NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5 - Arithmetic Progressions

# NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5 - Arithmetic Progressions

Edited By Ramraj Saini | Updated on Nov 17, 2023 04:52 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2

NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5 Arithmetic Progressions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This class 10 ex 5.2 the consists of 20 questions solved using the formula of the sum of n terms of the Arithmetic Progression. It also has some word problems too to enhance the understanding of this concept.

NCERT solutions for Exercise 5.2 Class 10 Maths Chapter 5 Arithmetic Progressions focus on the Arithmetic Progression’s basic notion, i.e., how an Arithmetic Progression is formed? Also, it stresses to clear the understanding of the nth term of the Arithmetic Progression. 10th class Maths exercise 5.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

##### PTE Registrations 2024

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

## Assess NCERT Solutions for Class 10 Maths chapter 5 exercise 5.2

Arithmetic Progressions Class 10 Chapter 5 Excercise: 5.2

 a d n $\small a_n$ (i) (ii) (iii) (iv) (v) 7 $\small -18$ $\small ...$ $\small -18.9$ $\small 3.5$ 3 $\small ...$ $\small -3$ $\small 2.5$ 0 8 10 18 $\small ...$ 105 $\small ...$ 0 $\small -5$ $\small 3.6$ $\small ...$

(i)
It is given that
$a=7, d = 3 , n = 8$
Now, we know that
$a_n = a+(n-1)d$
$a_8 = 7+(8-1)3= 7+7\times 3 = 7+21 = 28$
Therefore,
$a_8 = 28$

(ii) It is given that
$a=-18, n = 10, a_{10} = 0$
Now, we know that
$a_n = a+(n-1)d$
$a_{10} = -18+(10-1)d$
$0 +18=9d$
$d = \frac{18}{9}=2$
(iii) It is given that
$d=-3, n = 18, a_{18} = -5$
Now, we know that
$a_n = a+(n-1)d$
$a_{18} = a+(18-1)(-3)$
$-5=a+17\times (-3)$
$a = 51-5 = 46$
Therefore,
$a = 46$

(iv) It is given that
$a=-18.9, d = 2.5, a_{n} = 3.6$
Now, we know that
$a_n = a+(n-1)d$
$a_{n} = -18.9+(n-1)2.5$
$3.6+18.9= 2.5n-2.5$
$n = \frac{22.5+2.5}{2.5}= \frac{25}{2.5}= 10$
Therefore,
$n = 10$

(v) It is given that
$a=3.5, d = 0, n = 105$
Now, we know that
$a_n = a+(n-1)d$
$a_{105} = 3.5+(105-1)0$
$a_{105} = 3.5$
Therefore,
$a_{105} = 3.5$

(A) $\small 97$ (B) $\small 77$ (C) $\small -77$ (D) $\small -87$

Given series is
$\small 10,7,4,...,$
Here, $a = 10$
and
$d = 7 - 10 = -3$
Now, we know that
$a_n = a+(n-1)d$
It is given that $n = 30$
Therefore,
$a_{30} = 10+(30-1)(-3)$
$a_{30} = 10+(29)(-3)$
$a_{30} = 10-87 = -77$
Therefore, $\small 30$ th term of the AP: $\small 10,7,4,...,$ is -77

(A) $\small 28$ (B) $\small 22$ (C) $\small -38$ (D) $\small -48\frac{1}{2}$

Given series is
$\small -3,-\frac{1}{2},2,...,$
Here, $a = -3$
and
$d =-\frac{1}{2} -(-3)= -\frac{1}{2} + 3 = \frac{-1+6}{2}= \frac{5}{2}$
Now, we know that
$a_n = a+(n-1)d$
It is given that $n = 11$
Therefore,
$a_{11} = -3+(11-1)\left ( \frac{5}{2} \right )$
$a_{11} = -3+(10)\left ( \frac{5}{2} \right )$
$a_{11} = -3+5\times 5 = -3+25 = 22$
Therefore, 11th term of the AP: $\small -3,-\frac{1}{2},2,...,$ is 22
Hence, the Correct answer is (B)

