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# NCERT Solutions for Class 6 Maths Chapter 11: Algebra - Exercise 11.4

Last updated date: 29th Feb 2024
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## NCERT Solutions for Class 6 Maths Chapter 11 (Ex 11.4)

Free PDF download of NCERT Solutions for Class 6 Maths Chapter 11 Exercise 11.4 (Ex 11.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 6 Maths Chapter 11 Algebra Exercise 11.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 6 Subject: Class 6 Maths Chapter Name: Chapter 11 - Algebra Exercise: Exercise - 11.4 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2023-24 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

## Access NCERT Solutions for Class 6 Maths Chapter 11- Algebra

Exercise 11.4

(a) Take Sarita’s present age to be $y$ years.

(i) What will be her age 5 years from now?

Ans: Given, the present age of Sarita is $y$ years.

${\text{After }}5{\text{ years Sarita's age = Sarita's present age + }}5$

${\text{After }}5{\text{ years Sarita's age = }}\left( {y{\text{ + }}5} \right){\text{ years}}$

(ii) What was her age 3 years back?

Ans: Given, the present age of Sarita is $y$ years.

$3{\text{ years back Sarita's age = Sarita's present age }} - 3$

$3{\text{ years back Sarita's age = }}\left( {y - 3} \right){\text{ years}}$

(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?

Ans: Given, the present age of Sarita is $y$ years and her grandfather’s age is 6 times her age.

(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?

Ans: Obtained age of Sarita’s grandfather is $6y$ years.

Given, her grandmother is 2 years younger than her grandfather.

(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

Ans: Given, the present age of Sarita is $y$ years.

$3\;{\text{times of her present age = Her present age}} \times 3$

$3\;{\text{times of her present age = }}y \times 3$

$3\;{\text{times of her present age = }}3y$

It is also given that her father’s age is 5 years more than 3 times her age.

${\text{Her father's age = }}3{\text{ times of her present age}} + 5$

${\text{Her father's age = }}\left( {3y + 5} \right){\text{ years}}$

(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is $b$ meters?

Ans: Given, the breadth of the hall is $b$ meters.

$3\;{\text{times of breadth of the hall = Breadth of the hall}} \times 3$

$3\;{\text{times of breadth of the hall = }}b \times 3$

$3\;{\text{times of breadth of the hall = }}3b$

It is also given that the length of the hall is 4 meters less than three times the breadth of the hall.

Therefore,

${\text{Length of the hall = }}3{\text{ times breadth of the hall}} - 4$

${\text{Length of the hall = }}\left( {3b - 4} \right){\text{ meters}}$

(c) A rectangular box has height $h$ cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.

Ans: Given, height of the box is $h$cm and length of the box is 5 times of the height.

Therefore,

${\text{Length of the box = Height of the box}} \times 5$

${\text{Length of the box = }}h \times 5$

${\text{Length of the box = }}\left( {5h} \right){\text{ cm}}$

Also it is given that the breadth of the box is 10 cm less than the length of the box.

Hence,

${\text{Breadth of the box = Length of the box}} - 10$

${\text{Breadth of the box = }}\left( {5h - 10} \right){\text{ cm}}$

(d) Meena, Beena and Leena are climbing the steps to the hilltop. Meena is at step $s$, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Leena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using $s$.

Ans: Given, Meena is at $s$ step and Beena is 8 steps behind her.

Hence,

${\text{Step at which Beena is = }}\left( {{\text{step at which Meena is}}} \right) + 8$

${\text{Step at which Beena is }} = s + 8$

It is also given that Leena is 7 steps behind Meena.

${\text{Step at which Leena is = }}\left( {{\text{step at which Meena is}}} \right) - 7$

${\text{Step at which Leena is }} = s - 7$

The total number of steps is also given by 10, less than 4 times what Meena has reached.

Therefore,

${\text{Total number of steps}} = 4{\text{ times of }}\left( {{\text{the step where Meena is}}} \right) - 10$

${\text{Total number of steps}} = 4 \times \left( {{\text{the step where Meena is}}} \right) - 10$

${\text{Total number of steps}} = 4 \times {\text{s}} - 10$

${\text{Total number of steps}} = 4{\text{s}} - 10$

(e) A bus travels at $v$ km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using $v$.

Ans: The distance $d$ is given by $d = s \times t$, where $s$ is the speed and $t$ is the time.

Given, the speed of the bus is $v$ km per hour and the travel time of the bus is 5 hours.

Therefore, $s = v$ and $t = 5$.

Hence, the distance travelled by the bus in 5 hours is as follows.

$d = v \times 5$

$d = 5v{\text{ km}}$

Since, Beespur is still 20 km away, the distance of Beespur is as follows,

${\text{Distance of Beespur from Daspur = }}d + 20$

${\text{Distance of Beespur from Daspur = }}\left( {5v + 20} \right){\text{ km}}$.

2. Change the following statements using expressions into statements in ordinary language.

(For example, Given Salim scores $r$ runs in a cricket match, Nalin scores $\left( {r + 15} \right)$ runs. In ordinary language – Nalin scores 15 runs more than Salim.)

(a) A notebook costs ₹ $p$. A book costs ₹ $3p$

Ans: Given, the cost of a notebook is ₹ $p$ where the cost of a book is ₹ $3p$.

