Find the square root of the following number by using their ones and tens digit ?(b) 1444 ,(c)3025 (d)4761

Test for divisibility by 11. Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.

So, for instance, 2728 has alternating sum of digits 2 – 7 + 2 – 8 = -11. Since -11 is divisible by 11, so is 2728.

Similarly, for 31415, the alternating sum of digits is 3 – 1 + 4 – 1 + 5 = 10. This is not divisible by 11, so neither is 31415.

By Euclid's division algorithm
b = aq + r,0 ≤ r < a [∵ dividend = divisor × quotient + remainder]
⇒ 117 = 65 × 1 + 52
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF (65,117) = 13             ...(i)     
Also, given that, HCF (65,117) = 65m - 117  ...(ii)
From Eqs.(i) and (ii),
65m - 117 = 13
⇒ 65m = 130
⇒ m = 2

S t e p - b y - s t e p e x p l a n a t i o n:

It is G iv e n 

i ) 

=

/* From (1) */

=

=

=

=---(1)

ii) 

= 

After cancellation, we get

= ---(2)

Therefore,

 

OR

Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

 

This theorem also says that the prime factorisation of a natural number is unique, except for the order of its factors.

 

For example 20 can be expressed as

2×2×5

Using this theorem the LCM and HCF of the given pair of positive integers can be calculated.

It's already deleted from our syllabus....

