Question 12 Marks
There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?
AnswerLet the number of mangoes in each smaller box be x
$\therefore$ Number of mangoes in 8 such boxes $=8 x$
According to the question,
Number of mangoes in each larger box $=8 x+4$
and each larger box contains 100 mangoes (given).
So, the equation becomes $8 x+4=100$
On transposing ( +4 ) from LHS to RHS, we get
$8x=100-4$
$\Rightarrow8x=96$
On dividing both sides by 8, we get
$\frac{8x}{8}=\frac{96}{8}$
$\Rightarrow x=12$
Hence, the required number of mangoes in each smaller
box is 12.
View full question & answer→Question 22 Marks
(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?
(ii) What is that number one-third of which added to 5 gives 8?
Answer(i) Let the number be $x$.
On multiplying the number by 6 , we get $=6 x$
On subtracting 5 from the product, we get $6 x-5$
According to the question, $6 x-5=7$
Now, on transposing ( -5 ) from LHS to RHS, we get
$
6 x=7+5 \Rightarrow 6 x=12
$
Again, dividing both sides by 6 , we get
$
\frac{6 x}{6}=\frac{12}{6} \Rightarrow x=2
$
Hence, the required number is 2.
(ii) Let the number be $x$.
Then, one-third of $x=\frac{x}{3}$
On adding 5 to one-third of number, we get
$\frac{x}{3}+5 $
$\Rightarrow \frac{x}{3}+5=8$
Now, transposing ( +5 ) from LHS to RHS, we get
$
\frac{x}{3}=8-5 \Rightarrow \frac{x}{3}=3
$
Again, multiplying both sides by 3, we get
$
\frac{x}{3} \times 3 =3 \times 3$
$\Rightarrow x=9
$
Hence, the required number is 9 .
View full question & answer→Question 32 Marks
If 45 is added to half a number, then result is triple the number. Find the number.
AnswerLet $x$ be the number. So, half of $x$ is $\frac{x}{2}$.
Then, $\frac{x}{2}+45=3 x$
$\Rightarrow \frac{x+90}{2}=3 x$
$\Rightarrow x+90=6 x$
$\Rightarrow
90=5 x$
$
\Rightarrow x=\frac{90}{5}=18
$
Hence, the number is 18.
View full question & answer→Question 42 Marks
Solve the following 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 the number be x.
Now, seven times of this number $=7 x$, then add a fifty to it, we get $7 x+50$
To reach a triple century, we need forty.
Therefore, we get the equation
$7 x+50+40 =300 $
$\Rightarrow 7 x+90 =300$
On transposing 90 from LHS to RHS, we get
$7 x =300-90$
$\Rightarrow 7 x =210
$
On dividing both sides by 7, we get
$\frac{7 x}{7} =\frac{210}{7}$
$\Rightarrow x =3$
Hence, the required number is 30 .
View full question & answer→Question 52 Marks
Solve the following.
People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted, if the number of non-fruit trees planted was 77 ?
AnswerLet the number of fruit trees planted be x, then three
times of fruits trees be 3x.
$\therefore$Number of non-fruit trees = 2 + Three times the
number of fruit trees = 2 + 3x
But number of non-fruit trees was 77. Therefore,
we get the equation 2+3x =77
On transposing 2 from LHS to RHS, we get
$
\begin{aligned}
3 x =77-2 \\
\Rightarrow 3 x =75
\end{aligned}
$
On dividing both sides by 3 , we get
$
\begin{array}{rlrl}
\frac{3 x}{3} =\frac{75}{3} \\
\Rightarrow x 25
\end{array}
$
Hence, the number of fruit trees planted was 25 .
View full question & answer→Question 62 Marks
Solve the following.
Laxmi's father is 49 yr old. He is 4 yr older than three times Laxmi's age. What is Laxmi's age?
AnswerLet Laxmi 's age be $x y r$, then 3 times of Laxmi's age be $3 x$.
$\therefore$ Laxmi's father age $=3 x+4$,
but Laxmi's father is 49 yr .
Therefore, we get the equation, $3 x+4=49$
On transposing 4 from LHS to RHS, we get
$
\begin{array}{ll}
3 x=49-4 \\
\Rightarrow
3 x=45
\end{array}
$
On dividing both sides by 3 , we get
$
x=15
$
Hence, Laxmi's age is 15 yr .