Therefore,

${\text{The cost of book = }}3p$

${\text{The cost of book }} = {\text{ }}3 \times p$

${\text{The cost of book = }}3{\text{ times }}p$

${\text{The cost of book = }}3{\text{ times cost of notebook}}$

Hence, the cost of a book is 3 times the cost of a notebook.

(b) Tony put $q$ marbles on the table. He has $8q$ marbles in his box.

Ans: Given, the number of marbles on the table is $q$and the number of marbles in the box is $8q$.

Therefore,

${\text{The number of marbles in the box }} = 8q$

${\text{The number of marbles in the box }} = 8 \times q$

${\text{The number of marbles in the box }} = 8{\text{ times }}q$

${\text{The number of marbles in the box }} = 8{\text{ times the number of marbles on the table}}$

Hence, the box contains 8 times the number of marbles on the table.

(c) Our class has $n$ students. The school has $20n$ students.

Ans: Given, the number of students in the class is $n$ whereas the number of students in the school is $20n$.

${\text{The number of students in the school}} = 20n$

${\text{The number of students in the school}} = 20 \times n$

${\text{The number of students in the school}} = 20{\text{ times }}n$

${\text{The number of students in the school}} = 20{\text{ times the number of students in the school}}$

Hence, the total number of students in the school is 20 times that of the number of students in our class.

(d) Jaggu is $z$ years old. His uncle is $4z$ years old and his aunt is $\left( {4z - 3} \right)$ years old.

Ans: Given, the age of Jaggu is $z$ years and his uncle is $4z$ years old.

${\text{His uncle's age }} = 4z$

${\text{His uncle's age }} = 4 \times z$

${\text{His uncle's age }} = 4{\text{ times }}z$

${\text{His uncle's age }} = 4{\text{ times Jaggu's age}}$

Hence, Jaggu’s uncle is 4 times older than him.

Given the age of his aunt is $\left( {4z - 3} \right)$years.

${\text{His aunt's age }} = 4z - 3$

${\text{His aunt's age }} = {\text{His uncle's age}} - 3$

Therefore, Jaggu’s aunt is 3 years younger than his uncle.

(e) In an arrangement of dots there are $r$ rows. Each row contains 5 dots

Ans: Given the number of rows are $r$ and each row contains 5 dots.

Therefore,

${\text{The total number of dots = 5 }} \times {\text{ the number of rows}}$

${\text{The total number of dots = 5 times the number of rows}}$

Hence, the total number of dots in the arrangement is 5 times the number of rows.

3. (a) Given Munnu’s age to be $x$ years, can you guess what $\left( {x - 2} \right)$ may show?

Can you guess what $\left( {x + 4} \right)$ may show? What $\left( {3x + 7} \right)$ may show?

Ans: Given, the age of Munnu is $x$ years.

Therefore, $\left( {x - 2} \right)$ represents 2 years less than $x$.

Hence, $\left( {x--2} \right)$ represents a person who is 2 years younger to Munnu.

Therefore, $\left( {x + 4} \right)$ represents 4 years greater than $x$.

Hence, $\left( {x + 4} \right)$ represents a person who is 4 years older than Munnu.

Therefore, $\left( {3x + 7} \right)$ represents 7 years greater than $3x$.

Since $3x = 3{\text{ times }}x$

Hence, $\left( {3x + 7} \right)$ represents 7 years greater than 3 times $x$ and consequently $\left( {3x + 7} \right)$represents the person whose age is 7 years more than the 3 times of the age of Munnu.

(b) Given Sara’s age today to be $y$ years. Think of her age in the future or in the past.

What will the following expression indicate? $\left( {y + 7} \right),{\text{ }}\left( {y--3} \right),{\text{ }}y + 4\dfrac{1}{2},{\text{ }}y - 2\dfrac{1}{2}$

Ans: Given, the age of Sara is $y$ years.

Therefore, $\left( {y + 7} \right)$ represents 7 years greater than $y$.

Hence, $\left( {y + 7} \right)$ represents a person who is 7 years older than Sara.

Therefore, $\left( {y--3} \right)$ represents 3 years less than $y$.

Hence, $\left( {y--3} \right)$ represents the person who is 3 years younger to Sara.

Therefore, $y + 4\dfrac{1}{2}$ represents $4\dfrac{1}{2}$ years greater than $y$.

Hence, $y + 4\dfrac{1}{2}$ represents the person who is $4\dfrac{1}{2}$ years older than Sara.

Therefore, $y - 2\dfrac{1}{2}$ represents $2\dfrac{1}{2}$ years less than $y$.

Hence, $y - 2\dfrac{1}{2}$ represents the person who is $2\dfrac{1}{2}$ years younger to Sara.

(c) Given $n$ students in the class like football, what may $2n$ show? What may $\dfrac{n}{2}$ show?

Ans: Suppose, swimming is another game which is liked by $2n$students.

Given, the number of students in the class who likes football is $n$.

Since $2n = 2{\text{ times }}n$

$2n = 2{\text{ times of the number of students who like football}}$

Therefore, the number of students who like swimming is twice the number of students who like football.

Suppose tennis is another game which is liked by $\dfrac{n}{2}$students.

Since $\dfrac{n}{2} = {\text{ half of }}n$,

$\dfrac{n}{2} = {\text{ half of the number of students who like football}}$

Therefore, the number of students who like tennis is half the number of students who like football.

## NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.4

Opting for the NCERT solutions for Ex 11.4 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.4 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 11 Exercise 11.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 6 Maths Chapter 11 Exercise 11.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 6 Maths Chapter 11 Exercise 11.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.