Two water taps can fill a tank in [ 9+(3/8) ] hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank

 cbse chapter 4 quadratic equations word problems / time - work problems

exercise 4.3

 

9. Two water taps can fill a tank in [ 9+(3/8) ] hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank

 

Let x = time taken by tap of larger diameter to fill the tank (in hours)

 

given that the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately

 

so  time taken by tap of smaller diameter to fill the tank =(x+10) hours

together they take [ 9+(3/8) ] hours or (75/8) hours by converting the mixed fraction

remeber to use reciprocals in time /work problems

[1/x] +[1/(x+10)] = 1/{75/8}

[1/x] +[1/(x+10)] =[8/75]

 

multiply each term with 75x(x+10)

 

75(x+10) +75x =8x(x+10)

75x+750 +75x =8(x^2)+80x

8(x^2) -70x -750=0 dividing by 2

4(x^2) -35x -375=0

 

we search for two numbers 

whose sum is (-35)  and 

product =4*(-375) =(-1500) 

the numbers are (-60) and 25

use this to split the middle term

4(x^2) -35x -375=0

4(x^2) -60x +25x -375=0

 

4x[x-15]+25[x-15]=0

[x-15][4x+25] =0

 

x=15  or x={(-25) /4} which is rejected

 

x=15 hours for the tap of larger diameter.

 

x+10 =15+10=25hours for the tap of smaller diameter.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop atintermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

x =speed of passenger train (in hours)
(x+11) =speed express train (in hours)

distance = 132 km

time = distance/speed

t1= 132/x [time for passenger train]

t2= 132/(x+11) [time for express train]

given express train takes 1 hour less than a passenger train to travel 132 km  

so

t1 -t2 =1 

using the above values

[132/x] -[132/(x+11)] =1

multiply x(x+11)

132(x+11)-132x = 1x(x+11)

1452= (x^2) + 11 x

 

(x^2) +11x -1452 =0

to factorise search for two numbers whose product is (-1452) and sum is 11 

numbers are 44 and (-33)

so (x+44)(x-33)=0

x=-44(which is rejected

x=33 km/hr

speed of passenger train =x = 33 km/hr

speed of express train =(x+11)=33+11 =44 km/hr

 

----------------------------------------------------------------------------------------

If the equations x^2 -ax+b=0 and x^2-ex+f =0 have a root in common and the second equation has equal roots show that ae =2(b+f) 

 

solution

solving a quadratic equation by completing the square

solve 2x² -4x -7 = 0

solution

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

 

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.2

 

 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

given t3=12 , t(50)  =106

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d with n=3 and n=50

t3 = a+(3-1)d =a+2d

t50 = a+(50-1)d = a+49d

a+2d   =12-----------------(1)

a+49d =106---------------(2)

--------------------------------subtracting

   (-47) d =( -94 )

d = (-94)/(-47)

d=2

 

substituting in (1)

a+2d   =12

a+2(2) =12

a+4=12

a=12-4

a=8 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d with n=29

t(29) = 8 + (29-1)(2) = 8+28*2 = 8+56 =64

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

given t11=38 , t(16)  =73

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d with n=11 and n=16

t(31)=a+(11-1)d =a+10d

t(16)=a+(16-1)d =a+15d

a+10d =38--------------(1)

a+15d=73--------------(2)

------------------------------------substracting

(-5)d  = (-35)

d=(-35)/(-5)

d=7

substitute in (1)

a+10(7)=38

a+70=38

a=38-70

a=(-32)

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d with n=31

 

t(31)=  a+ 30d = (-32)+ 30*7=178

 

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

 solution

 

19.

