Important Questions for CBSE Class 6 Maths Chapter 11 - Algebra

Very Short Answer Questions 1 Marks

1. Express algebraically: $\mathbf{3}$ more than $\mathbf{x}$.

Ans: The algebraic expression of the given statement is $x+3$.

2. What is the exponential form of $\mathbf{x}\,\times \,\mathbf{x}\,\times \,\mathbf{x}\,\times \,\mathbf{x}\,\times ....\mathbf{15}\ \mathbf{times}$?

Ans: The given expression $x\times x\times x\times x\times ...15\ times$.

Recall that, $x\times x={{x}^{2}}$,

$x\times x\times x={{x}^{3}}$,

$x\times x\times x\times x={{x}^{4}}$, etc.

Therefore, the number of times $x$ is multiplied with itself equals $x$ to the power that number.

Thus, $x\times x\times x\times x\times ...15\ times={{x}^{15}}$.

3. What is the product form of ${{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}^{\mathbf{7}}}$?

Ans: The given algebraic expression is ${{x}^{2}}{{y}^{7}}$.

Recall that, the number of times $x$ is multiplied with itself equals $x$ to the power that number.

Therefore, ${{x}^{2}}=x\times x$.

Similarly, ${{y}^{7}}=y\times y\times y\times y\times y\times y\times y$.

Thus, ${{x}^{2}}{{y}^{7}}=x\times x\times y\times y\times y\times y\times y\times y\times y$.

4. What is the coefficient of $\mathbf{b}$ in $\mathbf{-8abc}$?

Ans: The given algebraic expression is $-8abc$.

Recall that, the coefficient of any term in an algebraic expression is the multiplication of other terms in that algebraic expression.

Therefore, the coefficient of the term $b$ in the expression $-8abc$ is $-8ac$.

5. What is the numerical coefficient of $\mathbf{4}{{\mathbf{a}}^{\mathbf{2}}}$?

Ans: The given expression is $4{{a}^{2}}$.

It is known that, the numerical coefficient in an algebraic expression is the number multiplied in that expression.

Therefore, the numerical coefficient of $4{{a}^{2}}$ is $4$.

6.  What is the constant term of $\mathbf{2}{{\mathbf{a}}^{\mathbf{2}}}-\mathbf{9}$?

Ans: The constant term of the expression $2{{a}^{2}}-9$ is $-9$.

7. If $\dfrac{\mathbf{y}}{\mathbf{5}}=\mathbf{1}$, then $\mathbf{y}=$?

Ans: The given equation is

$\dfrac{y}{5}=1$.

Multiply $5$ both sides of the equation.

$5\times \dfrac{y}{5}=1\times 5$

$\Rightarrow y=5$.

Short Answer Questions 2 Marks

1. Express algebraically the following statement:

$\mathbf{8}$ times a number $\mathbf{x}$ is less than a number $\ge $.

Ans: The required algebraic expression of the given statement is

$8x=z-y$.

2. What are the terms in the following algebraic expression?

$\mathbf{4a}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{+3}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{-5ab+9}$.

Ans: The terms of in the given expression are 

$4a{{b}^{2}},\,3{{c}^{2}},-5ab,$ and $9$.

3. Solve the following equation and evaluate $\mathbf{x}$. Also, justify the result.

$\mathbf{x}-\mathbf{8}=\mathbf{17}$.

Ans: The given equation is

$x-8=17$

Add $8$ both sides of the equation.

$ x-8+8=17+8 $

$ \Rightarrow x=25$ 

Justification:

$ LHS=x-8 $

$=25-8$

$ =17$ 

$ =RHS $

Thus, Left-hand-side $=$ Right-hand-side.

4. Solve the following equation and evaluate $\mathbf{x}$. Also, justify the result.

$\dfrac{\mathbf{4}}{\mathbf{6}}\mathbf{x=18}$.

Ans: The given equation is 

$\dfrac{4}{6}x=18$

Multiply $6$ both sides of the equation.

$6\times \dfrac{4}{6}x=6\times 18$

$\Rightarrow 4x=18\times 6$

Divid both sides of the equation by $4$.

$\dfrac{4x}{4}=\dfrac{18\times 6}{4}$

$\Rightarrow x=27$.

Justification:

$ LHS=\dfrac{4}{6}x $

$ =\dfrac{4}{6}\times 27 $

$ =18 $

$=RHS $

Hence, Left-hand-side $=$ Right-hand-side.

Long Answer Questions 3 Marks

1. Sanjit scored $\mathbf{70}$ marks in mathematics and $\mathbf{''x''}$ marks in English. Find the total score in both the subjects?