Dear it' s not in syllabus

Ex1.1

1. 1/365 2. 364/365

No it is not possible because ratio can not be negative

Yes

In which of the following situations, does the list of numbers involved make an 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. (ii) The amount of air present in a cylinder when a vacuum pump removes one fourth of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter. (iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum. Ans. (i)Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23 Taxi fare after 3 km = 23 + 8 = Rs 31 Taxi fare after 4 km = 31 + 8 = Rs 39 Therefore, the sequence is 15, 23, 31, 39... It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, ...) (ii)Let amount of air initially present in a cylinder = V Amount of air left after pumping out air by vacuum pump =  Amount of air left when vacuum pump again pumps out air =  So, the sequence we get is like  Checking for difference between consecutive terms ...  Difference between consecutive terms is not equal. Therefore, it is not an arithmetic progression. (iii) Cost of digging 1 meter of well = Rs 150 Cost of digging 2 meters of well = 150 + 50 = Rs 200 Cost of digging 3 meters of well = 200 + 50 = Rs 250 Therefore, we get a sequence of the form 150, 200, 250... It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50...) Here, difference between any two consecutive terms which is also called common difference is equal to 50. (iv)Amount in bank after Ist year = … (1) Amount in bank after two years = … (2) Amount in bank after three years = … (3) Amount in bank after four years = … (4) It is not an arithmetic progression because (2) − (1) ≠ (3) − (2) (Difference between consecutive terms is not equal) Therefore, it is not an Arithmetic Progression. 2. Write first four terms of the AP, when the first term a and common difference d are given as follows: (i)a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3 (iv) a = -1, d =  (v) a = -1.25, d = -0.25 Ans. (i) First term = a = 10, d = 10 Second term = a + d = 10 + 10 = 20 Third term = second term + d = 20 + 10 = 30 Fourth term = third term + d = 30 + 10 = 40 Therefore, first four terms are: 10, 20, 30, 40 (ii) First term = a = –2 , d = 0 Second term = a + d = –2 + 0 = –2 Third term = second term + d = –2 + 0 = –2 Fourth term = third term + d = –2 + 0 = –2 Therefore, first four terms are: –2, –2, –2, –2 (iii) First term = a = 4, d =–3 Second term = a + d = 4 – 3 = 1 Third term = second term + d = 1 – 3 = –2 Fourth term = third term + d = –2 – 3 = –5 Therefore, first four terms are: 4, 1, –2, –5 (iv) First term = a = –1, d =  Second term = a + d = –1 +  = − Third term = second term + d = −+ = 0 Fourth term = third term + d = 0 + = Therefore, first four terms are: –1, −, 0, (v) First term = a = –1.25, d = –0.25 Second term = a + d = –1.25 – 0.25 = –1.50 Third term = second term + d = –1.50 – 0.25 = –1.75 Fourth term = third term + d = –1.75 – 0.25 = –2.00 Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00 3. For the following APs, write the first term and the common difference. (i) 3, 1, –1, –3 … (ii) –5, –1, 3, 7... (iii)  (iv) 0.6, 1.7, 2.8, 3.9 ... Ans. (i) 3, 1, –1, –3... First term = a = 3, Common difference (d) = Second term – first term = Third term – second term and so on Therefore, Common difference (d) = 1 – 3 = –2 (ii) –5, –1, 3, 7... First term = a = –5 Common difference (d) = Second term – First term = Third term – Second term and so on Therefore, Common difference (d) = –1 – (–5) = –1 + 5 = 4 (iii) First term = a =  Common difference (d) = Second term – First term = Third term – Second term and so on Therefore, Common difference (d) =  (iv) 0.6, 1.7, 2.8, 3.9... First term = a = 0.6 Common difference (d) = Second term – First term = Third term – Second term and so on Therefore, Common difference (d) = 1.7 − 0.6 = 1.1 4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16... (ii) 2, , 3, ... (iii) −1.2, −3.2, −5.2, −7.2... (iv) −10, −6, −2, 2... (v)  (vi) 0.2, 0.22, 0.222, 0.2222... (vii) 0, −4, −8, −12... (viii)  (ix) 1, 3, 9, 27... (x) a, 2a, 3a, 4a... (xi)  (xii)  (xiii)  (xiv)  (xv)  Ans. (i) 2, 4, 8, 16... It is not an AP because difference between consecutive terms is not equal. As4 – 2 ≠ 8 − 4 (ii)2, , 3, ... It is an AP because difference between consecutive terms is equal.  Common difference (d) =  Fifth term =  Sixth term = 4 + ½ =  Seventh term =  Therefore, next three terms are 4, and 5. (iii)−1.2, −3.2, −5.2, −7.2... It is an AP because difference between consecutive terms is equal.  −3.2 − (−1.2) = −5.2 − (−3.2) = −7.2 − (−5.2) = −2 Common difference (d) = −2 Fifth term = −7.2 – 2 = −9.2Sixth term = −9.2 – 2 = −11.2 Seventh term = −11.2 – 2 = −13.2 Therefore, next three terms are −9.2, −11.2 and −13.2 (iv) −10, −6, −2, 2... It isan AP because difference between consecutive terms is equal.  −6 − (−10) = −2 − (−6) = 2 − (−2) = 4 Common difference (d) = 4 Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10 Seventh term = 10 + 4 = 14 Therefore, next three terms are 6, 10 and 14 (v)  It is an AP because difference between consecutive terms is equal.     Common difference (d) =  Fifth term =  Sixth term =  Seventh term =  Therefore, next three terms are  (vi) 0.2, 0.22, 0.222, 0.2222... It is not an AP because difference between consecutive terms is not equal. 0.22 − 0.2 ≠ 0.222 − 0.22 (vii) 0, −4, −8, −12... It is an AP because difference between consecutive terms is equal.  −4 – 0 = −8 − (−4) = −12 − (−8) = −4 Common difference (d) = −4 Fifth term = −12 – 4 =−16 Sixth term = −16 – 4 = −20 Seventh term = −20 – 4 = −24 Therefore, next three terms are −16, −20 and −24 (viii)  It is an AP because difference between consecutive terms is equal.   Common difference (d) = 0 Fifth term = Sixth term =  Seventh term =  Therefore, next three terms are  (ix) 1, 3, 9, 27... It is not an AP because difference between consecutive terms is not equal. 3 – 1 ≠ 9 − 3 (x) a, 2a, 3a, 4a... It is an AP because difference between consecutive terms is equal.  2a – a = 3a − 2a = 4a − 3a = a Common difference (d) = a Fifth term = 4a + a = 5a Sixth term = 5a + a = 6a Seventh term = 6a + a = 7a Therefore, next three terms are 5a, 6a and 7a (xi) a, a2, a3, a4... It is not an AP because difference between consecutive terms is not equal. a2 – a ≠ a3 − a2 (xii)    It is an AP because difference between consecutive terms is equal.   Common difference (d) =  Fifth term =  Sixth term =  Seventh term =  Therefore, next three terms are  (xiii)  It is not an AP because difference between consecutive terms is not equal.  (xiv)  It is not an AP because difference between consecutive terms is not equal.  (xv)   1, 25, 49, 73... It is an AP because difference between consecutive terms is equal.   = = 24 Common difference (d) = 24 Fifth term = 73 + 24 = 97 Sixth term = 97 + 24 = 121 Seventh term = 121 + 24 = 145 Therefore, next three terms are 97, 121 and 145 Exercise 1

Which exercise?

