Exercise 11.1
1. Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
Solution
a) Let us write the letter n for the number of T’s.
If one T is made, n = 1; if two T’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 2 matchsticks are required for making one letter T. So,
For n = 1, the number of matchsticks required is 2 × 1 = 2. [T]
For n = 2, the number of matchsticks required is 2 × 2 = 4. [TT]
For n = 3, the number of matchsticks required is 2 × 3 = 6. [TTT]
…
For n number of T’s, the number of matchsticks requires is = 2 × n = 2n
b) Let us write the letter n for the number of Z’s.
If one Z is made, n = 1; if two Z’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 3 matchsticks are required for making one letter Z. So,
For n = 1, the number of matchsticks required is 3 × 1 = 3. [Z]
For n = 2, the number of matchsticks required is 3 × 2 = 6. [ZZ]
For n = 3, the number of matchsticks required is 3 × 3 = 9. [ZZZ]
…
For n number of Z’s, the number of matchsticks requires is = 3 × n = 3n
c) Let us write the letter n for the number of U’s.
If one U is made, n = 1; if two U’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 3 matchsticks are required for making one letter U. So,
For n = 1, the number of matchsticks required is 3 × 1 = 3. [U]
For n = 2, the number of matchsticks required is 3 × 2 = 6. [UU]
For n = 3, the number of matchsticks required is 3 × 3 = 9. [UUU]
…
For n number of U’s, the number of matchsticks requires is = 3 × n = 3n
d) Let us write the letter n for the number of V’s.
If one V is made, n = 1; if two V’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 2 matchsticks are required for making one letter V. So,
For n = 1, the number of matchsticks required is 2 × 1 = 2. [V]
For n = 2, the number of matchsticks required is 2 × 2 = 4. [VV]
For n = 3, the number of matchsticks required is 2 × 3 = 6. [VVV]
…
For n number of V’s, the number of matchsticks requires is = 2 × n = 2n
e) Let us write the letter n for the number of E’s.
If one E is made, n = 1; if two E’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 5 matchsticks are required for making one letter E. So,
For n = 1, the number of matchsticks required is 5 × 1 = 5. [E]
For n = 2, the number of matchsticks required is 5 × 2 = 10. [EE]
For n = 3, the number of matchsticks required is 5 × 3 = 15. [EEE]
…
For n number of E’s, the number of matchsticks requires is = 5 × n = 5n
f) Let us write the letter n for the number of S’s.
If one S is made, n = 1; if two S’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 5 matchsticks are required for making one letter S. So,
For n = 1, the number of matchsticks required is 5 × 1 = 5. [S]
For n = 2, the number of matchsticks required is 5 × 2 = 10. [SS]
For n = 3, the number of matchsticks required is 5 × 3 = 15. [SSS]
…
For n number of S’s, the number of matchsticks requires is = 5 × n = 5n
g) Let us write the letter n for the number of A’s.
If one A is made, n = 1; if two A’s are made, n = 2 and so on; thus, n can be any natural number 1, 2, 3, 4, 5 …
We see that 6 matchsticks are required for making one letter A. So,
For n = 1, the number of matchsticks required is 6 × 1 = 6. [A]
For n = 2, the number of matchsticks required is 6 × 2 = 12. [AA]
For n = 3, the number of matchsticks required is 6 × 3 = 18. [AAA]
…
For n number of A’s, the number of matchsticks requires is = 6 × n = 6n
2. We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution
(a) And (d), that is, letters T and V have the same pattern rule as for letters L, C and F as the number of matchsticks required to make the letter is 2.
3. Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution
Number of rows = n
n is a variable that can take any value 1, 2, 3, 4, …
Number of cadets in a row = 5
Total number of cadets is given by the rule = 5 × number of rows
= 5n
4. If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution
Number of boxes is given by the variable b.
Number of mangoes in one box = 50
So, total number of mangoes is given by the rule = 50 × number of boxes
= 50b
5. The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution
Number of students is given by the variable s.