Ans: He got marks in mathematics $=70$.

He got marks in English $=x$.

Therefore, the total marks in mathematics and English $=x+70$.

2. Let $\mathbf{p}=\mathbf{1},\,\,\mathbf{q}=-\mathbf{1},$ and $\mathbf{r}=\mathbf{3}$. What is the value of ${{\mathbf{p}}^{\mathbf{3}}}\mathbf{+}{{\mathbf{q}}^{\mathbf{3}}}\mathbf{+}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{-3pqr}$?

Ans: The given algebraic expression is 

${{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr$                                             …… (i)

Substituting $p=1,q=-1,$ and $r=3$ into the equation (i) gives

$ {{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr$

$ ={{\left( 1 \right)}^{3}}+{{\left( -1 \right)}^{3}}+{{\left( 3 \right)}^{3}}-3\left( 1 \right)\left( -1 \right)\left( 3 \right) $

 $=1-1+27+9$ 

 $=27+9 $ 

 $ =36 $

3. Choose the similar terms from the following expressions.

$\mathbf{xyz,}\,\,\mathbf{x}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{z,}\,\,\mathbf{xz}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{,}\,\,{{\mathbf{z}}^{\mathbf{2}}}\mathbf{xy,}\,\,{{\mathbf{y}}^{\mathbf{2}}}\mathbf{xz,}\,\,{{\mathbf{x}}^{\mathbf{2}}}\mathbf{yz,}\,\,\mathbf{zx}{{\mathbf{y}}^{\mathbf{2}}}$

Ans:

The like terms are those which have exactly one square term involved.

Therefore, the similar terms are

$x{{y}^{2}}z,\,\,xz{{y}^{2}},\,\,{{y}^{2}}xz,$ and $zx{{y}^{2}}$.

4. Solve the equation $\mathbf{12x}-\mathbf{30}=\mathbf{6}$.

Ans:

The given equation is 

$12x-30=6$

Add $30$ both sides of the equation.

$12x-30+30=6+30$

$\Rightarrow 12x=36$

Divide $12$ both sides of the equation.

$\dfrac{12x}{12}=\dfrac{36}{12}$

$\Rightarrow x=3$.

Hence, the solution is $x=3$.

5. Verify whether $\mathbf{y}=\mathbf{4}$ satisfy the equation $\mathbf{3y}-\mathbf{5}=\mathbf{7}$.

Ans: The given equation is $3y-5=7$.

Then,

$ LHS=3\left( 4 \right)-5 $

$=12-5$

$ =7 $

$ =RHS $

Thus, Left-hand-side $=$ Right-hand-side.

Hence, $y=4$ satisfy the given equation.

6. Solve the equation $\mathbf{6x}+\mathbf{10}=\mathbf{26}-\mathbf{2x}$ and evaluate $\mathbf{x}$ by using the method of transposition. Then justify the answer.

Ans: The given equation is

$6x+10=26-2x$

$\Rightarrow 6x=26-2x-10$, by the method of transposition.

$\Rightarrow 6x=16-2x$

$\Rightarrow 6x+2x=16$, by the method of transposition.

$\Rightarrow 8x=16$

Divide $8$ both sides of the equation.

$\dfrac{8x}{8}=\dfrac{16}{8}$

$\Rightarrow x=2$.

Justification:

$ LHS=6x+10$

$=6\times 2+10 $ 

$ =12+10 $

$=22$

Also,

$ RHS=26-2x $

 $=26-2\times 2 $

$ =26-4$ 

 $=22 $

Thus, Left-hand-side $=$ Right-hand-side.

Hence, $x=2$ is the solution of the given equation.

Long Answer Questions 4 or 5 Marks 

1. Find the algebraic expression of the statements given below.

(a) $\mathbf{6}$ less than the quotient of $\mathbf{x}$ and $\mathbf{y}$.

Ans. The required algebraic expression is 

$\dfrac{x}{y}-6$.

(b) The sum of the quotient of $\mathbf{p}$, $\mathbf{q}$ and the product of $\mathbf{p},\,\,\mathbf{q}$.

Ans: The quotient of $p,q$ is $\dfrac{p}{q}$.

The product of $p,q$ is $pq$.

Therefore, the required algebraic expression is given by

$\dfrac{p}{q}+pq$.

(c) One fourth of the sum of $\mathbf{x}$ and $\mathbf{y}$.

Ans: The sum of $x$ and $y$ is $x+y$.

Therefore, the required algebraic expression is given by

$\dfrac{1}{4}\left( x+y \right)$.

(d) $\mathbf{8}$ removed from twice the term $\mathbf{x}$.

