Class 6 Maths NCERT Solutions
Chapter 11 Algebra
Exercise 11.1
Q.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
Ans:
(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.
Q.2: We already know the rule for the pattern of letters L, C and F. 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?
Ans:
It is known that L
requires only two matchsticks. Therefore, the pattern for L is 2n. Among
all the letters given in the question, only T and V are the two letters which
require two matchsticks.
Hence, (a) and (d)
Q.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.)
Ans:
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
Q.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.)
Ans:
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
Q.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.)
Ans:
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
Q.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.)
Ans:
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
Q.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?
Ans:
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
Q.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.
Ans:
Let Radha’s age
be x years.
Leela’s age = Radha’s
age − 4
= (x −
4) years
Q.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?
Ans:
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
Q.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?
Ans:
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
Q.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.
Ans:
(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.
Exercise 11.2
Q.1: The side of an equilateral triangle is shown by l. Express
the perimeter of the equilateral triangle using l.
Ans:
Side of equilateral triangle
= l
Perimeter = l + l + l =
3l
Q.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.)
Ans:
Side of regular
hexagon = l
Perimeter = 6l
Q.3: A cube is a three-dimensional 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.
Ans:
Length of edge
= l
Number of edges = 12
Total length of the
edges = Number of edges × Length of one edge
= 12l
Q.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).
Ans:
Diameter = AB = AC +
CB = r + r = 2r
d = 2r
Q.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.
Ans:
For any three whole
numbers a, b, and c,
(a + b)
+ c = a + (b + c)
Exercise 11.3
Q.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.)
Ans:
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
Q.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
Ans:
It can be observed
that the expressions in alternatives (c) and (d) are formed by using numbers
only.
Q.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
Ans:
(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.
Q.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
Ans:
(a) p +
7
(b) p −
7
(c) 7p
(d) 
(e) − m −
7
(f) − 5p
(g) 
(h) − 5p
Q.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
Ans:
(a) 2m +
11
(b) 2m −
11
(c) 5y +
3
(d) 5y −
3
(e) − 8y
(f) − 8y +
5
(g) 16 − 5y
(h) − 5y +
16
Q.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.
Ans:
(a) t +
4, t − 4, 4t, 
, 
,
4 − t, 4 + t
(b) 2y +
7, 2y − 7, 7y + 2,…
Exercise 11.4
Q.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.
Ans:
(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
Q.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.
Ans:
(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.
Q.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).
Ans:
(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.
Exercise 11.5
Q.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
Ans:
(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
Q.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 |
- |
Ans:
(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.
Q.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)
Ans:
(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
Q.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 |
− |
− |
− |
− |
− |
− |
− |
− |
− |
− |
Ans:
(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
Q.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!
Ans:
(i)There are 4 corners
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.
End of Questions