Sponsor Area
Use Euclid’s division algorithm to find the HCF of:
(i)135 and 225 (ii) 196 and 38220.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Let a be any odd positive integer and b = 6. Let q be a quotient and r be remainder.
Therefore, applying division lemma, we have 
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Given integers are 32 and 616.
Clearly 616 > 32. Therefore, applying Euclid’s division lemma to 616 and 32, we get
Since, the remainder 8 ≠ 0, we apply the division lemma, to get
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8.
Therefore, the maximum number of columns in which both 616 members (army contingent) and 32 members (army band) can march is 8.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b
Taking b = 3, we get
a = 3q + r; where 0 ≤ r < 3
When, r = 0 = ⇒ a = 3q
When, r = 1 = ⇒ a = 3q + 1
When, r = 2 = ⇒ a = 3q + 2
Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m.
⇒ Squares of 3q = (3q)2
= 9q2 = 3(3q)2 = 3 m where m is some integer.
Square of 3q + 1 = (3q + 1)2
= 9q2 + 6q + 1 = 3(3q2 + 2 q) + 1
= 3m +1, where m is some integer
Square of 3q + 2 = (3q + 2)2
= (3q + 2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1)+ 1
= 3m + 1 for some integer m.
∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Let a and b be two positive integergs such that a is greater than b; then :
a = bq + r;
where q and r are positive integers 0 ≤ r < b.
Taking b = 3, we get
a = 3q + r ; where 0 ≤ r < 3.
⇒ Different values of integer a are 3q, 3q + 1 or 3q + 2.
Cube of 3q = (3q)3
= 27q3 = 9(3q3) = 9m ;
where m is some integer.
Cube of 3q + 1 = (3q + 1)3
= (3q + 3(3q)2 × 1 + 3(3q) × 12+ l3
[∵ (a + b)3 = a3 + 3a2b + 3ab2 + 1]
= 27q3 + 27q2 + 9q + 1
= 9(3q3 + 3q2 + q) + 1
= 9m + 1; where m is some integer.
Cube of 3q + 2 = (3q + 2)3
= (3q)3 + 3(3q)2 × 2 + 3 × 3q × 22 + 23
= 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8; where m is some integer.
∴ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.
Express each number as a product of its prime factors: (i) 140
(i) 140 
So, 140 = 2 × 2 × 5 × 7
= 22 × 5 × 7.
Express each number as a product of its prime factors: (ii) 156
(ii) 156
So, 156 = 2 × 2 × 3 × 13
= 22 × 3 × 13
Express each number as a product of its prime factors: (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
(iii) 3825
So, 3825 = 3 × 3 × × 5 × 5 × 17
= 32 × 52 × 17.
Express each number as a product of its prime factors: (v) 7429
(v) 7429
So, 7429 = 17 × 19 × 23.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91
So, 26 = 2 × 13
and 91 = 7 × 13
∴ H.C.F. (26, 91) = 13
and L.C.M. (26, 91) = 2 × 7 × 13 = 182
Verification :
L.C.M. × H.C.F. = 182 × 13 = 2366
Product of the two nos. = 26 × 91 = 2366
∴ L.C.M. × H.C.F. = Product of the two nos.
Verified.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (ii) 510 and 92
(ii)
So, 510 = 2 × 3 × 5 × 17
and 92 = 2 × 2 × 23
∴ H.C.F. (510, 92) = 2
and L.C.M. (510, 92) = 2 × 2 × 3 × 5 × 17 × 23
= 23460
Verification :
L.C.M. × H.C.F. = 23460 × 2 = 46920
Product of the two nos. = 510 × 92 = 46920
L.C.M. × H.C.F. = Product of the two nos.
Verified.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (iii) 336 and 54
(iii)
So, 336 = 24 × 3 × 7
and 54 = 2 × 33
∴ H.C.F. = 2 × 3 = 6
L.C.M. = 24 × 33 × 7 = 3024
= 23460
Verification :
L.C.M. × H.C.F. = 3024 × 6 = 18144
Product of the two nos. = 336 × 54 = 18144
L.C.M. × H.C.F. = Product of the two nos.
Verified.
Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21
(i) 12, 15 and 21 
Now, 12 = 2 × 2 × 3= 22 × 3,
15 = 3 × 5, 21 = 3 × 7
Here, 3 is the smallest common prime factor.
∴ H.C.F. (12, 15, 21) = 3
And 22, 31, 51 and 71 are the greatest powers of prime factors.
∴ L.C.M. (12, 15,21)= 22 × 31 × 51 × 71 =420.
Find the LCM and HCF of the following integers by applying the prime factorisation method. (ii) 17, 23 and 29
(ii) 17, 23 and 29 
Now, 17 = 1 × 17, 23 = 1 × 23, 29 = 1 × 29
Here, 1 is the smallest common factor.
∴ H.C.F. (17, 23, 29)= 1
And 171, 231 and 291 are the greatest powers of prime factors.
∴ L.C.M. (17, 23, 29) = 17 × 23 × 29 = 11339.
Find the LCM and HCF of the following integers by applying the prime factorisation method. (iii) 8, 9 and 25
(iii) 8, 9 and 25
Now, 8 = 2 × 2 × 2 = 23,
9 = 3 × 3 = 32
and 25 = 5 × 5 = 52
Since, there is no smallest common factor.
∴ H.C.F. (8, 9, 25) = 1.
And 23, 32, and 52 are greatest powers of prime factors.
∴ L.C.M. (8, 9, 25) = 23 × 32 × 52 = 1800.
Sponsor Area
Check whether 6n can end with the digit 0 for any natural number n
If the number 6n, for any natural number n, ends with digit 0, then it is divisible by 5. That is, the prime factorisation of 6n contains the prime 5. This is not possible because the primes in the factorisation of 6n are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6n.
So, there is no natural number n for which 6n ends with digit zero.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Any positive number, which has more than two factors, is called a composite number.
Since,
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)
= 13 × 78
= 13 × 13 × 3 × 2
that is, the given number has more than two factors and it is a composite number.
Similarly,
7 × 6 × 5 × 4 × 3 + 5
= 5 × (7 × 6 × 4 × 3 + 1)
= 5 × 505
= 5 × 5 × 101
⇒ The given number is a composite number.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Required number of minutes is the LCM of 18 and 12.
We have,
18 = 2 × 32
and 12 = 22 × 3
∴ LCM of 18 and 12 is 22 × 32 = 36.
Hence, Ravi and Sonia will meet again at the starting point after 36 minutes.
Prove that 
is irrational.
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is rational.
Squaring on both sides, we get
5b2 = a2
Therefore, 5 divides a2
Therefore, 5 divides a
So, we can write
a = 5c for some integer c.
Substituting for a, we get
5b2 = 25c2
⇒ b2 = 5c2
This means that 5 divides b2, and so 5 divides b.
Therefore, a and b have at least 5 ts a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 
is rational.
So, we conclude that 
is irrational.
Prove that 
is irrational.
is rational.
That is, we can find coprime a and b (b ≠ 0)
Since a and b are integers, we get 
is rational, and so 
is rational.
But this contradicts the fact that 
is irrational.
This contradiction has arisen because of our incorrect assumption that 
is rational
So, we conclude that 
is irrational.
(i) 
Let us assume, to the contrary, that 
is rational.
So, we can find coprime integers a and b (≠ 0) such that
Since, a and b are integers, 
si rational, and so 
is irrational
So, we conclude that 
is irrational.
(ii) 
Let us assume to the contrary, that 
is rational.
So we can find coprime integers a and b (
0 ) such that
Since, a and b are integers, 
is rational. and so, 
is rational
But this contradicts the fact that 
is irrational.