 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on . In how many rows are the 200 logs placed and how many logs are in the top row?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . .  What is the total length of such a spiral made up of thirteen consecutive semicircles.

solution 

 

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

16. A sum of Rs.700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each of the prizes. 

solution

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs.300 for the third day, etc., the penalty for each succeeding day being Rs.50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? 

solution

 

14. Find the sum of the odd numbers between 0 and 50.

 solution

  13. Find the sum of the first 15 multiples of 8.

solution

 12. Find the sum of the first 40 positive integers divisible by 6.

solution

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

solution

10.  Show that a1 , a2 , . . ., an , . . . form an AP where a n is defined as below :
 an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

solution

 

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

  solution

 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

solution

 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

 solution

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

solution

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

solution 

 

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

solution 

 

 3.

given a = 5, d = 3, an = 50, find n and Sn

solution

 (ii) given a = 7, a13 = 35, find d and  S13

solution

(iii) given a(12) = 37, d = 3, find a and S(12 )

solution

(iv) given a3 = 15, S(10) = 125, find d and a(10)

 solution 

(v) given d = 5, S9 = 75, find a and a9 .

solution 

(vi) given a = 2, d = 8, Sn = 90, find n and an .

solution 

 vii) given a = 8, an = 62, Sn = 210, find n and d

solution

 (vii) given an = 4, d = 2, Sn = –14, find n and a.

solution 

 ix) given a = 3, n = 8, S = 192, find d.

solution

(x) given L= 28, S = 144, and there are total 9 terms. Find a.

solution

find the sums given below :

7 + [10 +(1/2) ] +14 + ...+84

solution

(ii) 34 + 32 + 30 + . . . + 10

 solution

(iii) –5 + (–8) + (–11) + . . . + (–230)

solution  

 Find the sum of the following APs:
 2, 7, 12, . . ., to 10 terms.

solution   

 (ii) –37, –33, –29, . . ., to 12 terms.

solution 

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

solution  

 (iv) (1/15) +(1/12) +(1/10) + .... 11terms

solution

 

Aruna saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs.20.75, find n.

solution

exercise 5.2

19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000?

solution  

 

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

solution  

 

17.Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

solution

 

 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

solution
 

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . 

and 3, 10, 17, . . . equal?

solution

 14. How many multiples of 4 lie between 10 and 250?

solution

 

13.How many three-digit numbers are divisible by 7?

solution   

 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

solution   

 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

solution 

 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

solution 

 

 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

solution

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

solution

6. Check whether (-150) is a term of the AP : 11, 8, 5, 2 . . .

solution  

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

 

 

5. Find the number of terms in each of the following APs :

 7,13,19, ....,205

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.2

 

6. Check whether (-150) is a term of the AP : 11, 8, 5, 2 . . .

 

 a = 11

d= 8-11=(-3)

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

tn=11+(n-1)(-3)

 

take tn = (-150)  and try to find n

(-150)=11+(n-1)(-3)

(-150) -11 =(n-1)(-3)

 

(-161)=(n-1)(-3)

 

n-1 =[161/3]

n=[161/3] +1 is not a natural number

so (-150) is not a term of the AP 

 

5. Find the number of terms in each of the following APs :

 7,13,19, ....,205

a=7

d=13-7=6

 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

tn=7+(n-1)(6) 

 

take tn = (205)  and try to find n

 

205 = 7+(n-1)(6) 

(n-1) =[205-7]/6

(n-1)=33

n=34

number of terms =34

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

 solution

 

19.

 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on . In how many rows are the 200 logs placed and how many logs are in the top row?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . .  What is the total length of such a spiral made up of thirteen consecutive semicircles.

solution 

 

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

16. A sum of Rs.700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each of the prizes. 

solution

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs.300 for the third day, etc., the penalty for each succeeding day being Rs.50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? 

solution

 

14. Find the sum of the odd numbers between 0 and 50.

 solution

  13. Find the sum of the first 15 multiples of 8.

solution

 12. Find the sum of the first 40 positive integers divisible by 6.

solution

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

solution

10.  Show that a1 , a2 , . . ., an , . . . form an AP where a n is defined as below :
 an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

solution

 

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

  solution

 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

solution

 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

 solution

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

solution

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

solution 

 

