## Online Gramin Dak Sevak Engagement

Extension(17.07.2017 to 22.07.2017) for apply online is provided for AP & Telangana candidates who has registered earlier and could not able to apply
2. Candidates who did not get OTP with old registered mobile number can change their mobile number Click Here
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## GDS Online Recruitment - Clarification regarding Marks & Grades and Extension of Date for online submission

GDS Online Recruitment - Clarification regarding Marks & Grades and Extension of Date for online submission : Andhra Pradesh Circle Press Release.
Extension Dates for other Circles.

## Extension of date for GDS Online Recruitment

Source: GDS Blog

ARTHMETIC PROGRESSION:
1The tenth term of an A.P is -15 and 31st term is -57 . Find the 15thterm?

Solution:    t10  = a+(n-1)d = a+9d and  t31 =a+30d By solving both          -21d=-42 ie d=-2 again by substituting d value in above a-18= -15 ie a  =3 then the 15th term is 3+(15-1)(-2) = -25

2. The first row of seating in an outdoor amphitheater contains 26 seats, the second row  contains 28 seats, the third row contains 30 seats, and so on. If there are 18 rows, what is the total seating capacity of the theater?

Solution: :Here the difference between any two successive terms is 2. The sequence is an arithmetic progression where a= 26 and = 2.

The no of seats in 18th row is a18 =a1+(n-1)d ie a18 =26+(18-1)2 =60
Sn =n(a1+an/2) =18(26+60)/2 =774

3. If a triangular stack of bricks has 37 bricks on the bottom row, 34 bricks on the second row and so on with one brick on top. How many bricks are in the stack?

Solution:  Here d = -3 then  37+(n-1)-3 =1 ie 37-3n+3=1 ie -3n=-39 ie n=13
Sn = 13(37+1)/2= 247

4. Which term of A.P 5,11,17 is 119

Solution : Here a=5, d=6 then tn =119=5+(n-1)6 ie 5+6n-6 =119 ie 6n =120 n=20

5. The common difference of an A.P is 3 and 15th term is 37 Find the first term ?

Solution : Here t15 = a+(15-1)*3 =37 ie a+42=37 then a=-5

6. Three numbers are in A.P the difference between the 1st and last is 8 and the product of this is 20 . Find the series.

Solution : The numbers are a,a+d,a+2d where a-(a+2d) =8 ie -2d=8 and d =-4 again a*a+2d = a2+2ad =20 ie a2+2a(-4) =20 ie a2-8a-20=0 ie by solving this a is 10 or -2 The series is 10,6,2 or -2,-6,-10

7. The first term of A.P is 10 and the last term is 50. The sum of all terms is 480 . Find the common difference ?

Solution: Here a =10 tn =50 and Sn =480 therefore Sn=n/2{2a+(n-1)d)
480= n/2{20+(n-1)d tn =50=10+(n-1)d ie d=10
480=n/2(20+(n-1)10) ie n=16 again 50=10+(15)d ie 15d =40 d=8/3

8. Find the first term of A.P where 8th term is 39 and 12th term is 59?

Solution: Given a8 = a1+7d = 39 and a12 =a1+11d =59 by solving above
-4d =-20 ie d=5 by substituting d=5 in a8 a1+35=39 ie a1=4

9. Find the lowest number in A.P such that the sum of all the terms is 105 and greater term is 6 times of the least?

Solution: Here Sn =105=n/2(a+L) and 6a=L ie 105=n/2(7a) ie 7an =210
an =210/7=30 by substituting in 30=6a then a=5

10. How many numbers between 300 to 500 are multiple of 7 ?

Solution: The first number divisible by 7 is 301 and last is 497

Here a1 =301 and L=497 and d=7
497= a+(n-1)d ie 301+(n-1)7 =7n-7+301 =497 ie 203=7n ie n =203/7 =29

11. The sum of the terms is 28 and the sum of terms except first and last is 14 in a 4 number series . Find first term and sum of all the first 16 terms?

Solution : The series are a,a+d,a+2d,a+3d
The sum of series =4a+6d=28 and sum of series except first and last term is 2a+3d =14   by solving this two we get d=2 and by substituting we get a=4. The sum of the series is Sn =n/2(8+30) =8*38=304

12.The first 5 even terms of a series is 15 the sum of first 3 terms is -3 Find the second term ?

Solution : The terms are a,a+d,a+2d,a+3d,a+4d ….
The first 5 even terms are a+d,a+3d,a+5d,a+7d,a+9d
The sum 5 even terms =5a+25d =15 ie a+5d=3
The sum of first 3 terms =3a+3d =-3 ie a+d =-1 which is the second term.  By solving 4d= 4 ie d=1 and a= -2

13. How many terms of an A.P is equal to 120 if its 3rd term is 9 and the difference between 7th term and 2nd term is 20?

Solution : The 3rd term a+2d =9 As given a+6d-20=a+d ie 5d=20 ie d=4 by substituting in a+2d=9 ie a+8=9 ie a=1

Sn = 120= n/2(2a+(n-1)d) ie 120= n/2 (1+(n-1)4) ie 2n2-n-120=0 by solving n= 8 or 15
See t8= 1+(7)4 =29

14. The sum of terms of a series is 2 n2 +5 for all values of n, then find its 7th term.

Solution: Given, Sn 2n2 5 n
Sn-1 = 2(n-1)2+5(n-1) = 2n2-4n+2+5n-5 ie 2n2+n-3
Sn-Sn-1 = 4n+3 t7 =28+3=31

15. The sum of the 1st term and 5th term is 26 and the product of 2nd and 4th term is 160.Find the sum of first 7 terms?

Solution : Given a+a+4d =26 ie 2a+4d=26 ie a+2d =13 ie 3rd term
Then (a+d)*(a+3d) =160 ie a2+4ad+3d2=160 ie a2+4ad+4d2-d2=160 ie (a+2d)2-d2=160 by substituting 132-d2=160 ie d2=9 ie d=3. then the series is 7,10,13,16 the 7th term is 7+6*3=25

16. The sum of all the terms of the A.P having 10 terms except first is 99 and the except 6th term is 89 . Find the third term?

Solution: Given 9a+45d =99 and 9a+40d=89 by splving this d=2. By substituting d=2 in 9a+45d =99 ie 9a+90=99 a=1. Third term a+2d =5

17. The first term of a series is 107 and the last term and 5th term is 253 Find the sum of the series ?

Solution : Given a=107, a5=253 ie a+4d=253 ie 107+4d=253 ie 4d=146
d=73/2 then Sum = 5/2{214+(5-1)(73/2) = 5/2(360) =900

18. The sum of first 3 terms is 15 and sum of its square is 83 Find the common difference and series ?

Solution : Given a+a+d=a+2d =15 ie 3a+3d=15 ie a+d=5 again a2+(a+d)2+(a+2d)2 =83 ie a2+52+(a+2d)2=83 ie 2a2+4ad+4d2 =58
ie a2+2ad+2d2 =29 = a2+2d(a+d)=29 = a2+2d(5) =a2+10d=29
a2+10(5-a) =29 ie a2-10a+21=0 by solving a =3 or 7 d=2. Therefore series is 3,5,7

19. Find the sum to 200 terms of the series 1,4,6,5,11,6 ?

Solution : There are two A.P one 1,6,11 … and another 4,5,6 ….

In first a=1 , d=5 n=100 Sn=100/2(2+99x5) =50x497 =24850
In 2nd a=4 ,d=1 ,n=100 Sn =100/2(8+99x1)=50x107 =   5350

Total     sum of the series                                         =30200

20. An A.P has 3 as its first term, also the sum of the first eight terms is twice the sum of the first five terms . Find the common difference ?

Solution : Given 3,3+d,3+2d,3+3d,3+4d,3+5d,3+6d,3+7d
The sum of all terms is 24+28d
The sum of first 5 terms is 15a+10d given 2(15+10d)=24+28d
ie 30+20d=24+28d ie 8d=6 ie d=3/4

21. The three digit number in A.P and their sum is 15 the number obtained by reversing the digits is 594 less than the original number. Find the number?

Solution: let the digit in unit place be a-d and digit in ten’s place is a and in 100th place is a+d

Sum of the digits  a-d+a+a+d =15 ie 3a =15 a=5
Original number (a-d)+10a+100(a+d)= 111a+99d=555+99d
Reverse a+d+10a+100a-100d =555-99d
Given 555+99d-(555-99d)=594 ie 198d=594 d=3
Hence the original number is 852

22. Find the 1st negative term in the series 20,19 1/4,18 1/2   etc

Solution: Given a=20 d=- ¾
tn = 20+(n-1)-3/4< 0 ie 80-3n+3 < 0 ie n=83/3 ie 27 2/3 Then the 28th term of the sequence is the first negative term.

