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.

 (a) A pattern of letter T as T

(b) A pattern of letter Z as Z

(c) A pattern of letter U as U

(d) A pattern of letter V as V

(e) A pattern of letter E as E

(f) A pattern of letter S as S

(g) A pattern of letter A as R

Solution:

(a) The given pattern is T

Two matchsticks are used in the given pattern.

∴ To create n such patterns, we need 2n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =2n

(b) The given pattern is Z

Three matchsticks are used in the given pattern.

∴ To create n such patterns, we need 3n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =3n

(c)The given pattern is U

Three matchsticks are used in the given pattern.

∴ To create n such patterns, we need 3n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =3n

(d) The given pattern is V

Two matchsticks are used in the given pattern.

∴ To create n such patterns, we need 2n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =2n

(e) The given pattern is E

Five matchsticks are used in the given pattern.

∴ To create n such patterns, we need 5n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =5n

(f) The given pattern is S

Five matchsticks are used in the given pattern.

∴ To create n such patterns, we need 5n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =5n

(g) The given pattern is R

six matchsticks are used in the given pattern.

∴ To create n such patterns, we need 6n matchsticks.

The rule which can give the number of matchsticks required to make the given pattern =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:

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: 

Given: The number of cadets in a row = 5

Let n be the number of rows

Total number of cadets = number of cadets in a row × number of rows

= 5n

Hence, total number of cadets = 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:

Let the number of boxes= b

Number of mangoes in a box = 50

Total number of mangoes = number of mangoes in a box × number of boxes

= 50b

Hence total 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? (Uses for the number of students.)

Solution:

The number of pencils distributed per student is 5.

Let the given number of students be s

Total number of pencils needed = Number of pencils distributed per student × Number of students

Total number of pencils needed = 5 × s = 5s.

6. A bird flies 1 kilometre 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:

Time taken by bird = t minutes

Speed of bird = 1 km per minute

Therefore, Distance covered by bird = speed x time = 1 x 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:

Given, number of dots in each row =9
Number of rows =r
∴ Total number of dots in r rows =9r

Case 1: When number of rows =8

Number of dots in r rows =9r

∴ Number of dots in 8 rows =9×8=72

Case 2: When number of rows =10

Number of dots in r rows =9r

∴ Number of dots in 10 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:

We have assumed Radha’s age to be x years.

As Leela is Radha’s younger sister her age must be less than that of Radha.

According to given condition, Leela is 4 years younger than Radha,

So, Leela’s age=(x-4) years

Hence, the Leela’s age can be expressed as (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:

Let the number of laddus given away be ′l′.
Number of laddus remaining = 5

Therefore, the total number of laddus made will be equal to the sum of laddus given away and laddus remaining.
Total number of laddus = l+5

11. 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:

Given, number of oranges in small box = x
Number of small boxes = 2
Therefore, total number of oranges in small boxes = x + x=2x
Remaining oranges = 10

Number of oranges in the large box will be the sum of oranges in the small boxes and the remaining oranges.
∴ Number of oranges in large box = 2x+10

12. (a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are not separate. Two neighboring 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) Fig 11.7 gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

Solution:

(a)

(a) 4 matchsticks

(b) 7 matchsticks

(c) 10 matchsticks

(d) 13 matchsticks

If we remove 1 matchstick from each then it forms a table of 3 i.e.,3,6,9, 12,

So, required equation = 3x+1, x= number of squares

(b)

(a) 3 matchsticks

(b) 5 matchsticks

(c) 7 matchsticks

(d) 9 matchsticks

If we remove 1 matchstick from each then it forms a table of 2 i.e., 2,4,6,8…………

So, required equation = 2x+1, x= number of triangles

 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 is a triangle with three equal sides.

Given, side of the equilateral triangle = l

Perimeter is the total sum of lengths of all the bounding sides of a planar object.

∴ Perimeter of an equilateral triangle =3×side length = 3l

2. The side of a regular hexagon (Fig 11.10) 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:

We will be using the formula for the perimeter of the regular hexagon

We know that the perimeter of a regular hexagon is 6 x side-length.

Given, length of each side of a hexagon = l units

Hence, the perimeter of the regular hexagon = 6 × l

Therefore, 6l units is the perimeter of the given regular hexagon.

3. A cube is a three-dimensional figure as shown in Fig 11.11. 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:

In a cube,

We have six identical squares on 6 faces

So, all the sides must be equal.

