Page No 226:
Question 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 A
Answer:
(a)
From the figure, it can be observed that it will require two matchsticks to make a T. Therefore, the pattern is 2n.
(b)
From the figure, it can be observed that it will require three matchsticks to make a Z. Therefore, the pattern is 3n.
(c)
From the figure, it can be observed that it will require three matchsticks to make a U. Therefore, the pattern is 3n.
(d)
From the figure, it can be observed that it will require two matchsticks to make a V. Therefore, the pattern is 2n.
(e)
From the figure, it can be observed that it will require five matchsticks to make an E. Therefore, the pattern is 5n.
(f)
From the figure, it can be observed that it will require five matchsticks to make a S. Therefore, the pattern is 5n.
(g)
From the figure, it can be observed that it will require six matchsticks to make an A. Therefore, the pattern is 6n.
Page No 227:
Question 2:
We already know the rule for the pattern of letters L, C and F. Some of the letters from some of the letters out of (a) T, (b) Z, (c) U, (d) V, (e) E, (f) S, (g) R give us the same rule as that given by L. Which are these? Why does this happen?
Answer:
It is known that L requires only two matchsticks. Therefore, the pattern for L is 2n. Among all the letters given above in question 1, only T and V are the two letters which require two matchsticks.
Hence, (a) and (d)
Page No 227:
Question 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.)
Answer:
Let number of rows be n.
Number of cadets in one row = 5
Total number of cadets = Number of cadets in a row × Number of rows
= 5n
Page No 227:
Question 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.)
Answer:
Let the number of boxes be b.
Number of mangoes in a box = 50
Total number of mangoes = Number of mangoes in a box × Number of boxes
= 50b
Page No 227:
Question 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.)
Answer:
Let the number of students be s.
Pencils given to each student = 5
Total number of pencils
= Number of pencils given to each student × Number of students
= 5s
Page No 227:
Question 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.)
Answer:
Let the flying time be t minutes.
Distance covered in one minute = 1 km
Distance covered in t minutes = Distance covered in one minute × Flying time
= 1 × t = t km
Page No 227:
Question 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?
Answer:
Number of dots in 1 row = 9
Number of rows = r
Total number of dots in r rows = Number of rows × Number of dots in a row
= 9r
Number of dots in 8 rows = 8 × 9 = 72
Number of dots in 10 rows = 10 × 9 = 90
Page No 227:
Question 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.
Answer:
Let Radha’s age be x years.
Leela’s age = Radha’s age − 4
= (x − 4) years
Page No 227:
Question 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?
Answer:
Number of laddus given away = l
Number of laddus remaining = 5
Total number of laddus = Number of laddus given away + Number of laddus
remaining
= l + 5
Page No 227:
Question 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?
Answer:
Number of oranges in one small box = x
Number of oranges in two small boxes = 2x
Number of oranges left = 10
Number of oranges in the large box = Number of oranges in two small boxes
+ Number of oranges left
= 2x + 10
Page No 227:
Question 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 given figure gives a matchstick pattern of triangles. Find the general rule that gives the number of matchsticks in terms of the number of triangles.
Answer:
(a) It can be observed that in the given matchstick pattern, the number of
matchsticks are 4, 7, 10, and 13, which is 1 more than thrice of the number of squares in the pattern.
Hence, the pattern is 3n + 1, where n is the number of squares.
(b) It can be observed that in the given matchstick pattern, the number of
matchsticks are 3, 5, 7, and 9, which is 1 more than twice of the number of triangles in the pattern.
Hence, the pattern is 2n + 1, where n is the number of triangles.
Page No 230:
Question 1:
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Answer:
Side of equilateral triangle = l
Perimeter = l + l + l = 3l
Page No 230:
Question 2:
The Side of a regular hexagon (see the given figure) is denoted by l. Express the perimeter of the hexagon using l.
(Hint: A regular hexagon has all its six sides equal in length.)
Answer:
Side of regular hexagon = l
Perimeter = 6l
Page No 230:
Question 3:
A cube is a threedimensional figure as shown in the given figure. 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.
Answer:
Length of edge = l
Number of edges = 12
Total length of the edges = Number of edges × Length of one edge
= 12l
Page No 231:
Question 4:
The diameter of a circle is a line which joins two points on the circle and also passed through the centre of the circle. (In the adjoining figure AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius(r).
Answer:
Diameter = AB = AC + CB = r + r = 2r
d = 2r
Page No 231:
Question 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.
Answer:
For any three whole numbers a, b, and c,
(a + b) + c = a + (b + c)
Page No 233:
Question 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.)
Answer:
Many expressions can be formed by using the three numbers 5, 7, and 8.
Some of these are as follows.
5 × (8 − 7)
5 × (8 + 7)
(8 + 5) ×7
(8 − 5) ×7
(7 + 5) ×8
(7 − 5) ×8
Page No 233:
Question 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
Answer:
It can be observed that the expressions in alternatives (c) and (d) are formed by using numbers only.
Page No 233:
Question 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)
(c) 2y + 17, 2y − 17 (d) 7m, − 7m + 3, − 7m − 3
Answer:
(a) Addition as 1 is added to z.
Subtraction as 1 is subtracted from z.
Addition as 17 is added to y.
Subtraction as 17 is subtracted from y.
(b) Multiplication as y is multiplied with 17.
Division as y is divided by 17.
Multiplication as z is multiplied with 5.
(c) Multiplication and addition
y is multiplied with 2, and 17 is added to the result.
Multiplication and subtraction
y is multiplied with 2, and 17 is subtracted from the result.
(d) Multiplication as m is multiplied with 7.
Multiplication and addition as m is multiplied with −7, and 3 is added
to the result.
Multiplication and subtractionas m is multiplied by −7, and 3 is
subtracted from the result.
Page No 233:
Question 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
Answer:
(a) p + 7
(b) p − 7
(c) 7p
(d)
(e) − m − 7
(f) − 5p
(g)
(h) − 5p
Page No 234:
Question 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
Answer:
(a) 2m + 11
(b) 2m − 11
(c) 5y + 3
(d) 5y − 3
(e) − 8y
(f) − 8y + 5
(g) 16 − 5y
(h) − 5y + 16
Page No 234:
Question 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.
Answer:
(a) t + 4, t − 4, 4t, , , 4 − t, 4 + t
(b) 2y + 7, 2y − 7, 7y + 2,…
Page No 235:
Question 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.
Answer:
(a) (i) Sarita’s age after 5 years from now = Sarita’s present age + 5
= y + 5
(ii) 3 years ago, Sarita’s age = Sarita’s present age − 3
= y − 3
(iii) Grandfather’s age = 6 × Sarita’s present age = 6y
(iv) Grandmother’s age = Grandfather’s present age − 2 = 6y − 2
(v) Father’s age = 5 + 3 × Sarita’s present age = 5 + 3y
(b) Length = 3 × Breadth − 4
l = (3b − 4) metres
(c) Length = 5 × Height
l = 5h cm
Breadth = 5 × Height − 10
b = (5h − 10) cm
(d) Step at which Beena is = (Step at which Meena is) + 8
= s + 8
Step at which leena is = (Step at which Meena is) − 7
= s − 7
Total steps = 4 × (Step at which Meena is) − 10 = 4s − 10
(e) Speed = v km/hr
Distance travelled in 5 hrs = 5 × v = 5v km
Total distance between Daspur and Beespur = (5v + 20) km
Page No 236:
Question 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 note book 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.
Answer:
(a) A book costs three times the cost 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.
Page No 236:
Question 3:
(a) Given Munnu’s age to be x years, can you guess what (x − 2) may show?
(Hint: Think of Mannu’s younger brother.)
Can you guess what (x + 4) may show? What (3x + 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?
(c) Given n students in the class like football, what may 2n show? What may show? (Hint: Think of games other than football).
Answer:
(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 + ) represents that the person, whose age is (y + ) years, is years elder to Sara.
(y −) represents that the person, whose age is (y − ) years, is 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.
Page No 240:
Question 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) (d) (7 × 3) − 19 = 8
(e) 5 × 4 − 8 = 2x (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)
(m) z + 12 > 24 (n) 20 − (10 − 5) = 3 × 5
(o) 7 − x = 5
Answer:
(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
Page No 241:
Question 2:
Complete the entries in the third column of the table.
S. No. 
Equation 
Value of variable 
Equation satisfied Yes/No 
(a) 
10y = 80 
y = 10 
 