23. How many term of A.P is equal to 120 if its 3rd term is 9 and the difference between 7th term and 2nd term is 20 ?

Solution : Series are a,a+d,a+2d …. Given a+2d= 9 and given

a+6d –(a+d) =20 ie 5d=20 ie d=4 by substituting a+2d=9 ie a+8=9 ie a=1

Sn =120= n/2(2+(n-1)4

24. If a+b+c =0 and a/b+c , b/c+a, c/a+b are in A.P then prove -1/b+c,1/c+a,1/a+b are also in A.P ?

Solution : Given b/c+a –a/b+c = c/a+b –b/c+a
= (b/c+a +1) –(a/b+c+1) =(c/a+b+1) –(b/c-a+1)
= a+b+c/c+a –a+b+c/b+a = a+b+c/c+b –a+b+c/c-a
= 1/c+a-1/b+c =1/c+b – 1/c+a ie in A.P

Geometric Progression :

25. Find the 6th term in G.P of 4,8,16… ?

Solution: Given a =4 r=2 then t6 =a(rn-1) =4*25 =4*32=128

26. Find the G.P where 4th term is 8 and 9th term is 256?

Solution : Given ar3 =8  and ar8=256 then ar8/ ar3 =256/8 ier5=32 r=2
By substituting in a.8=8 a=1 then the series is 1,2,4,8,16,32

27.Which term of G.P 5,-10,20,-40 is 320?

Solution :  Given a =5 r=-2 tn =320= arn-1ie 5*(-2)n-1 ie (-2)n-1=64=26
n-1=6 hence n=7

28.A certain ball bounces back to two-thirds of the height it fell from. If this ball is initially dropped from 27 feet, approximate the total distance the ball travels.

Solution: We can calculate the height of each successive bounce:
27*2/3=18
18*2/3=12
12*2/3=8
The series like 27,18,12,8..
The total distance that the ball travels is the sum of the distances the ball is falling and the distances the ball is rising. The distances the ball falls forms a geometric series,where a1 = 27 and = 2/3
Because is a fraction between −1 and 1, this sum can be calculated as
S∞ =a/1-r ie 27/1-2/3 ie 81
The distances the ball rises forms a geometric series,as 18,12,8
S∞ =a/1-r ie 18/1-2/3=54

29.The number of cells in a culture of a certain bacteria doubles every 4 hours. If 200 cells are initially present, write a sequence that shows the population of cells after every nth 4-hour period for one day. Write a formula that gives the number of cells after any 4-hour period.

Solutions : 400 cells; 800 cells; 1,600 cells; 3,200 cells; 6,400 cells; 12,800 cells;then the formula be pn = 400(2)n−1 cells

30.Three numbers are in G.P their sum is 43 and its product is 216. Find the numbers ?

Solution: Given a+ar+ar2 =43 and a3r3 =216 ie ar =6 ie 2nd term is 6 by substituting a+6+36=43 ie a=1 The series 1,6,36

31.Find the sum of G.P 0.6,0.06,0.006,0.0006… to nth term ?

Solution: A=6/10 r=1/10 r< 1 hence Sn = a(1-rn)/1-r ie 6/10(1-(1/10)n)/1-1/10 =6/9(1-1/10n)

32.Find the sum of G.P of 10th term where a=6 and r=2 ?

Solution: Given a=6 r=2 Sn =a(rn-1)/r-1 ie 6*210-1=6138

33. The sum of infinite G.P is 3 and sum of its first two terms is 8/3 Find the first term ?

Solution: Given Sn = 3 Let a be the first term and r is the common ratio a+ar=8/3 or 3a(1=r) =8 but Sn =a/1-r  ie 3.3(i-r)(1+r)=8 ie 1-r2 =8/9 ie r2=1/9 ie r=1/3

34. A G.P series 8,16,32,64 having sum 8184 find the term ?

Solution : Given a=6 r=2 Sn =a(rn-1)/r-1ie 8184 =8*(2n-1)/2-1
ie 8184/8=2n-1 ie 1023=2n-1 ie 1024=2ie 210 hence n=10

35. Find all terms between a1=-5 and a4 =-135 of a G.P ?

Solution : Given a1=-5 and  ar3 =-135 ie r3= -135/-5 =27 then r=3
Then the series is -5,-15,-45,-135

36. Find the series where a2=-2 and a5=2/125 ?

Solution : Given a2 =ar =-2 and ar4 =2/125 By dividing 2/125 by -2 r3=-1/125 ie r=-1/5 then the series 10,-2,2/5

37. Find the sum of 10th terms of the given series 4,-8,16,-32,64 ?

Solution : Given a=4 r=-2 the 10th term =ar9 =4(-2)9 =-2048 the sum is given as Sn =4(1-(-2)n)/1-(-2) ie 4*(1-1024)/3 ie 4*-1023/3= -1364

38.   Suppose you agreed to work for  30 days. You will earn 1 rupee on the first day, 2 rupees the second day, 4 rupees the third day, and so on. How many total rupees will you have earned at the end of the 30 day period?

Solution: Given 1,2,4,8,16,32,64,128,256,512 ….. n30
Here a=1 r=2 the sum is s30 =a(rn-1)/r-1 ie Sn =1*(230-1)2-1 ie 230 -1
ie 1073741824-1 = Rs 1,07,37,41,823

39. The 7th term of G.P is 8 times of 4th term find the 5th term is 48 ?

Solution : Given ar6=8ar3 ie r3 =8 ie r=2 and ar4 =48 ie a.16 =48 ie a=3
The series is 3,6,12,24,48

40. The sum of first and 3rd term is 20 and the sum of first three terms is 26. Find the series?

Solution : Given a+ar2 =20 a+ar+ar2 =26 ie ar =6 ie the 2nd term is 6 ie r=6/a by substituting a+a(6/a)2 =20 ie a+36a/a2 ie 36/a +a =20 ie 36+a2=20a ie a2+30-20a=0  a=18 or 2  r=3 or 1/3 Then series is2,6,18,

41. The sum of first and 3rd term is 40 and 2nd term and 4th term is 80. Find the series ?

Solution : Given a+ar2=40 ie a(1+r2)=40—1 and ar+ar3 =ar(1+r2)=80---2
Divide 2 by 1 then r=2 by substituting 5a=20 a=4 then the required series is 8,16,32,64….

42.Find the common ratio and 10th term of series 2,√2,1 ?

Solution : a=2 , r= √2/2 ie 1/√2 the 10th term ar9 =2.(1/√2)9=1/8√2

43. Find the sum of n terms of the series x+y/x-y,1, x-y/x+y ?

Solution : a =x+y/x-y r=x-y/x+y then Sn = a(rn-1)/r-1
= x+y/x-y[(x-y/x+y)n-1]/x-y/x+y -1

44.Find the sum of nth term of G,P x+y, x2+2xy+y2,x3+3x2y+3xy2+y3 ?

Solution : Given x+y,(x+y)2,(x+y)3 ie a=x+y = x+y the Sn =a(rn-1)/r-1
=x+y[(x+y)n-1]/x+y-1

45. The product of 3 successive term of G.P is 216 and sum of their square is 133, Find the number ?

Solution: Let the number is a/r, a, ar . The product of number a3=216 ie a=6 2ndterm is 6 .The number is 469

46. Find the sum of first 24 terms of G.P t1,t2,t3 if it is know that t1+t5+t10+t15+t20+t24 =225

Solution : (t1+t24)+(t5+t20)+(t10+t15) =225
3(t1+t24) =225 ie t1+t24 =225/3 =75
Sn = 24/2(t1+t24) =7512=900

47. If 1st term of a G.P is a 3rd term is b and (n+1)th term is c then 2n+1 th term of a G.P is prove c2/a ?

Solution: Let common ratio =r  b=ar2 = r =√b/a  then c =arn =rn =c/a
t2n+1 = a(rn)2 =a(c/a)2 =c2/a

48. The 4th term is 10 and 7th term is 80 and which term is 2560.?

Solution : Given ar6/ar3 =80/10 ie r3 =8 r=2 by substituting in ar3
8a =10 ie a= 5/4 tn = 5/4(2n-1)=2560 = 2560*4/5 =2048
ie 2n-1 =211 ie n-1 =11 ie n=10

49. Find the 3 digit number in G.P if 792 is subtract from this number it will be reversed in digits. Now if we inverse the 2nd digit of the required number by 2 the resulting digits will be in A.P ?

Solution: Let the digits are a,ar, ar2
=100a+10ar+ar2-792 = 100ar2+10ar+a ie 99a((1-r2)=792 ie 9a(1-r2)=72
= a(1-r2)=8 and a,ar+2,ar2 in A.P then 2(ar+2) =a+ar2 =a(r2-2r+1) =4
=a(r-1)2 =4= r+1/r-1=-2 ie r=1/3 a=9 then the number is 931

50.The lengths of the radii of circles form an infinite geometric series. The length of the first circle is 10cm. The  length  of the  radius   of each of the circles is_4/5 of the length of the previous circle. Determine the total area of all of the circles formed in this way.?

Solution: Area Circle 1: π x (10)2 = 100 π
Area Circle 2: π x (10 x  4/5 ) 2 = π x (10)2 x ( 4/5 ) 2
Area Circle 3: π x (10 x 4/5 x 4/5 ) 2 = π [10 x( 4/5 ) 2] 2 = π (10)2 x ( 4/5 ) 4
∴ Series: π x (10)2 + π (10)2 x ( 4/5 ) 2 + π (10)2 x ( 4/5 )4 + …
∴ = 100π and = (4/5 ) 2 = 16/25
∴ S∞ = 9/1 − 100 π/1-16/25
100 π/9/25
= 100 π  25/9
2500/9 π cm2

51. A square is drawn by joining the mid-points of the sides of a given square. A third square is drawn inside the second square in the same way and this process continues indefinitely. If a side of the first square is 16 cm, determine the sum of the areas of all the squares.

