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FA – IV
Class - 10
Arithmetic Progression
Page No: 99
Exercise 5.1
1. In which of the following situations, does the list of numbers involved make as arithmetic
progression and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
Answer
It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in
the cylinder at a time.
Answer
Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of
air remaining in the cylinder at a time. In other words, after every stroke, only 1 - 1/4 = 3/4th part of air
will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3...
Clearly, it can be observed that the adjacent terms of this series do not have the same difference
between them. Therefore, this is not an A.P.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and
rises by Rs 50 for each subsequent metre.
Answer
Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound
interest at 8% per annum.
Answer
We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be
Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this
is not an A.P.
2. Write first four terms of the A.P. when the first term a and the common differenced are given as
follows
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = - 3
(iv) a = -1 d = 1/2
(v) a = - 1.25, d = - 0.25
Answer
(i) a = 10, d = 10
Let the series be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1 + d = 10 + 10 = 20
a3 = a2 + d = 20 + 10 = 30
a4 = a3 + d = 30 + 10 = 40
a5 = a4 + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = - 2, d = 0
Let the series be a1, a2, a3, a4 …
a1 = a = -2
a2 = a1 + d = - 2 + 0 = - 2
a3 = a2 + d = - 2 + 0 = - 2
a4 = a3 + d = - 2 + 0 = - 2
Therefore, the series will be - 2, - 2, - 2, - 2 …
First four terms of this A.P. will be - 2, - 2, - 2 and - 2.
(iii) a = 4, d = - 3
Let the series be a1, a2, a3, a4 …
a1 = a = 4
a2 = a1 + d = 4 - 3 = 1
a3 = a2 + d = 1 - 3 = - 2
a4 = a3 + d = - 2 - 3 = - 5
Therefore, the series will be 4, 1, - 2 - 5 …
First four terms of this A.P. will be 4, 1, - 2 and - 5.
(iv) a = - 1, d = 1/2
Let the series be a1, a2, a3, a4 …a1 = a = -1
a2 = a1 + d = -1 + 1/2 = -1/2
a3 = a2 + d = -1/2 + 1/2 = 0
a4 = a3 + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) a = - 1.25, d = - 0.25
Let the series be a1, a2, a3, a4 …
a1 = a = - 1.25
a2 = a1 + d = - 1.25 - 0.25 = - 1.50
a3 = a2 + d = - 1.50 - 0.25 = - 1.75
a4 = a3 + d = - 1.75 - 0.25 = - 2.00
Clearly, the series will be 1.25, - 1.50, - 1.75, - 2.00 ……..
First four terms of this A.P. will be - 1.25, - 1.50, - 1.75 and - 2.00.
3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, - 1, - 3 …
(ii) -5, - 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 …
Answer
(i) 3, 1, - 1, - 3 …
Here, first term, a = 3
Common difference, d = Second term - First term
=1-3=-2
(ii) - 5, - 1, 3, 7 …
Here, first term, a = - 5
Common difference, d = Second term - First term
= ( - 1) - ( - 5) = - 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ....
Here, first term, a = 1/3
Common difference, d = Second term - First term
= 5/3 - 1/3 = 4/3
(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term - First term
= 1.7 - 0.6
= 1.1
4. Which of the following are APs? If they form an A.P. find the common difference d and write three
more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ....
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, - 6, - 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, - 4, - 8, - 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ....
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a2, a3, a4 …
(xii) √2, √8, √18, √32 ...
(xiii) √3, √6, √9, √12 ...
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …
Answer
(i) 2, 4, 8, 16 …
Here,
a2 - a1 = 4 - 2 = 2
a3 - a2 = 8 - 4 = 4
a4 - a3 = 16 - 8 = 8
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(ii) 2, 5/2, 3, 7/2 ....
Here,
a2 - a1 = 5/2 - 2 = 1/2
a3 - a2 = 3 - 5/2 = 1/2
a4 - a3 = 7/2 - 3 = 1/2
⇒ an+1 - an is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a5 = 7/2 + 1/2 = 4
a6 = 4 + 1/2 = 9/2
a7 = 9/2 + 1/2 = 5
(iii) -1.2, - 3.2, -5.2, -7.2 …
Here,
a2 - a1 = ( -3.2) - ( -1.2) = -2
a3 - a2 = ( -5.2) - ( -3.2) = -2
a4 - a3 = ( -7.2) - ( -5.2) = -2
⇒ an+1 - an is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a5 = - 7.2 - 2 = - 9.2
a6 = - 9.2 - 2 = - 11.2
a7 = - 11.2 - 2 = - 13.2
(iv) -10, - 6, - 2, 2 …
Here,
a2 - a1 = (-6) - (-10) = 4
a3 - a2 = (-2) - (-6) = 4
a4 - a3 = (2) - (-2) = 4
⇒ an+1 - an is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a5 = 2 + 4 = 6
a6 = 6 + 4 = 10
a7 = 10 + 4 = 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a2 - a1 = 3 + √2 - 3 = √2
a3 - a2 = (3 + 2√2) - (3 + √2) = √2
a4 - a3 = (3 + 3√2) - (3 + 2√2) = √2
⇒ an+1 - an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a5 = (3 + √2) + √2 = 3 + 4√2
a6 = (3 + 4√2) + √2 = 3 + 5√2
a7 = (3 + 5√2) + √2 = 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2 - a1 = 0.22 - 0.2 = 0.02
a3 - a2 = 0.222 - 0.22 = 0.002
a4 - a3 = 0.2222 - 0.222 = 0.0002
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(vii) 0, -4, -8, -12 …
Here,
a2 - a1 = (-4) - 0 = -4
a3 - a2 = (-8) - (-4) = -4
a4 - a3 = (-12) - (-8) = -4
⇒ an+1 - an is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a5 = -12 - 4 = -16
a6 = -16 - 4 = -20
a7 = -20 - 4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ....
Here,
a2 - a1 = (-1/2) - (-1/2) = 0
a3 - a2 = (-1/2) - (-1/2) = 0
a4 - a3 = (-1/2) - (-1/2) = 0
⇒ an+1 - an is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a5 = (-1/2) - 0 = -1/2
a6 = (-1/2) - 0 = -1/2
a7 = (-1/2) - 0 = -1/2
(ix) 1, 3, 9, 27 …
Here,
a2 - a1 = 3 - 1 = 2
a3 - a2 = 9 - 3 = 6
a4 - a3 = 27 - 9 = 18
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(x) a, 2a, 3a, 4a …
Here,
a2 - a1 = 2a - a = a
a3 - a2 = 3a - 2a = a
a4 - a3 = 4a - 3a = a
⇒ an+1 - an is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a5 = 4a + a = 5a
a6 = 5a + a = 6a
a7 = 6a + a = 7a
(xi) a, a2, a3, a4 …
Here,
a2 - a1 = a2 - a = (a - 1)
a3 - a2 = a3 - a2 = a2 (a - 1)
a4 - a3 = a4 - a3 = a3(a - 1)
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(xii) √2, √8, √18, √32 ...
Here,
a2 - a1 = √8 - √2 = 2√2 - √2 = √2
a3 - a2 = √18 - √8 = 3√2 - 2√2 = √2
a4 - a3 = 4√2 - 3√2 = √2
⇒ an+1 - an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a5 = √32 + √2 = 4√2 + √2 = 5√2 = √50
a6 = 5√2 +√2 = 6√2 = √72
a7 = 6√2 + √2 = 7√2 = √98
(xiii) √3, √6, √9, √12 ...
Here,
a2 - a1 = √6 - √3 = √3 × 2 -√3 = √3(√2 - 1)
a3 - a2 = √9 - √6 = 3 - √6 = √3(√3 - √2)
a4 - a3 = √12 - √9 = 2√3 - √3 × 3 = √3(2 - √3)
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(xiv) 12, 32, 52, 72 …
Or, 1, 9, 25, 49 …..
Here,
a2 − a1 = 9 − 1 = 8
a3 − a2 = 25 − 9 = 16
a4 − a3 = 49 − 25 = 24
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2 − a1 = 25 − 1 = 24
a3 − a2 = 49 − 25 = 24
a4 − a3 = 73 − 49 = 24
i.e., ak+1 − ak is same every time.
⇒ an+1 - an is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a5 = 73+ 24 = 97
a6 = 97 + 24 = 121
a7 = 121 + 24 = 145
Page No: 105
Exercise 5.2
1. Fill in the blanks in the following table, given that a is the first term, d the common difference
and an the nth term of the A.P.
a
d
n
an
(i)
7
3
8
…...
(ii)
− 18
…..
10
0
(iii)
…..
−3
18
−5
(iv)
− 18.9
2.5
…..
