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Frank Learning Maths Class 6 Solutions Chapter 7 Algebra
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 7 Algebra. Here students can easily find step by step solutions of all the problems for Algebra. Here students will find solutions for Exercise 7.1 and 7.2 and also solution for Chapter Check-up questions. Exercise wise proper solutions. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Algebra Exercise 7.1 Solution
Question no – (1)
Solution :
| No. of rows | 1 | 2 | 3 | ….. | n |
| No. of cadets | 7 | 14 | 21 | ….. | 7n |
Question no – (2)
Solution :
| No. of boxes | 1 | 2 | 3 | ….. | n |
| No. of apples | 60 | 120 | 180 | ….. | 60n |
Question no – (3)
Solution :
Shinky scores 100 marks in Mathematics
Marks in Science = x
∴ Total score in Mathematics and Science,
= (100 + x)
Question no – (4)
Solution :
As per the given question,
Nishu is 4 years younger than Rishu.
Nishu’s age in terms of Rishu’s age
∴ Nishu’s age,
= (z – 4) years
Question no – (5)
Solution :
Given, Ravi gets x as pocket money from his father every month.
Now,
| No. of months | 1 | 2 | 3 | ….. | 12 |
| Total pocket money | x | 2x | 3x | ….. | 12x |
So, Ravi’s pocket money for a year will be 12x.
Question no – (6)
Solution :
Let’s take the number = x
Now, Multiply it by 7 we get,
= 7x
Now, Add 5 to the result and subtract 35 from it we get,
= 7x + 5
= 7x – 30
Question no – (7)
Solution :
As per the question we get,
Cost of 12 notebooks = 12x rupees
Cost of 6 textbooks = 6y rupees
∴ Total cost,
= 12x + 6y rupees
Question no – (8)
Solution :
According to the question,
In a company, the manager’s salary is 5000 more than that of the sales executive.
If the salary of the sales executive is g,
∴ The salary of the manager in terms of g,
= (g + 5000) rupees
Question no – (9)
Solution :
| Number of symbols | 1 | 2 | 3 | …… | 10 |
| Number of dots | 15 | 30 | 45 | …… | 150 |
Therefore, he has to join 150 dots if he draws 10 pictures.
Question no – (10)
Solution :
| Number of handsets | 1 | 2 | 3 | …… | M |
| Number of keys | 16 | 32 | 48 | …… | 16m |
Question no – (11)
Solution :
There are p students and each student paid
= 50 rupees
∴ Total amount,
= 50p
∴ Amount of money that is left after payment,
= (50p – 1800)
Question no – (12)
Solution :
There are 8 tanks and each tank collected x litres.
∴ Total amount,
= 8x litre
and there was 100 litres already.
∴ Total amount,
= (8x + 100) litres
Question no – (13)
Solution :
(a) What will be the length (in cm) of the aluminium strip required to frame the board, if 10 cm extra strip is required to fix it properly?
= The circumference of the board is 2(r + t) cm
∴ The length of required strip is –[2 (r + t) + 10] cm
(b) If x nails are used to repair one board, how many nails will be required to repair 15 such boards?
= X nails are required to repair one board
∴ 15x nails are required to repair 15 boards.
(c) If 500 sq cm extra cloth per board is required to cover the edges, what will be the total area of the cloth required to cover 8 such boards?
= Required cloth for one board is (rt + 500) cm2
∴ Required cloth for 8 board is [8 (rt + 500] cm2
(d) What will be the expenditure for making 23 boards, if the carpenter charges x per board.
= Carpenter charges x rupees per board
∴ Carpenter charges 23x rupees per 23 boards.
Try These 7 (i) :
Question no – (1)
Solution :
Given, t, 5 and 12 and any two number operations. Every expression must have t in it.
Therefore,
= (12 × t) + 5
= 12t + 5.
Question no – (2)
Solution :
(a) The product of b and 100
Algebraic Expression : = 100b
(b) Two times a number added to 9
Algebraic Expression : = 2x + 9
(c) A number is dived by 15, then add 6 with the quotient
Algebraic Expression : = (d ÷ 15) + 6
(d) 7 subtracted from p
Algebraic Expression : = p – 7
(e) 5 times the sum of x and y
Algebraic Expression : = 5(x + y)
(f) Substract 7 from 11 times of a number
Algebraic Expression : = 11y – 7
(g) Multiply 5 with a number then add 1/2 to the result
Algebraic Expression : = 5y + 1/2
(h) d is multiplied by – 4 and the result is added to 81
Algebraic Expression : = {d × (-4)} + 81
(i) The product of p and q added to the quotient of p and q
Algebraic Expression : = pq + p/q
(j) Sum of 3 consecutive numbers
Algebraic Expression : = a +(a + 1) + (a + 2)
(k) One-fourth of y multiplied by the sum of a and 7
Algebraic Expression : = y/4 (a + 7)
Question no – (3)
Solution :
(a) Perimeter of square with side x unit.