24:19

Frustum is removed from syllabus

If three 3⃣ sides of a triangle are equal to the three corresponding sides of other triangle, then triangle??? are congruent.

Total cost of computer = 32000
total cost of microwave= 6500
loss on computer=5%
               s.p.=((100-L%)/100)Xc.p.
               s.p.=95/100X32000
               S.P. = 30400
profit on microwave=15%
s.p.=115/100X6500
S.P.=7475
                 total  c.p.=(38500
                          s.p.=37875
 p%=  (625/38500) X 100
 =1.62% gain percent 

 (i) We can prove irrational by contradiction.

Let us suppose that  is rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

=ab

⇒  … (1)

R.H.S of (1) is rational but we know that  is irrational.

It is not possible which means our supposition is wrong.

Therefore,  cannot be rational.

Hence, it is irrational.

Answer:

The value of k is 3.

Step-by-step explanation:

Given the equations of two lines 4x+6y-1=0 & 2x+ky-7=0

we have to find the value of k for which the above two lines are parallel.

The two lines are

Also the slope of parallel lines are equal

⇒ 

Cross-multiplying both sides

Hence, the value of k is 3.

A N S W E R : 
HCF ( 52,91 ) = greatest possible speed.
2∣52                                                              7∣91
___________                                             _____________
2∣26                                                             13∣13
___________                                             ______________
13∣13                                                                  1
___________  
     1
So 13 km/hr is the greatest possible speed.

(I) 1/2 (2) 3/4 (3) 1 (4) 1/2

Thanks to all of you

On putting n = 1, 2,3 ,…. in eq (1), For n = 1 (an) = - 4n + 15 1 = - 4(1) + 15   a1 = 11   For n = 2 a2 = - 4(2) + 15   a2 = - 8 + 15   a2 = 7   Common Difference , d = a2 - a1   d = 7-11   d = - 4   By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d] S20 = (20/2)[2 × 11 + (20 -1)(-4)] S20 = 10 [22 + 19 × - 4[ S20 = 10 [22 - 76] S20 = 10 × - 54 S20 = - 540   Hence,  the sum of first 20 terms of AP is  - 540.

nth term of an A.P, (a_n) = - 4n + 15……..(1)

On putting n = 1, 2,3 ,…. in eq (1),

For n = 1

(a_n) = - 4n + 15

1 = - 4(1) + 15  

a₁ = 11

For n = 2

a₂ = - 4(2) + 15  

a₂ = - 8 + 15  

a₂ = 7

Common Difference , d = a₂ - a₁  

d = 7-11  

d = - 4

By using the formula ,Sum of nth terms , S_n = n/2 [2a + (n – 1) d]

S₂₀ = (20/2)[2 × 11 + (20 -1)(-4)]

S₂₀ = 10 [22 + 19 × - 4[

S₂₀ = 10 [22 - 76]

S₂₀ = 10 × - 54

S₂₀ = - 540  

Hence,  the sum of first 20 terms of AP is  - 540.

 

Kis book ka h

Given :  

nth term of an A.P, (an) = - 4n + 15……..(1)

 

On putting n = 1, 2,3 ,…. in eq (1),

For n = 1

(an) = - 4n + 15

1 = - 4(1) + 15  

a1 = 11

 

For n = 2

a2 = - 4(2) + 15  

a2 = - 8 + 15  

a2 = 7

 

Common Difference , d = a2 - a1  

d = 7-11  

d = - 4

 

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

S20 = (20/2)[2 × 11 + (20 -1)(-4)]

S20 = 10 [22 + 19 × - 4[

S20 = 10 [22 - 76]

S20 = 10 × - 54

S20 = - 540  

Hence,  the sum of first 20 terms of AP is  - 540.

It 's deleted from syllabus sister

Division algorithm is deducted from syllabus

10 dafny

Which class?

HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.

Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.

The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.

CBSE stated in its notification that the deleted portions of the CBSE syllabus should not be considered for internal assessments or for the year-end board exams.

Deleted syllabus of CBSE Class 10 Mathematics

 

You could easily know which question is in syllabus if you are through with the syllabus

Exampler has mixed questions so we need to aware which topics have been deleted. It is not possible to tell like this

Syallbus of standard maths and basic maths is same

 

HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.

Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.

The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.

CBSE stated in its notification that the deleted portions of the CBSE syllabus should not be considered for internal assessments or for the year-end board exams.

Deleted syllabus of CBSE Class 10 Mathematics