Therefore 
is irrational.
is rational.
is rational.
is rational and so, 
is rational and so, 
is rational.
is irrational.
is irrational.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) 
has a terminating decimal expansion.
has a terminating decimal expansion
[By NCERT Theorem 1.6]
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Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
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has a non-terminating decimal expansion.Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
has a terminating decimal expansion.
[By NCERT Theorem 1.6]
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
has a non-terminating repeating decimal expansion.
[By NCERT Theorem 1.7]
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
has a terminating decimal expansion.
[By NCERT Theorem 1.6]
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
[By NCERT Theorem 1.6]
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
has a non-terminating repeating decimal.
[By NCERT Theorem 1.7]
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form , p q what can you say about the prime factors of q?
(i) 43.123456789
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form , p q what can you say about the prime factors of q?
(ii) 0.120120012000120000. .
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form , p q what can you say about the prime factors of q?
Sponsor Area
Required distance is the LCM of 84 cm, 90 cm and 120 cm.
Thus,
Since, 84 = 2 × 2 × 3 × 7 = 22 × 3 × 7
90 = 2 × 3 × 3 × 5 = 2 × 32 × 5
and 120 = 2 × 2 × 2 × 3 × 5
= 23 × 3 × 5
∴ LCM = Product of each prime factor with highest power
= 23 × 32 × 5 × 7 = 2520
= 25 m 20 cm
Problems Based on Irrational Numbers
∵ HCF × LCM
= Product of the numbers
= 100 × 190
= 19000 Ans.
Given two positive integers a and b, there exists a unique pair of integers q and r such that
a = bq + r, 0 ≤ r < b.
We know, if any number ends with the digit zero it is always divisible by 5.
⇒ If 12n ends with the digit zero, it must be divisible by 5.
This is possible only if prime factorisation of 12n contains the prime number 5.
Now, 12 = 2 × 2 × 3 = 22 × 3
⇒ 12n = (22 × 3)n = 22 × 3n
i.e., prime factorisation of 12' does not contain the prime number 5.
⇒ There is no value of n ∊ N for which 12n ends with the digit zero.
Since 6 = 2 × 3
6n = 2n × 3n
⇒ The prime factorisation of given number 6n does not contain the prime number 7.
⇒ 6n is not divisible by 7.
L.C.M. (150, 100) = 300
⇒ L.C.M. of 150 and 100 = 300
And, the product of numbers 150 and 100 = 150 × 100
Now, H.C.F. (150, 100)
Since, the product of two numbers
= Their H.C.F. × their L.C.M
⇒ a × b = H.C.F. × L.C.M. [where a and b are two numbers]
Other No. 
What is the HCF of 52 and 130.
i.e.
Prime factors of
52 = 2 × 2 × 13
and, Prime factors of
130 = 2 × 5 × 13
∴ HCF (52, 130) = 2 × 13 = 26
is a terminating or a non-terminating repeating decimal.
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will terminate after how many places of decimals?
will terminate after places of decimals.
q is of the form of 2m × 5n, where m, n are negative integers.
i.e.
960 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5
= 26 × 3 × 5
432 = 2 × 2 × 2 × 2 × 3 × 3 × 3
= 24 × 33
HCF (960, 432) = 24 × 3
= 16 × 3
= 48
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#7 {main}</pre>](/application/zrc/images/qvar/MAEN10041235.png)
will have a terminating decimal or a non 
864 = 3 × 3 × 3 × 3 × 7
= 34 × 7
Hence, exponent of 3 in the prime factorization of 864 is 4.
Since 7 × 11 ×13 + 13 = (7 ×11 + 1) × 13
and 7 × 6 × 5 × 4 × 3 × 2 × 1 × 5 = (7 × 6 × 4 × 3 × 2 × 1) × 5
Use Euclid’s division to find HCF of:
(i) 196 and 38220 (ii) 392 and 267540 (iii) 455 and 42
(iv) 81 and 237 (v) 867 and 255.
State whether 
has terminating or non terminating repeating decimal expansion.
Use Euclid’s division algorithm to find the HFC of 867 and 255
Given integers are 867 and 255.
Clearly 867 > 255
Therefore, by applying Euclid’s division lemma to 867 and 255, we get
II. Since the remainder 102 ≠ 0, we apply division lemma to get,
III. We consider the new divisor 102 and new remainder 51 and apply division lemma to get,
The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 51 is the HCF of 867 and 255.
Given integers are 4052 and 12576, clearly 12576 > 4052.
Therefore, by applying Euclid's division lemma to 4052 and 12576, we get
I. 12576 = 4052 × 3 + 420
II. Since the remainder 420 ≠ 0, we apply division lemma to 4052 and 420 to get
III. We consider the new divisor 420 and new remainder 272 and apply division lemma to gel 
IV. We consider the new divisor 272 and new remainder 148 and apply division lemma to get
V. We consider the new divisor 148 and new remainder 124 and apply division lemma to get
VI. We consider the new divisor 124 and new remainder 24 and apply division lemma to get
VII.We consider the new divisor 24 and new remainder 4 and apply division lemma to get
The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e. 4 is the HCF of 4052 and 12576.
Sponsor Area
It is given that required number which divides 245 and 1029, the remainder is 5 in each case.
⇒ 245 - 5 = 240 and 1029 - 5 = 1024 are completely divisible by the required number.
Since, it is given that the required number is the largest number.
Therefore, it is the HCF of 240 and 1024.
Now, finding HCF by Euclid’s division algorithm.
Given integers are 240 and 1024.
Clearly 1024 > 240.
Therefore, it is the HCF of 240 and 1024 and 240, we get
II. Since, the remainder 64 ≠ 0, we apply division lemma to get
III. We consider the new divisor 64 and remainder 48 and apply division lemma to get
IV. We consider the new divisor 48 and new remainder 16 to get
V. The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 16 is the HCF of 245 and 1029.
It is given that required number when divides 615 and 963, the remainder is 6 in each case.
⇒ 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the required number.
Since, it is given that the required number is the largest number. Given integers are 957 and 609 clearly 957 > 609.
Therefore, it is the HCF of 609 and 957.
Now, finding HCF by using Euclid’s division lemma to 609 and 957, we get
II. Since the remainder 348 ≠ 0, we apply division lemma to 348 and 609 to get
III. We consider the new divisor 348 and new remainder 261 and apply division lemma to get
IV. We consider the new divisor 261 and new remainder 87 and apply division lemma to get
The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 87 is the HCF of 615 and 963.
Clearly the required number is the HCF of the following numbers
626 - 1 = 625, 3127 - 2 = 3125 and
15628 - 3 = 15625
Case I. Finding the HCF of 625 and 3125 by applying Euclid’s division lemma.
I. 3125 = 625 × 5 + 0
Since, the remainder at this stage is zero, so the divisor i.e., 625 at this stage is the HCF of 625 and 3125.
Case II. Finding the HCF of 625 and third number 15625 by applying Euclid’s division lemma.
Now, the remainder at this stage is zero. So the divisor i.e., 625 at this stage is the HCF of 625 and 15625.
Hence, HCF of (626, 3127, 15628) is 625.
In order to arrange the books as required, we have to find the largest number that divides 105, 140 and 175 exactly.
Clearly, required number is the HCF of 105, 140 and 175.
Case I :
I. Finding HCF of 105 and 140 by applying Euclid’s division lemma, we get
II. Since, the remainder 35 ≠ 0, we apply division lemma to get
Since, the remainder at this stage is zero, so the divisor i.e., 35 at this stage is the HCF of 105 and 140.
Case II :
I. Finding the HCF of 35 and 175 by applying Euclid’s division lemma, we get
II. Since the remainder at this stage is zero, so the divisor i.e., 35 at this stage is the HCF of 35 and 175.
Thus, HCF of 105, 140) and 175 is 35.