Ans: Twice the term $x$ is $2x$.

Therefore, the required algebraic expression is $2x-8$.

2. Classify monomials, binomials, and trinomials from the following algebraic expressions.

$\mathbf{y}+\mathbf{5},\,\,\,\mathbf{6}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{y},\,\,\,-\mathbf{8},\,\,\,-\mathbf{3pqr},\,\,\,{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{z}}^{\mathbf{2}}}$

Ans: Recall that, sum of two terms is called binomials, sum of three terms is called trinomials and single algebraic term is called monomials.

The following table shows the required classifications.

3. Determine the solution of $\mathbf{3x}=\mathbf{12}$, by using the method of trial and errors.

Ans: The given equation is $3x=12$.

First substitute $x=1$ into the given equation.

$3x=3\times 1=3\ne 12$.

So, $x\ne 1$.

Then, substitute $x=2$ into the equation.

$3x=3\times 2=6\ne 12$.

So, $x\ne 2$.

Again, substitute $x=3$ into the given equation.

$3x=3\times 3=9\ne 12$.

Therefore, $x\ne 3$.

Now, substitute $x=4$ into the given equation.

Then $3x=3\times 4=12$.

Thus, $x=4$ is the solution of the given equation.

4. Solve the following equation and evaluate $\mathbf{x}$ by using the method of isolation.

$\mathbf{a}-\mathbf{7}=\mathbf{5}+\dfrac{\mathbf{a}}{\mathbf{2}}$.

Ans: The given equation is 

$a-7=5+\dfrac{a}{2}$

Add $7$ on both sides of the equation.

$a-7+7=5+7+\dfrac{a}{2}$

$\Rightarrow a=12+\dfrac{a}{2}$

Add $-\dfrac{a}{2}$  on both sides of the equation.

$a+\left( -\dfrac{a}{2} \right)=12+\dfrac{a}{2}+\left( -\dfrac{a}{2} \right)$

$ \Rightarrow a-\dfrac{a}{2}=12+\dfrac{a}{2}-\dfrac{a}{2} $

$ \Rightarrow \dfrac{2a-a}{2}=12 $ 

$\Rightarrow \dfrac{a}{2}=12 $

Multiply $2$ on both sides of the equation.

$\dfrac{a}{2}\times 2=12\times 2$

$\Rightarrow a=24$.

Hence, the required solution is $a=24$.

5. Solve the following equation and justify the result.

$\mathbf{3}\left( \mathbf{a+3} \right)\mathbf{+3}\left( \mathbf{a-1} \right)=\mathbf{5}\left( \mathbf{a+5} \right)$.

Ans: The given equation is 

$\text{3}\left( \text{a+3} \right)\text{+3}\left( \text{a-1} \right)=\text{5}\left( \text{a+5} \right)$

$\Rightarrow 3a+9+3a-3=5a+25$, multiplying the terms.

$\Rightarrow 6a+6=5a+25$, adding the like terms.

$\Rightarrow 6a-5a=25-6$, by the method of transposition.

$\Rightarrow a=19$.

Justification:

$ LHS=3\left( a+3 \right)+3\left( a-1 \right) $

$=3\left( 19+3 \right)+3\left( 19-1 \right)$

$ =3\times 22+3\times 18 $

$ =66+54$

$=120 $

Also, 

$ RHS=5\left( a+5 \right) $

$ =5\left( 19+5 \right) $ 

$=5\times 24 $ 

$=120 $

Thus, Left-hand-side $=$ Right-hand-side.

Hence, the required solution of the given equation is $a=19$.

6. Determine two numbers so that one is greater by $\mathbf{21}$ than another and sum of them is $\mathbf{91}$.

Ans: Let $x$ be one number.

Then, another number is $x+21$.

By the given conditions,

$x+\left( x+21 \right)=91$

$\Rightarrow 2x+21=91$

Add $-21$ on both sides of the equation.

$2x+21+\left( -21 \right)=91+\left( -21 \right) $ 

$\Rightarrow 2x+21-21=91-21 $ 

$ \Rightarrow 2x=70 $

Divide $2$ on both sides of the equation.

$ \dfrac{2x}{2}=\dfrac{70}{2} $

$\Rightarrow x=35 $

Hence, the required numbers are $35$ and $35+21$, that is $35,\,\,56$.

Important Questions for Class 6 Maths Chapter 11

CBSE Class 6 Maths Chapter 11 - Algebra PDF Free Download

To prepare for any mathematics concept, the best way to do it is to practice. You must have a set of important questions for class 6 maths chapter 11 to get your concepts cleared about the chapter. The students can download the PDF version of this question set from the Vedantu website onto their system for free. They need to have an account on the website. They can also enroll in some tuition classes to get accustomed to the topics. The students can download these class 6 maths chapter 11 important questions and use them in their preparation before any examinations.