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

solution 

 

 3.

given a = 5, d = 3, an = 50, find n and Sn

solution

 (ii) given a = 7, a13 = 35, find d and  S13

solution

(iii) given a(12) = 37, d = 3, find a and S(12 )

solution

(iv) given a3 = 15, S(10) = 125, find d and a(10)

 solution 

(v) given d = 5, S9 = 75, find a and a9 .

solution 

(vi) given a = 2, d = 8, Sn = 90, find n and an .

solution 

 vii) given a = 8, an = 62, Sn = 210, find n and d

solution

 (vii) given an = 4, d = 2, Sn = –14, find n and a.

solution 

 ix) given a = 3, n = 8, S = 192, find d.

solution

(x) given L= 28, S = 144, and there are total 9 terms. Find a.

solution

find the sums given below :

7 + [10 +(1/2) ] +14 + ...+84

solution

(ii) 34 + 32 + 30 + . . . + 10

 solution

(iii) –5 + (–8) + (–11) + . . . + (–230)

solution  

 Find the sum of the following APs:
 2, 7, 12, . . ., to 10 terms.

solution   

 (ii) –37, –33, –29, . . ., to 12 terms.

solution 

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

solution  

 (iv) (1/15) +(1/12) +(1/10) + .... 11terms

solution

 

Aruna saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs.20.75, find n.

solution

exercise 5.2

19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000?

solution  

 

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

solution  

 

17.Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

solution

 

 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

solution
 

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . 

and 3, 10, 17, . . . equal?

solution

 14. How many multiples of 4 lie between 10 and 250?

solution

 

13.How many three-digit numbers are divisible by 7?

solution   

 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

solution   

 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

solution 

 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

solution 

 

6. Check whether (-150) is a term of the AP : 11, 8, 5, 2 . . .

solution  

 

5. Find the number of terms in each of the following APs :

 7,13,19, ....,205

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

 

 

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.2

 

 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

n=17, n=10

t(17) =a+(17-1)d

t(17) =a+16d

similarly

t(10) =a+9d

17th term of an AP exceeds its 10th term by 7

means

t(17) = t(10) + 7

a+16d = [a+9d] +7

16d -9d = 7

7d=7

d=1

 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

n=3, n=9

 

t3 = a+(3-1)d

t3=a+2d

 

t9=a+8d

 

given

3rd and the 9th terms of an AP are 4 and – 8 respectively

 

so 

a+2d =4--------------------(1)

a+8d=(-8)--------------------(2)

-----------------------------------------------subtracting

-6d =12

d=12 /(-6)

d=(-2)

 

use (1)

 

a+2(-2) =4

a-4=4

a=4+4

a=8

 

 tn= a+ (n-1)d

tn=8+(n-1)(-2)

 

now when tn=0 , n=?

0= 8+(n-1)(-2)

(-8) =(n-1)(-2)

 

(-8)/(-2) = (n-1)

4=(n-1)

n=4+1

n=5 

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

 solution

 

19.

 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on . In how many rows are the 200 logs placed and how many logs are in the top row?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . .  What is the total length of such a spiral made up of thirteen consecutive semicircles.

solution 

 

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

16. A sum of Rs.700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each of the prizes. 

solution

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs.300 for the third day, etc., the penalty for each succeeding day being Rs.50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? 

solution

 

14. Find the sum of the odd numbers between 0 and 50.

 solution

  13. Find the sum of the first 15 multiples of 8.

solution

 12. Find the sum of the first 40 positive integers divisible by 6.

solution

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

solution

10.  Show that a1 , a2 , . . ., an , . . . form an AP where a n is defined as below :
 an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

solution

 

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

  solution

 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

solution

 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

 solution

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

solution

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

solution 

 

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

solution 

 

 3.