Therefore, Length of all side of cube = l

We need to find total side of cube 

Total length of all sides = AB + BC + CD + AD + BE + EF + CF + GF + DG + GH + HE + AH

= l + l + l + l + l + l + l + l + l + l + l + l

= 12 l

4. The diameter of a circle is a line which joins two points on the circle and also passes through the center of the circle. (In the adjoining figure (Fig 11.12) AB is a diameter of the circle; C is its center.) Express the diameter of the circle (d) in terms of its radius (r).

Solution:

In the given figure, AB is the diameter and CP is the radius.

AB = AC + CB

AC= CB =CP [Radii of the same circle]

⇒AB = CP+CP

⇒AB = 2CP

Given, diameter AB= d and radius CP = r

∴ d =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:

Associative property of addition can be expressed as follows:

(a + b) + c = a + (b + c)

The sum of two numbers ‘a’ and ‘b’ can be added to a third number ‘c’ or the sum of two numbers ‘b’ and ‘c’ can be added to a third number ‘a’. In both the cases, the final answer will be the same which is (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:

(a) (8 x 5) – 7
(b) (8 + 5) – 7
(c) (8 x 7) – 5
(d) (8 + 7) – 5
(e) 5 x (7 + 8)
(f) 5 + (7 x 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)
(i) 5- (7 – 8)
(j) 5 + (7 – 8)

Similarly, more expressions could be made.

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:

(a) y + 3 has a variable y
(b) (7 × 20) – 8z has a variable z
(c) 5 (21 – 7) + 7 × 2 is an expression that is independent of variables and has only numerical constants. Thus, this is an expression with numbers only.
(d) 5 is an expression that is independent of variables and has only a numerical constant. Thus, this is an expression with numbers only.
(e) 3x has a variable x
(f) 5 – 5n has a variable n
(g) (7 × 20) – (5 × 10) – 45 + p has a variable p

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) z + 1, z – 1, y + 17, y – 17

z + 1

=> 1 is added to z, hence it is addition.

z – 1

=> 1 is subtracted from z, hence it is subtraction.

y + 17

=> 17 is added to y, hence it is addition.

y – 17

=> 17 is subtracted from y , hence it is subtraction.

(b) 17y , y / 17, 5z

17y

=> y is multiplied by 17 , hence it is multiplication.

y / 17

=> y is divided by 17 , hence it is division.

5z

=> z is multiplied by 5, hence it is multiplication.

(c) 2y + 17, 2y – 17

2y + 17

=> y is multiplied by 2 and 17 is added to the result, hence it is multiplication and addition.

2y – 17

=> y is multiplied by 2 and 17 is subtracted from the result, hence it is multiplication and subtraction.

(d) 7m, -7m + 3, -7m – 3

7m

=> m is multiplied by 7, hence it is multiplication.

-7m + 3

=> m is multiplied by -7 and 3 is added to the result, hence it is multiplication and addition

-7m – 3

=> m is multiplied by -7 and 3 is subtracted from the result, hence it is multiplication and subtraction.

Hence, we have identified all the operations used in the given expressions.

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) p + 7 (b) p – 7 (c) 7 p (d) 7 p (e) – m – 7 (f) – 5p (g) – 5 p (h) – 5 p

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) 2m + 11 (b) 2m – 11 (c) 5y + 3 (d) 5y – 3 (e) – 8y (f) – 8 y + 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:

•  Different expressions using t and 4, with number not more than one number are listed below.
• (t + 4)
• (t – 4)
• 4t
• t/4
• 4/t
• (4 – t)
• The following expressions can be formed using y, 2 and 7:
• 2y+7
•  2y−7
• 7y+2
• 7y−2
• 2y7 

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 are 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) Given, Sarita’s present age is y years.

(i) To get Sarita’s age five years from now, we should add 5 to her present age.

∴ Sarita’s age 5 years from now =y+5

(ii)

To get Sarita’s age 3 years back, we should subtract 3 from her present age.

∴ Sarita’s age 3 years back =y−3

(iii) Since, Sarita’s age is y, her grandfather’s age will be 6×y =6y

(iv) Sarita’s grandfather’s age =6y
It is given that Sarita’s Grandmother is 2 years younger than her grandfather

∴ Sarita’s grandmother’s age =6y−2

(v) Given, Sarita’s age =y

3 times Sarita’s age = 3×y =3y

Since, Sarita’s father’s age is 5 years more than 3 times her age, therefore his age =3y+5

(b) Given, breadth = b meters

∴ 3 times the breadth = 3×breadth = 3×b = 3b

Since the length is 4 meters less than 3 times the breadth, length

= (3b − 4) meters.