(b) 
10y = 80 
y = 8 
 
(c) 
10y = 80 
y = 5 
 
(d) 
4l = 20 
l = 20 
 
(e) 
4l = 20 
l = 80 
 
(f) 
4l = 20 
l = 5 
 
(g) 
b + 5 = 9 
b =5 
 
(h) 
b + 5 = 9 
b = 9 
 
(i) 
b + 5 = 9 
b = 4 
 
(j) 
h − 8 = 5 
h = 13 
 
(k) 
h − 8 = 5 
h = 8 
 
(l) 
h − 8 = 5 
h = 0 
 
(m) 
p + 3 = 1 
p = 3 
 
(n) 
p + 3 = 1 
p = 1 
 
(o) 
p + 3 = 1 
p = 0 
 
(p) 
p + 3 = 1 
P = − 1 
 
(q) 
p + 3 = 1 
P = − 2 
 
Answer:
(a) 10y = 80
y = 10 is not a solution to the given equation because for y = 10,
10y = 10 × 10 = 100, and not 80
(b) 10y = 80
y = 8 is a solution to the given equation because for y = 8,
10y = 10 × 8 = 80 and hence, the equation is satisfied.
(c) 10y = 80
y = 5 is not a solution to the given equation because for y = 5,
10y = 10 × 5 = 50, and not 80
(d) 4l = 20
l = 20 is not a solution to the given equation because for l = 20,
4l = 4 × 20 = 80, and not 20
(e) 4l = 20
l = 80 is not a solution to the given equation because for l = 80,
4l = 4 × 80 = 320, and not 20
(f) 4l = 20
l = 5 is a solution to the given equation because for l = 5,
4l = 4 × 5 = 20 and hence, the equation is satisfied.
(g) b + 5 = 9
b = 5 is not a solution to the given equation because for b = 5,
b + 5= 5 + 5 = 10, and not 9
(h) b + 5 = 9
b = 9 is not a solution to the given equation because for b = 9,
b + 5= 9 + 5 = 14, and not 9
(i) b + 5 = 9
b = 4 is a solution to the given equation because for b = 4,
b + 5= 4 + 5 = 9 and hence, the equation is satisfied.
(j) h − 8 = 5
h = 13 is a solution to the given equation because for h = 13,
h − 8= 13 − 8 = 5 and hence, the equation is satisfied.
(k) h − 8 = 5
h = 8 is not a solution to the given equation because for h = 8,
h − 8= 8 − 8 = 0, and not 5
(l) h − 8 = 5
h = 0 is not a solution to the given equation because for h = 0,
h − 8= 0 − 8 = −8, and not 5
(m) p + 3 = 1
p = 3 is not a solution to the given equation because for p = 3,
p + 3= 3 + 3 = 6, and not 1
(n) p + 3 = 1
p = 1 is not a solution to the given equation because for p = 1,
p + 3= 1 + 3 = 4, and not 1
(o) p + 3 = 1
p = 0 is not a solution to the given equation because for p = 0,
p + 3= 0 + 3 = 3, and not 1
(p) p + 3 = 1
p = −1 is not a solution to the given equation because for p = −1,
p + 3= −1 + 3 = 2, and not 1
(q) p + 3 = 1
p = −2 is a solution to the given equation because for p = −2,
p + 3= − 2 + 3 = 1 and hence, the equation is satisfied.
Page No 241:
Question 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) (7, 2, 10, 14)
(e) r − 4 = 0 (4, − 4, 8, 0)
(f) x + 4 = 2 (− 2, 0, 2, 4)
Answer:
(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
Page No 241:
Question 4:
(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16.

m
1
2
3
4
5
6
7
8
9
10
…
m + 10
−
−
−
−
−
−
−
−
−
−
−
(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

t
3
4
5
6
7
8
9
10
11
…
5t
−
−
−
−
−
−
−
−
−
−
(c) Complete the table and find the solution of the equation z/3 = 4 using the table.

z
8
9
10
11
12
13
14
15
16
...
3
−
−
−
−
−
−
−
(d) Complete the table and find the solution to the equation m − 7 = 3

m
5
6
7
8
9
10
11
12
13
…
m − 7
−
−
−
−
−
−
−
−
−
−
Answer:
(a) For m + 10, the table can be constructed as follows.

m
m + 10
1
1 + 10 = 11
2
2 + 10 = 12
3
3 + 10 = 13
4
4 + 10 = 14
5
5 + 10 = 15
6
6 + 10 = 16
7
7 + 10 = 17
8
8 + 10 = 18
9
9 + 10 = 19
10
10 + 10 = 20
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.

t
5t
3
5 × 3 = 15
4
5 × 4 = 20
5
5 × 5 = 25
6
5 × 6 = 30
7
5 × 7 = 35
8
5 × 8 = 40
9
5 × 9 = 45
10
5 × 10 = 50
11
5 ×11 = 55
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.

z
8
9
10
11
12
13
14
15
16
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.

m
m − 7
5
5 − 7 = − 2
6
6 − 7 = − 1
7
7 − 7 = 0
8
8 − 7 = 1
9
9 − 7 = 2
10
10 − 7 = 3
11
11 − 7 = 4
12
12 − 7 = 5
13
13 − 7 = 6
By inspection, we can find that m = 10 is the solution of the above equation as for m = 10, m − 7 = 10 − 7 = 3
Page No 242:
Question 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!
(ii) For each day of the week
Make an upcount 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!
Answer:
(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.
View NCERT Solutions for all chapters of Class 10