given a = 5, d = 3, an = 50, find n and Sn

solution

 (ii) given a = 7, a13 = 35, find d and  S13

solution

(iii) given a(12) = 37, d = 3, find a and S(12 )

solution

(iv) given a3 = 15, S(10) = 125, find d and a(10)

 solution 

(v) given d = 5, S9 = 75, find a and a9 .

solution 

(vi) given a = 2, d = 8, Sn = 90, find n and an .

solution 

 vii) given a = 8, an = 62, Sn = 210, find n and d

solution

 (vii) given an = 4, d = 2, Sn = –14, find n and a.

solution 

 ix) given a = 3, n = 8, S = 192, find d.

solution

(x) given L= 28, S = 144, and there are total 9 terms. Find a.

solution

find the sums given below :

7 + [10 +(1/2) ] +14 + ...+84

solution

(ii) 34 + 32 + 30 + . . . + 10

 solution

(iii) –5 + (–8) + (–11) + . . . + (–230)

solution  

 Find the sum of the following APs:
 2, 7, 12, . . ., to 10 terms.

solution   

 (ii) –37, –33, –29, . . ., to 12 terms.

solution 

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

solution  

 (iv) (1/15) +(1/12) +(1/10) + .... 11terms

solution

 

Aruna saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs.20.75, find n.

solution

exercise 5.2

19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000?

solution  

 

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

solution  

 

17.Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

solution

 

 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

solution
 

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . 

and 3, 10, 17, . . . equal?

solution

 14. How many multiples of 4 lie between 10 and 250?

solution

 

13.How many three-digit numbers are divisible by 7?

solution   

 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

solution   

 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

solution 

 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

solution 

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probability formulae

 Let S be the sample space and A a subset of S

P(A)=n(A)/n(S)

 

P(S)=1

P( {} )  = 0,  {} stands for null set

 

let A' denote the event that "A does not happen"

P(A) +P(A') =1

 

P(A')  = 1-P(A)

P(A) = 1-P(A')

 

addition theorem

P[A ∪ B] = P(A) +P(B) - P[A∩B] 

P[A or B] = P(A) +P(B) - P[A and B] 

 

Bayes theorem

Total probability

 

 

problem 2
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

solution to problem 2 of  bayes theorem

problem 3
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostelier?

solution to problem 3 of  bayes theorem


problem 4
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let ( 3/4 ) be the probability that he knows the answer and ( 1/4) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability ( 1/4 ). What is the probability that the student knows the answer given that he answered it correctly?
solution to bayes theorem problem 4 for ncert cbse 12th mathematics

problem 5

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?
solution to bayes theorem problem 5 for ncert cbse 12th mathematics

problem 6

There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

solution to problem 6 of bayes theorem for cbse ncert mathematics

problem 7

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

solution to problem 7 of bayes theorem for cbse ncert mathematics 

problem 8
 A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further,2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
solution of problem 8 on bayes theorem for cbse mathematics 

problem 9
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
solution of problem 9 on bayes theorem for cbse mathematics

problem 10
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
solution of problem10 on bayes theorem for ncert cbse mathematics

problem 11
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
solution of problem11 on bayes theorem for ncert cbse mathematics

problem 12
A card from a pack of 52 cards is lost. From the remaining cards of the pack,two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
 solution to bayes theorem problem 12 for ncert cbse mathematics probability

problem 13
Probability that a man speaks truth is (  4 / 5 ). A coin is tossed and the man reports that a head appeared.Find the probability that actually there was a head.
 solution to bayes theorem problem 13for ncert cbse mathematics probability


miscellaneous exercise problem 14
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of a certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
solution to miscellaneous exercise problem 14 on bayes theorem in  ncert cbse 12th mathematics

miscellaneous exercise problem 15
Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
solution to miscellaneous exercise problem 15on bayes theorem in  ncert cbse 12th mathematics


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bayes theorem is a topic for cbse / ncert  / scert of 12th standard in India. These are some of the important questions from the topic of bayes theorem from various cbse ncert textbooks and old question paper

How many three-digit numbers are divisible by 7?