(c) Given that,

Height of the box h cm

Now,

Length of the box = 5 * (Height of the box)

= 5h cm

And

Breadth of the box = Length of the box – 10

= (5h – 10) cm.

(d)  Given that,

Meena’s position = s step

Beena’s position= meena’s position + 8

= (s + 8) step

Leena’s position = meena’s position – 7

= (s – 7) step 

Given that

The total number of steps to the hill top is 10 less than 4 times what meena has reached

Total number of steps = 4 * (meena’s position) – 10

= 4s – 10 

(e) Given,

Speed of bus = v km/hr

Now,

Distance = speed * time

Distance travelled by the bus in 5 hours

= v * 5 km

= 5v km.

Distance from Daspur to Beespur

= Distance travelled by bus in 5 hour + 20 km

= (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 ` p. A book costs ` 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)A book costs 3 times the costs of a notebook.

(b) Tony’s box contains 8 times the number of marbles on the table

(c) Total number of students in the school is 20 times that of our class

(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle

(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 +4 1 2, y – 2 1 2 .

(c) Given n students in the class like football, what may 2n show? What may 2 n show? (Hint: Think of games other than football).

Solution:

(a)(x − 2) represents that the person, whose age is (x − 2) years, is 2 years younger to Munnu.

(x + 4) represents that the person, whose age is (x + 4) years, is 4 years elder to Munnu.

(3x + 7) represents that the person, whose age is (3x + 7) years, is elder to Munnu and his age is 7 years more than three times of the age of Munnu.

(b) In future

After n years from now, Sara’s age will be (y + n) years.

In past

n years ago, Sara’s age was (y − n) years.

(y + 7) represents that the person, whose age is (y + 7) years, is 7 years elder to Sara.

(y − 3) represents that the person, whose age is (y − 3) years, is 3 years younger to Sara.

(y + 4(1/2)) represents that the person, whose age is (y +4(1/2)) years, is 4(1/2) years elder to Sara.

(y − 2(1/2)) represents that the person, whose age is (y − 2(1/2)) years, is 2(1/2) years younger to Sara.

(c) 2n may represent the number of students who like either football or some other game such as cricket whereas   represents the number of students who like cricket, out of the total number of students who like football.

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.

(a) 17 = x + 7

(b) (t – 7) > 5

(c) 4 2 = 2

(d) (7 × 3) – 19 = 8

(e) 5 × 4 – 8 = 2 x

(f ) x – 2 = 0

(g) 2m < 30

(h) 2n + 1 = 11

(i) 7 = (11 × 5) – (12 × 4)

( j) 7 = (11 × 2) + p

(k) 20 = 5y

( l) 3 2 q < 5

(m) z + 12 > 24

(n) 20 – (10 – 5) = 3 × 5

(o) 7 – x = 5

Solution:

(a) An equation with variable x

(b) An inequality

(c) No, it is a numerical equation.

(d) No, it is a numerical equation.

(e) An equation with variable x

(f) An equation with variable x

(g) An inequality

(h) An equation with variable n

(i) No, it is a numerical equation.

(j) An equation with variable p

(k) An equation with variable y

(l) An inequality

(m) An inequality

(n) No, it is a numerical equation.

(o) An equation with variable x

2. Complete the entries in the third column of the table.

Solution:

(a) No (b) Yes (c) No (d) No (e) No (f) Yes (g) No (h) No (i) Yes (j) Yes (k) No (l) No (m) No (n) No (o) No (p) No (q) Yes

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.

 (a) 5m = 60            (10, 5, 12, 15)

 (b) n + 12 = 20       (12, 8, 20, 0)

(c) p – 5 = 5             (0, 10, 5 – 5)

(d) q /2 = 7                (7, 2, 10, 14)

(e) r – 4 = 0              (4, – 4, 8, 0)

(f) x + 4 = 2               (– 2, 0, 2, 4)

Solution:

(a) 5m = 60

m = 12 is a solution to the given equation because for m = 12,

5m = 5 × 12 = 60 and hence, the equation is satisfied.