$\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26$

Given AP series is
$\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26$
Here, $a = 2 , n = 3 \ and \ a_3 = 26$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_3 =2+(3-1)d$
$\Rightarrow 26 -2=(2)d$
$\Rightarrow d = \frac{24}{2}= 12$
Now,
$a_2= a_1+d$
$a_2= 2+12 = 14$
Therefore, the missing term is 14

$\small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3$

Given AP series is
$\small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3$
Here, $a_2 = 13 , n = 4 \ and \ a_4 = 3$
Now,
$a_2= a_1+d$
$a_1= a = 13 - d$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_4 =13-d+(4-1)d$
$\Rightarrow 3-13=-d+3d$
$\Rightarrow d = -\frac{10}{2}= -5$
Now,
$a_2= a_1+d$
$a_1= a = 13 - d= 13-(-5 ) = 18$
And
$a_3=a_2+d$
$a_3=13-5 = 8$
Therefore, missing terms are 18 and 8
AP series is 18,13,8,3

$\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}$

Given AP series is
$\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}$
Here, $a = 5 , n = 4 \ and \ a_4 = 9\frac{1}{2}= \frac{19}{2}$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_4 =5+(4-1)d$
$\Rightarrow \frac{19}{2} -5=3d$
$\Rightarrow d = \frac{19-10}{2\times 3} = \frac{9}{6} = \frac{3}{2}$
Now,
$a_2= a_1+d$
$a_2 = 5+\frac{3}{2} = \frac{13}{2}$
And
$a_3=a_2+d$
$a_3=\frac{13}{2}+\frac{3}{2} = \frac{16}{2} = 8$
Therefore, missing terms are $\frac{13}{2}$ and 8
AP series is $5,\frac{13}{2}, 8 , \frac{19}{2}$

$\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6$

Given AP series is
$\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6$
Here, $a = -4 , n = 6 \ and \ a_6 = 6$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_6 =-4+(6-1)d$
$\Rightarrow 6+4 = 5d$
$\Rightarrow d = \frac{10}{5} = 2$
Now,
$a_2= a_1+d$
$a_2 = -4+2 = -2$
And
$a_3=a_2+d$
$a_3=-2+2 = 0$
And
$a_4 = a_3+d$
$a_4 = 0+2 = 2$
And
$a_5 = a_4 + d$
$a_5 = 2+2 = 4$
Therefore, missing terms are -2,0,2,4
AP series is -4,-2,0,2,4,6

$\small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22$

Given AP series is
$\small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22$
Here, $a_2 = 38 , n = 6 \ and \ a_6 = -22$
Now,
$a_2=a_1+d$
$a_1=a =38-d \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_6 =38-d+(6-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$\Rightarrow -22-38-=-d+5d$
$\Rightarrow d = -\frac{60}{4} = - 15$
Now,
$a_2= a_1+d$
$a_1 = 38-(-15) = 38+15 = 53$
And
$a_3=a_2+d$
$a_3=38-15 = 23$
And
$a_4 = a_3+d$
$a_4 = 23-15 = 8$
And
$a_5 = a_4 + d$
$a_5 =8-15 = -7$
Therefore, missing terms are 53,23,8,-7
AP series is 53,38,23,8,-7,-22

Given AP is
$\small 3,8,13,18,...,$
Let suppose that nth term of AP is 78
Here, $a = 3$
And
$d = a_2-a_1 = 8 - 3 = 5$
Now, we know that that
$a_n = a + (n-1)d$
$\Rightarrow 78 = 3 + (n-1)5$
$\Rightarrow 78 -3 = 5n-5$
$\Rightarrow n = \frac{75 +5}{5}= \frac{80}{5} = 16$
Therefore, value of 16th term of given AP is 78

$\small 7,13,19,...,205$

Given AP series is
$\small 7,13,19,...,205$
Let's suppose there are n terms in given AP
Then,
$a = 7 , a_n = 205$
And
$d= a_2-a_1 = 13-7 = 6$
Now, we know that
$a_n =a + (n-1)d$
$\Rightarrow 205=7 + (n-1)6$
$\Rightarrow 205-7 = 6n-6$
$\Rightarrow n = \frac{198+6}{6} = \frac{204}{6} = 34$
Therefore, there are 34 terms in given AP