View full question & answer→Question 72 Marks
Solve the following.
Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
AnswerLet Parmit has $x$ marbles, then the five times of $x$ be $5 x$.
$\therefore$ Number of marbles Irfan has $=5 x+7$
But Irfan has 37 marbles, then we get the equation
$
5 x+7=37
$
On transposing (+7) from LHS to RHS, we get $5 x$
$
=37-7 \Rightarrow 5 x=30
$
On dividing both sides by 5 , we get $\frac{5 x}{5}=\frac{30}{5}$
$
\Rightarrow
x=6
$
Hence, Parmit has 6 marbles.
View full question & answer→Question 82 Marks
Solve the following.
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 be x.
Then, runs scored by Sachin=Twice of runs scored by Rahul = 2x
$\therefore$Sum of their runs =x + 2x = 3x
Since, the sum of their runs be two short of a double century.
Therefore, we get the equation
$
3 x+2=200
$
On transposing ( +2 ) from LHS to RHS, we get
$
\begin{array}{ll}
3 x=200-2 \\
\Rightarrow 3 x=198
\end{array}
$
On dividing both sides by 3 , we get
$
\frac{3 x}{3}=\frac{198}{3} \Rightarrow x=66
$
Hence, the runs scored by Rahul is 66 and by Sachin
$
=2 \times 66=132
$
View full question & answer→Question 92 Marks
Solve the following.
In an isosceles triangle, the base angles are equal. The vertex angle is $40^{\circ}$. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180^{\circ}$ ).
AnswerLet $A B C$ be an isosceles triangle whose base angles are equal and let measure of base angle be $x$. Also, vertex angle is $40^{\circ}$.
Since, the sum of three angles of a triangle is $180^{\circ}$.
$
\begin{array}{lr}
\therefore \angle A+\angle B+\angle C=180^{\circ} \\
\Rightarrow 40^{\circ}+x+x=180^{\circ} \\
\Rightarrow 40^{\circ}+2 x=180^{\circ}
\end{array}
$
which is the required equation.
On transposing (+40) from LHS to RHS, we get
$\begin{aligned} & 2 x=180^{\circ}-40^{\circ} \\ \Rightarrow \quad & 2 x=140^{\circ}\end{aligned}$
On dividing both sides by 2, we get
$\frac{2 x}{2}=\frac{140^{\circ}}{2}=70^{\circ}$
$\Rightarrow \quad x=70^{\circ}$
Hence, the base angles of the triangle are of measure 70°.
View full question & answer→Question 102 Marks
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. What is the lowest score?
AnswerLet the lowest score in the class be $x$. Then, twice of the lowest marks be $2 x$.
$\therefore$ Highest marks obtained by a student
$=$ Twice the lowest marks $+7=2 x+7$
But the highest score is 87 .
So, we get the equation $2 x+7=87$
On transposing ( +7 ) from LHS to RHS, we get
$
\begin{aligned}
2 x & =87-7 \\
\Rightarrow 2 x & =80
\end{aligned}
$
On dividing both sides by 2 , we get
$\frac{2x}{2}=\frac{80}{2}$
$\Rightarrow$x=40
Hence, the lowest score in the class is 40.
View full question & answer→Question 112 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
Anwar thinks of a number. If he takes away 7 from $\frac{5}{2}$ of the number, then the result is 23.
AnswerLet Anwar thinks the number be $x$.
$\therefore \frac{5}{2}$ of the number $=\frac{5 x}{2}$
On subtracting 7 from $\frac{5}{2}$ of the number, we get
$
\frac{5 x}{2}-7
$
According to the question, $\frac{5}{2} x-7=23$
Now, on transposing ( -7 ) from LHS to RHS, we get
$
\frac{5}{2} x=23+7 \Rightarrow \frac{5}{2} x=30
$
On multiplying both sides by 2 , we get
$
\frac{5}{2} x \times 2=30 \times 2 \Rightarrow 5 x=60
$
Again, dividing both sides by 5, we get
$
\frac{5 x}{5}=\frac{60}{5} \Rightarrow x=12
$
Hence, the required number is 12 .
View full question & answer→Question 122 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5 , then she will get 8 .