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.2

 

 13.How many three-digit numbers are divisible by 7?

 

smallest three digit number divisible by 7 is 105

next is 112, 119 and so on

 biggest three digit number divisible by 7 is 994

 

so the required numbers are 105,112,119, ...994

which is an AP with a=105, d =7

tn=994 , n=?

 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

994=105 +(n-1)7

994-105= (n-1)7

889=(n-1)7

889/7 =(n-1)

127=n-1

n=127+1

n=128

 

 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

 

a=3

d=15-3=12

 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

n=54

t(54) =3 +(54-1)(12) =3+53*12=639

 

132 more than its 54th term means

132+639=771

 

tn=771, n=?

 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d

771= 3 +(n-1)12

[771-3] /12  =(n-1)

768/12=(n-1)

64=n-1

n=64+1

n=65

 

 

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

 solution

 

19.

 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on . In how many rows are the 200 logs placed and how many logs are in the top row?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . .  What is the total length of such a spiral made up of thirteen consecutive semicircles.

solution 

 

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

16. A sum of Rs.700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each of the prizes. 

solution

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs.300 for the third day, etc., the penalty for each succeeding day being Rs.50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? 

solution

 

14. Find the sum of the odd numbers between 0 and 50.

 solution

  13. Find the sum of the first 15 multiples of 8.

solution

 12. Find the sum of the first 40 positive integers divisible by 6.

solution

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

solution

10.  Show that a1 , a2 , . . ., an , . . . form an AP where a n is defined as below :
 an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

solution

 

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

  solution

 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

solution

 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

 solution

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

solution

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

solution 

 

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

solution 

 

 3.

given a = 5, d = 3, an = 50, find n and Sn

solution

 (ii) given a = 7, a13 = 35, find d and  S13

solution

(iii) given a(12) = 37, d = 3, find a and S(12 )

solution

(iv) given a3 = 15, S(10) = 125, find d and a(10)

 solution 

(v) given d = 5, S9 = 75, find a and a9 .

solution 

(vi) given a = 2, d = 8, Sn = 90, find n and an .

solution 

 vii) given a = 8, an = 62, Sn = 210, find n and d

solution

 (vii) given an = 4, d = 2, Sn = –14, find n and a.

solution 

 ix) given a = 3, n = 8, S = 192, find d.

solution

(x) given L= 28, S = 144, and there are total 9 terms. Find a.

solution

find the sums given below :

7 + [10 +(1/2) ] +14 + ...+84

solution

(ii) 34 + 32 + 30 + . . . + 10

 solution

(iii) –5 + (–8) + (–11) + . . . + (–230)

solution  

 Find the sum of the following APs:
 2, 7, 12, . . ., to 10 terms.

solution   

 (ii) –37, –33, –29, . . ., to 12 terms.

solution 

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

solution  

 (iv) (1/15) +(1/12) +(1/10) + .... 11terms

solution

 

Aruna saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs.20.75, find n.

solution

exercise 5.2

19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000?

solution  

 

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

solution  

 

17.Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

solution

 

 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

solution
 

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . 

and 3, 10, 17, . . . equal?

solution

 14. How many multiples of 4 lie between 10 and 250?

solution

 

13.How many three-digit numbers are divisible by 7?

solution   

 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

solution   

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

 

values of trigonometric functions of particular angles

 taking the angle in degree measure

 

 sin(0)=0

cos(0)=1

tan(0)=0

 

sin(90)=1

cos(90)=0

tan(90)=infinity

 

sin(30)=1/2

cos(30)=[( sqrt(3) ) / 2]

tan(30)=1 / {sqrt(3)}

 

 

sin(60)=[( sqrt(3) ) / 2]

cos(60)=1/2

tan(60)= {sqrt(3)}

 

sin(45) ={ 1 / [  sqrt(2) ] }

cos(45) ={ 1 / [  sqrt(2) ] }

tan(45)=1

 

  trigonometry identities

 

1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work