$\small 18,15\frac{1}{2},13,...,-47$

Given AP series is
$\small 18,15\frac{1}{2},13,...,-47$
suppose there are n terms in given AP
Then,
$a = 18 , a_n = -47$
And
$d= a_2-a_1 = \frac{31}{2}-18 = \frac{31-36}{2} = -\frac{5}{2}$
Now, we know that
$a_n =a + (n-1)d$
$\Rightarrow -47=18 + (n-1)\left ( -\frac{5}{2} \right )$
$\Rightarrow -47-18= -\frac{5n}{2}+\frac{5}{2}$
$\Rightarrow -\frac{5n}{2}= -65-\frac{5}{2}$
$\Rightarrow -\frac{5n}{2}= -\frac{135}{2}$
$\Rightarrow n = 27$
Therefore, there are 27 terms in given AP

Given AP series is
$\small 11,8,5,2...$
Here, $a = 11$
And
$d = a_2-a_1 = 8-11 = -3$
Now,
suppose -150 is nth term of the given AP
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow -150 = 11+(n-1)(-3)$
$\Rightarrow -150- 11=-3n+3$
$\Rightarrow =n = \frac{161+3}{3}= \frac{164}{3} = 54.66$
Value of n is not an integer
Therefore, -150 is not a term of AP $\small 11,8,5,2...$

It is given that
$\small 11$ th term of an AP is $\small 38$ and the $\small 16$ th term is $\small 73$
Now,
$a_{11} =38= a+ 10d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_{16} =73= a+ 15d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= -32 \ \ \ and \ \ \ d = 7$
Now,
$a_{31} = a+30d = -32 + 30\times 7 = -32+210 = 178$
Therefore, 31st terms of given AP is 178

It is given that
AP consists of $\small 50$ terms of which $\small 3$ rd term is $\small 12$ and the last term is $\small 106$
Now,
$a_3 = 12=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_{50} = 106=a+49d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= 8 \ \ \ and \ \ \ d = 2$
Now,
$a_{29} = a+28d=8+28\times 2 = 8 +56 = 64$
Therefore, 29th term of given AP is 64

It is given that
$\small 3$ rd and the $\small 9$ th terms of an AP are $\small 4$ and $\small -8$ respectively
Now,
$a_3 = 4=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_{9} = -8=a+8d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= 8 \ \ \ and \ \ \ d = -2$
Now,
Let nth term of given AP is 0
Then,
$a_{n} = a+(n-1)d$
$0 = 8+(n-1)(-2)$
$2n = 8+2= 10$
$n = \frac{10}{2} = 5$
Therefore, 5th term of given AP is 0

It is given that
$\small 17$ th term of an AP exceeds its $\small 10$ th term by $\small 7$
i.e.
$a_{17}= a_{10}+7$
$\Rightarrow a+16d = a+9d+7$
$\Rightarrow a+16d - a-9d=7$
$\Rightarrow 7d=7$
$\Rightarrow d = 1$
Therefore, the common difference of AP is 1

Given AP is
$\small 3,15,27,39,...$
Here, $a= 3$
And
$d= a_2-a_1 = 15 - 3 = 12$
Now, let's suppose nth term of given AP is $\small 132$ more than its $\small 54$ th term
Then,
$a_n= a_{54}+132$
$\Rightarrow a+(n-1)d = a+53d+132$
$\Rightarrow 3+(n-1)12 = 3+53\times 12+132$
$\Rightarrow 12n = 3+636+132+12$
$\Rightarrow 12n = 636+132+12$
$\Rightarrow n = \frac{780}{12}= 65$
Therefore, 65th term of given AP is $\small 132$ more than its $\small 54$ th term

It is given that
Two APs have the same common difference and difference between their $\small 100$ th terms is $\small 100$
i.e.
$a_{100}-a'_{100}= 100$
Let common difference of both the AP's is d
$\Rightarrow a+99d-a'-99d=100$
$\Rightarrow a-a'=100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - (i)$
Now, difference between 1000th term is
$a_{1000}-a'_{1000}$
$\Rightarrow a+999d -a'-999d$
$\Rightarrow a-a'$
$\Rightarrow 100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i) )$
Therefore, difference between 1000th term is 100