3.6
(v)
3.5
0
105
…..
Answer
(i) a = 7, d = 3, n = 8, an = ?
We know that,
For an A.P. an = a + (n − 1) d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, an = 28
(ii) Given that
a = −18, n = 10, an = 0, d = ?
We know that,
an = a + (n − 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2
(iii) Given that
d = −3, n = 18, an = −5
We know that,
an = a + (n − 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51
a = 51 − 5 = 46
Hence, a = 46
(iv) a = −18.9, d = 2.5, an = 3.6, n = ?
We know that,
an = a + (n − 1) d
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5
(n - 1) = 22.5/2.5
n-1=9
n = 10
Hence, n = 10
(v) a = 3.5, d = 0, n = 105, an = ?
We know that,
an = a + (n − 1) d
an = 3.5 + (105 − 1) 0
an = 3.5 + 104 × 0
an = 3.5
Hence, an = 3.5
Page No: 106
Choose the correct choice in the following and justify
(i) 30th term of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) −77 (D.) −87
(ii) 11th term of the A.P. -3, -1/2, ,2 .... is
(A) 28 (B) 22 (C) - 38 (D)
Answer
(i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a2 − a1 = 7 − 10 = −3
We know that, an = a + (n − 1) d
a30 = 10 + (30 − 1) (−3)
a30 = 10 + (29) (−3)
a30 = 10 − 87 = −77
Hence, the correct answer is option C.
(ii) Given that A.P. is -3, -1/2, ,2 ...
First term a = - 3
Common difference, d = a2 − a1 = (-1/2) - (-3)
= (-1/2) + 3 = 5/2
We know that, an = a + (n − 1) d
a11 = 3 + (11 -1)(5/2)
a11 = 3 + (10)(5/2)
a11 = -3 + 25
a11 = 22
Hence, the answer is option B.
3. In the following APs find the missing term in the boxes.
Answer
(i) For this A.P.,
a=2
a3 = 26
We know that, an = a + (n − 1) d
a3 = 2 + (3 - 1) d
26 = 2 + 2d
24 = 2d
d = 12
a2 = 2 + (2 - 1) 12
= 14
Therefore, 14 is the missing term.
(ii) For this A.P.,
a2 = 13 and
a4 = 3
We know that, an = a + (n − 1) d
a2 = a + (2 - 1) d
13 = a + d ... (i)
a4 = a + (4 - 1) d
3 = a + 3d ... (ii)
On subtracting (i) from (ii), we get
- 10 = 2d
d=-5
From equation (i), we get
13 = a + (-5)
a = 18
a3 = 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For this A.P.,
a2 = 13 and
a4 = 3
We know that, an = a + (n − 1) d
a2 = a + (2 - 1) d
13 = a + d ... (i)
a4 = a + (4 - 1) d
3 = a + 3d ... (ii)
On subtracting (i) from (ii), we get,
- 10 = 2d
d=-5
From equation (i), we get,
13 = a + (-5)
a = 18
a3 = 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iv) For this A.P.,
a = −4 and
a6 = 6
We know that,
an = a + (n − 1) d
a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d=2
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v)
For this A.P.,
a2 = 38
a6 = −22
We know that
an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d ... (i)
a6 = a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
4. Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer
3, 8, 13, 18, …
For this A.P.,
a=3
d = a2 − a1 = 8 − 3 = 5
Let nth term of this A.P. be 78.
an = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16th term of this A.P. is 78.
5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,
, 13,...., -47
Answer
(i) For this A.P.,
a=7
d = a2 − a1 = 13 − 7 = 6
Let there are n terms in this A.P.
an = 205
We know that
an = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.
(ii) For this A.P.,
a = 18
Let there are n terms in this A.P.
an = 205
an = a + (n − 1) d
-47 = 18 + (n - 1) (-5/2)
-47 - 18 = (n - 1) (-5/2)
-65 = (n - 1)(-5/2)
(n - 1) = -130/-5
(n - 1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Answer
For this A.P.,
a = 11
d = a2 − a1 = 8 − 11 = −3
Let −150 be the nth term of this A.P.
We know that,
an = a + (n − 1) d
-150 = 11 + (n - 1)(-3)
-150 = 11 - 3n + 3
-164 = -3n
n = 164/3
Clearly, n is not an integer.
Therefore, - 150 is not a term of this A.P.
7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Answer
Given that,
a11 = 38
a16 = 73
We know that,
an = a + (n − 1) d
a11 = a + (11 − 1) d
38 = a + 10d ... (i)
Similarly,
a16 = a + (16 − 1) d
73 = a + 15d ... (ii)
On subtracting (i) from (ii), we get
35 = 5d
d=7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a31 = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31st term is 178.
8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer
Given that,
a3 = 12
a50 = 106
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a + 2d ... (i)
Similarly, a50 = a + (50 − 1) d
106 = a + 49d ... (ii)
On subtracting (i) from (ii), we get
94 = 47d
d=2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore, 29th term is 64.
9. If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.
Answer
Given that,
a3 = 4
a9 = −8
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
4 = a + 2d ... (i)
a9 = a + (9 − 1) d
−8 = a + 8d ... (ii)
On subtracting equation (i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = a + 2 (−2)
4=a−4
a=8
Let nth term of this A.P. be zero.
an = a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2n = 10
n=5
Hence, 5th term of this A.P. is 0.
10. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Answer
We know that,
For an A.P., an = a + (n − 1) d
a17 = a + (17 − 1) d
a17 = a + 16d
Similarly, a10 = a + 9d
It is given that
a17 − a10 = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d=1
Therefore, the common difference is 1.
11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?
Answer
Given A.P. is 3, 15, 27, 39, …
a=3
d = a2 − a1 = 15 − 3 = 12
a54 = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let nth term be 771.
an = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.
Or
Let nth term be 132 more than 54th term.
n = 54 + 132/2
= 54 + 11 = 65th term
12. Two APs have the same common difference. The difference between their 100th term is 100, what
is the difference between their 1000th terms?
Answer
Let the first term of these A.P.s be a1 and a2 respectively and the common difference of these A.P.s
be d.
For first A.P.,
a100 = a1 + (100 − 1) d
= a1 + 99d
a1000 = a1 + (1000 − 1) d
a1000 = a1 + 999d
For second A.P.,
a100 = a2 + (100 − 1) d
= a2 + 99d
a1000 = a2 + (1000 − 1) d
= a2 + 999d
Given that, difference between
100th term of these A.P.s = 100
Therefore, (a1 + 99d) − (a2 + 99d) = 100
a1 − a2 = 100 ... (i)
Difference between 1000th terms of these A.P.s
(a1 + 999d) − (a2 + 999d) = a1 − a2
From equation (i),
This difference, a1 − a2 = 100
Hence, the difference between 1000th terms of these A.P. will be 100.
13. How many three digit numbers are divisible by 7?
Answer
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first
term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5.
Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d=7
an = 994
n=?
an = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
Or
Three digit numbers which are divisible by 7 are 105, 112, 119, ....994 .
These numbers form an AP with a = 105 and d = 7.
Let number of three-digit numbers divisible by 7 be n, an = 994
⇒ a + (n - 1) d = 994
⇒ 105 + (n - 1) × 7 = 994
⇒7(n - 1) = 889
⇒ n - 1 = 127
⇒ n = 128
14. How many multiples of 4 lie between 10 and 250?
Answer
First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common
difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
a = 12
d=4
an = 248
an = a + (n - 1) d
248 = 12 + (n - 1) × 4
236/4 = n - 1
59 = n - 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
Or
Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be n, an = 248
⇒ a + (n - 1) d = 248
⇒ 12 + (n - 1) × 4 = 248
⇒4(n - 1) = 248
⇒ n - 1 = 59
⇒ n = 60
15. For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Answer
63, 65, 67, …
a = 63
d = a2 − a1 = 65 − 63 = 2
nth term of this A.P. = an = a + (n − 1) d
an= 63 + (n − 1) 2 = 63 + 2n − 2
an = 61 + 2n ... (i)
3, 10, 17, …
a=3
d = a2 − a1 = 10 − 3 = 7
nth term of this A.P. = 3 + (n − 1) 7
an = 3 + 7n − 7
an = 7n − 4 ... (ii)
It is given that, nth term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer
a3 = 16
a + (3 − 1) d = 16
a + 2d = 16 ... (i)
a7 − a5 = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d=6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a=4
Therefore, A.P. will be
4, 10, 16, 22, …
Page No: 107
17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Answer
Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
a20 = 253 + (19) (−5)
a20 = 253 − 95
a = 158
Therefore, 20th term from the last term is 158.
18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the
first three terms of the A.P.
