marks
Solve the following equation and check your answer:
$5(x-2)+3(x+1)=25$
View full question & answer→Question 563 Marks
Solve the following equation and check your answer:
$\frac{x-3}{5}-2=-1$
View full question & answer→Question 573 Marks
Solve the following equation and check your answer:
$5(2-3 x)-17(2 x-5)=16$
View full question & answer→Question 583 Marks
Solve the following equation and check your answer:
$x+9=13$
View full question & answer→Question 593 Marks
Solve the following equation and check your answer:
$6(1-4 x)+7(2+5 x)=53$
View full question & answer→Question 603 Marks
Solve the following equation and check your answer:
$8(2 x-5)-6(3 x-7)=1$
View full question & answer→Question 613 Marks
Solve the following equation and check your answer:
$3(x+2)-2(x-1)=7$
View full question & answer→Question 623 Marks
Solve the following equation and check your answer:
$3(x+6)=24$
View full question & answer→Question 633 Marks
Solve the following equation and check your answer:
$\frac{1}{3}-2 x=0$
View full question & answer→Question 643 Marks
Solve the following equation and check your answer:
$\frac{x}{4}=\frac{7}{8}$
View full question & answer→Question 653 Marks
Solve the following equation and check your answer:
$3(x+2)=15$
View full question & answer→Question 663 Marks
Solve the following equation and check your answer:
$14=\frac{7 x}{10}-8$
View full question & answer→Question 673 Marks
Solve the following equation and check your answer:
$2 y-\frac{1}{2}=-\frac{1}{3}$
View full question & answer→Question 683 Marks
Solve the following equation and check your answer:
$\frac{4}{5}-x=\frac{3}{5}$
View full question & answer→Question 693 Marks
Solve the following equation and check your answer:
$x-3=5$
View full question & answer→Question 703 Marks
Solve the following equation by trial-and-error method:
$\frac{x}{18}=20$
View full question & answer→Question 713 Marks
Solve the following equation by trial-and-error method:
$\frac{15}{x}=3$
View full question & answer→Question 723 Marks
Solve the following equation by trial-and-error method:
$\frac{x}{4}=12$
View full question & answer→Question 733 Marks
Solve the following equation by trial-and-error method:
$2 x+4=3 x$
View full question & answer→Question 743 Marks
Solve the following equation by trial-and-error method:
$\frac{x}{2}+7=11$
View full question & answer→Question 753 Marks
Solve the following equation by trial-and-error method:
$4 x=28$
View full question & answer→Question 763 Marks
Solve the following equation by trial-and-error method:
$x-7=10$
View full question & answer→Question 773 Marks
Solve the following equation by trial-and-error method:
$x+3=12$
View full question & answer→