= 4x unit
(b) Perimeter of △ with sides as a, b, c units.
= (a + b + c) units
(c) Perimeter of rectangle with length = x and breadth = y.
= 2(x + y) units
(d) Area of square with side s.
= s² square units
(e) Area of rectangle with length = l and breadth = b.
= lb square units
Try These 7 (ii) :
Question no – (1)
Solution :
(a) Her brother is 2 years younger.
Expression = (x – 2)
(b) Her father’s age exceeds her age by 35 years.
Expression = (x + 35)
(c) Mother’s age is 3 years less than that of her father.
Expression = (x + 35 – 3) = (x + 32)
(d) Her grandfather’s age is 8 times of her age.
Expression = 8x
Question no – (2)
Solution :
The rule that is expressed by this formula in words is,
= Perimeter is a rectangle is two times the sum of its length and breath.
Question no – (3)
Solution :
| Position of term | 1st | 2nd | 3rd | 4th | …. | 100th |
| Term | 13 | 26 | 39 | 52 | …. | 1300 |
Question no – (4)
Solution :
| Term | 7 | 12 | 17 | 22 | …. | 5n + 2 |
| Formation of term | 5(1) + 2 | 5(2) + 2 | 5(3) + 2 | 5(4) + 2 | ….. | 5(n) + 2 |
| Position of term | 1 | 2 | 3 | 4 | ….. | n |
Question no – (5)
Solution :
| Position of term | 1 | 2 | 3 | 4 | 5 | 6 |
| Formation of term | 19 × 1 | 19 × 2 | 19 × 3 | 19 × 4 | 19 × 5 | 19 × 6 |
| Term | 19 | 38 | 57 | 76 | 95 | 114 |
Algebra Exercise 7.2 Solution
Question no – (1)
Solution :
Given, The side of an equilateral triangle is = L.
∴ Perimeter of equilateral triangle,
= 3L
So, the Perimeter is 3L
Question no – (2)
Solution :
Given, The side of a regular hexagon is = y.
∴ The perimeter of the hexagon,
= 6y
Therefore, the Perimeter is 6y.
Question no – (4)
Solution :
Smallest natural number is 1.
x times of 3 = 3x
x times of 3 is added to the smallest natural number = 3x + 1
Question no – (5)
Solution :
Smallest two digit number is 10.
6 times q = 6 x q = 6q
6 times q is subtracted from the smallest two digit number = 1 – 6q.
Question no – (6)
Solution :
A cube is a three-dimensional figure having six identical faces.
The length of an edge of a cube is given by l
We know, There are 12 edges of a cube.
∴ The total length of the edges is
= 12l
Therefore, total length of the edges is 12l
Question no – (7)
Solution :
(a) What will be her age 10 years from now?
= x + 10 years
(b) What was her age 3 years ago?
= x – 3 years
(c) Isha’s father is 5 years more than twice her age. What is her father’s age?
Her father’s age,
= 2x + 5 years
Question no – (8)
Solution :
Let, the breath be b.
∴ Then length,
= (4b – 6) metres
Hence, the length is (4b – 6) metres.
Question no – (9)
Solution :
(a) 2, 4, 6, 8, 10, 12, ….
| Term | 2 | 4 | 6 | 8 | …… | 2n |
| formation | 2 × 1 | 2 × 2 | 2 × 3 | 2 × 4 | …… | 2 × n |
The general term is 2n
(b) 5, 10, 15, 20, 25, 30,…
| Term | 5 | 10 | 15 | 20 | …… | 5n |
| formation | 5 × 1 | 5 × 2 | 5 × 3 | 5 × 4 | …… | 5 × n |
The general term is 5n
(c) 3,5,7,9, 11, 13, 15,…
| Term | 3 | 5 | 7 | 9 | …… | 2n + 1 |
| formation | 2 × 1 + 1 | 2 × 2+ 1 | 2 × 3 + 1 | 2 × 4 + 1 | …… | 2 × n + 1 |
The general term is 2n + 1
(d) 6, 11, 16, 21, 26, 31,…
| Term | 6 | 11 | 16 | 21 | …… | 5n + 1 |
| formation | 5 × 1 + 1 | 5 × 2 + 1 | 5 × 3 + 1 | 5 × 4 + 1 | …… | 5 × n + 1 |
The general term is 5n + 1
Question no – (10)
Solution :
As per the given question,
Raghu’s age is 18 years less than two times his cousin’s age
∴ Raghu’s age,
= (2 × x) – 18
= 2x – 18 years
Question no – (11)
Solution :
(a) What will be her age 15 years from now?
= x + 15 years
(b) What was her age 8 years back?