We know that the first three digit number divisible by 7 is 105 and last three-digit number divisible by 7 is 994
Therefore,
$a = 105 , d = 7 \ and \ a_n = 994$
Let there are n three digit numbers divisible by 7
Now, we know that
$a_n = a+ (n-1)d$
$\Rightarrow 994 = 105 + (n-1)7$
$\Rightarrow 7n = 896$
$\Rightarrow n = \frac{896}{7} = 128$
Therefore, there are 128 three-digit numbers divisible by 7

We know that the first number divisible by 4 between 10 to 250 is 12 and last number divisible by 4 is 248
Therefore,
$a = 12 , d = 4 \ and \ a_n = 248$
Let there are n numbers divisible by 4
Now, we know that
$a_n = a+ (n-1)d$
$\Rightarrow 248 = 12 + (n-1)4$
$\Rightarrow 4n = 240$
$\Rightarrow n = \frac{240}{4} = 60$
Therefore, there are 60 numbers between 10 to 250 that are divisible by 4

Given two AP's are
$\small 63,65,67,...$ and $\small 3,10,17,...$
Let first term and the common difference of two AP's are a , a' and d , d'
$a = 63 \ , d = a_2-a_1 = 65-63 = 2$
And
$a' = 3 \ , d' = a'_2-a'_1 = 10-3 = 7$
Now,
Let nth term of both the AP's are equal
$a_n = a'_n$
$\Rightarrow a+(n-1)d=a'+(n-1)d'$
$\Rightarrow 63+(n-1)2=3+(n-1)7$
$\Rightarrow 5n=65$
$\Rightarrow n=\frac{65}{5} = 13$
Therefore, the 13th term of both the AP's are equal

It is given that
3rd term of AP is $\small 16$ and the $\small 7$ th term exceeds the $\small 5$ th term by $\small 12$
i.e.
$a_3=a+2d = 16 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_7=a_5+12$
$a+6d=a+4d+12$
$2d = 12$
$d = 6$
Put the value of d in equation (i) we will get
$a = 4$
Now, AP with first term = 4 and common difference = 6 is
4,10,16,22,.....

Given AP is
$\small 3,8,13,...,253$
Here, $a = 3 \ and \ a_n = 253$
And
$d = a_2-a_1=8-3 = 5$
Let suppose there are n terms in the AP
Now, we know that
$a_n = a+(n-1)d$
$253= 3+(n-1)5$
$5n = 255$
$n = 51$
So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting
Therefore,
$a_{32} = a+31d$
$a_{32} = 3+31\times 5 = 3+155 = 158$
Therefore, 20th term from the of given AP is 158

It is given that
sum of the < img alt="\small 4" class="fr-fic fr-dii" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%204"> th and $\small 8$ th terms of an AP is $\small 24$ and the sum of the $\small 6$ th and $\small 10$ th terms is $\small 44$
i.e.
$a_4+a_8=24$
$\Rightarrow a+3d+a+7d=24$
$\Rightarrow 2a+10d=24$
$\Rightarrow a+5d=12 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_6+a_{10}=44$
$\Rightarrow a+5d+a+9d=44$
$\Rightarrow 2a+14d=44$
$\Rightarrow a+7d=22 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= -13 \ and \ d= 5$
Therefore,first three of AP with a = -13 and d = 5 is
-13,-8,-3

It is given that
Subba Rao started work at an annual salary of Rs 5000 and received an increment of Rs 200 each year
Therefore, $a = 5000 \ and \ d =200$
Let's suppose after n years his salary will be Rs 7000
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow 7000=5000+(n-1)200$
$\Rightarrow 2000=200n-200$
$\Rightarrow 200n=2200$
$\Rightarrow n = 11$
Therefore, after 11 years his salary will be Rs 7000
after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years
Thus , 1995+10 = 2005