Now,
Number of stacks of Physics books
Number of stacks of Chemistry books
Number of stacks of Biology books
Now, HCF (56, 96)= 8
Applying Euclid’s division algorithm on 8 and 404, we get
Now, HCF (404, 8) = 4
Hence, H.C.F. of 56, 96 and 404 is 4. Ans.
Problems Based on Fundamental theorem of Arithmetic
and 20 = 2 × 2 × 5
∴ LCM = Product of each prime factor with highest powers
= 22 × 3 × 5 = 60
i.e., LCM (6,20) = 60
HCF = Product of common prime factors with lowest powers = 2
i.e., H.C.F. (6, 20) = 2.
Since,
1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 24 × 34
and 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7
= 23 × 32 × 5 × 7
∴ LCM = Product of each prime factor with highest powers
= 24 × 34 × 5 × 7 = 45360
i.e., LCM (1296, 2520)= 45360
H.C.F. = Product of common prime factors with lowest powers
= 23 × 32 = 8 × 9 = 72
i.e., H.C.F. (1296, 2520) = 72.
Find the LCM and HCF of the following integers by applying the prime factorisation method :
12, 15 and 21
Since, 12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
and 21 = 3 × 7
∴ LCM = Product of each prime factor with highest power
= 22 × 3 × 5 × 7 = 420
i.e., LCM (12, 15, 21) = 420
HCF = Product of common prime factors with lowest powers
= 3
i.e., HCF (12, 15,21) = 3.
Since 96 = 2 × 2 × 2 × 2 × 2 × 3
= 2s × 3 and 404 = 2 × 2 × 101 = 22 × 101
∴ LCM = Product of each prime factor with highest power
= 2s × 3 × 101 = 9696
i.e., LCM (96, 404) = 9696
HCF Product of common prime factors with lowest powers
= 22 = 4
We know that,
HCF (a. b) × LCM (a,b) = a × b
⇒ HCF (96, 404) × LMC (96, 404) = 96 × 404
⇒ LCM (96, 404)![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041321-1.png)
Using prime factorization method, Find the LCM of
(i) 12, 15, 20, 27 (ii) 21, 28, 36, 45.
Since, 12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
20 = 2 × 2 × 5 = 22 × 5
and 27 = 3 × 3 × 3 = 33
∴ LCM = Product of each prime factor with highest power
= 22 × 33 × 5 = 540
i.e., LCM (12, 15, 20, 27) = 540.
(ii) 21, 28, 36, 45
Since, 21 = 3 × 7
28 = 2 × 2 × 7 = 22 × 7>
36 = 2 × 2 × 3 × 3 = 22 × 32
and 45 = 5 × 3 × 3 = 5 × 32
∴ LCM = Product of each prime factor with highest power
= 32 × 7 × 22 × 5
= 1260
i.e. LCM (21, 28, 36, 45)= 1260.
Since, 18 = 2 × 3 × 3 = 2 × 32
and 12 = 2 × 2 × 3 = 22 × 3
∴ LCM = Product of each prime factor with highest power
= 22 × 32 = 36
Hence, Ravi and Priya will meet again at the starting point after 36 minutes.
Since, 12 = 2 × 2 × 3 = 22 × 3
20 = 2 × 2 × 5 = 22 × 5
and 30 = 2 × 3 × 5
HCF = Product of common prime factors with lowest powers
= 2
So, there must be 2 books in each stack.
∴ Number of stacks of Physics books
Number of stacks of Chemistry books![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041326-2.png)
Number of stacks of Mathematics books
Required distance is the LCM of 84 cm, 90 cm and 120 cm.
Thus,
Since, 84 = 2 × 2 × 3 × 7 = 22 × 3 × 7
90 = 2 × 3 × 3 × 5 = 2 × 32 × 5
and 120 = 2 × 2 × 2 × 3 × 5
= 23 × 3 × 5
∴ LCM = Product of each prime factor with highest power
= 23 × 32 × 5 × 7 = 2520
= 25 m 20 cm
Problems Based on Irrational Numbers
Prove that 
is irrational.
is is rational. We know that, rational number can be written as 
where ‘a’ and ‘b’ are integers and b ≠ 0. i.e.. Assume that
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⇒ a2 is divisible by 2
⇒ a is divisible by 2
Let a = 2c for some integer c.
Putting a = 2c in (i), we get
2 b2 = (2c)2
⇒ 2b2 = 4c2
⇒ b2 — 2c2
⇒ b2 is divisible by 2
⇒ b is divisible by 2.
Thus, 2 is a common factor of a and b. But, this contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.
The contradiction arises by assuming that 
is rational.
Hence, 
is irrational.
Prove that 
is irrational.
Let us assume that 
is rational.
We know that rational number can be written as 
where ‘a’ and ‘b’ are integers and b ≠ 0.
i.e., assume that
Squaring both side, we get
⇒ a2 is divisible by 3
⇒ a is divisible by 3
Let a = 3c for some integer ‘c’.
Putting a = 3c in (i)
a2 = 3b2
⇒ (3c)2 = 3b2
⇒ 9c2 = 3b2
⇒ b2 = 3c2
⇒ b2 is divisible by
⇒ b is divisible by 3
Thus, 3 is a common factor of ‘a’ and ‘b’.
But this contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.
The contradiction arises by assuming that 
is rational.
Hence, 
is irrational.
Show that ![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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is an irrational number.
Let ![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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is rational
i.e., it can be expressed as 
whereas ‘a’ and ‘b’ both are integers and b ≠ 0.
Thus, 
Now 
is rational and w'e know that 2 is also rational.
is also rational
[∵ Difference, sum and product aftwo rational numbers are always rational]
Comparing it with result (i), we get 
is rational, which is not true as 
is an irrational number.
∴ Our assumption that ![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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is rational is not correct.
Find the LCM and HCF of the following integers by applying the prime factorisation method :
(i) 40. 36 and 126 (ii) 24, 15 and 36
where p and q are co-prime and also express q in the form 22 × 5m.
we find that 2 and 5 cannot be factors of p = 2317.
Thus, p = 23 17 and q = 10000 are co-prime.
Here, q = 24 × 54
p=73 and q = 99 are co-prime.
Here, q=32 X 11.
Prove that 
is an irrational number.
is rational i.e. we can find coprimes a and b (b ≠ 0) such that
is rational, and so 
is rational
is irrational.
This contradiction has arises because of our incorrect assumption that ![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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is an rational.
Hence, we conclude that 
is an irrational number.
Sollution is not provided.
is an irrational number.
Sollution is not provided.
Sponsor Area
If 
is a rational number (q = 0), what is the condition on q so that the decimal reprsentation of the 
is terminating?
2, (Sollution not provided)
(Sollution not provided)
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will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
(Sollution not provided)
(Sollution not provided)
(Sollution not provided)
is a
A.
terminating decimal
is a terminating decimal what can you say about q?
D.
q must be in the form 2n.5m; where n and m are non-negative integers.
C.
non-terminating repeating decimalcomposite number
composite number
A.
composite number
C.
both (a) and (b)A.
a = bq + r, 0 ≤ r < b; a, b, q, r ∊ Integers0
, impossible event1
, sureor certain eventIf a die is thrown once, the possible outcomes are
S = {1, 2, 3, 4,5, 6} i.e. n(S) = 6
Let A be the favourable outcomes of getting add number. Then
A = {1,3,5} i.e. n(S) = 3
Therefore, P(A) = 
Thus, the given statement is correct.
Total possible ways of visiting shop by them = 5 x 5 = 25
(i) They can visit the shop on a 11 week days Tuesday to Saturday.
Let A be the favourable outcomes of visiting shop by them on the same day = 5 Then, n(A) = 5
(ii) Let Be be the favourable outcomes of visiting shop on the different days by them = 25 – 5 = 20 days.
i.e. n(B) = 20
Therefore, P(B) = 
(iii) Shyam T W Th. F
Ekta W Th. F S
Ekta T W Th. F
Shyam W Th. F S
Let A be the favourable outcomes of visiting shop on consecutive days.