= x – 8 years
(c) Vidushi’s mother’s age is three times her age. What is her mother’s age?
= 3x years
(d) Vidushi’s father is 4 years older than her mother. What is her father’s age?
= 3x + 4 years
(e) Vidushi’s grandfather’s age is 4 years more than 6 times her age. What is her grandfather’s age?
= (6 × x) + 4
= (6x + 4) years
Question no – (12)
Solution :
According to the question,
Akshay is at step = s.
Arjun is at step = (s + 6)
Abhay is at step = (s – 7)
∴ The total numbers of steps
= 3 × s – 9
= 3s – 9
Question no – (13)
Solution :
(a) A pen costs Rs. C. A notebook costs Rs. 5C.
In ordinary language, you can say :
= A notebook costs 5 times of the cost of a pen
(b) Shruti scores x marks. Preeti scores 2x + 1 marks.
In ordinary language, you can say :
= Preeti scores 1 more than 2 times the marks that Shruti scores.
(c) Class VII has n boys. The number of girls in class is
In ordinary language, you can say :
= In class vii, the number of girls is half of the number of boys.
(d) Gautam scores r runs in a cricket match. Sachin scores 3r – 2 runs.
In ordinary language, you can say :
= Sachin scores 2 less than 3 times of Goutam’s score
(e) Monu is y years old. Her sister is y-6 years old and her father is 4y years old.
In ordinary language, you can say :
= Momi’s sister is 6 years younger than Monu.
= Monu’s father is 4 times older than Monu.
Algebra Chapter Check-up Solution
Question no – (1)
Solution :
(a) 6p + q
Variables = p, q
(b) x + y
Variables = x, y
(c) 7a + 5
Variable = a
(d) 8x – 7
Variable = x
Question no – (2)
Solution :
(a) Statement = 5 added to thrice of x
Algebraic expression = 3x + 5
(b) Statement = 5 subtracted from twice of y
Algebraic expression = 2y – 5
(c) Statement = Sum of p and q multiplied by 5
Algebraic expression = 5 (x + y)
(d) Statement = Half of the addition of 5 and thrice of x
Algebraic expression = 1/2 (3x + 5)
(e) Statement = X subtract from twice of y
Algebraic expression = 2x – x
(f) Statement = One-fifth of the summation of p and q
Algebraic expression = (p + q)/5
(g) Statement = X divided by y minus 7
Algebraic expression = x/y – 7
(h) Statement = P divided by 11 added to 10
Algebraic expression = p/11 + 10
| Statement | Algebraic expression | |
| (a) | 5 added to thrice of x | 3x + 5 |
| (b) | 5 subtracted from twice of y | 2y – 5 |
| (c) | Sum of p and q multiplied by 5 | 5 (x + y) |
| (d) | Half of the addition of 5 and thrice of x | 1/2 (3x + 5) |
| (e) | X subtract from twice of y | 2x – x |
| (f) | One-fifth of the summation of p and q | (p + q)/5 |
| (g) | X divided by y minus 7 | x/y – 7 |
| (h) | P divided by 11 added to 10 | p/11 + 10 |
Question no – (3)
Solution :
(a) A pencil costs p and a pen costs ₹ 5p.
In ordinary language,
= A pen costs five time the cost of a pencil
(b) The maximum temperature on a day in Delhi was p° C The minimum temperature was (p – 10) ° C.
In ordinary language,
= The maximum temperature in Delhi was 10° Celsius more than the maximum temperature on the same day.
(c) Sharad used to take p cups of tea in a day. After having some health problem, he takes p-2 cups of tea a day.
In ordinary language,
= After having health problem, Sharad has reduced his tea consumption by 2 cups.
(d) The number of students dropping out of school last year was m, number of students dropping out of school this year is m – 30.
In ordinary language,
= The number of dropping out students has reduced this year by 30.
(e) Khader’s monthly salary was ₹ P in the year 2015. His salary in the 2016 was ₹ (P + 1000).
In ordinary language,
= Khader’s monthly salary has increased 1000 in 2016
(f) The number of girls enrolled in a school last year was g. The number of girls enrolled this year in the school is 3g – 10.
In ordinary language,
= The number of girls enrolled this year is 10 less than twice the number of girls enrolled last year.
Question no – (4)
Solution :
(a) 0 + 10 in simple statement,
= Increased by 15
(b) p – 34 in simple statement,
= Reduced by 34
(c) 18q + 2 in simple statement,
= Increased is times then added 2
(d) r/21 in simple statement,
= Divided by 21
(e) – 6s – 6 in simple statement,
= Multiplied by (- 6) then reduced by 6
(f) t/2 + 3 in simple statement,
= lessen by half then 3 is added
(g) 1/2 (w + x) in simple statement,
= Half of the summation.
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