Solution. Let be the side length of square, then
AB BC CD DA a
EF,Gare the mid-points of ABBC,CD and DA respectively.
EF= FG =GH =HE=a/√2
and IJKare the mid-points of EFFG,GHHE respectively.
IJ =JK= KL =LI=a/2

Similarly, MN= NO= OP= PM =a/2√2
and QR= RS= ST =TQ=a/4
sum of areas
ABCD EFGH IJKL MNOP QRST ....
=a2+ (a/√2)2+(a/2)2+( a/2√2)2+(a/4)2….
=a2(1+1/2+1/4+1/8…) =a2(1/1-1/2)=2a2=2*162=512sqcm

52. 5,x,y is in A.P and x,y,81 is in G.P find the x ,y ?

Solution : Given a=5 a+d =x a+2d=y  ie 5+d=x d=x-5 substituting this in a+2d =y ie a+2(x-5) =y ie 2x-y =5 again
Given x,2x-5,81 ie 2x-5/x=81/2x-5 by solving this (4x-1)(x-25)=0 ie x=25 and y=45

HARMONIC PROGRESSION:

53. If a,b,c are in H.P show that a-b/b-c=a/c ?

Solution : Since a,b,c are in H.P then 1/a,1/b,1/c is in A.P

=1/b-1/a =1/c-1/b = a-b/ab=b-c/bc or a-b/b =a/c

54. Find the 1st term of H.P whose second term is 5/4 and the 3rd term is ½ ?

Solution : Let the series a,5/4,1/2 then a/5/4=5/4/1/2=a/1/2 ie 4a-5/5-2=2a ie a=-5/2

55. The sum of three numbers in H.P is 37 and the sum of their reciprocal is ¼ then find the number ?

Solution : Let the numbers be 1/a-d,1/a,1/a+d  the sum of 1/a-d+1/a+1/a+d =37 and the sum of reciprocal a-d+a+a+d= 1/4  ie 3a=1/4 a=1/12 by substitute 12/1-12d+12+12/1+12d=37 ie 12/1-12d+12/1+12d =25 ie d2=1/25*144=1/60 then the nubers 1/15,1/12,1/10

LINES

56.Find the slope of the line through the points (—2, 5) and (7,1).

Solution : Remember that the slope of the line through two points (x1, y1) and (x2, y2) is given by the equation m= y2-y1/x2-x1 ie 1-5/7+2 =-4/9

57.Write a point-slope equation of the line through the points (1,2) and (1,3).

Solution : The line through (1,2) and (1,3) is vertical and, therefore, does not have a slope. Thus, there is no point-slope equation of the line.

58.Find a point-slope equation of the line going through the point (1,3) with slope 5.

Solution : y -3 = 5(x 1). ie y-3=5x-5 ie y-5x=-2

59. Find the slope of the line having the equation y - 7 = 2(x - 3) and find a point on the line.

Solution : y — 7 = 2(x - 3) is a point-slope equation of the line. Hence, the slope m = 2, and (3, 7) is a point on the line.

60.Find the slope-intercept equation of the line through the points (2,4) and (4,8).

Solution : Remember that the slope-intercept equation of a line is y = mx + b,where is the slope and is the y-intercept (that is, the v-coordinate of the point where the line cuts the y-axis). In this case, the slope m = (8-4)/(4-2) |=2. A point-slope equation of the line is y-8 = 2(x — 4). This is equivalent to y-8 = 2x — 8, or, finally, to y = 2x.

61.Find the slope and y-intercept of the line given by the equation 7x + 4y = 8.

Solution : If we solve the equation 7x + 4y = 8 for y, we obtain the equation y = —7/4 x + 2, which is the slope-intercept equation. Hence, the slope = — 7/4and the y-intercept b = 2.

62.Find an equation of the line L through (-1,4) and parallel to the line M with the equation 3x + 4y = 2.

Solution : Remember that two lines are parallel if and only if their slopes are equal. If we solve 3x + 4y = 2 for y,namely, y = — 3/4x +1/2, we obtain the slope-intercept equation for M. Hence, the slope of M is —3/4 and,therefore, the slope of the parallel line L also is -3/4. So, L has a slope-intercept equation of the form y=-3/4x + b. Since L goes through (-1,4), 4= -3/4(-1) + b, and, therefore,b=4-3/4=13/4. Thus, the equation of L is y = - 3/4x + 13/4

63.Find an equation of the line through (2, 3) and parallel to the line with the equation 4x — 2y = 7.

Solution: The required line must have an equation of the form 4x - 2y = E. Since (2, 3) lies on the line, 4(2) - 2(3) = E. So, E = 8-6 = 2. Hence, the desired equation is 4x - 2y = 2.

64.Find an equation of the line through (2,3) and parallel to the line with the equation y = 5.

Solution: Since y = 5 is the equation of a horizontal line, the required parallel line is horizontal. Since it passes through (2, 3), an equation for it is y = 3.

65.Find an equation of the line through the points (0,2) and (3,0).

Solution : The y-intercept is b = 2 and the x-intercept is a = 3. So, the equation of the line is x/3+y/2 =1

66.If the point (2, k) lies on the line with slope m = 3 passing through the point (1, 6), find k.

Solution: A point-slope equation of the line is y — 6 = 3(x — 1). Since (2, k) lies on the line, k — 6 = 3(2-1).Hence, k = 9.

67.Does the point (-1, -2) lie on the line L through the points (4,7) and (5,9)?

Solution :The slope of L is (9 - 7)/(5 - 4) = 2. Hence, a point-slope equation of L is y-7 = 2(x- 4). If we substitute —1 for and -2 for in this equation, we obtain —2 — 7 = 2(-l — 4), or —9 = -10, which is false. Hence, (—1, —2) does not lie on L.

68.Find the slope-intercept equation of the line through (1,4) that is perpendicular to the line L with equation 2x - 6y = 5.

Solution: Solve 2x - 6y = 5 for y, obtaining = 2/3— 5/6 . So, the slope of L is 2/3 Recall that two lines with slopes m1 and m2 are perpendicular if and only if m1m2 = —1, or, equivalently, m, = —1 /m2. Hence, the slope of is the negative reciprocal of 3, that is, -3. The slope-intercept equation of has the form -3x + b.Since (1,4) is on M, 4 = — 3 - 1 + fe. Hence, b = 7, and the required equation is y = -3x + 1.

69.Determine whether the lines 3x + 6y = 7 and 2x + 4y = 5 are parallel.

Solution: The coefficients of and are proportional: 2/3 = 4/6. Hence, the lines are parallel.

70.Use slopes to determine whether the points A(4,1), 5(7, 3), and C(3,9) are the vertices of a right triangle.

Solution: The slope m1 of line AB is (3 - l)/(7-4) = 2/3 . The slope m2 of line BC is (9 -3)/(3 -7) = -3/2 .Since m2 is the negative reciprocal of m1, the lines ABand BC are perpendicular. Hence, ABC has a right angle at B.

71.Determine so that the points A(7,5), B(-l, 2), and C(k, 0) are the vertices of a right triangle with right angle at B.

Solution: The slope of line AB is (5-2)/[7-(-1) = 3/8 The slope of line BC is (2-0)/(-l - k) = -2/(l +k).The condition for A ABC to have a right angle at is that lines AB and BC are perpendicular, which holds when and only when the product of their slopes is —1, that is (3/8)[—2/(l + k) = —1. This is equivalent to 6 = 8(1 + k), or 8k = -2, or k=-1/4.

72.Use slopes to show that the points A(5, 4), B(-4, 2), C(-3, -3), andD(6, -1) are vertices of a parallelogram.

Solution: The slope of AB is (4-2)/[5 - (-4)] =2/9 and the slope of CD is [-3 - (-l)/(-3 -6) = 2/9; hence,AB and CD are parallel. The slope of BC is (-3 - 2)/[-3 - (-4)] = -5 and the slope of AD is (-1 - 4)/(6 — 5) = -5, and, therefore, BC and ADare parallel. Thus, ABCD is a parallelogram.

73.For what value of will the line kx + 5y - 2k have slope 3?

Solution : y=kx/5-2k/5  This is the slope-intercept equation. Hence, the slope m = —k/5 = 3. So,k=-15

74.Find the midpoint of the line segment between (2, 5) and (—1, 3).

Solution; By the midpoint formula, the coordinates of the midpoint are the averages of the coordinates of the endpoints.
In this case, the midpoint (x, y) is given by ([2 + (-l)]/2, (5 + 3)/2) = (1, 4).

75.Find the intersection of the line through (1, 2) and (5, 8) with the line through (2,2) and (4, 0).

Solution: The slope of L is (8 — 2)/(5 — 1) =3/2. Its slope-intercept equation has the form y =3/2x + b. Since it passes through (1,2), 2=3/2(1) + b, and, therefore, b=1/2. So, L has equation y=3/2x+1/2.  Similarly,we find that the equation of M is y = -x + 4. So, we must solve the equations y = -x + 4 and y=3/2x +1/2simultaneously. Then, -x + 4=3/2x+1/2, -2x + 8 = 3x + 1, 7 = 5x,x=7/5 When x=7/5, y = -x +4 = - 7/5 + 4 = 13/5. Hence, the point of intersection is (7/5,13/5)•

76.Find the distance from the point (1, 2) to the line 3x - 4y = 10.