Then, n(A) = 8
Therefore, P(A) = 
.
Number of possible outcomes = 36
i.e. n(S) = 36
(i) Let A be the favourable outcomes that the total score is even. Then n(A) = 18
Therefore, P(A) = 
(ii) Let B be favourable outcomes that the total score is 6. Then
n(B) = 4
Therefore. P(B) = 
(iii) Let C be the favourable outcomes that total score is at least 6. Then n(C) = 15
Therefore, P(C) = 
.
Number of red balls in the bag = 5 Let number of blue balls in the bag = x Total number of balls in the bag = x + 5 i.e., n(S) = x + 5
Let A be the favourable outcomes of getting red balls, then
n(A) = 5
Therefore, P(A) = 
Let B be the favourable outcomes of getting blue balls, then
n(B) = x
Therefore, P(B) = 
According to equation, P(B) = 2 P(A)
Hence, the number of blue balls in the bag is 10.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
i.e., n(S) = 12
(0 Let A be the favourable outcomes of getting black ball, then
n(A) = x
Therefore,
P(A) = 
(ii) Number of white balls in the box = x + 6 Total number of balls in the box = 12 + 6 = 18. i.e., n(S) = 18
Let B be the favourable outcomes of getting new white balls, then
n(B) = x + 6
Therefore,
P(B) = 
Now according to question
P(B) = 2P(A)
Hence, the number of white balls = 3.
Find the number of blue balls in the jar.
Total number of marbles in the jar = 24
i.e. n(S) = 24
Let number of blue balls in the jar be x.
Then, the number of green balls in the jar = 24 – x Let A be the favourable outcomes of getting blue balls. Then
n(A) = x
Therefore, P(A) = 
Let B be favourable outcomes of getting green balls. Then
n(B) = 24 – x
Therefore, P(B) = 
According to the given condition :
P(B) = 
Hence, the number of blue balls in the jar = 8.
Total number of possible outcome(s) are 16 i.e., n(S) = 16
Let A be the favourable outcomes of getting getting a number which is perfect square. Then A = (9,16) i.e., n(A) = 2
Therefore,
When a die is thrown oner then possible outcome(s) are 6
i.e., n(S) = 6
Let A be the favourable outcomes of setting a number less than 3. then
A = (1,2) i.e., n(A) = 2
Therefore,
When a die is thrown once, then possible outcome(s) are 6
i.e., n(S) = 6
Let ‘A’ be the favourable outcomes of getting a number greater than 5. Then A = (6) i.e., n(A) = 1
Therefore,
Here, we have S = {1, 2, 3,....., 15}
⇒ n(s) = 15
Let A be the favourable av + causes of getting a multiple 4, then A = {4,8,12}
⇒ n (A) = 3
Therefore, P(A) = 
2000 is a leap year. So, their birthday can be any day of 366 days in the year. We assume that these 366 outcomes are equally likely. Because both the friends have the same birthday.
∴ Favourable outcomes for their birthday = 1 Hence, P (both have the same birthday)
As we know probabilty of event can not be less than O and greater than 1.
i.e. O ≤ P ≤ 1
∶ (B) –1.5 is not possible.
If a card is drawn from a well shuffled deck of 52 cards, then possible outcomes (3) are 52 i.e. n(s) = 52
Let A be the favourable outcomes of setting a diamond then.
A = {13} i.e. (A) = 13
Therefore,
.When a Card is drawn of random from a well shoffled pack of cards, then possible outcomes are 52.
i.e. n(s) = 52
Let A be the favourbale outcomes of setting a black queen, then n(A) = 2 Therefore,
P(A) = 
Total no. of possible outcome(s) are 18. i.e., n(S) =18
Let A be the favourable outcomes of getting the number on the card which is an even number, then
A = (4,6,8,10,12,14,16,18,20) i.e., m(A) = 9
Therefore,
P(A) = 
If a die is thrown once, then
S = {1,2,3,4,5,6} i.e. n(s) = 6
Let A be favourable outcomes of setting a prime number, then
A = {2,3,5} i.e. n(4) = 3
Therefore, P(A) = 
If a die is thrown once, then S = {1,2,3,4,5,6} i.e. n( s) = 6
Let A be the favourable outcomes of setting number divisible by 2. then
A = {2,4,5} i.e. n(A) = 3
Therefore, P(A) = 
A die thrown once. Find the probability of getting
(i) an even prime number
(ii) a multiple of 3
If a die thrown once then possible outcomes are
S = {1, 2, 3, 4, 5, 6} i.e. n(S) = 6
(i) Let A be the favourable outcomes of getting an even prime number, then
A = [2] i.e. n(A) = 2
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of getting a multiple of three, then
B = {3,6} i.e. n(B) = 2
Therefore, P(B) = 
[(RR), (RW), (WR), (WW)]
Solution not provided.
Ans. 1
Solution not provided.
Ans. 66%
Solution not provided.
Ans. 
Solution not provided.
Ans. 0.481
Solution not provided.
Ans. 
A bag contain 5 yellow and 3 red ribbons. What is the probability that a child take out one ribbon at random.
(a) A red ribbon
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 0.113
Solution not provided.
Ans. 
Two coins ar tossed once. Write the probability of getting.
(a) At least one head
Solution not provided.
Ans. 
If we toss a coin then we get either a head or a tail. Thus number of possible outcomes(s) are two–Head(H) and Tail(T).
i.e., S = |HT|
⇒ n(S) = 2
Let E be the favourable outcomes of getting head, then
E = {H}
⇒ n(E) = 1
Therefore, P(E) P(Head) = 
Let F is the favourable outcomes of getting Tail, then
F = {T}
⇒ n(F) = 1
Therefore, P(F) = P(Tail) = 
Note : For any event E, P(E) + P(Ē) = 1.
Where Ē stands for‘note’E and Ē are called complementary favourable events.
If we toss a pair of coins twice then we get following possible outcomes (S).
i.e., S = { HT, TH, HH, TT }
⇒ n(S) = 04
Let E be the favourable outcomes of getting one head, then
E = { HT, TH }
⇒ n(E) = 2
Therefore, P(E) = 
Let F is the favourable outcomes of getting two tails, then
F = { T, T }
⇒ n(F) = 1
Therefore, P(F) = 
Two coins are tossed simultaneously. Find the probability of getting.
(i) two heads, (ii) at least one head, (iii) no head.
If we toss two coins simultaneously, then possible out comes (s), are
S = { HT, TH, HH, TT }
⇒ n( S) = 4
Let E be the favourable outcomes of getting two heads, then
E = { H H }
⇒ n(E) = 1
Therefore, P(E) = 
Let F be the favourable outcomes of getting at least one head, then
F = { HH, HT, TH }
⇒ n(F) = 3
Therefore, P(F) = 
Let G be the favourable outcomes of getting no head then
⇒ n (G) = 1
Therefore, P(G) = 
If we toss two unbiased coins, then possible outcomes(s), are
S = { HH, TH, TT, HT }
⇒ n(S) = 4
Let A be the favourable outcomes of getting exactly one head, then
A = {TH, HT } i.e., n( A) = 2
Therefore, P(A) = 
If we throw a die once then possible outcomes (s) are
S = {1, 2, 3, 4, 5, 6}
⇒ n(S) = 6
Let E be the favourable outcome of getting a prime number then
E = 12, 3, 5}
⇒ irt(E) = 3
Therefore, P(E) = 
If we roll a die once, then possible outcomes (s), are
S = (1, 2, 3, 4, 5, 6)
⇒ n( S) = 6
Let E be the favourable outcomes of getting a number greater than 2, then E = { 3, 4, 5, 6, }
⇒ n(E) = 4
Therefore, P(E) = 
If we throw a die once, then possible outcomes (s), are
S = {1, 2, 3, 4, 5, 6}
⇒ n(S) = 6
Let E be the favourable outcomes of getting a number 5 or 6, then
E = (5, 6}
⇒ n (E) = 2
Therefore, P(E) = 
A die is thrown once. Find the probability of getting
(i) an even number.