AnswerLet Ibenhal thinks the number $=x$
On adding 19 to $x$, we get $x+19$
On dividing the sum by 5 , we get $\frac{x+19}{5}$
According to the question, $\frac{x+19}{5}=8$
Now, on multiplying both sides by 5 , we get
$
\frac{x+19}{5} \times 5=8 \times 5 \Rightarrow x+19=40
$
Again, on transposing +19 from LHS to RHS, we get
$
\begin{array}{l}
x=40-19 \\
\Rightarrow \qquad x=21
\end{array}
$
Hence, the required number is 21 .
View full question & answer→Question 132 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
AnswerLet the number of notebooks $=x$
$\therefore$ Thrice of the number $=3 x$
On subtracting thrice the number of notebooks from 50 , we get
$
50-3 x
$
According to the question,
$
50-3 x=8
$
On transposing 50 from LHS to RHS, we get
$
-3 x=8-50 \Rightarrow-3 x=-42
$
Again, dividing both sides by ( -3 ), we get
$
\frac{-3 x}{-3}=\frac{-42}{-3} \Rightarrow x=14
$
Hence, the required number of notebooks is 14.
View full question & answer→Question 142 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
When I subtracted 11 from twice a number, the result was 15.
AnswerLet the number be $x$.
$\therefore$ Twice of the number $=2 x$
Now, on subtracting 11 from twice the number, we get
$
2 x-11
$
According to the question,
$
2 x-11=15
$
On transposing ( -11 ) from LHS to RHS, we get
$
2 x=15+11 \Rightarrow 2 x=26
$
Again, dividing both sides by 2 , we get
$
\frac{2 x}{2}=\frac{26}{2} \Rightarrow x=13
$
Hence, the required number is 13.
View full question & answer→Question 152 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
If I take three-fourths of a number and add 3 to it, then I get 21.
AnswerLet the number be $x$.
$\therefore$ Three-fourth of the number $=\frac{3}{4} x$
On adding 3 to it, we get
$
\frac{3 x}{4}+3
$
According to the question,
$
\frac{3}{4} x+3=21
$
On transposing (+3) from LHS to RHS, we get
$
\frac{3}{4} x=21-3 \Rightarrow \frac{3}{4} x=18
$
On multiplying both sides by 4 , we get
$
\frac{3}{4} x \times 4=18 \times 4 \Rightarrow 3 x=72
$
Again, dividing both sides by 3 , we get
$
\frac{3 x}{3}=\frac{72}{3} \Rightarrow x=24
$
Hence, the required number is 24 .
View full question & answer→Question 162 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
One-fifth of a number minus 4 gives 3 .
AnswerLet the number be $x$.
$\therefore$ One-fifth of the number $=\frac{x}{5}$
Now, on subtracting 4 from one-fifth of a number, we get
$
\frac{x}{5}-4
$
According to the question,
$
\frac{x}{5}-4=3
$
On transposing ( -4 ) from LHS to RHS, we get
$
\frac{x}{5}=3+4 \Rightarrow \frac{x}{5}=7
$
On multiplying both sides by 5 , we get
$
\frac{x}{5} \times 5=7 \times 5 \Rightarrow x=35
$
Hence, the required number is 35 .
View full question & answer→Question 172 Marks
Set up equations and solve them to find the unknown numbers in the following cases.
Add 4 to eight times a number, you get 60 .
AnswerLet the number be $x$.
$\therefore$ Eight times of the number $=8 x$
Now, add 4 to eight times a number, we get $8 x+4$
According to the question,
$
8 x+4=60
$
Now, we find the value of $x$.
We have, $8 x+4=60$
On transposing (+4) from LHS to RHS, we get
$
8 x=60-4 \Rightarrow 8 x=56
$
On dividing both sides by 8 , we get
$
\frac{8 x}{8}=\frac{56}{8} \Rightarrow x=7
$
Hence, the required number is 7.
View full question & answer→Question 182 Marks
Solve the following equations.
$10 p+10=100$
AnswerWe have, $10 p+10=100$
On subtracting 10 from both sides, we get
$10 p+10-10 =100-10 \\$
$\Rightarrow 10 p =90$
On dividing both sides by 10 , we get
$
\frac{10 p}{10}=\frac{90}{10} \Rightarrow p=9
$
Hence, $p=9$ is the solution of the given equation.
View full question & answer→Question 192 Marks
Solve the following equations.
$10 p=100$
AnswerWe have, $10 p=100$
On dividing both sides by 10 , we get
$
\frac{10 p}{10}=\frac{100}{10} \Rightarrow p=10
$
Hence, $p=10$ is the solution of the given equation.