Solution: Remember that the distance from a point (x1, y1) to a line Ax + By+C = 0 is \Ax1 + By1 + C1l√A2 + B2. In our case, A = 3, B = -4, C=10, and √A2 + B2 =25 = 5. So, the distance is |3(1)-4(2)-10|/5=3.

77.Find equations of the lines of slope —3/4 that form with the coordinate axes a triangle of area 24 square units.

Solution: The slope-intercept equations have the form y = - 3/4x + b. When y =0, x = 4/3b. So, the x-intercept a is 4/3. Hence, the area of the triangle is 1/2ab = ½*4/3b*b= 2/3b2 = 24. So, b2 = 36, b = ±6, and the desired
equations are y=-3/4x± 6; that is, 3x + 4y = 24 and 3x + 4y = -24.

78.Find the intersection of the lines x - 2y and 3* + 4y = 6.

Solution: We must solve x - 2y = 2 and 3x + 4y = 6 simultaneously. Multiply the first equation by 2, obtaining 2x - 4y = 4, and add this to the second equation. The result is 5x = 10, .v = 2. When x = 2, substitution
in either equation yields y = 0. Hence, the intersection is the point (2, 0).

79. Find the intersection of the line y = 8x - 6 and the parabola 2x2.

Solution : Solve y = 8x-6 and y = 2x2 simultaneously. 2x2 = 8x - 6, x2 = 4x--3, x2-4x + 3 = 0,(x -3)(x- 1) = 0, A-= 3 or x =1l. When x = 3, y = 18, and whenx=1. y = 2. Thus, the intersection consists of (3, 18) and (1,2).

80. Find the intersection of the line y = x - 3 and the hyperbola xy = 4.

Solution: We must solve y = x - 3 and xy = 4 simultaneously. Then x(x - 3) = 4,x2 - 3x - 4 = 0,(x-4)(x + 1 ) = 0, x = 4 or x = -1. When x = 4, y = 1, and when x=-l, = -4. Hence, the intersection consists of the points (4,1) and (-1, -4).

CIRCLE

81. Write the standard equation for the circle with center (3,5) and radius 4.

Solution: (x-3)2+(y-5)2 =16

82. Find the standard equation of the circle with center at (1, -2) and passing through the point (7, 4).

Solution: The radius of the circle is the distance between (1, -2) and (7, 4):
√(7-1)2+(4+2)2=√72. Thus, the standard equation is: (x - I)2 + ( + 2)2 = 72.

83. Identify the graph of the equation x2 + y2 - I2x + 20y + 15 = 0.

Solution: Complete the square in and in y: (x - 6)2 + (y + 10)2 + 15 = 36 + 100.Simplifying, we obtain (x - 6)2 + (y + 10)2 = 121, the standard equation of a circle with center at (6, -10) and radius 11.

84.Find the center and radius of the circle passing through the points (3, 8), (9,6), and (13, -2).

Solution: The circle has an equation x2 + y2 + Dx + Ey + F = 0. Substituting the values (x, y) at each of the three given points, we obtain three equations: 3D +8E + F= -73, 9D + 6E + F= -117,13D - 2E + F = —173. By factoring we get D=-6,E=4 and F =-87. Hence, we obtain x2 + y2 -6x + 4y - 87 = 0. Complete the square: (x - 3)2 +(y +2)2-87 = 9 + 4, or, (x - 3)2 + (y + 2)2 = 100. Hence, the center is (3,-2) and the radius is 10

85.For what value(s) of does the circle (x - k)2 + (y - 2k)2 = 10 pass through the point (1,1)?

Solution : (1 - k)2 + (1 -2k)2 = 10. Squaring out and simplifying, we obtain 5k2 – 6k - 8 = 0. The left side factors into (5k+ 4)(k-2). Hence, the solutions are k = -4/5 and k = 2.

86.Determine the value of so that x2 + y2 — 8x + 10+ k = 0 is the equation of a circle of radius 7.

Solution : Complete the square: (x - 4)2 + (y + 5)2 + = 16 + 25. Thus, (x -4)2 + (y + 5)2 = 41 - k. So, 41-k=49 ie k =8

87.Find the standard equation of the circle which has as a diameter the segment joining the points (5, -1) and (-3,7)

Solution : The center is the midpoint (1, 3) of the given segment. The radius is the distance between (1, 3) and (5,-1): √(5-1)2+(-1-3)2=32 ie (x-1)2+(y-3)2=32.

88. Find the standard equation of a circle with radius 13 that passes through the origin, and whose center has abscissa -12.

Solution : Let the center be (-12, b). The distance formula yields 132=√144+b2 ie 144+b2 = 169, b2 = 25, and = ±5. Hence, there are two circles, with equations (x + 12)2 + (y - 5)2 =169 and (x + 12)2 + (y + 5)2 = 169.

89.Find the standard equation of the circle with center at (1, 3) and tangent to the line 5x - I2y -8 = 0.

Solution : The radius is the perpendicular distance from the center (1,3) to the line:5(1)-12(3)-8/13=3 standard equation is (x - I)2 + ( y - 3)2 = 9.

90.Find the center and radius of the circle passing through (2,4) and (-1,2) and having its center on the line x-3y=8

Solution : Let (a, b) be the center. Then the distances from (a, b) to the given points must be equal, and, if we square those distances, we get (a - 2)2 + (b -4)2 = (a + I)2+ (b - 2)2, -4a + 4 - 8b + 16 = 2a + 1 - 4b + 4, 15 =6a + 4b.Since (a, b) also is on the line x - 3y = 8, we have a - 3b = 8. If we multiply this equation by —6 and add the result to 6a + 4b = 15, we obtain 22b = -33, b = —3/2. Then a =7/2, and the center is (7/2 , -3/2). The radius is the distance between the center and (-1, 2) is √130/2

91.Find the points of intersection (if any) of the circles x2 + (y -4)2 = 10 and (x — 8)2 + y2 = 50.

Solution : The circles are x2 + y2 - 8y + 6 = 0 and x2 – 16y + y2 + 14 = 0. Subtract the second equation from the first: 16x – 8y - 8 = 0, 2x-y -1 = 0, y = 2x -1. Substitute this equation for in the second equation:
(x - 8)2 + (2x - I)2 = 50, 5x2 – 20x + 15 = 0, x2-4x + 3 = 0, (x - 3)(x - 1) = 0,x = 3 or x = 1.Hence, the points of intersection are (3,5) and (1,1).

92.Find an equation of the circle that contains the point (3,1) and passes through the points of intersection of the two circles x2 + y2-x-y-2 = Qand x2 + y2 + 4x - 4y - 8 = 0.

Solution:  substitute (3,1) for (x, y) in the equation (x2 + y2 - x - y - 2) + k(x2 + y2 + 4x -4y — 8) = 0. Then 4 + 10= 0, k = —4/10. So, the desired equation can be written as 5(x2+y2-x-y-2)-2(x2+y2+4x-4y-8)=0 or 3x2+3y2-13x+3y+6=0

93.Determine the locus of a point that moves so that the sum of the squares of its distances from the lines 5x + 12y -4 = 0 and 12x - 5y + 10 = 0 is 5.

Solution: [Note that the lines are perpendicular.]
Let (x, y) be the point. The distances from the two lines are |5x+12y-4|/13 and |12x-5y+10|/13.Hence,(5x+12y-4/13)2 +(12x-5y+10)2=5 by simplifying
169x2+169y2+200x-196y-729 = 0, the equation of a circle.

94.Find the locus of a point the sum of the squares of whose distances from (2, 3) and (-1, -2) is 34.

Let (x, y) be the point. Then (x -2)2 + (y -3)2 + (x + I)2 + (y + 2)2 = 34. Simplify: x2 + y2 - x -- 8 = 0, the equation of a circle.

95. Find the locus of a point (x, y) the square of whose distance from (-5,2) is equal to its distance from the line 5x + 12y - 26 = 0.

Solution: (x+5)2+(y-2)2 =|5x+12y-26|/13

Simplifying, we obtain two equations 13x2 + 13y2 + 125x - 64y + 403 = 0 and 13x2 + I3y2 + 135x - 40y +351=0

ELLIPSE

96. Find the standard form of the equation of the ellipse having foci at  (0.1)and (4,1) and a major axis of length 6.

Solution : The centre of foci is 0+4/2,1+1/2 =(2,1) and the distance from foci from the centre is c=2 . Given 2a=6 ie a=3. b=√a2-c2 =√5. The standard equation is (x-2)2/32+(y-1)2/√5 ie(x-2)2/9+(y-1)2/5=1

97. The given equation x2+4y2+6x-8y+9=0 find the foci and centre ?

Solution : (x2+6x+  )+4(y2-2y+  ) = -9 ie (x2+6x+9)+4(y2-2y+1)=-9+9+4 ie (x+3)2+4(y-1)2=4 ie (x+3)2/4+(y-1)2/1 =1 in the standard form (x+3)2/22+(y-1)2/12=1. Hence centre is (-3,1) and a=2 and b=1 then c2=√a2-b2=√3 foci is (-3-√3,1) and (-3+√3,1)

98. Find the center, vertices, and foci of the ellipse 4x2+y2-8x+4y-8=0

Solution : The given equation 4x2+y2-8x+4y-8=0
= 4(x2-2x+   )+ (y2+4y+   ) =8 ie 4(x2-2x+1)+(y2+4y+4)=8+8
=4(x-1)2+(y+2)2 =16 ie (x-1)2/4+(y+2)/16 =1 ie (x-1)2/22+(y+2)2/42=1 this is a standard equation of an eclipse with centre(1,-2) and a=2 b=4 and c=√16-4 =2√3.Hence centre (1,-2) vertex(1,2) and (1,-6) foci(1,-2-2√3) and (1,-2+2√3)

99.The moon travels about Earth in an elliptical orbit with Earth at one focus, as shown in Figure 10.26. The major and minor axes of the orbit have lengths of 768,800 kilometers and 767,640 kilometers, respectively. Find the greatest and smallest distances (the apogee and perigee), respectively from Earth’s center to the moon’s center.