m = 10 is not a solution to the given equation because for m = 10,

5m = 5 × 10 = 50, and not 60

m = 5 is not a solution to the given equation because for m = 5,

5m = 5 × 5 = 25, and not 60

m = 15 is not a solution to the given equation because for m = 15,

5m = 5 × 15 = 75, and not 60

(b) n + 12 = 20

n = 8 is a solution to the given equation because for n = 8,

n + 12 = 8 + 12 = 20 and hence, the equation is satisfied.

n = 12 is not a solution to the given equation because for n = 12,

n + 12 = 12 + 12 = 24, and not 20

n = 20 is not a solution to the given equation because for n = 20,

n + 12 = 20 + 12 = 32, and not 20

n = 0 is not a solution to the given equation because for n = 0,

n + 12 = 0 + 12 = 12, and not 20

(c) p − 5 = 5

p = 10 is a solution to the given equation because for p = 10,

p − 5 = 10 − 5 = 5 and hence, the equation is satisfied.

p = 0 is not a solution to the given equation because for p = 0,

p − 5 = 0 − 5 = −5, and not 5

p = 5 is not a solution to the given equation because for p = 5,

p − 5 = 5 − 5 = 0, and not 5

p = −5 is not a solution to the given equation because for p = −5,

p − 5 = − 5 − 5 = −10, and not 5

(d) 

q = 14 is a solution to the given equation because for q = 14,

and hence, the equation is satisfied.

q = 7 is not a solution to the given equation because for q = 7,

, and not 7

q = 2 is not a solution to the given equation because for q = 2,

, and not 7

q = 10 is not a solution to the given equation because for q = 10,

, and not 7

(e) r − 4 = 0

r = 4 is a solution to the given equation because for r = 4,

r − 4 = 4 − 4 = 0 and hence, the equation is satisfied.

r = −4 is not a solution to the given equation because for r = −4,

r − 4 = − 4 − 4 = −8, and not 0

r = 8 is not a solution to the given equation because for r = 8,

r − 4 = 8 − 4 = 4, and not 0

r = 0 is not a solution to the given equation because for r = 0,

r − 4 = 0 − 4 = −4, and not 0

(f) x + 4 = 2

x = −2 is a solution to the given equation because for x = −2,

x + 4 = − 2 + 4 = 2 and hence, the equation is satisfied.

x = 0 is not a solution to the given equation because for x = 0,

x + 4 = 0 + 4 = 4, and not 2

x = 2 is not a solution to the given equation because for x = 2,

x + 4 = 2 + 4 = 6, and not 2

x = 4 is not a solution to the given equation because for x = 4,

x + 4 = 4 + 4 = 8, and not 2

4. (a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35

(c) Complete the table and find the solution of the equation z/3 =4 using the table.

(d) Complete the table and find the solution to the equation m – 7 = 3

Solution:

(a) For m + 10, the table can be constructed as follows.

By inspection, we can find that m = 6 is the solution of the above equation as for m = 6, m + 10 = 6 + 10 = 16

(b) For 5t, the table can be constructed as follows.

By inspection, we can find that t = 7 is the solution of the above equation as for t = 7, 5t = 5 × 7 = 35

(c) For  , the table can be constructed as follows.

By inspection, we can find that z = 12 is the solution of the above equation as for z = 12,   = 4

(d) For m − 7, the table can be constructed as follows.

By inspection, we can find that m = 10 is the solution of the above equation as for m = 10, m − 7 = 10 − 7 = 3

• 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!

(ii) For each day of the week Make an up count from me If you make no mistake, you will get twenty-three!

(iii) I am a special number Take away from me a six! A whole cricket team You will still be able to fix!

(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:

(i)There are 4corners in a square.

Thrice the number of corners in the square will be 3 × 4 = 12

When this result, i.e. 12, is added to the number, it comes to be 34. Therefore, the number will be the difference of 34 and 12 i.e., 34 − 12 = 22

(ii) 23 was the result when the old number was up-counted on Sunday.

22 was the result when the old number was up-counted on Saturday.

21 was the result when the old number was up-counted on Friday.

20 was the result when the old number was up-counted on Thursday.

19 was the result when the old number was up-counted on Wednesday.

18 was the result when the old number was up-counted on Tuesday.

17 was the result when the old number was up-counted on Monday.

Therefore, number taken at the start = 17 − 1 = 16

(iii) In a cricket team, there are 11 players. Hence, the number is such that when 6 is subtracted from it, the result is 11. Therefore, the number is 11 + 6 = 17

(iv) The number is such that when it is subtracted from 22, the result is again the number itself. The number is 11, which again gives 11 when it is subtracted from 22.