(ii) a number greater than 3.
(iii) a number between 3 and 6.
If we throw a die once, then possible outcomes (s), are
S = { 1, 2, 3, 4, 5, 6 }
⇒ n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3
Therefore, P(E) = 
(ii) Let F be the favourable outcomes of getting a number greater than 3, then
F = { 4, 5, 6 }
⇒ n( F) = 3
Therefore, P(E) = 
(iii) Let G be the favourable outcomes of getting a number between 3 and 6.
G = { 4, 5 }
⇒ n(G) = 2
Therefore, P(G) = 
A die is thrown once. Find the probability of getting
(i) an odd number.
(ii) a number greater than 4.
(iii) seven.
If we throw a die once, then possible outcomes (s) are,
S = { 1, 2, 3, 4, 5, 6 )
n(S) = 6
(i) Let F be the favourable outcomes of getting an odd number, then
E = { 1, 3, 5 }
⇒ n(E) = 3
Therefore, P(E) = 
(ii) Let F be the favourable outcomes of getting a number greater than 4, then F = {5, 6}
⇒ n(F) = 2
Therefore, P(F) = 
(iii) Let G be the favourable outcome of getting seven, then
G = { 0 }
⇒ n(G) = 0
Therefore, P(G) = 
(i) If a die is thrown once, then possible outcomes are
S {1, 2, 3, 4, 5, 6} i.e. n(S) = 6 Let A be the favourable outcomes of setting a prime number, than A = {2, 3, 5} i.e. n(A) = 3 Therefore,
P(A) = 
(ii) Let B be the favourbale outcomes of setting a number divisibel by 2, then B = {2, 4, 6} i.e. n(B) - 3
Therefore, P(B) = 
Let A be the favourable outcomes of getting the sum of the two numbers appearing on the top is less than or equal to 10, then
A = (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)
(6.1), (6,2), (6,3)/ (6,4) i.e., n( A) = 33
Therefore, 
A pair of dice is thrown once. Find the probability of getting a total of 5 on two dice.
If a pair of dice is thrown once, then possible outcomes(s) are 36. i.e., n(S) = 36
Let E be the favourable outcomes of getting a total of 5 on two dice, then
E = {(1, 4), (2, 3), (3, 2), (4,1)}
⇒ n(E) = 4
Therefore, P(E) = 
When a pair of dice is thrown once then possible outcome(s) are 36. i.e., n( S) = 36
Let A be the favourable outcomes of getting the same number of each dice, then
S= ((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)}
⇒ (A) = {(11), (2,2), (3,3), (4,4), (5,5), (6,6))
i.e. n( A)= 6
Therefore,
Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them?
(ii) 5 will come up on at least one?
(iii) 5 will come at both dice?
(ii) Let B be the favourable outcomes that 5 will come up on at least one dice. Then
B = {(1,5), (2, 5), (3, 5), (4, 5), (5,1),
(5,2), (5, 3) (5, 4), (5,5), (5, 6), (6,5)}
i.e., n(B) = 11
Therefore, P(B) = 
(iii) Let C be the event that 5 will come up on both the dice.
C = {5, 5}, i.e., n(C) = l
Therefore, P(C)=
Two dice are thrown simultaneously. Find the probability of getting :
(i) a sum less than 6 (ii) a sum less than 7
(iii) a sum more than 7 (iv) 8 as the sum
If two dice are thrown simultaneously, then possible outcomes(s) are 36. i.e., n(S)= 36
(i) Let E be the favourable outcomes of getting a sum less than 6, then
E= {(1,1),(1, 2), (l, 3), (1,4), (2, 1), (2, 2), (2, 3), (3, l), (3, 2), (4, l)}
⇒ n(E)= 10 Therefore,
P(E) = 
(ii) Let F be the favourable outcomes of getting a sum less than 7.
F= {(1, 1), (1, 2), (1,3), (1,4), (1,5). (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4,1), (4, 2), (5, 1)}
⇒ n( F)=15 Therefore,
P(F) = 
(iii) Let G be the favourable outcomes of getting a sum more than 7, then
G = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)1
⇒ n(G) = 15 Therefore,
P(G) = 
(iv) Let H be the favourable outcomes of getting 8 as the sum. then
H = |(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
⇒ n( H) = 6 Therefore.
P(H) = 
If a deck of 52 cards is shuffled then sample space is 52. i.e., n(S) = 52
Let E be the favourable outcomes of getting Jack, then
n(E) = 4 Therefore,
A card is drawn from a pack of 52 playing cards. What is the probability that it is
(i) an ace
(ii) a face card
(iii) any card numbered from 2 to 10?
Total number of all possible outcomes = 52
i.e., n( S) = 52
(i) Let E be the favourable outcomes of getting an ace, then
n(E) = 4 Therefore,
P(E) = 
(ii) Let F be the favourable outcomes of getting a face card, then
n(F) = 12 Therefore,
P(F) = 
(iii) Let G be the favourable outcomes of getting any card numbered from 2 to 10, then n(G)= {36}
Therefore,
P(G) = 
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is :
(i) red (ii) either red or king
(iii) red and a king (iv) a red face card
(v) ‘2’ of spades (vi) ‘10’ of a black suit.
Total number of all possible outcomes = 52.
i,e,. n(S)=52
(i) Let B be the favourable outcomes of getting red,then
n (B) = 26.
Therefore, P(B) = 
(ii) Let C be the favourable outcomes of getting a card either red or king, then n(C)= 28.
Therefore, P(C) = 
(iii) Let D be the favourable outcomes of getting red and a king, then n(D)=2 Therefore,
P(D) = 
(iv) Let E be the favourable outcomes of getting a red face card, then
n(E) = 6
Therefore, P(E) = 
(v) Let F be the favourable outcomes of getting 2 of spades, then
E = {1}
⇒ n(E) = 1
Therefore, P(E) = 
(vi) Let G be the favourable outcomes of getting 10 of black suit then ⇒
n(G) = 2
Therefore, P(G) = 
f one card is drawn from a well-shuffled deck of playing card, then possible outcome(s) are (52)
i.e., n(S) = 52
I. Let A be the favourable outcomes of getting a face card, then
i.e., n(A) = 12
Therefore,
II. Let B be the favourable outcomes of getting card which is neither a king nor a red card, then i.e., n(B) = 24
Therefore,
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 blackface card.
(iv) a Jack of hearts.
(v) a spade
(vi) a queen of diamond.
Total number of all possible outcomes = 52
i.e., n(S) = 52
(i) Let E be the favourable outcomes of getting a king of red colour, then n(E)= 2 Therefore,
P(E) = 
(it) Let F be the favourable outcomes of getting a face card, then
n(F) = 12
Therefore,
P(F) = 
(iii) Let G be the favourable outcomes of a red face card, then
n(G) = 6
P(G) = 
(iv) Let H be the favourable outcomes of getting a jack from the hearts, then n( H) = 1
Therefore,
P(H) = 
(v) Let I be the favourable outcomes of getting a spade, then
n(I) = 13
Therefore:
P(l) = 
(vi) Let J be the favourable outcomes of getting a queen of diamonds, then n(J) = 1
Therefore,
P(J) = 
Five cards - the ten, jack, queen, king and ace of diamonds are well-shuffled with their face down wards. One card is then picked up at random.
(i) What is the probability that the card is a 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.