Solution: Because 2a =768800 and 2b =767640 then a=384400 and b=383820 which implies that c2=√a2-b2 ie √3844002-3838202 =21108. So the greatest distance between earth and moon is a+c =405508KM and the least distance between earth and moon is a-c =363292

100. Find the centre, foci,vertex for the following equation x2/4+y2/9 =1

Solution: The equation is x2/22+y2/32 =1 ie a=2,b=3 and centre (0,0) then vertices (2,0) and (-2,0) and foci(1,0),(-1,0)

101. Find the centre, foci,vertex for the following equation (x+3)2/16+(y-5)2/25 =1

Solution : From the above equation centre is (-3,5) and a2 =16 ie a=4 and b2=25 ie b=5 and c= √25-16 =√9 =3. Foci =(0,5) and (-6,5) and vertices is (-3+5,5) ie (2,5) and (-8,5)

102.Find the equation of the ellipse with vertices(5,0) and (-5,0) and the eccentricity e =3/5 ?

Solution: =e = c/a =3/5 where c=3 and a=5 b= √25-9=4 vertices (5,0)and (-5,0) means centre (0,0) foci = (3,0) and (-3,0) and the equation of the eclipse is x2/25+y2/16 =1

103. Find the equation of the ellipse with vertices(-7,-3) and (13-3) and foci (-5,-3),(11,-3) ?

Solution : Centre is the mid-point of vertices ie (-7+13/2,-3+(-3)/2) =(3,-3) ie (h,k) is (3,-3) and 2a=20 ie a=10 and 2c=16 ie c=8 and b=√36 =6. Hence the equation is (x-3)2/100+(y+3)2/36=1

104.Find the equation of an ellipse having foci(-10,8) and (14,8) and length of major axis is 30 ?

Solution : The centre is the mid-point the foci =(-10+14/2,8+8/2) ie (2,8) and the length of the major axis =30 ie 2a =30 ie a=15 and the distance between foci is 2c=24 ie c=12 then b= 9 then the required equation is
(x-2)2/225+(y-8)2/81 =1

105. The eccentricity 0.60 and the length is 36 find the equation of ellipse ?

Solution :  Given e=c/a =0.60 and 2a =36 a=18 ie c=a*e ie 18*0.60=10.8
Then b = 14.4 then the required equation is x2/324+y2/207=1

106. Determine the standard form for the equation of an ellipse given the following information with  Center (3, 4) with = 5and = 2.

Solution : The standard formula for ellipse is (x-3)2/25+(y-4)2/4=1

107. Find the x- and y-intercepts (x−3)2 /4 + (y−22/9 = 1

Solution : To find x-intercept y=0 (x-3)2/4 +4/9=1 ie (x-3)2/4=1-4/9 =5/9
(x-3)/2=√5/3 ie x-3=2(√5/3) ie x= 2√5/3-3 and y-intercept ia 0

108.Ellipse whose major axis has vertices (2, 9) and (2, −1) and minor axis has vertices (−2, 4) and (6, 4.

Solution: The mid-point of vertices =(2,4) ie the centre is (2,4)The distance between vertices in major axis 2a= 10 and 2b in minor axix =8 hence the equation is (x-2)2/25+(y-4)2/16=1

109. Ellipse whose major axis has vertices (−8, −2) and (0, −2) and minor axis has a length of 4 units.

Solution: The mid-point is (-4,-2) and 2b=4 given. 2a=8 a=4 then the equation is (x+4)2/16+(y+2)2/4=1

Parabolas
110. Determine the vertex and axis of symmetry in the following equation given y=2(x-4)2-3

Solution: The equation is in the form y=a(x-h)2+k ie h=4 and k =-3 and a=2 in this example a > 0 concave upward . The vertex (4,-3) the axis of symmetry of x=4

111. Determine the vertex and axis of symmetry in the following equation given x=(y-3)2+5

Solution: The equation is in the form x=a(y-k)2+h therefore a=1,k=3 and h=5. The vertex (5,-3) the axis of symmetry of y=3. it is a horizontal parabola.

112. Write an equation of a vertical parabola with a vertex of (2,6) and passing through the point (-1,4)

Solution: Since the vertex is (2,6) it follows that h=2 and k=6. Also, we know that the point (-1,4) is the solution of the equation. By substituting the value in the equation y=a(x-h)2+k the equation is 4=a(-1-2)2+6 ie 4=a(-3)2+6 ie 4=9a+6 then a= -2/9 now the equation is y=-2/9(x-2)2+6

113. A concrete bridge is designed with an arch in the shape of a parabola. The road over the bridge is 120 feet long and the maximum height of the arch is 50 feet. Write an equation for the parabolic arch.

Solution : The vertex of the parabola is located at its maximum height, or(0,50) The x-intercepts are (0,60),(0,-60) and We need only one x-intercept to find an equation, so we will use Substitute 0 for h, 50 fork, 60 for x, and 0 for in the equation for a vertical parabola, and solve for a.

y=a(x-h)2+k ie 0=a(60-0)2+50 ie 0=3600a+50 then a=-1/72 the equation of parabola is y=-1/72(x-o)2+50 ie y=-1/72x2+50

114. A satellite dish receiver is in the shape of a parabola. A cross section of the dish shows a diameter of 13 feet at a distance of 2.5 feet from the vertex of the parabola. Write an equation for the parabola.

Solution: The horizontal parabola has a vertex at the origin. If the parabola is 13 feet
in diameter at a distance along the axis of 2.5 feet from the vertex, the radius is 6.5 feet. We can thuslabel two points on the parabola as (2.5,6.5),(2.5,-6,5) .Substitute 0 for h, 0 for k, 2.5 for x, and 6.5 for y
in the equation for a horizontal parabola, and solve for a.

x= a(y-k)2+h ie 2.5=a(6.5-0)2+0 ie 2.5=42.25a ie a=10/169 then the equation is x=10/169y2.

115. Find the equation of vertical parabola having vertex (3,-2) and passes through (5,2)

Solution: Given vertex(3,-2) ie h=3 and k=-2 it passes through (5,2)let the equation be y=a(x-h)2+k by substituting (5,2) ie 2=a(5-3)2-2 ie 4a=4 a=1
Hence the equation is y=(x-3)2-2

116. Find the equation of horizontal parabola having vertex (4,3) and passes through (6,2)

Solution: Given vertex(4,3) ie h=4 and k=-3 it passes through (6,2)let the equation be x=a(y-k)2+h by substituting (6,2) ie 6=a(2-3)2-4 ie a=2
Hence the equation is x=(y-3)2-4

HYPERBOLAS

117.Equation of a hyperbola having centre(0.0) and foci(c,0),(-c,0) and constant difference 2a ?

Solution: The equation is x2/a2-y2/b2=1

118.Find the equation of the asymptotes for x2/36-y2/9=1  ?

Solution: The x-intercept are (6,0) and (-6,0) and y intercept (0,3) and (0,-3) and the asymptotes is (1/2,-1/2) ie equation y=1/2x or y=-1/2x

119. Find the centre ,vertex,foci and equation of the asymptotes for the following equation y2/9-x2/16=1

Solution : Centre is (0,0) and a=4 and b=3 vertices (0,3) and (0,-3) and foci is c=√25 =5 ie (0,5) and (0,-5) asymptotes +or -b/a ie 3/4x

120. Find the centre ,vertex,foci and equation of the asymptotes for the following equation (x-2)2/4-(y+3)2/9=1

Solution: Centre is (2,-3) and a=2 b=3 vetex (0,-3) and (4,-3) and foci is (2+√13,-3) and (2-√13,-3) asymptotes is y=3/2(x-2)-3

121. Find the centre ,vertex,foci and equation of the asymptotes for the following equation (x)2/3-(y-5)2/12=1

Solution: Centre is (0,5) and a=√3 b=√12 vetex (√3,5) and (-√3,5) and foci is (√15,5) and (-√15,5) asymptotes is y=2x+5

122. Find the stndrd equation of the equation 4x2-5y2-40x-20y+160=0

Solution : 4(x2-10x+ 25  )-5(y2+4y+ 4 ) +160=0 +100-20

= 4(x-5)2-5(y+2)2 =-80 ie (x-5)2/20-(y+2)2/16=-1 ie (y+2)2/16-(x-5)2/20=1 Here centre (5,-2) a=√20 b=4

LINEAR SYSTEM WITH TWO VARIABLES and three variable

123.The sum of 4 times a larger integer and 5 times a smaller integer is 7. When twice the smaller integer is subtracted from 3 times larger , the result is 11. Find the integers ?