Answer
We know that,
an = a + (n − 1) d
a4 = a + (4 − 1) d
a4 = a + 3d
Similarly,
a8 = a + 7d
a6 = a + 5d
a10 = a + 9d
Given that, a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 ... (i)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ... (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d=5
From equation (i), we get
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a2 = a + d = − 13 + 5 = −8
a3 = a2 + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.
19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs
200 each year. In which year did his income reach Rs 7000?
Answer
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every
year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Therefore, an = a + (n − 1) d
7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.
20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If
in the nth week, her week, her weekly savings become Rs 20.75, find n.
Answer
Given that,
a=5
d = 1.75
an = 20.75
n=?
an = a + (n − 1) d
20.75 = 5 + (n - 1) × 1.75
15.75 = (n - 1) × 1.75
(n - 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n-1=9
n = 10
Hence, n is 10.
Page No: 112
Exercise 5.3
1. Find the sum of the following APs.
(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15, 1/12, 1/10, ...... , to 11 terms
Answer
(i) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a=2
d = a2 − a1 = 7 − 2 = 5
n = 10
We know that,
Sn = n/2 [2a + (n - 1) d]
S10 = 10/2 [2(2) + (10 - 1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 245
(ii) −37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a2 − a1 = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
Sn = n/2 [2a + (n - 1) d]
S12 = 12/2 [2(-37) + (12 - 1) × 4]
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6(-30) = -180
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that,
Sn = n/2 [2a + (n - 1) d]
S12 = 50/2 [1.2 + (99) × 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505
(iv) 1/15, 1/12, 1/10, ...... , to 11 terms
For this A.P.,
2. Find the sums given below
(i) 7 +
+ 14 + .................. +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer
(i) For this A.P.,
a=7
l = 84
d = a2 − a1 =
- 7 = 21/2 - 7 = 7/2
Let 84 be the nth term of this A.P.
l = a (n - 1)d
84 = 7 + (n - 1) × 7/2
77 = (n - 1) × 7/2
22 = n − 1
n = 23
We know that,
Sn = n/2 (a + l)
Sn = 23/2 (7 + 84)
= (23×91/2) = 2093/2
=
(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a2 − a1 = 32 − 34 = −2
l = 10
Let 10 be the nth term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
Sn = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
(iii) (−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
a = −5
l = −230
d = a2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930
3. In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.
Answer
(i) Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
⇒ 50 = 5 + (n - 1) × 3
⇒ 3(n - 1) = 45
⇒ n - 1 = 15
⇒ n = 16
Now, Sn = n/2 (a + an)
Sn = 16/2 (5 + 50) = 440
(ii) Given that, a = 7, a13 = 35
As an = a + (n − 1)d,
⇒ 35 = 7 + (13 - 1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, Sn = n/2 (a + an)
S13 = 13/2 (7 + 35) = 273
(iii)Given that, a12 = 37, d = 3
As an = a + (n − 1)d,
⇒ a12 = a + (12 − 1)3
⇒ 37 = a + 33
⇒a=4
Sn = n/2 (a + an)
Sn = 12/2 (4 + 37)
= 246
(iv) Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d ... (i)
Sn = n/2 [2a + (n - 1)d]
S10 = 10/2 [2a + (10 - 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d ... (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d ... (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9 = 8
(v) Given that, d = 5, S9 = 75
As Sn = n/2 [2a + (n - 1)d]
S9 = 9/2 [2a + (9 - 1)5]
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
a = -35/3
an = a + (n − 1)d
a9 = a + (9 − 1) (5)
= -35/3 + 8(5)
= -35/3 + 40
= (35+120/3) = 85/3
(vi) Given that, a = 2, d = 8, Sn = 90
As Sn = n/2 [2a + (n - 1)d]
90 = n/2 [2a + (n - 1)d]
⇒ 180 = n(4 + 8n - 8) = n(8n - 4) = 8n2 - 4n
⇒ 8n2 - 4n - 180 = 0
⇒ 2n2 - n - 45 = 0
⇒ 2n2 - 10n + 9n - 45 = 0
⇒ 2n(n -5) + 9(n - 5) = 0
⇒ (2n - 9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a5 = 8 + 5 × 4 = 34
(vii) Given that, a = 8, an = 62, Sn = 210
As Sn = n/2 (a + an)
210 = n/2 (8 + 62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8 + 5d
⇒ 5d = 62 - 8 = 54
⇒ d = 54/5 = 10.8
(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n ... (i)
Sn = n/2 (a + an)
-14 = n/2 (a + 4)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n2 + 10n
2n2 − 10n − 28 = 0
n2 − 5n −14 = 0
n2 − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8
(ix) Given that, a = 3, n = 8, S = 192
As Sn = n/2 [2a + (n - 1)d]
192 = 8/2 [2 × 3 + (8 - 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d=6
(x) Given that, l = 28, S = 144 and there are total of 9 terms.
Sn = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a=4
Page No: 113
4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Answer
Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 − a1 = 17 − 9 = 8
As Sn = n/2 [2a + (n - 1)d]
636 = n/2 [2 × a + (8 - 1) × 8]
636 = n/2 [18 + (n- 1) × 8]
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
n = (-53/4) or n = 12
n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12
only.
5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the
common difference.
Answer
Given that,
a=5
l = 45
Sn = 400
Sn = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
d = 40/15 = 8/3
6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how
many terms are there and what is their sum?
Answer
Given that,
a = 17
l = 350
d=9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Sn = n/2 (a + l)
S38 = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer
d=7
a22 = 149
S22 = ?
an = a + (n − 1)d
a22 = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a=2
Sn = n/2 (a + an)
= 22/2 (2 + 149)
= 11 × 151
= 1661
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
Given that,
a2 = 14
a3 = 18
d = a3 − a2 = 18 − 14 = 4
a2 = a + d
14 = a + 4
a = 10
Sn = n/2 [2a + (n - 1)d]
S51 = 51/2 [2 × 10 + (51 - 1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610
9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer
Given that,
S7 = 49
S17 = 289
S7
= 7/2 [2a + (n - 1)d]
S7 = 7/2 [2a + (7 - 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)
Similarly,
S17 = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d=2
From equation (i),
a + 3(2) = 7
a+6=7
a=1
Sn = n/2 [2a + (n - 1)d]
= n/2 [2(1) + (n - 1) × 2]
= n/2 (2 + 2n - 2)
= n/2 (2n)
= n2
10. Show that a1, a2 … , an , … form an AP where an is defined as below
(i) an = 3 + 4n
(ii) an = 9 − 5n
Also find the sum of the first 15 terms in each case.
Answer
(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term
as 7.
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(7) + (15 - 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525
(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first
term as 4.
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(4) + (15 - 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 - 70)
= 15/2 (-62)
= 15(-31)
= -465
11. If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum
of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.
Answer
Given that,
Sn = 4n − n2
First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3
Sum of first two terms = S2
= 4(2) − (2)2 = 8 − 4 = 4
Second term, a2 = S2 − S1 = 4 − 3 = 1
d = a2 − a = 1 − 3 = −2
an = a + (n − 1)d
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a3 = 5 − 2(3) = 5 − 6 = −1
a10 = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are −1, −15, and 5
− 2n respectively.
12. Find the sum of first 40 positive integers divisible by 6.
Answer
The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a=6
d=6
S40 = ?
Sn = n/2 [2a + (n - 1)d]
S40 = 40/2 [2(6) + (40 - 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920
13. Find the sum of first 15 multiples of 8.
Answer
The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d=8
S15 = ?
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960
14. Find the sum of the odd numbers between 0 and 50.
Answer
The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a=1
d=2
l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
Sn = n/2 (a + l)
S25 = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas
follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty
for each succeeding day being Rs. 50 more than for the preceding day. How much money the
contractor has to pay as penalty, if he has delayed the work by 30 days.
Answer
It can be observed that these penalties are in an A.P. having first term as 200 and common difference
as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S30
= 30/2 [2(200) + (30 - 1) 50]
= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.
16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall
academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of
the prizes.
Answer
Let the cost of 1st prize be P.
Cost of 2nd prize = P − 20
And cost of 3rd prize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and
first term as P.
a=P
d = −20
Given that, S7 = 700
7/2 [2a + (7 - 1)d] = 700
a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs
40.
17. In a school, students thought of planting trees in and around the school to reduce air pollution. It
was decided that the number of trees, that each section of each class will plant, will be the same as
the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will
plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be
planted by the students?
Answer
It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
Sn = n/2 [2a + (n - 1)d]
S12 = 12/2 [2(1) + (12 - 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.