Number of pencils given to one student = 5
So, total number of pencils is given by the rule = 5 × number of students
= 5s
6. A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution
Flying time in minutes is given by the variable t.
Distance fled by the bird in one minute is 1 km
Distance covered by the bird in minutes is given by the rule = distance covered in 1 min × flying time = 1 × t = t km.
7. Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Solution
Number of dots in a row = 9
Number of rows is a variable r.
Total number of dots in r rows = number of dots in one row × number of rows
= 9r
Total number of dots in 8 rows = number of dots in one row × number of rows
= 9 × 8 = 72
Total number of dots in 10 rows = number of dots in one row × number of rows
= 9 × 10 = 90
8. Leela is Radha's younger sister. Leela is 4 years younger than Radha. Can you write Leela's age in terms of Radha's age? Take Radha's age to be x years.
Solution
Radha's age = x years
Leela is 4 years younger than Radha, i.e. Radha’s age – 4
Leela’s age = x – 4 years
9. Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution
Number of laddus mother gave away = l.
Number of laddus remaining = 5
Total number of laddus she made = l + 5.
10. Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Solution
Number of oranges in the small box = x
When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside.
That is, Number of oranges in the larger box = Number of oranges in two small boxes + 10 oranges
= x + x + 10 = 2x + 10
11. (a) Look at the following matchstick pattern of squares. The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)
(b) The figure below gives a matchstick pattern of triangles. As in 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.
Solution
a)
1 square 
- 4 matchsticks – 3 × 1 + 1
- 4 matchsticks – 3 × 1 + 1
2 squares 
- 7 matchsticks - 3 × 2 + 1
- 7 matchsticks - 3 × 2 + 1
3 squares 
- 10 matchsticks - 3 × 3 + 1
- 10 matchsticks - 3 × 3 + 1
4 squares 
- 13 matchsticks - 3 × 4 + 1
- 13 matchsticks - 3 × 4 + 1
So, the pattern for n number of squares is 1 more than 3 times the number of squares, n, OR 3n + 1.
b)
1 triangle 
- 3 matchsticks – 2 × 1 + 1
- 3 matchsticks – 2 × 1 + 1
2 triangles 
- 5 matchsticks - 2 × 2 + 1
- 5 matchsticks - 2 × 2 + 1
3 triangles 
- 7 matchsticks - 2 × 3 + 1
- 7 matchsticks - 2 × 3 + 1
So the pattern for n number of triangles is 1 more than twice the number of triangles, n, OR 2n + 1
Exercise 11. 2
1. The side of an equilateral triangle is shown by l. express the perimeter of the equilateral triangle using l.
Solution
An equilateral triangle has three equal sides, l and its perimeter is the sum of the three sides, 3l.
2. The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l. (Hint : A regular hexagon has all its six sides equal in length.)
Solution
Perimeter of the hexagon is = 6l
3. A cube is a three-dimensional figure as shown below. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.
Solution
Length of one edge of a cube = l
Total number of edges in a cube = 12
So, total length of the edges of a cube = 12l
4. The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. AB is a diameter of the circle; C is its centre. Express the diameter of the circle (d) in terms of its radius (r).
Solution
C is the centre of the circle. AC and CB are the radius(r) of the circle.
So, Diameter AB = AC + CB = r + r = 2r
5. To find sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.
Solution
For any three whole numbers a, b, and c
(a + b) + c = a + (b + c)
Exercise 11.3
1. Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication. (Hint: Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 × 8) + 7; make the other expressions.)
Solution
Some expressions that can be formed using 5, 8 and 7 are given below.
5 – (8 + 7), 5 + (8 + 7), 5 × (8 + 7), 5 – (8 × 7), 5 + (8 × 7), 5 – (8 – 7)…
[Many such expressions can be formed using the numbers]
2. Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) (7 × 20) – (5 × 10) – 45 + p
Solution
Expression (c) and (d) are expressions with only numbers.
3. Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
(a) z +1, z – 1, y + 17, y – 17
(b) 17y, y/17, 5 z
(c) 2y + 17, 2 y – 17
(d) 7 m, – 7 m + 3, – 7 m – 3
Solution
a) 1 is added to z, 1 is subtracted from z, 17 is added to y and 17 is subtracted from y. So, operations involved are addition, subtraction, addition, subtraction.
b) 17 and y are multiplied, y is divided by 17, 5 is multiplied to z. So, the operations involved are multiplication , division and multiplication.
c) y is multiplied by 2 and is added to 17, y is multiplied by 2 and 17 is subtracted from the product. So the operations involved are multiplication and addition, multiplication and subtraction.
d) m is multiplied to 7, m is multiplied to -7 and 3 added, m is multiplied to -7 and 3 subtracted. So, the operations involved are multiplication, multiplication and addition, multiplication and subtraction.
4. Give expressions for the following cases.
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) – p multiplied by 5
(g) – p divided by 5
(h) p multiplied by – 5
Solution
(a) 7 + p
b) p – 7
c) 7 p
d) p /7
e) –m – 7
f) -5p
g) –p/5
h) -5p
5. Give expressions in the following cases.
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to 16.
Solution
a) 11 + 2m
b) 2m – 11
c) 5y + 3
d) 5y – 3
e) -8y
f) -8y + 5
g) 16 – 5y
h) -5y + 16
6. (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution
a) t + 4, t – 4, 4t, t/4, 4/t, 4 – t
b) 2y + 7, 2y – 7, 2y/7, 7/2y, 2 + 7y …
[Many expressions can be formed]
Exercise 11.4
1. Answer the following:
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother's age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father's age?
(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?
(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.
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena
is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? 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.
(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.
Solution
a)
i) Sarita’s present age is y years.
Her age 5 years from now will be her present age + 5 = y + 5 years.
ii) Her age 3 years back will be her present age – 3 years = y – 3 years
iii) Sarita’s grandfather is 6 times her age = 6y years
iv) Grandmother’s age = Grandfather’s age – 2 = 6y – 2 years
v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age = 5 + 3y
b) breadth of the rectangular hall = b meters
length of the hall = 4 meters less than 3 times the breadth = 3b – 4 meters
c) height of the rectangular box = h cm
length = 5 times the height = 5h cm
breadth = 10 cm less than the length = 5h – 10 cm
d) Step at which Meena is = s
Step at which Beena is = Meena’s step + 8 steps = s + 8
Step at which Leena is = Meena’s step – 7 = s – 7
Total number of steps = 4s – 10
e) Distance travelled by the bus in 1 hour = v km
Distance travelled in 5 hours = 5v km
So, total distance from Daspur to Beespur = Distance travelled in 5 hours + 20 = 5v + 20 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 (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)
(a) A notebook costs Rs p. A book costs Rs 3 p.
(b) Tony puts q marbles on the table. He has 8 q marbles in his box.
(c) Our class has n students. The school has 20 n students.
(d) Jaggu is z years old. His uncle is 4 z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution
a) The cost of a book is three times the cost of a notebook.
b) Tony has 8 times the number of marbles in his box than on the table.
c) Total number of students in the class is 20 times that of our class.
d) Jaggu’s uncle is 4 times the age of Jaggu and his aunt is 3 less than 4 times the age of Jaggu.
e) The total number of dots is 5 times the number of rows.
3. (a) Given Munnu’s age to be x years, can you guess what (x – 2) may show? (Hint: Think of Munnu’s younger brother.)
Can you guess what (x + 4) may show? What (3 x + 7) may show?
(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? y + 7, y – 3, y + 4½ , y – 2½
(c) Given n students in the class like football, what may 2n show? What may n/2 show? (Hint: Think of games other than football).
Solution
a) If Munnu’s age is x years, then (x – 2) may show the age is 2 years younger than Munnu.
(x + 4) shows that the age is 4 years older than Munnu and 3x + 7 shows that the age considered is 7 more than thrice the age of Munnu.
b) Sara’s present age = y years
y + 7 represents her future age, i.e., her age after 7 years.
y – 3 represents her past age, i.e. her age 3 years ago.
y + 4½ represents her future age, i.e her age after 4½ years
y - 2½ represents her past age, i.e. her age 2½ years ago.
c) 2n represents twice the number of students who like football. n/2 represents half the number of students who like football.