Solution: Let x is the larger integer and y is the smaller integer
=4x+5y =7  and 3x-2y =11 By solving this equation by elimination method the eqn1x3= 12x+15y =21 and 2x4 =12x-8y =44 by subtracting 23y=-23 ie y=-1 and by substituting the value of y in eqn1 x=3

124.An integer is 1 less  than twice the other. If their sum is 20 Find the integers?

Solution: 2x-1=y and x+y =20 eqn1 2x-y=1 by using elimination method 3x=21 x=7 by substituting in eqn2 7+y=20 then y =13.

125.An executive traveled a total of 4 hours and 875 miles by car and by plane.Driving to the airport by car, she averaged 50 miles per hour. In the air, the plane averaged 320 miles per hour. How long did it take her to drive to the airport?

Solution : Travel by car   50    x     50x ( x time taken )
Travel by air   320   y    320y (y time taken)
Total distance traveled 50x+320y=875
Total time taken x+y =4 by solving this equation by elimination method 50x+320y =875 and 50x+50y =200 by subtracting 270y = 675 ie y=5/2 ie 2 1/2 then x= 4-21/2 =1 1/2hrs.

126.The sum of two integers is 45. The larger integer is 3 less than twice the smaller. Find the two integers.

Solution : The large integer is x and smaller is y the given  x+y=45 and 2y-3=x ie x-2y=-3 then 3y =48 ie y =16 then x=45-16=29

127.The difference of a smaller integer and twice a larger is 0. When 3 times the larger integer is subtracted from 2 times the smaller, the result is −5. Find the integers.

Solution : The large integer is x and smaller is y given 2x-y=0 and 3x-2y=-5 then by solving these 4x-2y=0 ie x=5 by substituting in 10-y=0 ie y=10

128.The length of a rectangle is 5 more than twice its width. If the perimeter measures 46 meters, then find the dimensions of the rectangle.

Solution : Given l-5=2w ie l-2w=5 and 2l+2w =46 ie l+w =23  by solving -3w =18 ie w=6 by substituting in l+6=23 ie l=17

129. A Rs5,200 principal is invested in two accounts, one earning 3% interest and another earning 6% interest. If the total interest for the year is Rs210, then how much is invested in each account?

Solution :Given the amount invested in 3% interest is x and y in 6% interest
Then x+y =5200 and 3x/100+6x/100=210 ie 3x+6y =21000 by multiplying x+y =5200 with 3 ie 3x+3y =15600 by solving 3y =5400 then y=1800 and x=3400

130. A small business invested Rs12000 in two accounts. The account earning 4% annual interest yielded twice as much interest as the account earning 3% annual interest. How much was invested in each account?

Solution : Given x+y =12000 and 2(3y/100)=4x/100 ie 6y/50=4x/100 ie 12y=4x ie 4x-12y=0 eqn1 by 4 4x+4y=48000 by solving 16y=48000 then y =3000 hence x= 9000

131. A community theater sold 140 tickets to the evening musical for a total of Rs1,540. Each adult ticket was sold for Rs12 and each child ticket was sold for Rs8.How many adult tickets were sold?

Solution : x+y =140 and 12x+8y =1540 multiply by 12 then 12x+12y=1680 by solving both -4y=-140 ie y=35 and x=105

132. The sum of three integers is 38. Two less than 4 times the smaller integer is equal to the sum of the others. The sum of the smaller and larger integer is equal to 2 more than twice that of the other. Find the integers.

Solution : let x the larger integer and z the smaller integer x+y+z =38 and 4z=x+y-2 ie x+y-4z=-2 and z+x-2=2y ie x-2y+z=2 by solving 1 and 2 5z= 40 ie z=8 by substituting z value in 2 and 3 x+y-32+2=0 ie x+y =30 and x-2y =2-8 =-6 by solving 3y=36 ie y=12 then x=18

133. Let the angles AB, and of a triangle . The larger angle is equal to twice the sum of the other two. Four times the smallest angle is equal to the difference of angle and B. Find the angles.

Solution : A+B+C =180 then given C =2(A+B) ie 2A+2B-C =0 and again 4A=C-B ie 4A-C+B=0 egn1 by 2 2A+2B+2C=360 by solving -3C=-360 ie C =120 then A+B =60  4A+B =120 by solving 3A=60 ie  A=20then B=40

Cramer’s Rule :

134.  Solve by using Cramers rule 3x-y=-2 and 6x+4y=2

Solution : |3 -1|             |-2  -1|           |3   -2|
|6  4|   = D    |  2   4 | =Dx   |6    2|  = Dy

=  D = 18   Dx =- 6    Dy = 18    then x = Dx/D =-6/18=-1/3 Dy/D =18/18=1

135. Solve by using Cramers rule 3x-5y=8 and 2x-7y=9

Solution: |3 -5|              |8- 5|            |3    8|
|2-7|   =D      |9  -7|   =Dx   |2    9|  =Dy

=D= -11   Dx = -11      Dy  = 11 then x= 11/11 =1 y=11/-11=-1

136. Solve by using Cramers rule x-y+2z=-3 and 3x+2y-z=13 and
-4x-3y+z=-18

Solution : |1 -1 2 |          |-3 -1  2 |         |1 -3  2|          |1 -1 -3|
|3  2 -1| = D   | 13 2  -1| =Dx  |3 13-1| =Dy  |3  2  13|= Dz
|-4-3  1|          | -18 -3 1|        |-4 -18 1|       |-4 -3 -18|

=D = 1(2-3)+1(3-4)+2(-9+8) =-1-1-2=-4
=Dx = -3(2-3)+1(13-18)+2(-39+36) = 3-5-6 =-8  x=2
=Dy = 1(13-18)+3(3-4)+2(-54+52)=-5-3-4 = -12 y=3
=Dz =1(-36+39)+1(-54+52)-3(-9+8)=3-2+3=-4 z=-1

137. Find (f+g) and (f-g) in (x) = 2 − 3+ 2(x) = 2 + 4− 7

Solution : (f+g) = x2-3x+2+x2+4x-7 = 2x2+x-5
(f-g) = x2-3x+2-(x2+4x-7) = -7x+9

138.  Given (x) = 3 + 22 − 8 and (x) = 22 − 3+ 5 , evaluate  (fg(−2)

Solution : (f+g)= x3+2x2-8 +2x2-3x+5 =x3+4x2-3x-3
Then (f+g)(-2)  = (-2)^3+4(-2)^2-3(-2)-3 = -8+16+6-3 =11

139. Find f*g of  (x) = 2(x) = 2 − 7+ 5

Solution : f*g = (2x2)*( x2-7x+5) = 2x4 -14x3+10x2

140. Find f/g of (x) = 3 + 42 + 3− 2(x) = + 2

Solution : f/g = x3+4x2+3x-2/x+2 = (x+2)(x2+2x-1)/x+2=x2+2x-1

Factoring polynomial

141. Factorize 12x4 − 16x3 + 4x2 by GCF method ?

Solution: GCF : 4x2(3x2-4x+1) again (3x2-3x-x+1) =3x(x-1)-1(x-1) then
=4x2(3x-1)(x-1)

142. Factorize by grouping 63 − 15x2 − 2+ 5

Solution : 6x3-15x2-2x+5 here 6x3-15x2 one group and -2x+5 another
= 3x2(2x-5)-1(2x-5) ie (3x2-1)(2x-1)

143. (2+ 5)2 − (− 3)2 by factor ?

Solution : (2x+5) =a (x-3) =b ie a2-b2 =(a+b)(a-b) then by factor
= (2x+5+x-3)(2x+5-x+3) =(3x+2)(x+8)

Factoring trinomials

144. Factor: y2 − 7xy + 12.

Solution: Factor x2y2 =(xy)2 ie (xy      )(xy       )
Factor 12  = (-3)(-4) shows -7 by adding  and +12  on multiply
= (xy-3)(xy-4)

145.  Factor: 6 − 3 − 42.

Solution : Factor x6 =(x3)2 ie  (x3     )(x3     )
Factor 42 =(-7)(+6) by add -1 by multiply -42
= (x3-7)(x3+6)

146. Factorize -18x6-69x4+12x2 ?

Solution : GCF is 3x2 then -3x2(6x4+23x2-4) by factorize 6x4+23x2-4
= first factor 3x2 and 2x2 24 = +24 and -1
= 6x4+24x2-x2-4 = 6x2(x2+4) -1(x2+4) ie -3x2(6x2-1)(x2+4)

147. Factorize 54x4-36x3-24x2+16x ?

Solution : GCF 2x then 2x(27x3-18x2-12x+8) by grouping 27x3-18x2 and -12x+8 ie 9x2(3x-2)-4(3x-2) ie (3x-2)(9x2-4) again difference of square 9x2-4 is factorized (3x+2)(3x-2) hence 2x(3x-2)(3x+2)(3x-2)

148. Factorize x6+6x3-16 ?

Solution: Factor x3.x3 then (x3+8 )(x3 -2 ) 16= +8 -2 again sum of cube x3+8 = (x+2)(x2-2x+4) hence  (x3-2)(x+2)(x2-2x+4)

149. Solve: 43 − 2 − 100+ 25 = 0.

Solution: x2(4x-1)-25(4x-1) =0 by grouping  ie (4x-1)(x2-25)
ie (4x-1)(x+5)(x-5) by difference in square ie 4x-1=0 x=1/4 and x+5 =0 ie x=-5 or 5 hence x=+5 or -5 or ¼

150. The profit in rupees generated by producing and selling bicycles per week is given by the formula (n) = −5n2 + 400− 6000. How many bicycles must be produced and sold to break even?