View full question & answer→Question 202 Marks
Set up an equation in the following cases.
In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be bin degrees. Remember that the sum of angles of a triangle is $180^*$.)
AnswerLet the base angle be $b$.
Since, the given triangle is an isosceles, so both base angles are $b$ and $b$.
Vertex angle $=2 \times$ Either base angle $=2 b$
We know that sum of all the angles of a triangle is $180^{\circ}$.
$
\begin{array}{l}
\therefore b+b+2 b=180^{\circ} \\
\Rightarrow 4 b=180^{\circ}
\end{array}
$
Hence, the required equation is $4 b=180^{\circ}$.
View full question & answer→Question 212 Marks
Set up an equation in the following cases.
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 1 )
AnswerLet the lowest marks (score) beL.
$\therefore$ Twice the lowest marks $=2 l$
According to the question,
Highest marks $=($Twice the lowest marks $+7$$)=2 l+7$
But the highest score is 87.
Hence, the required equation is $2 l+7=87$.
View full question & answer→Question 222 Marks
Set up an equation in the following cases.
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 $y yr$.
$\therefore 3$ times of Laxmi's age $=3 yyr$
According to the question,
Age of Laxmi's father $=(3$ times Laxmi's age +4$)$
$
=(3 y+4) y z
$
But Laxmi's father is 49 yr old.
Hence, the required equation is $3 y+4=49$
View full question & answer→Question 232 Marks
Set up an equation in the following cases.
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.
$\therefore$ Five times of Parmit's marbles $=5 m$
According to the question,
Number of marbles Irfan has
$=7$ more than five times $m$
$=5 m+7$
But, Irfan has 37 marbles$\qquad$[given]
Hence, the required equation is $5 m+7=37$.
View full question & answer→Question 242 Marks
Write equations for the following statements.
If you add 3 to one-third of $z$, you get 30 .
AnswerAccording to the question,
one-third of $z=\frac{1}{3} z$
Now, add 3 to one-third of $z=\frac{1}{3} z+3$
and the result $=30$
Hence, the required equation is $\frac{1}{3} z+3=30$.
View full question & answer→Question 252 Marks
Write equations for the following statements.
If you take away 6 from 6 times $y$, you get 60 .
AnswerAccording to the question,
Six times of $y=6 y$
Now, take away 6 from 6 times $y=6 y-6$
and the result $=60$
Hence, the required equation is $6 y-6=60$.
View full question & answer→Question 262 Marks
Write equations for the following statements.
One-fourth of a number $x$ minus 4 gives 4 .
AnswerAccording to the question,
One-fourth of a number $x=\frac{1}{4} \times x=\frac{x}{4}$
Now, on subtracting $4=\frac{x}{4}-4$ and the result $=4$
Hence, the required equation is $\frac{1}{4} x-4=4$.
View full question & answer→Question 272 Marks
Write equations for the following statements.
Seven times $m$ plus 7 gets you 77 .
AnswerAccording to the question,
Seven times of $m=7 m$
Now, seven times $m$ plus $7=7 m+7$
and the result $=77$
Hence, the required equation is $7 m+7=77$.
View full question & answer→Question 282 Marks
Write equations for the following statements.
Three-fourth of $t$ is 15 .
AnswerAccording to the question,
Three-fourth of $t=\frac{3}{4} t$ and three-fourth of $t=15$
Hence, the required equation is $\frac{3}{4} t=15$.
View full question & answer→Question 292 Marks
Write equations for the following statements.
The number b divided by 5 gives 6 .
AnswerAccording to the question,
Number $b$ divided by $5=\frac{b}{5}$ and the quotient $=6$
Hence, the required equation is $\frac{b}{5}=6$.
View full question & answer→Question 302 Marks
Write equations for the following statements.
Ten times $a$ is 70 .
AnswerAccording to the question,
Ten times of $a=10 a$ and ten times $a=70$
Hence, the required equation is $10 a=70$.
View full question & answer→Question 312 Marks
Write equations for the following statements.
2 subtracted from $y$ is 8 .
AnswerAccording to the question,
Subtract 2 from $y$, we get $=y-2$
and the difference $=8$
Hence, the required equation is $y-2=8$.
View full question & answer→Question 322 Marks
The age of Sohan Lal is four times that of his son Amit. If the difference of their ages is 27 yr , find the age of Amit.