18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with
centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of
such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Answer
perimeter of semi-circle = πr
= π(0.5) = π/2 cm
P2 = π(1) = π cm
P1
= π(1.5) = 3π/2 cm
P1, P2, P3 are the lengths of the semi-circles
P3
π/2, π, 3π/2, 2π, ....
P1 = π/2 cm
P2 = π cm
d = P2- P1 = π - π/2 = π/2
First term = P1 = a = π/2 cm
Sn = n/2 [2a + (n - 1)d]
Therefor, Sum of the length of 13 consecutive circles
S13 = 13/2 [2(π/2) + (13 - 1)π/2]
= 13/2 [π + 6π]
=13/2 (7π)
= 13/2 × 7 × 22/7
= 143 cm
Page No: 114
19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in
the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the
top row?
Answer
It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a2 − a1 = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
Sn = 200
Sn = n/2 [2a + (n - 1)d]
S12 = 12/2 [2(20) + (n - 1)(-1)]
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n − n2
n2 − 41n + 400 = 0
n2 − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
an = a + (n − 1)d
a16 = 20 + (16 − 1) (−1)
a16 = 20 − 15
a16 = 5
Similarly,
a25 = 20 + (25 − 1) (−1)
a25 = 20 − 24
= −4
Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which
is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16 th row is 5.
20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other
potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the
bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the
same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a
competitor is 2 × 5 + 2 ×(5 + 3)]
Answer
The distances of potatoes from the bucket are 5, 8, 11, 14…
Distance run by the competitor for collecting these potatoes are two times of the distance at which the
potatoes have been kept. Therefore, distances to be run are
10, 16, 22, 28, 34,……….
a = 10
d = 16 − 10 = 6
S10 =?
S10 = 12/2 [2(20) + (n - 1)(-1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.
Constructions
Page No: 219
Exercise 11.1
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Answer
Steps of Construction:
Step I: AB = 7.6 cm is drawn.
Step II: A ray AX making an acute angle with
AB is drawn.
Step III: After that, a ray BY is drawn parallel to
AX making equal acute angle as in the
previous step.
Step IV: Point A1, A2, A3, A4 and A5 is marked
on AX and point B1, B2.... to B8 is
marked on BY such that AA1 = A1A2 =
A2A3 =....BB1= B1B2 = .... B7B8
Step V: A5 and B8 is joined and it intersected AB at point C diving it in the ratio 5:8.
AC : CB = 5 : 8
Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are
2/3 of the corresponding sides of the first triangle.
Answer
Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4
cm, an arc is draw.
Step III: Again, with B as a centre and radius
equal to 5 cm an arc is drawn on same
side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute
angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3
=5) is marked on AX as A1 A2....A5
Step VII: A5B is joined. A2B' is drawn parallel
to A5B and B'C' is drawn parallel to
BC.
ΔAB'C' is the required triangle
Justification:
∠A(Common)
∠C = ∠C' and ∠B = ∠ B' (corresponding angles)
Thus ΔAB'C' ~ ΔABC by AAA similarity condition
From the figure,
AB'/AB = AA2/AA5 = 2/3
AB' =2/3 AB
AC' = 2/3 AC
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.
Answer
Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray BX making an acute angle with BC is drawn
opposite to vertex A.
Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked
on BX.
Step IV; Point B5 is joined with C to draw B5C.
Step V: B7C' is drawn parallel to B5C and C'A' is parallel to
CA.
Thus A'BC' is the required triangle.
Justification
ΔAB'C' ~ ΔABC by AAA similarity condition
∴ AB/A'B = AC/A'C' = BC/BC'
and BC/BC' = BB5/BB7 = 5/7
∴A'B/AB = A'C'/AC = = BC'/BC = 7/5
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle
whose sides are 1.5 times the corresponding sides of the isosceles triangle.
Answer
Steps of Construction:
Step I: BC = 5 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersect BC at O.
Step III: At a distance of 4 cm, a point A is marked on perpendicular
Step IV: AB and AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to
Step VI: 3 points B1 B2 and B3 is marked BX.
Step VII: B2 is joined with C to form B2C.
bisector of BC.
vertex A.
Step VIII: B3C' is drawn parallel to B2C and C'A' is drawn parallel to
CA.
Thus, A'BC' is the required triangle formed.
Justification:
ΔAB'C' ~ ΔABC by AA similarity condition.
∴ AB/AB' = BC/B'C' = AC/AC'
also,
AB/AB' = AA2/AA3 = 2/3
⇒ AB' = 3/2 AB, B'C' = 3/2 BC and AC' = 3/2 AC
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle
whose sides are 3/4 of the corresponding sides of the triangle ABC.
Answer
Steps of Construction:
Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an
∠ABC = 60° with BC.
Step III: AC is joined to form ΔABC.
Step IV: A ray BX is drawn making an acute angle with BC
Step V: 4 points B1 B2 B3 and B4 at equal distance is
Step VII: B3 is joined with C' to form B3C'.
Step VIII: C'A' is drawn parallel CA.
Thus, A'BC' is the required triangle.
Justification:
∠A = 60° (Common)
∠C = ∠C'
ΔAB'C' ~ ΔABC by AA similarity condition.
∴ AB/AB' = BC/B'C' = AC/AC'
opposite to vertex A.
marked on BX.
also,
AB/AB' = AA3/AA4 = 4/3
⇒ AB' = 3/4 AB, B'C' = 3/4 BC and AC' = 3/4 AC
6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose
sides are 4/3 times the corresponding sides of Δ ABC.
Answer
Sum of all side of triangle = 180°
∴∠A + ∠B + ∠C = 180°
∠C = 180° - 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of 45° with BC.
Step III: At C, a ray making an angle of 30° with
BC is drawn intersecting the previous
ray at A. Thus, ∠A = 105°.
Step IV: A ray BX is drawn making an acute
angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal
distance is marked on BX.
Step VI: B3C is joined and B4C' is made parallel
to B3C.
Step VII: C'A' is made parallel CA.
Thus, A'BC' is the required triangle.
Justification:
∠B = 45° (Common)
∠C = ∠C'
ΔAB'C' ~ ΔABC by AA similarity condition.
∴ BC/BC' = AB/A'B' = AC/A'C'
also,
BC/BC' = BB3/BB4 = 34
⇒ AB = 4/3 AB', BC = 4/3 BC' and AC = 4/3 A'C'
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then
construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Answer
Steps of Construction:
Step I: BC = 3 cm is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is
intersecting it at point A.
Step IV: AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC
Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is
Step VII: B3C is joined B5C' is made parallel to B3C.
Step VIII: A'C' is joined together.
Thus, ΔA'BC' is the required triangle.
made on previous ray
opposite to vertex A.
marked on BX.
Justification:
As in the previous question 6.
Page No. 221
Exercise 11.2
In each of the following, give also the justification of the construction:
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of
tangents to the circle and measure their lengths.
Answer
Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is
Step II: A point P at a distance of 10 cm from the centre O is
Step III: Perpendicular bisector OP is drawn and let it
drawn.
taken. OP is joined.
intersected at M.
Step IV: With M as a centre and OM as a radius, a circle is
circle at Q and R.
Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.
drawn intersecting previous
Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm
and measure its length. Also verify the measurement by actual calculation.
Answer
Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is
Step II: With O as a centre and radius equal to 6 cm, a
Step III: P be any point on the circle of radius 6 cm and OP is
Step IV: Perpendicular bisector of OP is drawn which cuts it at
Step V: With M as a centre and OM as a radius, a circle is
circle of radius 4 cm at Q
and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.
Measurement:
OQ = 4 cm (Radius of the circle)
PQ = 6 cm ( Radius of the circle)
∠PQO = 90° (Angle in the semi circle)
Applying Pythagoras theorem in ΔPQO,
PQ2 + QO2 = PO2
⇒ PQ2 + 42= 62
⇒ PQ2 + 16 = 36
⇒ PQ2 = 36 - 16
⇒ PQ2 = 20
⇒ PQ = 2√5
drawn.
concentric circle is drawn.
joined.
M
drawn which intersect the the
Justification:
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.
3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a
distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Answer
Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a
circle is drawn.
Step II: The diameter of the circle is extended both sides
and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn
and x and y be its mid-point.
Step IV: With O as a centre and Ox be its radius, a circle
is drawn which intersected the
previous circle at
M and N.
Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI: PM and PN are joined also QR and QT are joined.
Thus, PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from
point Q.
Justification:
∠PMO = 90° (Angle in the semi circle)
∴ OM ⊥ PM
Therefor, OM is the radius of the circle then PM has to be a tangent of the circle.
Similarly, PN, QR and QT are tangents of the circle.
4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Answer
We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O) = 360° - (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and OQ
the circle.
Step II: Draw another radius OR making an angle equal to 120° with
is joined. OQ is radius of
the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined
which is perpendicular
OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.