It is given that the total number of cards = 5
i.e. n(S) = 5
(i) Let E be the favourable outcomes of getting a queen, then
E = {1}
⇒ n(E) = 1
Therefore,
P(E) = 
(ii) If the Queen is drawn and put aside then possible outcomes become 4 i.e., n(S) = 4
(a) Let A be the favourable outcomes that the picked up card is an ace, then n(A) = {1}
Therefore,
P(A) = 
(b) Let B be the favourable outcomes that the picked up card is a queen, then n(B) = 0
Therefore,
P(B) = 
The king, queen and jack of clubs are removed from a deck of 52 cards and then well shuffled. One card is selected from the remaining cards. Find the probability of getting
(i) a heart (ii) a kitty (iii) a club (iv) the 1O' of heart.
If the king, queen and jack of clubs are removed from a deck of 52 cards, then possible outcomes become 49.
i.e., n(S) = 49 [52 –3 = 49]
(i) Let E be the favourable outcomes of getting a heart, then
n (E) = 13 Therefore,
P(E) = 
(ii) Let F be the favourable outcomes of getting a king then,
n(F) = 3
Therefore,
P(F) = 
(iii) Let G be the favourable outcomes of getting a club, then
n(G) = 10
Therefore,
P(G) = 
(iv) Let H be the favourable outcomes of getting the 10 of hearts, then n(H) = 1
Therefore,
P(H) = 
If 2 black kings and 2 red aces are removed from a deck of 52 cards, find the probability of getting
(i) ati ace of heart
(ii) a king
(iii) a ace
(iv) a heart
(v) a red card
If 2 black kings and 2 red aces are removed from a deck of 52 cards, then possible outcomes are 52 – 4 = 48
i.e. n (S) = 48
(i) Let E be the favourable outcomes of getting an ace, then
n(E) = 0
Therefore,
P(E) = 
(ii) Let F be the favourable outcomes of getting an ace, then
n(F) = 2
Therefore,
P(F) = 
(iii) Let G be the favourable outcomes of getting a king, then
n(G) = 2
Therefore,
(v) Let 1 be the favourable outcomes of getting a red card, then,
n(I) = 24
Therefore,
P(I) = 
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is
(i) n black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king.
Total number of all possible outcomes = 52.
i.e., n( S) = 52
(i) Let A be the favourable outcomes of getting a black king, then
n( A) = 2
Therefore,
P(A) = 
(ii) Let B be the favourable outcomes of getting either a black card or a king, then n(B) = 28
Therefore,
P(B) = 
(iii) Let C be the favourable outcomes of getting black and a king, then n(C) = 2
Therefore,
P(C) = 
(iv) Let D be tine favourable outcomes of getting a jack, queen or a king then n(D) = 12
Therefore,
P(D) = 
(v) Let E be the favourable outcomes of getting neither a Heart nor a king, then n(E) = 36
Therefore,
P(E) = 
(vi) Let F be the favourable outcomes of getting spade or an ace, then
n(F) = 16
Therefore,
P(F) = 
(vii) Let G be the favourable outcomes of getting neither an ace nor a king, then n(G) = 44
Therefore,
P(G) = 
When a king, queen and jack of diamonds are removed from a pack of 52 cards, then possible outcome(s)are
52–3 = 49 i.e., n(S) = 49
I. Let A be the favourable outcomes of getting a card of diamonds. Then i.e., n( A) = 10
Therefore,
II. Let B be the favourable outcomes of getting a jack, then
i.e., n(B) = 3
Therefore,
All the three cards of spades are removed from a well-shuffled pack of 52 cards. A card is drawm at random from the remaining pack. Find the probability of getting.
(a) a black face card (b) a queen (c) a black card.
If all the three face cards of spades are removed, then possible outcome(s) are 49. i.e., n(S) = 49
(a) Let A be the favourable outcomes of getting a black face card, then n (A) = 3
Therefore,
P(A) = 
(b) Let B be the favourable outcomes of getting a queen, then
n(B) = 3
Therefore, P(B) = 
(c) Let C be the favourable outcomes of getting a black card, then
n(C) = {23}
Therefore, P(C) = 
Red kings, queens and jacks are removed from a deck of 52 playing cards and then well shuffled. A card is drawn from the remaining cards. Find the probability of getting
(i) a king, (ii) a red card, (iii) a spade.
If Red kings, queens and jacks are removed from a deck at 52 cards, then n(S) = 52
(i) Let ‘A’ be the favourable outcomes of getting a king, then
n(A) = 02
Therefore, P(A) = 
(ii) Let 'B' be the favourable outcomes of getting 'a red card' their
n(B) = 20
Therefore, P(B) = 
(iii) Let 'C' be the favourable outcomes of getting 'a spade'. Then
n(C) = 11
Therefore, P(C) = 
(a) a face card
(b) not a face card
Out of 52 cards all cards of ace, jack and queen are removed.
So, remaining cards = 52 – 12 = 40 i.e., n(S) = 40
(a) Let E be the favourable outcomes of getting face cards, then
E= {total face cards - face cards which are removed}
E = {12-8} i.e., n(E) = 4
Now, P (getting face cards)
(b) Let 'F' be the favourable outcomes of getting 'not a face card', then
F = {40 – 4} i.e., n( F) = 36
Now, P (getting not a face card)
No. of red colour balls = 5
No. of blue colour balls = 4
No. of green colour balls = 3
Total no. of balls = 12
i.e., n( S) = 12
I. Let ‘A’ be the favourable outcomes of getting ‘of red colour’. Then,
i.e., n( A) = 5
Therefore,
II. Let 'B' be the favourable outcomes of getting not of green colour. Then
i.e., n( B) = 9
Therefore,
No. of red balls = 4
No. black balls = 4
No. of yellow balls = 3
Total no. of balls (4+5+3) = 12 n(S) = 12
I. Let ‘A’ be the favourable outcomes of getting the ball taken out is of ‘yellow colour’ 1 hen
n( A) = 3
Therefore,
II. Let ‘B’ be the favourable outcomes of getting the ball taken out is of ‘not of red colour’ I hen,
i.e., n(B) = 8
Therefore,
No. of red balls in the bag = 5
No. of white balls in the bag = 8
No. of black balls in the bag = 7
No. of green balls in the bag = 4
Total no. of balls in the bag = 24
i.e., n(S) = 24
(i) Let ‘A’ be die favourable outcomes of getting black ball, then
i(A) = 7
Therefore,
P(A) = 
(ii) Let B' be the favourable outcomes of getting red balls, then
n(B) = 5
Therefore,
P(B) = 
(iii) Let ‘C’ be the favourable outcomes of getting a ball ‘not green’ then n(C) = 20
Therefore,
P(C) = 
Number of red balls in the bag = 5
Let number of blue balls in the bag = x
Now, total number of balls in the bag = x + 5
i.e., n( S) = n + 5
(i) Let ‘A’ be the favourable outcomes of getting blue balls, then
n( A) = x
Therefore,
P(A) = 
(i) Let 'B' be the favourable outcomes of getting red balls, then
n(B) = 5
Therefore,
P(B) = 
According to question :
P(A) = 4 P (B)
Hencem the number of blue balls in the bag is 20.
A box contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is
(a) White (b) neither red nor white
Number of red balls in the box = 5
Number of green balls in the box = 4
Number of white balls in the box = 7
Total number of balls in the box = 16
i.e., n (S) = 16
(i) Let A be the favourable outcomes of getting white balls, then
n(A) = 7
Therefore,
P(A) = 
(ii) Let B be the favourable outcomes of getting neither red nor white balls, then
n(B) = 1 - P (red or white)
= 1 - [P (red) + P (white)]
(i) white ball or a green ball.
(ii) neither a green ball not a red ball.