Solution : By solving 5n2-400n+6000 ie 5n2+300n+100n-6000 ie 5n(n-60)+100(n-60) ie (5n+100)(n-60) then n=60

151. A uniform border is to be placed around an 8 × 10 inch picture. If the total area including the border must be 168 square inches, then how wide should the border be?

Solution : The area is given as 168 sq inches
= (10+2x)(8+2x)=168 ie 80+20x+16x+4x2=168 ie 4x2+36x-88 =0
= x2+9x-22 =0 ie x2+11x-2x-22 =0 ie (x+11)(x-2)=0 ie x=2

152. Simplify 4-12/x  +9/x2  by 2-5/x +3/x2

Solution : LCD for both Dr and Nr is x2 in the fractions 1,x,x2
= Multiply by x2 ie 4x2-12x+9 diveded by 2x2-5x+3 by factorize Dr and Nr
= 2x(2x-3)-3(2x-3) ie (2x-3)(2x-3) then x(2x-3)-1(2x-3) ie (2x-3)(x-1)
=(2x-3)(2x-3) /(2x-3)(x-1) = 2x-3/x-1

153. Solve 1/x+2/x2 =x+9/2x2

Solution : Find LCD as 2x2 Multiply both side by LCD ie 2x+4 =x+9
= 2x-x=9-4 ie x=5

154. Solve 3(x+2)/x-4 - x+4/x-2 = x-2/x-4

Solution : LCD (x-4)(x-2) Multiply by LCD on bothside
=3(x+2)(x-2)-(x+4)(x-4) =(x-2)(x-2) ie  3(x2-4)-(x2-16)=x2-4x+4
= 3x2-12-x2+16=x2-4x+4 ie x2+4x =0 x(x+4) =0 x=-4

155. Solve 6+x -1 =x -2

Solution : 6+1/x=1/x2 Here LCD is x2 multiply by x2 on both side
= 6x2+x=1 ie 6x2+3x-2x-1=0 ie 3x(2x+1)-1(2x+1) ie (2x+1)(3x-1)=0
= 2x=-1 or x=-1/2 or 3x=1 or x=1/3

156. When 2 is added to 5 times the reciprocal of a number, the result is 12. Find the number.?

Solution : let be x then given  5/x+2 =12  ie 5+2x=12x = 10x =5 ie x=1/2

157. When 1 is subtracted from 4 times the reciprocal of a number, the result is 11Find the number ?

Solution : let be x then given 4/x-1=11 ie 4-x=11x ie -12x=-4 x=1/3

158. The sum of the reciprocals of two consecutive odd integers is 12/35 Find the integers.

Solution : Let be x then given 1/x+ 1/x+2 =12/35 ie 35(2x+2)=12(x2+2x)
= 70x+70 =12x2+24x ie 12x2-46x-70=0 by factorizing 12x(x-5)-14(x-5)
=x-5=0 ie x=5 the integer are 5,7

159. If 3 times the reciprocal of the smaller of two consecutive integers is subtracted from 7 times the reciprocal of the larger, then the result is12. Find the two integers.

Solution : Let be x and x+1 then given 3(1/x=1)-2(1/x)=1/2  ie 3/x-1-2/x= ½ ie 2(3x-2x+2)=x2+2x ie 2x+4 =x2+2x ie x2 =4 x=2 the answer 2,3

160. Solve 5x2-27x+10 by Bluma’s method?

Solution : It is in the form of ax2+bx+c multiply ac as x2-27x+50=0
Then factorize x2-25x-2x+50 =0 ie (x-25)(x-2) =o x=25/5 or 2/5

161. Find two numbers whose difference multiplied by the difference of their square is 160 annd where sum multiplied by the sum of their square gives 580. ?

Solution : Let x+y and x-y may be the numbers then given difference 2y and difference of square x2+2xy+y2-(x2-2xy+y2)=4xy ie 2y(4xy)=160 or xy2=20 –eqn1 then given 2x{2(x2+y2)}=580 ie x(x2+y2) =145 eqn2 by substituting xy2 =20 in eqn2 x3+20=145 ie x3=125 x=5 then 2y2(20)=160 ie y2 =4 y=2 the numbers are 7,3

162. What two numbers are those whose sum multiplied by the greater is 204 and whose difference multiplied by lesser is 35 ?

Solution: Let the greater is x and lesser is y . Given (x+y)x=204 eqn1 then given (x-y)y =35 eqn2 . By adding 1 & 2 x2+xy+xy-y2=204+35=239  by subtracting x2+xy-xy+y2 =204-35 =169 ie x2+y2=169 ie x2-y2=239-2xy  again

163. Sankar traveled 15 KM on the bus and then another 72 KM on a train. The train was 18 KM per hour faster than the bus, and the total trip took 2 hours.what was the average speed of the train?

Solution : Let x be the speed of the bus . The train speed is x+18 then given  15/x+72/x+18=2 .Multiply by both side by LCD x(x+18) as shown below
15(x+18)+72x =2x(x+18) ie 87x+270=2x2+36x ie 2x2-51x-270=0 by factorizing 2x2-60x+9x-270=0 ie 2x(x-30)+9(x-30) =0 hence x=30KM The speed of the train is 30+18=48KM

164. Gopal can paint a typical room in 2 hours less time than Mani. If gopal and Mani can paint 5 rooms working together in a 12 hour shift, how long does it take each to paint a single room?

Solution : Let x be the time taken by Mani to paint a single room.Then Gopal can do the same in x-2 hrs then 1/x+1/x-2=5/12 then 12/x+12/x-2 =5 Multiply by LCD x(x-2) by both side we get 12(x-2)+12(x)=5(x)(x-2) ie 12x-24+12x=5x2-10x ie 5x2-34x+24=0 by factorizing 5x2-30x-4x+24=0 ie 5x(x-6)-4(x-6) =0 ie x=6 then answer is Mani can in 5hrs and gopal in 3 hrs

LOGARITHM

165. Which of the following statements is not correct?

A.log10 10 = 1      B.log (2 + 3) =log (2 x 3)   C.log10 1 = 0   D.log (1 + 2 + 3) = log 1 + log 2 + log 3

Solution : B  log(2+3)=log 5 log (2x3)=log 6

166. If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:

Solution : log 5 512 = log 5 (29) =log (29)/log 10/2 =9 log2 /log10-log 2=9*3010/1-0.3010=3.867

167. If log 27 = 1.431, then the value of log 9 is:

Solution : log 33 =1.431 ie 3log3 =1.431 ie log3 =0.0477 log9 =2log3 ie 0.0477x2=0.0954

168. If log a/b +log b/a = log (a + b), then:

Solution : log (a+b) =log(a/b*b/a) =log 1 =1

169. If log10 7 = a, then log10 (1/70) is equal to:

Solution : log10(1/ 70) = log10  1-(log10 7+log10 10) = –(a +1)

170.If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, then x is equal to:

Solution : log10 (5(5x + 1) =log10(x+5)+log10 10 ie log25x+5=log10 (10(x+5))=10x+50 then 15x =45 ie x=3

171.The value of log3(1/60) + log4(1/60) + log5(1/60) is:

Solution : log60 3+log60 4+log60 5 ie log60 (3*4*5) =log60 60 =1

172. If logx y = 100 and log2 x = 10, then the value of y is:

Solution : log 2 x =10 => x= 2 10   log x y =100 y=x 100  ie y =2 1000

173. If logx 9/16=-1/2, then x is equal to:

Solution : logx 9/16 =x-1/2=9/16=√x =16/9=x=256/81

174. If log10000 x = ¼ then x equal to

Solution : log 10 x =1/4 ie 4log 10 x =1/4

175. If log4 + log2 = 6, then is equal to:

Solution : 2log2 x +log2 x =6  ie 3log2 x =6  x=2

176. If log5 (x2 + x) - log5 (+ 1) = 2, then the value of is:

Solution: log5(x2+x)/x+1=2  ie x2+x/x+1=5 ie x(x+1)/x+1 =x=25

177. Expand using the binomial theorem: (+ 2)5 .

Solution: Use binomial theorem n=5 and  y=2
=  5                5                   5                5                5                5
(x5)(20)+     (x4)(21) +    (x3)(22) +    (x2)(23) +    (x1)(24) + x025
0                1                  2                3                 4                5
=x5 +5x4*2+10x3*4+10x2*8+5x*16+32
=x5+10x4+40x3+80x2+80x+32

178. Expand using the binomial theorem: (2x -5)4

Solution : Use binomial theorem n=4 and  y=5
=  4                4                   4                4                4
(2x)4(-50)+    (2x)3(-51) +   (2x)2(-52) +  (2x)(-53) +  (x0)(-55)
0                1                  2                3                 4
=16x4 +32x3*(-5)+24x2*25+8x*(-125)+625
=16x4-160x3+600x2-1000x+625

180. Sum of two integers is 10 and the sum of their reciprocal is 5/12 Find the integers?

Solution :Let the integers are x and y then given x+y=10 and 1/x+1/y=5/12
= x+y/xy =5/12 substituting x+y ie 10/xy=5/12 ie 5xy=120 xy =24 then x2+2xy+y2 =100 ie x2+y2=100-48=52 then (x-y)2 =x2+y2-2xy =52-48=4
x-y =2 by solving x+y=10 and x-y =2 then  x= 6 and y =4

181. The multiplication of two numbers is 399 and its sum is 40 Find the numbers ?

Solution : Let the numbers be x and y . Given xy =399 and x+y=40 Then (x+y)2 = 402 = x2+y2+2xy =x2+y2+798 then x2+y2 =1600-798=802Then (x-y)2 =x2+y2-2xy =802-798=4 ie x-y =2 then by solving with x+y=40 we get x=21 and y=19

182. A certain numbers between 10-100 is 8 times the sum of its digits and if 45 is subtracted from it the digits will be reversed. Find the numbers?