Justification:
Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° - 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.
5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking
B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of
the other circle.
Answer
Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4
cm, a circle is drawn which cut the
line at point O.
Step III: With B as a centre and radius equal to
3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a
radius, a circle is drawn which
intersect the previous two circles at P,
Q and R, S.
Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.
Justification:
∠BPA = 90° (Angle in the semi circle)
∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle.
Similarly, AQ, BR and BS are tangents of the circle.
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular
from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Answer
Steps of Construction:
Step I: A ΔABC is drawn.
Step II: Perpendicular to AC is drawn to point B
which intersected it at D.
Step III: With O as a centre and OC as a radius, a
circle is drawn. The circle through B, C,
D is drawn.
Step IV: OA is joined and a circle is drawn with
diameter OA which intersected the
previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.
Justification:
∠OEA = 90° (Angle in the semi circle)
∴ OE ⊥ AE
Therefor, OE is the radius of the circle then AE has to be a tangent of the circle.
Similarly, AB is tangent of the circle.
Probability
Page No: 308
Exercise 15.1
1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________ .
(ii) The probability of an event that cannot happen is __________. Such an event is called ________ .
(iii) The probability of an event that is certain to happen is _________ . Such an event is called
_________ .
(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .
(v) The probability of an event is greater than or equal to and less than or equal to __________ .
Answer
(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure or certain
event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Answer
(i) It does not have equally likely outcomes as it depends on various reasons like mechanical
problems, fuels etc.
(ii) It does not have equally likely outcomes as it depends on the player how he/she shoots.
(iii) It has equally likely outcomes.
(iv)It has equally likely outcomes.
3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the
beginning of a football game?
Answer
Yes, tossing of a coin is a fair way of deciding which team should get the ball at the beginning of a
football game because it has only two outcomes either head or tail. A coin is always unbiased.
4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) -1.5 (C) 15%
(D) 0.7
Answer
The probability of an event is always greater than or equal to 0 and less than or equal to 1.
Thus, (B) -1.5 cannot be the probability of an event.
5. If P(E) = 0.05, what is the probability of ‘not E’?
Answer
P(E) = 0.05
also, P(E) + P(not E) = 1
⇒ P(not E) = 1 - P(E)
⇒ P(not E) = 1 - 0.05
⇒ P(not E) = 0.95
6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the
bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Answer
(i) Since the bag contains only lemon flavoured.
Therefor, No. of orange flavoured candies = 0
Probability of taking out orange flavoured candies = 0/1 = 0
(ii) The bag only have lemon flavoured candies.
Probability of taking out lemon flavoured candies = 1/1 = 1
7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is
0.992. What is the probability that the 2 students have the same birthday?
Answer
Let E be the event of having the same birthday.
P(E) = 0.992
⇒ P(E) + P(not E) = 1
⇒ P(not E) = 1 – P(E)
⇒ 1 - 0.992 = 0.008
The probability that the 2 students have the same birthday is 0.008
8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the
probability that the ball drawn is (i) red? (ii) not red?
Answer
No. of red balls = 3
No. of black balls = 5
Total no. of balls = 5+3 = 8
(i) Probability of drawing red balls = No. of red balls/Total no. of balls = 3/8
(ii) Probability of drawing black balls = No. of black balls/Total no. of balls = 5/8
9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the
box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not
green?
Answer
No. of red marbles = 5
No. of white marbles =8
No. of green marbles = 4
Total no. of balls = 5+8+4 = 17
(i) Favourable no. of elementary events = 5
Probability of taking out red marble = 5/17
(ii) Favourable no. of elementary events = 8
Probability of taking out red marble = 8/17
(iii) Favourable no. of elementary events = 4
Probability of taking out red marble = 4/17
Page No: 309
10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is
equally likely that one of the coins will fall out when the bank is turned upside down, what is the
probability that the coin (i) will be a 50 p coin ? (ii) will not be a ₹5 coin?
Answer
No. of 50p coins = 100
No. of ₹1 coins = 50
No. of ₹2 coins = 20
No. of ₹5 coins = 10
Total no. of coins = 100 + 50 + 20 + 10 = 180
(i) Favourable no. of elementary events = 100
Probability that it will be 50p coins = 100/180 = 5/9
(ii) Favourable no. of elementary events = 100+50+20 = 170
Probability that it will be 50p coins = 170/180 = 17/18
11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from
a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish
taken out is a male fish?
Answer
No. of male fish in the tank = 5
no. of female fish in the tank = 8
Total number of fish in the tank = 5 + 8 = 13
Favourable number events = 5
Probability of taking out a male fish = 5/13
12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the
numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the
probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Answer
Possible no. of events = 8
(i) Favourable number of events = 1
Probability that it will point at 8 = 1/8
(ii) Odd numbers = 1, 3, 5 and 7
Favourable number of events = 4
Probability that it will be an odd number = 4/8 = 1/2
(iii) Numbers greater than 2 = 3, 4, 5, 6, 7 and 8
Favourable number of events = 6
Probability that a number greater than 4 = 6/8 = 3/4
(iv) Numbers less than 9 = 1,2,3,4,5,6,7,8
Favourable number of events = 8
Probability that a number less than 9 = 8/8 = 1
13. A die is thrown once. Find the probability of getting
(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Answer
Possible numbers of events on throwing a dice = 6
Numbers on dice = 1,2,3,4,5 and 6
(i) Prime numbers = 2, 3 and 5
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2
(ii) Numbers lying between 2 and 6 = 3, 4 and 5
Favourable number of events = 3
Probability that a number between 2 and 6 = 3/6 = 1/2
(iii) Odd numbers = 1, 3 and 5
Favourable number of events = 3
Probability that it will be an odd number = 3/6 = 1/2
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the
queen of diamonds
Answer
Possible numbers of events = 52
(i) Numbers of king of red colour = 2
Probability of getting a king of red colour = 2/52 = 1/26
(ii) Numbers of face cards = 12
Probability of getting a face card = 12/52 = 3/13
(iii) Numbers of red face cards = 6
Probability of getting a king of red colour = 6/52 = 3/26
(iv) Numbers of jack of hearts =1
Probability of getting a king of red colour = 1/526
(v) Numbers of king of spade = 13
Probability of getting a king of red colour = 13/52 = 1/4
(vi) Numbers of queen of diamonds = 1
Probability of getting a king of red colour = 1/52
15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face
downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an
ace? (b) a queen?
Answer
Total numbers of cards = 5
(i) Numbers of queen = 1
Probability of picking a queen = 1/5
(ii) When queen is drawn and put aside then total numbers of cards left is 4
(a) Numbers of ace = 1
Probability of picking an ace = 1/4
(a) Numbers of queen = 0
Probability of picking a queen = 0/4 = 0
16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen
and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the
probability that the pen taken out is a good one.
Answer
Numbers of defective pens = 12
Numbers of good pens = 132
Total numbers of pen = 132 + 12 = 144 pens
Favourable number of events = 132
Probability of getting a good pen = 132/144 = 11/12
17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the
probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at
random from the rest. What is the probability that this bulb is not defective?
Answer
(i) Total numbers of bulbs = 20
Numbers of defective bulbs = 4
Probability of getting a defective bulb = 4/20 = 1/5
(ii) One non defective bulb is drawn in (i) then the total numbers of bulb left is 19
Total numbers of events = 19
Favourable numbers of events = 19 - 4 = 15
Probability that the bulb is not defective = 15/19
18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the
box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number
divisible by 5.
Answer
Total numbers of discs = 50
(i) Total numbers of favourable events = 81
Probability that it bears a two-digit number = 81/90 = 9/10
(ii) Perfect square numbers = 1, 4, 9, 16, 25, 36, 49, 64 and 81
Favourable numbers of events = 9
Probability of getting a perfect square number = 9/90 = 1/10
(iii) Numbers which are divisible by 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85
and 90
Favourable numbers of events = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5
Page No: 310
19. A child has a die whose six faces show the letters as given below:
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Answer
Total numbers of events = 6
(i) Total numbers of faces having A on it = 2
Probability of getting A = 2/6 = 1/3
(ii) Total numbers of faces having D on it = 1
Probability of getting A = 1/6
20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the
probability that it will land inside the circle with diameter 1m?