Why is it Important to Practice Important Questions for Algebra?

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More Questions Mean More Practice

Mathematics has always been one of the rare subjects which require a lot of practice. Unlike other subjects where reading the concepts can help the students in their preparation, they have to first know about the concepts and then practice in maths. To help the students in their practice, the set of important questions for class 6 maths chapter 11 has been published. In this set of questions, all the basic principles of algebra are covered. The students will start from scratch and gradually can soak up the increasing complexity of the subject.

The Proper Introduction to a New Topic

Secondly, algebra will be a new chapter for the students in class 6. Chapter 11 will pose another issue for the students. The students will be confused about the type of questions that they will encounter in the examination. By solving these class 6 maths chapter 11 important questions, the students will know the type of questions. They can consider these questions to be the blueprint of what is coming in the examination and plan their preparation accordingly.

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Fine-Tuning the Preparation Process

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Last-Minute Exam Preparation

For any other subject, the last-minute exam preparation will involve revision of the important topics. However, Math is such a subject that can only be revised through practice. For the last-minute preparation, you would prefer to look at the problems that are most likely to come in the examination. You might also prefer to solve them to revise the techniques. These questions will be ideal for last-minute preparation before any examination.

Important Questions on Maths Chapter 11 Algebra

Q. What is the algebraic representation of six more than x?

Ans. The problem mentioned above can be best represented as x + 6.

Q. What is the algebraic representation of nine less than three times of x?

Ans. The problem mentioned above can be best represented as 3x - 9. 

Q. What is the algebraic representation of four more than half of x?

Ans. The problem mentioned above can be best represented as (x/2) + 4.

Q. In a horticultural garden, rose plants are arranged in rows. There are seven rose plants in each row. If the total number of rows dedicated to role plants is equal to n, how many rose plants are there in the garden?

Ans. In the problem mentioned above, the following information is obtained.

The number of rose plants in a single row = 7.

The number of rows in the horticultural garden dedicated to rose plants = n.

The total number of rose plants in the horticultural garden = total number of rose plants in each row X total number of rows dedicated to rose plants in the horticultural garden.

Therefore, the total number of rose plants in the horticultural garden = 7n.

Q. In a fruit store, there are different boxes designated for different fruits. There are n boxes of mangoes, each containing 15 mangoes. Apples are present in 4 boxes with x number of apples in each box. Additionally, there are 6 boxes of guavas, each containing 20 guavas. How many fruits are present in the store in total?

Ans. In the problem mentioned above, the following information is obtained.

Number of mangoes in each box = 15 mangoes

Number of boxes containing mangoes = n boxes

The total number of mangoes in the fruit store = number of mangoes in each box X Number of boxes containing mangoes.

Therefore, the total number of mangoes in the fruit store = 15n.

Number of apples in each box = x apples

Number of boxes containing apples = 4 boxes

The total number of apples in the fruit store = number of apples in each box X Number of boxes containing apples.

Therefore, the total number of apples in the fruit store = 4x.

Number of guavas in each box = 20 guavas

Number of boxes containing guavas = 6 boxes

The total number of guavas in the fruit store = number of guavas in each box X Number of boxes containing guavas.

Therefore, the total number of guavas in the fruit store = 120 guavas.

Finally, the total number of fruits present in the fruit store = total number of mangoes + total number of apples + the total number of guavas

Total number of fruits in the fruit store = 15n + 4x +120.

Q. The speed of a car is x km/ hr. If the car covers a distance of y km by going with that constant speed, how much time will it take to complete the distance?

Ans. In the problem mentioned above, the following information is obtained.

Speed of the car = x km/hr.

Distance covered by the car at that constant speed = y km.

Time-taken by the car to cover the distance = Distance covered by the car at that constant speed/ Speed of the car.

Therefore, Time-taken by the car to cover the distance = y/x.

Q. Manoj has x number of chocolates. Rahim gave him 4 more. How many chocolates does Manoj have now?

Ans. In the problem mentioned above, the following information is obtained.

The number of chocolates Manoj has = x chocolates.

The number of chocolates Rahim gave to Manoj = 4 chocolates.

Total number of chocolates that Manoj have = number of chocolates Manoj had previously + number of chocolates Rahim gave to Manoj,

Therefore, the total number of chocolates Manoj has now = x+4.

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Important Related Links for CBSE Class 6 Maths