Question

please give some algebra problems for class 6​

Answer 1

Step-by-step explanation:

Ch-11 Algebra

Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder. She has 10 dots in a row. How many dots will her Rangoli have for r rows?

10 + r 

10r

10 – r

r

Which of the following is expression with one variable?

x + y + z

y + 1

1

x + y – 5

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?

12b

3b – 4

None of these

3b + 4

The _______ of the variable in an equation which satisfies the equation is called a solution to the equation.

value

factor

term

None of these

The teacher distributes 4 pencils per student. Can you tell how many pencils are needed for given number of students? (Use s for the number of students.)

4 – s

4+s

s

4s

Match the following:

Column AColumn B(a) 3 times y added to 13(p) 5y – 8(b) 8 subtracted from 5 times y(q) 3x – 5(c) 5 reduced from 3 times x(r) 2x + 5(d) 5 added to double of x(s) 3y + 13

Fill in the blanks:

The value of 2x – 12 is zero, when x = ________.

The product of 2 and x is being added to the product of 3 and y is expressed as ________.

The numerical coefficient of the terms 12xy212xy2 is _________.

The no. of terms in the expression 3x2y–4x2y2+12xy2–5x3x2y–4x2y2+12xy2–5x is ______.

State whether the following statements are true or false:

The parts of an algebraic exponent which are connected by + or – sign are called its terms.

5 times x subtracted from 8 times y is 5x-8y.

A number having fixed value is called variable.

The numerical coefficient of -2x2y is -2.

Write which letters give us the same rule as that given by L.

Rearrange the terms of the following expressions in ascending order of powers of x:

5x2, 2x, 4x4, 3x3, 7x5

Give expressions for the following

7 added to

7 subtracted from

p multiplied by

p divided by

7 subtracted

– p multiplied by

– p divided by

p multiplied by – 5.

The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students ? (Use s for number of students.)

Form expressions using y, 2 and 7. Every expression must have y in it. use only two number operations. These should be different.

Find the value of the expression 2x – 3y + 4z, if x = 10, y = -12 and z = 11.

Deepak’s present age is one-third his mother’s present age. If the mother’s age was five times his age 6 years ago, what are their present ages?

Ch-11 Algebra

Answer

10r, Explanation: Let the total number of rows be ‘r’.

As, No. Of dots in a row =10.

So, the dots needed for 10 rows = r x 10= 10r.

y + 1, Explanation: The equation has one variable as “y” whose value is not known. therefore, the equation is in one variable.

3b – 4, Explanation: breadth of a rectangular hall = b meters

let length of a rectangular hall be ‘l’ meter

according to the question, l = 3 times the breadth – 4 = 3b – 4

value, Explanation: It is correct because the value of the variable must satisfy the equation.

(d) 4s, Explanation: Let the number of pencils be ‘s’.

As, the number of pemcils distributed to each student= 4

Thus, No. of pencils for ‘s’ students = 4 x s= 4s.

→→ (s)

→→ (p)

→→ (q)

→→ (r)

6;

2x + 3y;

1212;

4

True

False

False

True

The other letters which give us the same rule as L are T, V and X because the number of matchsticks required to make each of them is 2.

If the given terms are arranged in the ascending order of powers of x, we get,2x, 5x2, 3x3, 4x4, 7x5.

p + 7

p – 7

7p

p7p7

– m – 7

-5p

−p5−p5

– 5p.

Number of pencils to be distributed to each student= 5And, let the number of students in class be ‘s’.

As per the logic, Number of pencils needed =(Number of students in the class) x. (Number of pencils to be distributed to one student )

So, Number of pencils needed= 5 x s =5s.

The different expressions that can formed are: 2y + 7, 2y – 7, 7y + 2, 7y-2, (y/2) – 7, (y/7)-2, y – (7/2), y + (7/2)

Given expression = 2x – 3y + 4z

If x = 10, y = -12 and z = 11,

The expression becomes, (2×10)–(3×–12)+(4×11)(2×10)–(3×–12)+(4×11)

= 20 – (-36) + 44

= 20 + 36 + 44

= 100.