Answer
Area of the rectangle = (3 × 2) m2 = 6m2
Area of the circle = πr2 = π(1/2)2 m2 = π/4 m2
Probability that die will land inside the circle = (π/4) × 1/6 = π/24
21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a
pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it
to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
Answer
Total numbers of pens = 144
Numbers of defective pens = 20
Numbers of non defective pens = 144 - 20 = 124
(i) Numbers of favourable events = 124
Probability that she will buy it = 124/144 = 31/36
(ii) Numbers of favourable events = 20
Probability that she will not buy it = 20/144 = 5/36
22. Refer to Example 13. (i) Complete the following table:
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
Answer
Events that can happen on throwing two dices are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total numbers of events : 6 × 6 = 36
(i) To get sum as 2, possible outcomes = (1,1)
To get sum as 3, possible outcomes = (1,2) and (2,1)
To get sum as 4, possible outcomes = (1,3); (3,1); and (2,2)
To get sum as 5, possible outcomes = (1,4); (4,1); (2,3); and (3,2)
To get sum as 6, possible outcomes = (1,5); (5,1); (2,4); (4,2); and (3,3)
To get sum as 7, possible outcomes = (1,6); (6,1); (5,2); (2,5); (4,3); and (3,4)
To get sum as 8, possible outcomes = (2,6); (6,2); (3,5); (5,3); and (4,4)
To get sum as 9, possible outcomes = (3,6); (6,3); (4,5); and (5,4)
To get sum as 10, possible outcomes = (4,6); (6,4) and (5,5)
To get sum as 11, possible outcomes = (5,6) and (6,5)
To get sum as 12, possible outcomes = (6,6)
Event:
2 3
4
5
6
7
8
9
10
11
12
Sum on 2 dice
Probability
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
(ii) No, i don't agree with the argument. It is already justified in (i).
23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins
if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the
probability that Hanif will lose the game.
Answer
Events that can happen in tossing 3 coins = HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
Total number of events = 8
Hinif will lose the game if he gets HHT, HTH, THH, TTH, HTT, THT
Favourable number of elementary events = 6
Probability of losing the game = 6/8 = 3/4
24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Answer
(i) Consider the following events.
A = first throw shows 5,
B = second throw shows 5
P(A) = 6/36, P(B) = 6/36 and P(notB) = 5/6
⇒ P(notA) = 1– 6/36 = 30/36 = 5/6
Required probability = 5/6 × 5/6 = 25/36
(ii) Number of events when 5 comes at least once = 11
Probability = 11/36
Page No: 311
25. Which of the following arguments are correct and which are not correct? Give reasons for your
answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or
one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore,
the probability of getting an odd number is 1/2
Answer
(i) The statement is incorrect
Possible events = (H,H); (H,T); (T,H) and (T,T)
Probability of getting two heads = 1/4
Probability of getting one of the each = 2/4 = 1/2
(ii) Correct. The two outcomes considered are equally likely.
Quadratic Equations
Page No: 73
Exercise 4.1
1. Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x - 3)
(ii) x2 - 2x = (-2)(3 - x)
(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
(iv) (x - 3)(2x + 1) = x(x + 5)
(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
(vi) x2 + 3x + 1 = (x - 2)2
(vii) (x + 2)3 = 2x(x2 - 1)
(viii) x3 - 4x2 - x + 1 = (x - 2)3
Answer
(i) (x + 2)2 = 2(x - 3)
⇒ x2 + 2x + 1 = 2x - 6
⇒ x2 + 7 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.
(ii) x2 - 2x = (-2)(3 - x)
⇒ x2 - 2x = -6 + 2x
⇒ x2 - 4x + 6 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.
(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
⇒ x2 - x - 2 = x2 + 2x - 3
⇒ 3x - 1 =0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(iv) (x - 3)(2x + 1) = x(x + 5)
⇒ 2x2 - 5x - 3 = x2 + 5x
⇒ x2 - 10x - 3 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.
(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
⇒ 2x2 - 7x + 3 = x2 + 4x - 5
⇒ x2 - 11x + 8 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.
(vi) x2 + 3x + 1 = (x - 2)2
⇒ x2 + 3x + 1 = x2 + 4 - 4x
⇒ 7x - 3 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(vii) (x + 2)3 = 2x(x2 - 1)
⇒ x3 + 8 + x2 + 12x = 2x3 - 2x
⇒ x3 + 14x - 6x2 - 8 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(viii) x3 - 4x2 - x + 1 = (x - 2)3
⇒ x3 - 4x2 - x + 1 = x3 - 8 - 6x2 + 12x
⇒ 2x2 - 13x + 9 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.
2. Represent the following situations in the form of quadratic equations.
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its
breadth. We need to find the length and breadth of the plot.
Answer
Let the breadth of the rectangular plot = x m
Hence, the length of the plot is (2x + 1) m.
Formula of area of rectangle = length × breadth = 528 m2
Putting the value of length and width, we get
(2x + 1) × x = 528
⇒ 2x2 + x =528
⇒ 2x2 + x - 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Answer
Let the first integer number = x
Next consecutive positive integer will = x + 1
Product of both integers = x × (x +1) = 306
⇒ x2 + x = 306
⇒ x2 + x - 306 = 0
(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now
will be 360. We would like to find Rohan's present age.
Answer
Let take Rohan's age = x years
Hence, his mother's age = x + 26
3 years from now
Rohan's age = x + 3
Age of Rohan's mother will = x + 26 + 3 = x + 29
The product of their ages 3 years from now will be 360 so that
(x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 - 360 = 0
⇒ x2 + 32x - 273 = 0
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it
would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer
Let the speed of train be x km/h.
Time taken to travel 480 km = 480/x km/h
In second condition, let the speed of train = (x - 8) km/h
It is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = (480/x + 3) km/h
Speed × Time = Distance
(x - 8)(480/x + 3) = 480
⇒ 480 + 3x - 3840/x - 24 = 480
⇒ 3x - 3840/x = 24
⇒ 3x2 - 8x - 1280 = 0
Page No: 76
Exercise 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x + 1/8 = 0
(v) 100x2 – 20x + 1 = 0
Answer
(i) x2 – 3x – 10
= x2 - 5x + 2x - 10
= x(x - 5) + 2(x - 5)
= (x - 5)(x + 2)
Roots of this equation are the values for which (x - 5)(x + 2) = 0
∴ x - 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
(ii) 2x2 + x – 6
= 2x2 + 4x - 3x - 6
= 2x(x + 2) - 3(x + 2)
= (x + 2)(2x - 3)
Roots of this equation are the values for which (x + 2)(2x - 3) = 0
∴ x + 2 = 0 or 2x - 3 = 0
⇒ x = -2 or x = 3/2
(iii) √2 x2 + 7x + 5√2
= √2 x2 + 5x + 2x + 5√2
= x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)
Roots of this equation are the values for which (√2x + 5)(x + √2) = 0
∴ √2x + 5 = 0 or x + √2 = 0
⇒ x = -5/√2 or x = -√2
(iv) 2x2 – x + 1/8
= 1/8 (16x2 - 8x + 1)
= 1/8 (16x2 - 4x -4x + 1)
= 1/8 (4x(4x - 1) -1(4x - 1))
= 1/8(4x - 1)2
Roots of this equation are the values for which (4x - 1)2 = 0
∴ (4x - 1) = 0 or (4x - 1) = 0
⇒ x = 1/4 or x = 1/4
(v) 100x2 – 20x + 1
= 100x2 – 10x - 10x + 1
= 10x(10x - 1) -1(10x - 1)
= (10x - 1)2
Roots of this equation are the values for which (10x - 1)2 = 0
∴ (10x - 1) = 0 or (10x - 1) = 0
⇒ x = 1/10 or x = 1/10
2. (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of
the number of marbles they now have is 124. Find out how many marbles they had to start with.
Answer
Let the number of John's marbles be x.
Therefore, number of Jivanti's marble = 45 - x
After losing 5 marbles,
Number of John's marbles = x - 5
Number of Jivanti's marbles = 45 - x - 5 = 40 - x
It is given that the product of their marbles is 124.
∴ (x - 5)(40 - x) = 124
⇒ x2 – 45x + 324 = 0
⇒ x2 – 36x - 9x + 324 = 0
⇒ x(x - 36) -9(x - 36) = 0
⇒ (x - 36)(x - 9) = 0
Either x - 36 = 0 or x - 9 = 0
⇒ x = 36 or x = 9
If the number of John's marbles = 36,
Then, number of Jivanti's marbles = 45 - 36 = 9
If number of John's marbles = 9,
Then, number of Jivanti's marbles = 45 - 9 = 36
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy
(in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the
total cost of production was Rs 750. Find out the number of toys produced on that day.
Answer
Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 - x)
It is given that, total production of the toys = Rs 750
∴ x(55 - x) = 750
⇒ x2 – 55x + 750 = 0
⇒ x2 – 25x - 30x + 750 = 0
⇒ x(x - 25) -30(x - 25) = 0
⇒ (x - 25)(x - 30) = 0
Either, x -25 = 0 or x - 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.