Total number of balls in the bag = 5 + 8 + 7 = 20
i.e. n( S) = 20
(i) Let A be the favourable outcomes of getting a white ball or a green ball. Then n( A) = 15
Therefore,
P(getting a white ball or a green ball)
(ii) Let B be the favourable outcomes of getting neither of a green ball nor a red ball. Then h(B) = 7
Therefore, P(getting neither a green ball nor a red ball)
Cards marked with the numbers 2 to 101 are placed in a box and mixed throughly. One card is drawn from this box. Find the probability that the number on the cards is
(i) an even number
(ii) a number less than 14.
(iii) a number which is a perfect square.
(iv) a prime n umber less than 20.
(i) Total no. of cards = 100
i.e., n( S)= 100
(ii) Let A be the favourable outcomes of getting an even number, then
A = {2,4,6,8,10........100}
⇒ n( A) = 50
Therefore,
(ii) Let B be the favourable outcomes of getting a number less than 14, then
B = {2,3,4,5,6...........13}
⇒ n(B) = 12
Therefore,
(iii) Let C be the favourable outcomes of getting a number which is a perfect square, then
C = {4,9,16,25,36,49,64,81,100}
⇒ n(C) = 9 Therefore,
(iv) Let D be the favourable outcomes of getting a prime number less than 20, then
D = {2,3,5,7,11,13,17,19}
⇒ n(D) = 8
Therefore,
18 Cards, numbered 1, 2, 3, ..., 18 are put in a box and mixed throughly. A card is drawn at random from the box. Find the probability that the Card drawn bears
(i) an even number
(ii) a number divisible by 2 or 3
Total number of cards 18 i.e., n( S) = 18
(i) Let A be the favourable outcomes of getting an even number, then
A= { 2. 4, 6, 8, 10, 12, 14, 16, 18 }
⇒ n(A) = 9 Therefore,
(ii) Let B be the favourable outcomes of getting a number divisible by 2 or 3 then, N(B) = 12
Therefore,
12 cards, numbered 1, 2,3......., 12 are put in a box and mixed throughly. A card is drawn at random from the box. Find the probability that the card drawn bears
(i) an even number
(ii) a number divisible by 2 or 3.
(i) Let A be the favourable outcomes of getting an even number, then
A = {2,4,6,8,10,12}
⇒ n( A) = 6
Therefore,
(it) Let B be the favourable outcomes of getting number divisible by 2 or 3, then n(B) ='8
Therefore,
TotaLnumbers of bulbs = 400 i.e. n{S) = 400
Let A be the favourable outcomes of getting 'non defective bulb', then
n(A) = 385
Therefore,
A leap year has 366 days, i.e., 52 weeks and 2 days.
The remaming 2 days can be
(i) Sunday and Monday.
(ii) Monday and Tuesday.
(iii) Tuesday and Wednesday.
(iv) Wednesday and Thursday.
(v) Thursday and Friday.
(vi) Friday and Saturday.
(vii) Saturday and Sunday.
Out of these seven possibilities only tw'o are favourable for Sunday.
Hence, probability of getting 53 Sundays in a leap year = 
Total no. of tickets =20 Le., n(S) = 20
I. Let ‘A’ be the favourable outcomes of getting a multiple of 7, then
A = (14,21,28)
i.e., n( A) = 3
Therefore,
II. Let ’B' be the favourable outcomes of getting the number greater than 15 and multiple of 5, then B = (20,25,30) i.e., n(B) = 3
Therefore,
Cards marked with numbers 3,4, 5,.....,50 are placed in a box and mixed throughly. One card is drawn at random from the box. Find the probability that number on the drawn card is
(i) divisible by 7
(ii) a number which is a square.
Total number of cards 48
i.e., n(S) = 48
(i) Let A be the favourable outcomes of getting a numbers divisible 7, then
A = {7,14,21,28,35,42,49}
⇒ n(A) = 7
Therefore,
P(A) = 
(ii) Let B be the favourable outcomes of getting number a which is a perfect square, then
B = {4, 9, 16, 25, 36, 49)
⇒ n(B) = 6
Therefore,
P(B) = 
Total no. of cards : 10 i.e., n(S) = 10
Let A be the favourable outcomes of getting a card with numbers less than 10, then
A = {6, 7, 8, 9}
i.e., n(A) = 4
Therefore,
P(A) = 
If two dice are thrown at the same time, then possible outcomes are 36.
i.e., n( S) = 36.
Let A be the favourable outcomes of getting the sum of two numbers appearing on the top of the dice is more than ? They
A = {(4,6), (5,5), (6,4), (5, 6), (6,5), (6,6)}
i.e., n( A) = 6
Therefore,
A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn from the bag at random. Find the probability that the number on the card, drawn from the box is
(i) an odd number,
(ii) a perfect square number,
(iii) a number divisible by 7.
Total no. of cards = 86, i.e., n(S) = 86
(i) Let A be the favourable outcomes of getting an odd number. Then,
A = {15,17,19,21,23,25,27,29,31,33,35, 37, 39,41,43,45,47,49, 51,53, 55,57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99}
n( A) = 43
Therefore,
(ii) Let B be the favourable outcomes of getting a perfect square number. Then,
B = {16, 25,36, 49, 64,81) i.e., n(B) = 6
Therefore,
(iii) Let C be the favourable outcomes of getting a number divisible by 7. Then
C = {14,21,28,35,42,49,56,63,70,77,84, 91,98}
i.e., n( C) = 13
Therefore,
Two unbiased coins are tossed simultaneously. Calculate the probability of getting
(i) exactly one tail.
(ii) exactly two heads.
Solution not provided.
Two unbiased coins are tossed. Calculate the probability of getting
(i) At least two tails.
(ii) At most two tails.
Solution not provided.
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. S = [HH, HT, TH, TT]
Two coins are tossed once. Find the probability of getting:
(i) Exactly one head.
(ii) At most one head.
Solution not provided.
“Three unbiased” coins are tossed together. Find the probability of getting :
(i) One head.
(ii) All heads.
Solution not provided.
Three unbiased coins are tossed together. Find the probability of getting :
(i) Two heads.
(ii) At least two heads.
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
An unbiased die is thrown once. Find the probability of getting
(i) an even number and a multiple of 3. (ii) an even number or a multiple of 3.
Solution not provided.
Ans. (i) 
(ii) 
Tips: -
Problems Based on DieAn unbiased die is thrown once. Find the probability of getting:
(i) a multiple of 2 or 3.
(ii) 2 or 4.
Solution not provided.
Ans. (i) 
(ii) 
Tips: -
Problems Based on DieSolution not provided.
Ans. 
Tips: -
Problems Based on DieSolution not provided.
Ans. 0
Solution not provided.
Ans. 
A die in the shape of tetrahedron has four faces on which 3, 4, 6 and 8 are written. The die is rolled once. Find the probability’ of getting
(i) a prime number.
(ii) a number less than 6.
(iii) a number greater than 8.
(iv) a number between 4 and 8.
Solution not provided.
Tips: -
Problems Based on DieA fair die is tossed. List the sample space. State the probability of the favourable event.
(i) a number greater than 3 appears.
(ii) a number 3 appears.
Solution not provided.
Ans. 
Tips: -
Problems Based on DieAn unbiased die is thrown. Find the probability of getting
(i) a number less than 5.
(ii) a number greater than 5.
Solution not provided.
Ans. 
Tips: -
Problems Based on DieA die is thrown once. What is the probability of getting a number other ghan 3.
Problems Based on Two Dice
Solution not provided.
Ans. 
Tips: -
Problems Based on DieSolution not provided.
Ans. 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on Die.Two dice are thrown simultaneously. Find the probability of getting:
(i) an even number as sum
(ii) an odd number as the sum
(iii) the product as a perfect square.
Solution not provided.
Ans. 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on Die.In a single throw of two dice, find the probability of:
(i) not getting the same number on the two dice.
(ii) getting a sum greater than 9.
Solution not provided.
Ans. 
Tips: -
Problems Based on Die.A die is thrown once. Find the probability of getting:
(i) a number greater than 6.