AnswerLet $x$ be the age of Amit.
So, age of Sohan Lal, the father of Amit $=4 x y r$
If the difference of their ages is 27 yr .
Then, $4 x-x=27 \Rightarrow 3 x=27$
On dividing both sides by 3 , we get
$\Rightarrow
x=\frac{27}{3}=9$
Hence, age of Amit is 9 yr.
View full question & answer→Question 332 Marks
In a class of 60 students, the number of girls is one-third the number of boys. Find the number of girls and boys in the class.
AnswerAs per the given information in the question,
total number of students in the class $=60$
Let $x$ be the number of boys in the class.
So, the number of girls in the class $=\frac{x}{3}$
$\therefore
x+\frac{x}{3}=60$
$\Rightarrow
\frac{3 x+x}{3}=60$
$\Rightarrow
\frac{4 x}{3}=60$
$4 x=60 \times 3$
$\Rightarrow
4 x=180$
$\Rightarrow x=\frac{180}{4}=45$
Hence, the number of boys in the class is 45 and number of girls in the class is $\frac{45}{3}=15$.
View full question & answer→Question 342 Marks
Anamika thought of a number. She multiplied it by 2 , added 5 to the product and obtained 17 as the result. What is the number she had thought of?
AnswerLet $x$ be the number thought by Anamika.
If she multiplied it by 2 , then the number will be $2 x$.
Also, added 5 to it and obtained 17 as the result.
$
\therefore 2 x+5=17
$
Subtract 5 from both sides, we get
$2 x+5-5=17-5$
$\Rightarrow 2 x=17-5$
$
2 x=12
$
Dividing both sides by 2 , we get
$\frac{2 x}{2}=\frac{12}{2}$
$\Rightarrow x=\frac{12}{2}=6$
Hence, the number 6 is thought by Anamika.
View full question & answer→Question 352 Marks
In a family, the consumption of wheat is 4 times that of rice. The total consumption of the two cereals is 80 kg . Find the quantities of rice and wheat consumed in the family.
AnswerAs per the given information in the question,
total consumption of the two cereals $=80 kg$
Let $x$ be the consumption of rice.
So, the consumption of wheat $=4 x$
$\therefore x+4 x=80$
$\Rightarrow 5 x=80$
$\Rightarrow x=\frac{80}{5}=16 kg$
$\therefore$ Consumption of wheat $=4 x=4 \times 16=64 kg$
Hence, the consumption of rice and wheat are 16 kg and 64 kg , respectively.
View full question & answer→Question 362 Marks
Radha got ₹ 17480 as her monthly salary and overtime. Her salary exceeds the overtime by ₹ 10000 . What is her monthly salary?
AnswerRadha's monthly salary and overtime =₹ 17480
Let $x$ be the monthly salary.
Then, overtime = ₹$(x-10000)$
$\therefore 17480-x=x-10000$
$2 x=27480$
$
\Rightarrow x=13740
$
Hence, her monthly salary is ₹ 13740 .
View full question & answer→Question 372 Marks
Karim and Ramesh are two close friends. Both invested some amount in interest giving plan of a bank. At the end of the plan, Karim received ₹ 13000 more than that of Ramesh. But Ramesh needs some more money for an urgent work. If Karim gives $\frac{1}{2}$ of interest earned by him to Ramesh, also the total interest received by them is ₹ 17000 . Find the interest received by Ramesh and what amount is given by Karim to Ramesh for his help. What type of value depict by Karim?
AnswerLet $x$ be the interest received by Ramesh.
So, interest received by Karim =₹$(x+13000)$
The total interest received by both of them = ₹ 17000
$\therefore x+x+13000=17000$
$\Rightarrow 2 x=17000-13000$
$2 x=4000$
$\Rightarrow x=\frac{4000}{2}=$₹ 2000
Interest received by Ramesh =₹ 2000
And Karim received interest = 2000 +13000 = ₹15000
$\therefore$ Karim gives $\frac{1}{2}$ th of his interest to Ramesh
$=\frac{15000}{2}=$₹ 7500
Karim is very helping to his friend.
View full question & answer→Question 382 Marks
The perimeter of a rectangle is 40 m . The length of the rectangle is 4 m less than 5 times its breadth. Find the length of the rectangle.
AnswerAs per the given information in the question,
the perimeter of a rectangle is 40 m.