It is given that
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs $\small 1.75$
Therefore, $a = 5 \ and \ d = 1.75$
after $\small n$ th week, her weekly savings become Rs $\small 20.75$
Now, we know that
$a_n = a +(n-1)d$
$\Rightarrow 20.75= 5+(n-1)1.75$
$\Rightarrow 15.75= 1.75n-1.75$
$\Rightarrow 1.75n=17.5$
$\Rightarrow n=10$
Therefore, after 10 weeks her saving will become Rs 20.75

## More about NCERT Solutions for Class 10 Maths Exercise 5.2

It's about the nth term of the Arithmetic Progression and deals with the number of terms and rank of the particular term in the Arithmetic Progression. In addition to this, it has word problems that give in-depth knowledge of the topic. Exercise 5.2 Class 10 Maths - Arithmetic Progression is the progression in which the difference between two consecutive terms is constant. Using this very concept nth term, a number of terms and rank of a particular term of the progression could be determined. The NCERT solutions for Class 10 Maths exercise 5.2 mainly focuses on the nth term, the validity of the term, and the number of terms in the Arithmetic Progression, and twenty questions are given in exercise 5.2 Class 10 Maths. Students can quickly go through the Arithmetic Progressions Class 10 Notes to revise all concepts all together.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 5.2

• NCERT solutions for Class 10 Maths exercise 5.2 helps solve and revise all these exercises’ questions.
• You would be able to gain more points if you go through Class 10 Maths chapter 5 NCERT solutions for exercise 5.2, and if you practice it thoroughly, it will help you score well in maths in examinations.
• Exercise 5.2 in chapter 5 of maths for Class 10 is based on fundamental characteristics of Arithmetic Progression, i.e., any two consecutive terms of the Arithmetic Progression differ by the constant numerical value.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. How do we find a particular progression is an Arithmetic Progression?

In Arithmetic Progression, any two consecutive terms differ by a constant numerical value.

2. How is the nth term of the Arithmetic Progression found?

It could be calculated easily by using the prop[erty that any two consecutive terms differ by a constant numerical value. To calculate it, just add the difference (n-1) times to the first term of the Arithmetic Progression.

3. How can we find that a particular term belongs to a certain Arithmetic Progression?

For this, just find the rank of that term in that Arithmetic Progression, and if the position comes out to be in fraction, then that term doesn’t belong to that Arithmetic Progression; otherwise, it is.

4. Can a whole Arithmetic Progression be determined by just any two non-consecutive terms of the Arithmetic Progression?

Yes, it could be done efficiently by using the nth term formula. There would be two unknowns, namely the first term and the common difference, and we would have two equations with us; solving them, we would get those. And once the first term and common difference are calculated then, Arithmetic Progression could be determined.

5. How can we reverse the Arithmetic Progression?

For this, just do two things i.e.

1. Take the last term to be the first term.

2. Reverse the sign of the common difference(if it was +2, do it -2 and vice versa)

6. Any term or the Common Difference can be in fraction or not?

Yes, It could be in a fraction. Only the number of terms in the Arithmetic Progression can't be in a fraction.

7. What is n in the Arithmetic Progression?

It's just the generalized way to represent any term of the Arithmetic Progression. Based on the requirement, it could define the first term, last term, etc.

8. How can we find the Common Difference of the Arithmetic Progression?

To find the Common Difference of the Arithmetic Progression, just differentiate (n-1)th term from the nth term.

9. According to the NCERT answers for Class 10 Maths chapter 5 exercise 5.2, what is the nth term?

According to this exercise, the nth term is any term of the Arithmetic Progression that could be calculated by adding Common Difference (n-1) times to the first term of the Arithmetic Progression. This very concept is required to frame the whole Arithmetic Progression.

10. What sorts of questions are addressed in the NCERT solutions for Class 10 Maths chapter 5 exercise 5.2?

The questions are based on the concept that the two consecutive terms of the Arithmetic Progression always differ by a constant numerical value. Based on this concept, there are word problems too available in this exercise.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

#### National Institute of Open Schooling 10th examination

Exam Date:20 September,2024 - 07 October,2024

#### National Institute of Open Schooling 12th Examination

Exam Date:20 September,2024 - 07 October,2024

Edx
1113 courses
Coursera
804 courses
Udemy
394 courses
Futurelearn
222 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD

### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9