(ii) a number between 4 and 6.
Solution not provided.
Ans. (i) 0 (ii) 
Tips: -
Problems Based on Die.Solution not provided.
Ans. 
Tips: -
Problems Based on CardsSolution not provided.
Ans. 
The King, Queen, and Jack of clubs are removed from a deck of 52 playing cards. One card is selected from the remaining cards. Find the probability of getting:
(i) a heart, (ii) a king, (iii) a club, (iv) the '10' of heart.
Solution not provided.
Ans. 
Tips: -
Problems Based on CardsSolution not provided.
Ans. (i) 0 (ii) 
Tips: -
Problems Based on CardsOut of a pack of 52 playing cards, two black kings and 4 red cards were lost. Find the probability that the card drawn is a :
(i) red card, (ii) king, (iii) black card, (iv) black Jack, (v) black Queen.
Solution not provided.
Ans. 
Tips: -
Problems Based on CardsSolution not provided.
Ans. 
Tips: -
Problems Based on CardsTwo black kings and two black jacks are removed from a pack of 52 cards. Find the probability of getting
(i) A card of hearts
(ii) A black card
(iii) Either a red nor a king
(iv) Neither an ace nor a king
(v) A jack, queen or a king.
Solution not provided.
Ans. 
Tips: -
Problems Based on CardsSolution not provided.
Ans. 10
Tips: -
Problems Based on BallsA box contains 12 balls out of which x are black. If one ball is drawn at randone from the box, what is the probability that it will be a black ball.
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x
Solution not provided.
Ans. 
Tips: -
Problems Based on BallsSolution not provided.
Ans. 
Tips: -
Solution not provided.
Ans. 8
(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.
Solution not provided.
Ans. (i) 
(ii) 
Solution not provided.
Ans. 0.7
The number of matchsticks in each of 30 boxes found to have the following distribution ?
|
No. of match stikcs |
No. of boxes |
|
46 47 48 49 50 |
2 8 10 3 7 |
What is the probability' that a box selected at random will contain
(i) 50 matchsticks
(ii) less than 48 matchsticks.
(iii) at least 46 matchsticks.
(iv) more than 50 matchsticks.
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution nor provided.
Ans. 
Solution not provided.
Ans. 0.005
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Five cards: ten, jack, queen, king and an ace of diamonds are shuffled face downwards. One card is picked at random.
(i) What is the probability that the card is a queen?
(ii) If a king is drawn first and put aside, what is the probability that the second card picked up is the ace?
Solution not provided.
Ans. 
Solution not provided.
Ans. 24
(i) A lot of 20 bulbs contains 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 not replaced. Now bulb is drawn at random from the rest. What is the probability that this bulb is not defective.
Solution not provided.
Ans. 
Solution not provided.
(i) 
(ii) 
A piggy bank contains hundred 50 paises coins, fifty Re 1 coins twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fallout when the bank is turned up side dawn, what is the probability that the coin
(i) Will be a 50 paise coins ?
(ii) Will not be a Rs 5 coins?
Solution not provided.
(i) 
(ii) 
Solution not provided.
(i) 
(ii) 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
A die is thrown once. Find the probability of getting
(a) A prime number
(b) A number divisble by 2
Solution not provided.
Ans. 
A die is thrown once. Find the probability of getting..
(a) An even prime number
(b) A multiple of 3
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Three different coins are tossed together. Find the probability of getting
(i) exactly two heads
(ii) at least two heads
(iii) at least two tails.
When three coins are tossed together, the possible outcomes are
HHH, HTH, HHT, THH, THT, TTH, HTT, TTT
therefore,
Total number of possible outcomes = 8
(i) Favourable outcomes of exactly two heads are HTH, HHT, THH
Total number of favourable outcomes= 3
Probability of getting exactly two heads = 3/8
(ii) Favourable outcomes of at least two heads are HHH, HTH, HHT, THH
Total number of favourable outcomes =4
Probability of getting at least two heads =4/8 = 1/2
(iii)Favourable outcomes of at least two tails are THT, TTH, HTT, TTT
Total number of favourable outcomes =4
Probability of getting at least two tails - 4/8 = 1/2
A number x is selected at random from the numbers 1, 2, 3, and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.
x is selected from 1,2,3 and 4
1,2,3,4
y is selected from 1,4,9 and 16
Let A {1,4,9,16,2,8,18,32,3,12,27,48,36,64} which consists
of elements that are product of x and y
The different dice are tossed together. Find the probability that the product of the two number on the top of the dice is 6.
When two dice are thrown simultaneously, the possible outcomes can be listed as:
|
|
1 |
2 |
3 |
4 |
5 |
6 |
|
1 |
(1, 1) |
(1, 2) |
(1, 3) |
(1, 4) |
( 1, 5) |
(1, 6) |
|
2 |
(2, 1) |
(2, 2) |
(2, 3) |
(2, 4) |
(2, 5) |
(2, 6) |
|
3 |
(3, 1) |
(3, 2) |
(3, 3) |
(3, 4) |
(3, 5) |
(3, 6) |
|
4 |
(4, 1) |
(4, 2) |
(4, 3) |
(4, 4) |
(4, 5) |
(4, 6) |
|
5 |
(5, 1) |
(5, 2) |
(5, 3) |
(5, 4) |
(5, 5) |
(5, 6) |
|
6 |
(6, 1) |
(6, 2) |
(6, 3) |
(6, 4) |
(6, 5) |
(6, 6) |
∴ Total number of possible outcomes = 36
The outcomes favourable to the event the product of the two number of the top of the dice is 6 denoted by E are (1, 6), (2, 3), (3, 2) and (6, 1)
∴ Number of favourable outcomes = 4
Let E1, E2 and E3 be the events of drawing a red, blue and orange ball, respectively
We Know
P(E1) = 1/4
P (E2) = 1/3
Therefore,
P(E3) = 1 - P (E2) - P (E1)
Total number of cards = 52
(i) Number of cards that are spades or aces = 13 + 3 = 16
Probability that the card drawn is a card of spade or an ace
= 
(ii) Number of black kings = 2
Probability that the card drawn is a black king
= 
(iii) Number of cards that are neither jacks nor kings = 52 - 8 = 44
Probability that the card drawn is neither a jack nor a king
= 
(iv) Number of cards that are kings or queens = 4 + 4 = 8
Probability that the card drawn is either a king or queen
= 
What is the HCF of the smallest prime number and the smallest composite number?
Smallest prime number is 2.
Smallest composite number is 4
therefore, HCF is 2
Given that is irrational, prove that is an irrational number.
Let us assume that is rational. Then there exist co-prime positive integers a and b such that
This contradicts that is irrational.
So our assumption is incorrect.
Hence, is an irrational number.
Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.
Using the factor tree for the prime factorization of
404 and 96, we have
404 = 22× 101 and 96 = 25 × 3
To find the HCF, we list common prime factors and their smallest exponent in 404 and 96 as under: Common prime factor = 2, Least exponent = 2
therefore, HCF = 22 =4
To find the LCM, we list all prime factors of 404 and 96 and their greatest exponent as follows:
| Prime factors of 404 and 96 | Greatest Exponent |
| 2 | 5 |
| 3 | 1 |
| 101 | 1 |
Therefore,
LCM = 25 x 31 x 1011
= 25 x 31 x 1011
= 9696
Now,
HCF x LCM = 9696 x 4= 38784
Product of two numbers = 404 x 96
Therefore, HCF x LCM = Product of two numbers.
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#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041205.png)
.![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041205-1.png)
![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041247.png)
has a terminating or non-terminating decimal expansion.![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041255.png)
as a non-terminating recurring decimal.
![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041357.png)
be a rational number such that the prime factorization of q is not of the form 2![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>](/application/zrc/images/qvar/MAEN10041372.png)
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