Let present age of mother = x years

Deepak’s present age =x3years=x3years

6 years ago, mother’s age = (x – 6) years

Deepak’s age =(x3–6)=(x3–6) years

According to the problem, 6 years ago, mother’s age is 5 times Deepak age.

i.e., (x – 6) =5×(x3–6)=5×(x3–6)

x–5x3=–30+

Answer 2

Answer:

Linear Equation:

1. $\frac{x}{2} +\frac{x}{4} =\frac{1}{8}$
2. $\frac{2x+5}{3} =3m - 10$
3. $\frac{4x}{5} - \frac{3x}{4} =4$
4. 3x + 3 (x + 1) = 69
5. 2 (x - 2) + 3 (4x - 1) = 0

Word Problems:

1. A number when added to its two thirds gives 55. Find the number.
2. The length of a rectangular field is twice its breadth. If the perimeter of the field is 150 m. Find the length and breadth.
3. Find two consecutive natural numbers whose sum is 63.
4. Find two consecutive positive odd integers whose sum is 76.
5. 2/3 of a number is less than the original number by 20. Find the number.
6. A number when added to its half gives 72. Find the number.

Answers:

Linear Equation:

(1) $\frac{x}{2} +\frac{x}{4} =\frac{1}{8}$

$\frac{2x}{4} +\frac{x}{4} =\frac{1}{8}$

$\frac{3x}{4} = \frac{1}{8}$

$(3x) 8 = (1) 4$ (Cross Multiplication)

$24x = 4$

$x = \frac{4}{24}$

$x = \frac{1}{6}$

(2) $\frac{2x+5}{3} =3m - 10$

$2x + 5 = (3m - 10)3$

$2x + 5=9m-30$

$2x - 9x = - 30 - 5$

$- 7x=-35$

$x = \frac{-35}{-7}$

$x=5$

(3) $\frac{4x}{5} - \frac{3x}{4} =4$

$\frac{16x-15x}{20} =4$

$\frac{x}{20} =4$

$x = 4$ × $20$

$x=80$

(4) 3x + 3 (x + 1) = 69

3x + 3x + 3 = 69

6x + 3 = 69

6x = 69 - 3

6x = 66

x = 66/6

x = 11

(5) 2 (x - 2) + 3 (4x - 1) = 0

2x - 4 + 12x - 3 = 0

2x + 12x - 4 - 3 = 0

14x - 7 = 0

14x = 0 + 7

14x = 7

x = 7/14

x = 1/2

Word Problems:

(1) x + $\frac{2x}{3}$ = 55

$\frac{3x+2x}{3}$ = 55

5x = 55 × 3

5x = 165

x = 33

(2) Breadth = x

Length = 2x

Perimeter = 150 m

2 (l + b) = 150

2 (2x + x) = 150

2 (3x) = 150

6x = 150

x = 25

Hence, length = 2x

= 2 × 25

= 50 m

Breadth = x

= 25 m

(3) 1st natural number = x

2nd natural number = x + 1

Sum = 63

x + (x + 1) = 63

x + x + 1 = 63

2x + 1 = 63

2x = 63 - 1

x = 62/2

x = 31

Hence, 31 + 32 = 63

(4) 1st integer = 2x + 1

2nd integer = 2x + 3

Sum = 76

(2x + 1) + (2x + 3) = 76

2x + 1 + 2x + 3 = 76

2x + 2x + 1 + 3 = 76

4x + 4 = 76

4x = 76 - 4

4x = 72

x = 72/4

x = 18

Hence 1st odd integer = 2x + 1

= 2 × 18 + 1

= 36 + 1

= 37

2nd odd integer = 2x + 3

= 2 × 18 + 3

= 36 + 3

= 39

Hence, 37 + 39 = 76

(5) $x - \frac{2x}{3} = 20$

×

$x = 60$

(6) x + $\frac{x}{2}$ = 72

$\frac{2x+x}{2} =72$

$\frac{3x}{2} = 72$

$3x=72$ × $2$

$3x = 144$

x = 48

Hope it helps