[ $\because$ perimeter of rectangle $=2$ length +2 breadth $]$
Let $x$ be the breadth of the rectangle.
Then, length of the rectangle $=5 x-4$
$\begin{aligned} \therefore 2 x+2(5 x-4) & =40 \\
\Rightarrow 2 x+10 x-8 & =40\end{aligned}$
On adding 8 to both sides, we get
$\Rightarrow 12 x=48$
$\Rightarrow x=4$
Hence, length of the rectangle
$=5 x-4=(5 \times 4)-4$
$=20-4=16 m$
View full question & answer→Question 392 Marks
A girl is 28 yr younger than her father. The sum of their ages is 50 yr . Find the ages of the girl and her father.
AnswerAs per the given information in the question,
the sum of ages of the girl and her father $=50 yr$
Let $x$ be the age of the girl.
So, the age of the father will be $=(x+28) yr$
$\therefore x+(x+28)=50$
$\Rightarrow 2 x+28=50$
$2 x=50-28$
$\Rightarrow 2 x=22$
$\Rightarrow x=\frac{22}{2}=11 yr$
Hence, age of the girl is 11 yr .
and her father's age is $=11+28=39 yr$.
View full question & answer→Question 402 Marks
In a school, the number of girls is 50 more than the number of boys. The total number of students is 1070. Find the number of girls.
AnswerAs per the given information in the question,
the total number of students is 1070 .
Let $x$ be the number of boys in the school.
So, the number of girls in the school will be $x+50$.
Then, $ x+(x+50)=1070$
$2 x+50=1070$
$\Rightarrow 2 x=1070-50$
$2 x=1020 \Rightarrow x=510$
Hence, the number of boys in the school is 510.
And the number of girls in the school will be = 510 + 50 = 560.
View full question & answer→Question 412 Marks
Subramaniam and Naidu donate some money in a Relief Fund. The amount paid by Naidu is ₹ 125 more than that of Subramaniam. If the total money paid by them is ₹ 975 , then find the amount of money donated by Subramaniam.
AnswerLet ₹ $x$ be the amount donated in a Relief Fund by Subramaniam.
So, the amount donated by Naidu will be ₹$(x+125)$.
If the sum of their money is ₹975.
Then, $x+(x+125)=975$
$2 x+125=975$
$\Rightarrow 2 x=975-125$
$2 x=850$
$\Rightarrow x=\frac{850}{2}$
$\Rightarrow x=$ ₹ 425
Hence, the amount of money donated by Subramaniam is 425.
View full question & answer→Question 422 Marks
The length of a rectangle is two times its width. The perimeter of the rectangle is 180 cm . Find the dimensions of the rectangle.
AnswerAs per the given information in the question,
the perimeter of the rectangle is 180 cm .
Let $x$ be the width of the rectangle.
So, the length of the rectangle will be $2 x$.
Perimeter of a rectangle $=2$ length +2 width
$\therefore 2 x+2(2 x)=180 \Rightarrow 6 x=180$
$x=\frac{180}{6}=30 cm$
Hence, width of the rectangle is 30 cm and length of the rectangle is $2 \times 30=60 cm$.
View full question & answer→Question 432 Marks
Express each of the given statements as an equation.
(i) If $1$ is subtracted from a number and the difference is multiplied by $\frac{1}{2}$, the result is 7 .
(ii) Mohan is 3 yr older than Sohan. The sum of their ages is 43 yr .
(iii) Five times a number increased by 7 is 27 .
Answer(i) Let the number be $x$.
Now, 1 is subtracted from a number, then we get $(x-1)$,
The difference is multiplied by $\frac{1}{2}$.
Here, $\frac{1}{2}(x-1)$ gives result 7.
$\therefore \frac{1}{2}(x-1)=7$
(ii) Let age of Sohan be $x$ yr. Then, the age of Mohan is $(x+3) y r$.
$\therefore$ Sum of their ages $=43$
$\Rightarrow x+(x+3)=43$
(iii) Let the number be $x$. Then, five times of number be $5 x$.
Since, it is increased by 7.
So, $5 x+7$ gives result 27.
Hence, the equation is $5 x+7=27$.
View full question & answer→Question 442 Marks
Ajay subtracts thrice the number of notebooks he has from 50, he finds the result to be 14 . Now, set up the equation on the basis of given information and solve it to find the unknown number.