3. Find two numbers whose sum is 27 and product is 182.
Answer
Let the first number be x and the second number is 27 - x.
Therefore, their product = x (27 - x)
It is given that the product of these numbers is 182.
Therefore, x(27 - x) = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 13x - 14x + 182 = 0
⇒ x(x - 13) -14(x - 13) = 0
⇒ (x - 13)(x -14) = 0
Either x = -13 = 0 or x - 14 = 0
⇒ x = 13 or x = 14
If first number = 13, then
Other number = 27 - 13 = 14
If first number = 14, then
Other number = 27 - 14 = 13
Therefore, the numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Answer
Let the consecutive positive integers be x and x + 1.
Therefore, x2 + (x + 1)2 = 365
⇒ x2 + x2 + 1 + 2x = 365
⇒ 2x2 + 2x - 364 = 0
⇒ x2 + x - 182 = 0
⇒ x2 + 14x - 13x - 182 = 0
⇒ x(x + 14) -13(x + 14) = 0
⇒ (x + 14)(x - 13) = 0
Either x + 14 = 0 or x - 13 = 0,
⇒ x = - 14 or x = 13
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other
two sides.
Answer
Let the base of the right triangle be x cm.
Its altitude = (x - 7) cm
From Pythagoras theorem, we have
Base2 + Altitude2 = Hypotenuse2
∴ x2 + (x - 7)2 = 132
⇒ x2 + x2 + 49 - 14x = 169
⇒ 2x2 - 14x - 120 = 0
⇒ x2 - 7x - 60 = 0
⇒ x2 - 12x + 5x - 60 = 0
⇒ x(x - 12) + 5(x - 12) = 0
⇒ (x - 12)(x + 5) = 0
Either x - 12 = 0 or x + 5 = 0,
⇒ x = 12 or x = - 5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 - 7) cm = 5
cm.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a
particular day that the cost of production of each article (in rupees) was 3 more than twice the number
of articles produced on that day. If the total cost of production on that day was Rs 90, find the number
of articles produced and the cost of each article.
Answer
Let the number of articles produced be x.
Therefore, cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90.∴ x(2x + 3) = 0
⇒ 2x2 + 3x - 90 = 0
⇒ 2x2 + 15x -12x - 90 = 0
⇒ x(2x + 15) -6(2x + 15) = 0
⇒ (2x + 15)(x - 6) = 0
Either 2x + 15 = 0 or x - 6 = 0
⇒ x = -15/2 or x = 6
As the number of articles produced can only be a positive integer, therefore, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15.
Page No: 87
Exercise 4.3
1. Find the roots of the following quadratic equations, if they exist, by the method of completing
the square:
(i) 2x2 – 7x +3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0
Answer
(i) 2x2 – 7x + 3 = 0
⇒ 2x2 – 7x = - 3
On dividing both sides of the equation by 2, we get
⇒ x2 – 7x/2 = -3/2
⇒ x2 – 2 × x × 7/4 = -3/2
On adding (7/4)2 to both sides of equation, we get
⇒ (x)2 - 2 × x × 7/4 + (7/4)2 = (7/4)2 - 3/2
⇒ (x - 7/4)2 = 49/16 - 3/2
⇒ (x - 7/4)2 = 25/16
⇒ (x - 7/4) = ± 5/4
⇒ x = 7/4 ± 5/4
⇒ x = 7/4 + 5/4 or x = 7/4 - 5/4
⇒ x = 12/4 or x = 2/4
⇒ x = 3 or 1/2
(ii) 2x2 + x – 4 = 0
⇒ 2x2 + x = 4
On dividing both sides of the equation, we get
⇒ x2 + x/2 = 2
On adding (1/4)2 to both sides of the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2
⇒ (x + 1/4)2 = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 - 1/4
⇒ x = ± √33-1/4
⇒ x = √33-1/4 or x = -√33-1/4
(iii) 4x2 + 4√3x + 3 = 0
⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0
⇒ (2x + √3)2 = 0
⇒ (2x + √3) = 0 and (2x + √3) = 0
⇒ x = -√3/2 or x = -√3/2
(iv) 2x2 + x + 4 = 0
⇒ 2x2 + x = -4
On dividing both sides of the equation, we get
⇒ x2 + 1/2x = 2
⇒ x2 + 2 × x × 1/4 = -2
On adding (1/4)2 to both sides of the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 - 2
⇒ (x + 1/4)2 = 1/16 - 2
⇒ (x + 1/4)2 = -31/16
However, the square of number cannot be negative.
Therefore, there is no real root for the given equation.
2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Answer
(i) 2x2 – 7x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -7 and c = 3
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = 7±√49 - 24/4
⇒ x = 7±√25/4
⇒ x = 7±5/4
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4
∴ x = 3 or 1/2
(ii) 2x2 + x - 4 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = -4
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒x = -1±√1+32/4
⇒x = -1±√33/4
∴ x = -1+√33/4 or x = -1-√33/4
(iii) 4x2 + 4√3x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 4, b = 4√3 and c = 3
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = -4√3±√48-48/8
⇒ x = -4√3±0/8
∴ x = √3/2 or x = -√3/2
(iv) 2x2 + x + 4 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = 4
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = -1±√1-32/4
⇒ x = -1±√-31/4
The square of a number can never be negative.
∴There is no real solution of this equation.
Page No: 88
3. Find the roots of the following equations:
(i) x-1/x = 3, x ≠ 0
(ii) 1/x+4 - 1/x-7 = 11/30, x = -4, 7
Answer
(i) x-1/x = 3
⇒ x2 - 3x -1 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -3 and c = -1
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = 3±√9+4/2
⇒ x = 3±√13/2
∴ x = 3+√13/2 or x = 3-√13/2
(ii) 1/x+4 - 1/x-7 = 11/30
⇒ x-7-x-4/(x+4)(x-7) = 11/30
⇒ -11/(x+4)(x-7) = 11/30
⇒ (x+4)(x-7) = -30
⇒ x2 - 3x - 28 = 30
⇒ x2 - 3x + 2 = 0
⇒ x2 - 2x - x + 2 = 0
⇒ x(x - 2) - 1(x - 2) = 0
⇒ (x - 2)(x - 1) = 0
⇒ x = 1 or 2
4. The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is 1/3.
Find his present age.
Answer
Let the present age of Rehman be x years.
Three years ago, his age was (x - 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1/3.
∴ 1/x-3 + 1/x-5 = 1/3
x+5+x-3/(x-3)(x+5) = 1/3
2x+2/(x-3)(x+5) = 1/3
⇒ 3(2x + 2) = (x-3)(x+5)
⇒ 6x + 6 = x2 + 2x - 15
⇒ x2 - 4x - 21 = 0
⇒ x2 - 7x + 3x - 21 = 0
⇒ x(x - 7) + 3(x - 7) = 0
⇒ (x - 7)(x + 3) = 0
⇒ x = 7, -3
However, age cannot be negative.
Therefore, Rehman's present age is 7 years.
5. In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks
more in Mathematics and 3 marks less in English, the product of their marks would have been 210.
Find her marks in the two subjects.
Answer
Let the marks in Maths be x.
Then, the marks in English will be 30 - x.
According to the question,
(x + 2)(30 - x - 3) = 210
(x + 2)(27 - x) = 210
⇒ -x2 + 25x + 54 = 210
⇒ x2 - 25x + 156 = 0
⇒ x2 - 12x - 13x + 156 = 0
⇒ x(x - 12) -13(x - 12) = 0
⇒ (x - 12)(x - 13) = 0
⇒ x = 12, 13
If the marks in Maths are 12, then marks in English will be 30 - 12 = 18
If the marks in Maths are 13, then marks in English will be 30 - 13 = 17
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30
metres more than the shorter side, find the sides of the field.
Answer
Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 - 60x - 2700 = 0
⇒ x2 - 90x + 30x - 2700 = 0
⇒ x(x - 90) + 30(x -90)
⇒ (x - 90)(x + 30) = 0
⇒ x = 90, -30
However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m.
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the
larger number. Find the two numbers.
Answer
Let the larger and smaller number be x and y respectively.
According to the question,
x2 - y2 = 180 and y2 = 8x
⇒ x2 - 8x = 180
⇒ x2 - 8x - 180 = 0
⇒ x2 - 18x + 10x - 180 = 0
⇒ x(x - 18) +10(x - 18) = 0
⇒ (x - 18)(x + 10) = 0
⇒ x = 18, -10
However, the larger number cannot be negative as 8 times of the larger number will be negative and
hence, the square of the smaller number will be negative which is not possible.
Therefore, the larger number will be 18 only.
x = 18
∴ y2 = 8x = 8 × 18 = 144
⇒ y = ±√44 = ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and - 12.