It may represent the number of students who like cricket, or basketball etc.
Exercise 11.5
1. State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
Solution
a) Is an equation with one variable, x.
b) Is not an equation, is an inequation.
c) Is not an equation, It is a numerical equation.
d) Is not an equation, It is a numerical equation.
e) Is not an equation, is an inequation.
f) Is an equation with one variable, x.
g) Is not an equation, is an inequation.
h) Is an equation with one variable, n.
i) Is not an equation, It is a numerical equation.
j) Is an equation with one variable, p.
k) Is an equation with one variable, y.
l) Is not an equation, is an inequation
m) Is not an equation, is an inequation.
n) Is not an equation, It is a numerical equation.
o) Is an equation with one variable, x.
2. Complete the entries in the third column of the table.
Solution
3. Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
Solution
a) 5m = 60
For m = 10, 5m = 5(10) = 50
For m = 5, 5m = 5(5) = 25
For m = 12, 5m = 5(12) = 60
For m = 15, 5m = 5(15) = 75
So, we see that m = 12 satisfy the equation and is the solution.
b) n + 12 = 20
For n = 12, n + 12 = 12 + 12 = 24
For n = 8, n + 12 = 8 + 12 = 20
For n = 20, n + 12 = 20 + 12 = 32
For n = 0, n + 12 = 0 + 12 = 12
So, we see that n = 8 satisfy the equation and is the solution.
c) p – 5 = 5
For p = 0, p – 5 = 0 – 5 = -5
For p = 10, p – 5 = 10 – 5 = 5
For p = 5, p – 5 = 5 – 5 = 0
For p = -5, p – 5 = -5 – 5 = -10
So, we see that p = 10 satisfy the equation and is the solution.
d) q/2 = 7
For q = 7, q/2 = 7/2
For q = 2, q/2 = 2/2 = 1
For q = 10, q/2 = 10/2 = 5
For q = 14, q/2 = 14/2 = 7
So, we see that q = 14 satisfy the equation and is the solution.
e) r – 4 = 0
For r = 4, r – 4 = 4 – 4 = 0
For r = -4, r – 4 = -4 – 4 = -8
For r = 8, r – 4 = 8 – 4 = 4
For r = 0, r – 4 = 0 – 4 = -4
So, we see that r = 4 satisfy the equation and is the solution.
f) x + 4 = 2
For x = -2, x + 4 = -2 + 4 = 2
For x = 0, x + 4 = 0 + 4 = 4
For x = 2, x + 4 = 2 + 4 = 6
For x = 4, x + 4 = 4 + 4 = 8
So, we see that x = -2 satisfy the equation and is the solution.
4. (a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.
Solution
We see that m = 6 is the solution of the equation.
(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.
Solution
We see that t = 7 is the solution of the equation.
(c) Complete the table and find the solution of the equation z/3 =4 using the table.
Solution
We see that z = 12 is the solution of the equation.
(d) Complete the table and find the solution to the equation m – 7 = 3.
Solution
We see that m = 10 is the solution of the equation.
5. Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!
Solution
A square has 4 sides.
“Counting every corner Thrice and no more!”
Implies we go around the square thrice, i.e. 12 times.
“Add the count to me To get exactly thirty four!”
That is, when 12 is added to a number, say x, we get 34.
12 + x = 34
x = 34 – 12 = 22
So, 22 is the number.
(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!
Solution
If 23 is the number for Sunday,
Then counting up, Saturday is 22, Friday is 21, Thursday is 20, Wednesday is 19, Tuesday is 18, Monday is 17 and Sunday is 16.
Therefore, the number considered is 16.
(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!
Solution
Let the number be x.
If we take away 6 from x, we get a cricket team. [A cricket team has 11 players].
x – 6 = 11
x = 11 + 6 = 17
The special number is 17.
(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
Solution
Let the number be x.
22 – x = x
22 = x + x = 2x
x = 22/2 = 11
The number is 11.
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