AnswerLet $x$ be the number of nootbooks.
As per the given information, we have
$
50-3 x=14
$
On subtracting 50 from both sides, we get
$\begin{aligned} \Rightarrow 50-3 x-50 & =14-50 \\
-3 x =-36\end{aligned}$
On dividing both sides by -3 , we get
$\frac{3 x}{3}=\frac{36}{3}$
$
\Rightarrow x=12
$
Hence, the required number is 12 .
View full question & answer→Question 452 Marks
Kritika thinks of a number. If she takes away 7 from $\frac{9}{2}$ of the number, the result is 26 . Now, set up the equation on the basis of given information and solve it to find the unknown number.
AnswerLet $x$ be the number.
As per the given information, we have
$\frac{9}{2} x-7=26$
On adding 7 to both the sides, we get
$\frac{9}{2} x-7+7=26+7 \Rightarrow \frac{9}{2} x=33$
Now, on dividing both sides by $\frac{9}{2}$, we get
$\frac{9 / 2 x}{9 / 2}=\frac{33}{9 / 2} \Rightarrow x=\frac{33}{1} \times \frac{2}{9}=\frac{11 \times 2}{3}$
So, $ x=\frac{22}{3}$
Hence, the required number is $\frac{22}{3}$.
View full question & answer→Question 462 Marks
When you subtracted 12 from twice a number the result was 16.Now, set up the equation on the basis of given information and solve it to find the unknown number.
AnswerLet $x$ be the number.
As per the given information, we have
$2 x-12=16$
On adding 12 to both the sides, we get
$
2 x-12+12=16+12 \Rightarrow 2 x=28
$
Now, on dividing both sides by 2 , we get
$\frac{2 x}{2}=\frac{28}{2} \Rightarrow x=14$
Hence, the required number is 14.
View full question & answer→Question 472 Marks
Solve the following equations
$\frac{P}{2}+6=12$
AnswerGiven, equation $\frac{P}{2}+6=12$
On subtracting 6 from both the sides, we get
$
\frac{P}{2}+6-6=12-6 \Rightarrow \frac{P}{2}=6
$
Now, on multiplying both sides by 2 , we get
$\frac{P}{2} \times 2=6 \times 2$
$
\therefore
P=12,
$
which is the required solution.
View full question & answer→Question 482 Marks
Solve the following equations
$2 y+\frac{7}{2}=\frac{35}{2}$
AnswerGiven, $2 y+\frac{7}{2}=\frac{35}{2}$
On subtracting $\frac{7}{2}$ from both the sides, we get
$2 y+\frac{7}{2}-\frac{7}{2}=\frac{35}{2}-\frac{7}{2}$
$\Rightarrow 2 y=\frac{35-7}{2}$
$2 y=\frac{28}{2}$
Now, on dividing both sides by 2 , we get
$\frac{2 y}{2}=\frac{28}{2 \times 2} \Rightarrow y=\frac{28}{4}$
$\therefore y=7$,
which is the required solution.
View full question & answer→Question 492 Marks
Solve the following equations
$5(2 m-6)=12$
AnswerGiven, $5(2 m-6)=12$
On dividing both sides by 5 , we get
$\frac{5(2 m-6)}{5}=\frac{12}{5} \Rightarrow 2 m-6=\frac{12}{5}$
Now, on adding 6 in the both sides, we get
$2 m-6+6=\frac{12}{5}+6 \Rightarrow 2 m=\frac{12}{5}+6$
$\Rightarrow 2 m=\frac{12+30}{5} \Rightarrow 2 m=\frac{42}{5}$
Again, dividing both sides by 2 , we get
$\frac{2 m}{2}=\frac{42}{5 \times 2}$
$\therefore m=\frac{21}{5}$,
which is the required solution.
View full question & answer→Question 502 Marks
Solve the following equations
$4(4 P+6)=16$
AnswerGiven equation is $4(4 P+6)=16$
On dividing both sides by 4 , we get
$\frac{4(4 P+6)}{4}=\frac{16}{4} \Rightarrow 4 P+6=4$
Now, on subtracting 6 from both the sides, we get
$
4 P+6-6=4-6 \Rightarrow 4 P=-2
$
Again, dividing both sides by 4 , we get
$
\frac{4 P}{4}=-\frac{2}{4} \Rightarrow P=-\frac{1}{2}
$
which is the required solution.
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