8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1
hour less for the same journey. Find the speed of the train.
Answer
Let the speed of the train be x km/hr.
Time taken to cover 360 km = 360/x hr.According to the question,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 - x + 1800-5/x = 360
⇒ x2 + 5x + 10x - 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x - 40) = 0
⇒ x = 40, -45
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h.
9. Two water taps together can fill a tank in
hours. The tap of larger diameter takes 10 hours less
than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the
tank.
Answer
Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x - 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x
Part of tank filled by larger pipe in 1 hour = 1/x - 10
It is given that the tank can be filled in
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x - 10) = 8x2 - 80x
⇒ 150x - 750 = 8x2 - 80x
⇒ 8x2 - 230x +750 = 0
⇒ 8x2 - 200x - 30x + 750 = 0
⇒ 8x(x - 25) -30(x - 25) = 0
⇒ (x - 25)(8x -30) = 0
⇒ x = 25, 30/8
= 75/8 hours by both the pipes together. Therefore,
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours. As in this case, the time taken by the
larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 - 10 =15
hours respectively.
10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and
Bangalore (without taking into consideration the time they stop at intermediate stations). If the average
speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of
the two trains.
Answer
Let the average speed of passenger train be x km/h.
Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger
train to cover the same distance.
⇒ 132 × 11 = x(x + 11)
⇒ x2 + 11x - 1452 = 0
⇒ x2 + 44x -33x -1452 = 0
⇒ x(x + 44) -33(x + 44) = 0
⇒ (x + 44)(x - 33) = 0
⇒ x = - 44, 33
Speed cannot be negative.
Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train
will be 33 + 11 = 44 km/h.
11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the
sides of the two squares.
Answer
Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and
4y respectively and their areas will be x2 and y2 respectively.
It is given that
4x - 4y = 24
x-y=6
x=y+6
Also, x2 + y2 = 468
⇒ (6 + y2) + y2 = 468
⇒ 36 + y2 + 12y + y2 = 468
⇒ 2y2 + 12y + 432 = 0
⇒ y2 + 6y - 216 = 0
⇒ y2 + 18y - 12y - 216 = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y - 12) = 0
⇒ y = -18, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.
Page No: 91
Exercise 4.4
1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x2 - 3x + 5 = 0
(ii) 3x2 - 4√3x + 4 = 0
(iii) 2x2 - 6x + 3 = 0
Answer
(i) Consider the equation
x2 - 3x + 5 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
= ( - 3)2 - 4 (2) (5) = 9 - 40
= - 31
As b2 - 4ac < 0,
Therefore, no real root is possible for the given equation.
(ii) 3x2 - 4√3x + 4 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b2 - 4ac
= (-4√3)2 - 4(3)(4)
= 48 - 48 = 0
As b2 - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be -b/2a and -b/2a.-b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.
(iii) 2x2 - 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac
= (-6)2 - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation:
2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x - 2) + 6 = 0
Answer
(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b2 - 4ac
= (k)2 - 4(2) (3)
= k2 - 24
For equal roots,
Discriminant = 0
k2 - 24 = 0
k2 = 24
k = ±√24 = ±2√6
(ii) kx(x - 2) + 6 = 0
or kx2 - 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = - 2k and c = 6
Discriminant = b2 - 4ac
= ( - 2k)2 - 4 (k) (6)
= 4k2 - 24k
For equal roots,
b2 - 4ac = 0
4k2 - 24k = 0
4k (k - 6) = 0
Either 4k = 0
or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x2' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.
3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is
800 m2? If so, find its length and breadth.
Answer
Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)= 2l2
2l2 = 800
l2 = 800/2 = 400
l2 - 400 =0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = 0, c = 400
Discriminant = b2 - 4ac
= (0)2 - 4 × (1) × ( - 400) = 1600
Here, b2 - 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be
designed.
l = ±20
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m
4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two
friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer
Let the age of one friend be x years.
then the age of the other friend will be (20 - x) years.
4 years ago,
Age of 1st friend = (x - 4) years
Age of 2nd friend = (20 - x - 4) = (16 - x) years
A/q we get that,
(x - 4) (16 - x) = 48
16x - x2 - 64 + 4x = 48
- x2 + 20x - 112 = 0
x2 - 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b2 - 4ac
= (-20)2 - 4 × 112
= 400 - 448 = -48
b2 - 4ac < 0
Therefore, there will be no real solution possible for the equations. Such type of condition doesn't exist.
5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and
breadth.
Answer
Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 - l
Area = l×b = l(40 - l) = 40l - l240l - l2 = 400
l2 - 40l + 400 = 0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b2 - 4ac
(-40)2 - 4 × 400
= 1600 - 1600 = 0
b2 - 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,l = -b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of park, l = 20 m
And breadth of park, b = 40 - l = 40 - 20 = 20 m.
Excercise: 9.1
Page No: 203
1. A circus artist is climbing a 20 m long rope, which is tightly stretched and
tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the
rope with the ground level is 30° (see Fig. 9.11).
Answer
Let AB be the vertical pole Ac be 20 m long rope tied to point C.
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the
ground making an angle 30° with it. The distance between the foot of the tree to the point where the
top touches the ground is 8 m. Find the height of the tree.
Answer
Let AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
3. A contractor plans to install two slides for the children to play in a park. For the children below the
age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle
of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m,
and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Answer
There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides.
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
A/q,
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.
Page No: 204
4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the
foot of the tower, is 30°. Find the height of the tower.
Answer
Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the
tower.
A/q,
In right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.
5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily
tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the
string, assuming that there is no slack in the string.
Answer
Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
A/q,
In right ΔABC,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.
6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from
his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the
distance he walked towards the building.
Answer
Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ - BZ = (30 - 1.5) = 28.5 m
A/q,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.
7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission
tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Answer
Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC - BC
A/q,
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.
8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of
elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the
pedestal is 45°. Find the height of the pedestal.
Answer
Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC - AB
A/q,
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = BC/CD
⇒ BC = CD.
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC - BC = 1.6 m
⇒BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.
9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of
elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the
height of the building.
Answer
Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
A/q,
In right ΔBCD,
tan 60° = CD/BC
⇒ √3 = 50/BC
⇒ BC = 50/√3
also,
In right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/BC
⇒ AB = 50/3
Thus, the height of the building is 50/3.
10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80
m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60°
and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Answer
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
A/q,
AB = CD,
OB + OD = 80 m
Now,
In right ΔCDO,
tan 30° = CD/OD
⇒ 1/√3 = CD/OD
⇒ CD = OD/√3 ... (i)
also,
In right ΔABO,
tan 60° = AB/OB
⇒ √3 = AB/(80-OD)
⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 - 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m
respectively.
11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite
the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this
point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower
is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.
Answer
Here, AB is the height of the tower.
CD = 20 m (given)
A/q,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(20+BC)
⇒ AB = (20+BC)/√3 ... (i)
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ AB = √3 BC ... (ii)
From eqn (i) and (ii)
AB = √3 BC = (20+BC)/√3
⇒ 3 BC = 20 + BC
⇒ 2 BC = 20 ⇒ BC = 10 m
Putting the value of BC in eqn (ii)
AB = 10√3 m
Thus, the height of the tower 10√3 m and the width of the canal is 10 m.
12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and
the angle of depression of its foot is 45°. Determine the height of the tower.
Answer
Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
A/q,
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC = DE + CD
= (7√3 + 7) m = 7(√3+1) m.
13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of
two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse,
find the distance between the two ships.
Answer
Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
A/q,
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3 m
The distance between the two ships = CD = BD - BC = (75√3 - 75) m = 75(√3-1) m.
Page No: 205
14. A 1.2 m tall girl spots a balloon moving with the wind in a
horizontal line at a height of 88.2 m
from the ground. The height of 88.2 m from the ground. The angle of elevation of the balloon from the
eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig.
9.13). Find the distance travelled by the balloon during the interval.
Answer
Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m - 1.2 m = 87 m
Distance travelled by the balloon =
DE = CE - CD
A/q,
In right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon = DE = CE - CD = (87√3 - 29√3) m = 29√3(3 - 1) m = 58√3 m.
15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a
car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed.
Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car
to reach the foot of the tower from this point.
Answer
Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
A/q,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 - AB/√3
⇒ CD = AB(√3 - 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.
16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the
base of the tower and in the same straight line with it are complementary. Prove that the height of the
tower is 6 m.
Answer
Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
A/q,
In right ΔABC,
tan x = AB/BC
⇒ tan x = AB/4
⇒ AB = 4 tan x ... (i)
also,
In right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x ... (ii)
Multiplying eqn (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.
****************