Solution : 8(x+y)= 10x+y ie -2x+7y =0 –eqn1 then again 10x+y-45=10y+x ie 9x-9y=45 ie x-y=5 ---eqn 2 by solving 1&2 5y=10 ie y=2 by substituting we get x=7 then the number is 72.

183. What numbers are those whose difference is 45 and the quotient of the greater by the lesser is 4 ?

Solution: Given x-y=45 and x/y=4 ie x=4y substitute x=4y in eqn1 ie 4y-y=45 ie 3y =45 y=15 and x=60

184. Find the two numbers such that the first to be added to 5times the second the sum is 52 if the second be added to 8 times of the first the sum is 65 ?

Solution: let the first is x and the second is y then given x+5y=52 and 8x+y=65 eqn1 by 8 then 8x+40y=416 by solving 39y =351 y=9 and x=7 then the number is 79.

185. Divide 20 into two parts such that the difference between their square is 160 . Find the numbers ?

Solution : Let one part is x and another is 20-x then given x2-(x-20)2 =160
Ie x2-(x2-40x-400)=160 ie 40x =560 x=14 then y=6

186. A person distribute Rs 100/- among 36 persons for men Rs3/- and for women Rs 2.50 each . Find each gender ?

Solution: Let men x and women y then Given x+y=36 and 3x+2.5y=100
= 6x+5y =200 by solving two equation 6x+6y=216 & 6x+5y =200 y=16
And x=20

187. Find two numbers whose difference multiplied by the difference of the square is 160, and whose sum multiplied by the sum of the square is 580?

Solution : Let the numbers be x+y and x-y then given 2y(4xy)=160 ie 8xy2=160 ie xy2=20  and (2x){2(x2+y2)} =580 ie 2x(2x2+2y2)= 4x3+4xy2= 580 ie x3+xy2=145 x3=145-20=125 x=5 substitute in xy2=20 we get y2=20/5=4 ie y=2 Hence the required number is 7 & 3

188. A mirror having 18cm x 12cm to be framed with the frame of equal width whose area is equal that of glass. Find the width of frame. ?

Solution : The width of frame is x. Then the area of the frame is to be given as (18+2x)(12+2x) = 2x18x12 =512 ie 216+36x+24x+4x2 =512 ie 4x2+60x-216 =0 ie 4x2-12x+72x-216 =0 ie 4x(x-3)+72(x-3)=0 The width of the frame is 3 cm.

189. A shirt was purchased and sold for Rs 171/- and gains exactly as much  percent the shirt costs. What is the cost of the shirt ?

Solution: Let x be the cost of shirt.Given 171-x =x*x/100 =x2-100x-17100= 0 ie x2+190x-90x-17100=0 ie x(x+190)-90(x+190) =0 ie x=90

190. A kilogram of Tea and 2 kilograms of sugar costs Rs 140/- and if tea increased by 50% and tea 20% then it costs Rs 158/- Find the cost of Tea and sugar ?

Solution : Let tea costs Rs x and sugar costs Rs y then given x+2y =140 and (x+x/2)+(y+y/5)=174 ie 3x/2+6y/5=158 ie 15x+12y=1740 by multiplying by 15 eqn1 15x+30y=2100 by solving 18y=360 y=20 and x=100

191.A man has to travel some distance, he has traveled 40KM he increases his speed 2 KM , if he had traveled with his increased speed during entire journey he would have arrived 40minutes earlier , but if he had continued at originl speed he would have arrived 20 minutes later. How far he had to travelled ?

Solution : Let x be the number of kilometers to be traveled and y is the speed . Given 40/y+x-40/y+2 hrs =80+xy/y(y+2) the time taken if he traveled in increased speed is x/y+2 and time taken in the original speed =x/y . then x/y+2=80+xy/y(y+2)-2/3 ---1 x/y=80+xy/y(y+2)+1/3---2
By solving 1 & 2 x(1/y-1/y+2)=1 ie 2x=y(y+2) from the equation 2 ie 3x(y+2)=3(80+xy)+y(y+2) ie 6x-240=2x then x=60.

192. A boat going up-stream 30KMs and down-stream 44KMs in 10hrs.It also goes up-stream  40KMs and downstreams 55KMs in 13hrs. Find the rate of stream and of the boat ?

Solution : Let boat’s speed is x and y is the speed of stream. Given 30/x-y+44/x+y =10—1 and 40/x-y+55/x+y =13----2 by multiplying 1x4 ie 120/x-y +176/x-y=40 –3 and 120/x-y+165/x+y =39 by subtracting 11/x+y=1 ie x+y=11 hence from 1 30/x-y=10-4=6 x-y=5 by solving x=8 and y=3

193. Solve x4-10x2+9=0

Solution : Let x2 =y then y2-10y+9=0 by factorizing (y-9)(y-1)=0 hence y=1 or 9 x2 = 1 or 9 x=1 or 3

194. Solve 25a4/x2+x2=26a2

Solution : Multiply  both side by x2 ie 25a4+x4 =26a2x2 ie x4-26a2x2+25a4 put x2 =y  ie y2-26a2y+25a4 ie (y-25a2)(y-a2)=0 ie y=a2 or 25a2 hence x2=a2 or 25a2 ie x=a or 5a

195.Solve (x2+3x)2-(x2+3x)-6=0

Solution : Let x2+3x= y ie  y2-y-6=0 by factorize y2-3y+2y-6=0  ie  (y-3)(y+2) ie y =3 or y=-2 hence x2+3x =3 or x2+3x=-2 x2+3x+2=0 by factorizing (x+2)(x+1) ie x=-1 or -2 in x2+3x-3=0 x=-3+√21/2  or -3-√21/2

196.Solve (x+2)(x+3)(x+4)(x+5)=24(x2+7x+7)

Solution: Rearrange (x+2)(x+5)(x+3)(x+4)=24(x2+7x+7)  to make 7x
(x2+7x+10)(x2+7x+12) =24(x2+7x+7) put x2+7x =y then the equation as (y+10)(y+12)=24(y+7) ie y2+22y+120=24y+168 ie y2-2y-48=0 hence (y+6)(y-8) =0 then y=-6 or y=8 but y=x2+7x ie x2+7x+8 or x2+7x-6 by factorizing (x+8)(x-1) or (x+6)(x+1) hence x=1 or -8 or -6 or -1

197. If a/b+c +c/a+b =2b/c+a prove that a2+c2=2b2 or a+b+c=0

Solution :Say a/b+c-b/c+a=-(c/a+b)+b/c+a LHS a(c+a)-b(b+c)/(b+c)(c+a)
=ac+a2-b2-bc/(b+c)(c+a) = c(a-b)-b2+a2/(b+c)(c+a) then the RHS like
=a(b-c)+b2-c2/(a+b)(c+a) ie (a-b)(c+a+b)/b+c =(b-c)(a+b+c)/a+b
=(a2-b2)(a+b+c)=(b2-c2)(a+b+c) ie a+b+c{(a2-b2)-(b2-c2)}
=(a+b+c)(a2+c2-2b2)=0 ie a+b+c=0 or a2+c2=2b2

198. If a+b+c =0 then show 2(a4+b4+c4)=(a2+b2+c2)2

Solution : a+b+c=0 ie a+b=-c by squaring a2+2ab+b2 =c2 ie a2+b2-c2=-2ab again by squaring (a2+b2-c2)2 =4a2b2 or a4+b4+c4+2a2b2-2b2c2-2c2a2 = 4a2b2 ie a4+b4+c4=2(a2b2+b2c2+c2a2) by adding a4+b4+c4
=2(a4+b4+c4) =a4+b4+c4+2(a2b2+b2c2+c2a2) =(a2+b2+c2)2

199. If a+b+c=0 then prove 1/b2+c2-a2 +1/c2+a2-b2 + 1/a2+b2-c2 =0

Solution : a+b=-c then a2+2ab+b2=c2 ie a2+b2-c2=-2ab similarly b2+c2-a2=-2bc and c2+a2-b2 =-2ac ie 1/-2bc+1/-2ac+1/-2ab =a+b+c/-2abc =0

200. Solve the equation x-11+24/x=0 and solve (y2-2y)-11+24/y2-2y =0 ?

Solution: x-11+24/x =0 ie x2-11x+24 =0 by factorizing x2-8x-3x+24=0 ie (x-8)(x-3)=0 hence x=3 0r 8 again in second equation put y2-2y = x and the equation become x-11+24/x ie x=3 or 8 so y2-2y=3 or 8 by factorizing y2-2y-3=0 factorize y(y+1)(y-3) =0 ie y=3,-1 if y2-2y-8=0 (y-4)(y+2)=0 here y=4 or -2 The value of y =3,4,-1,-2

201. Solve x= √19-2x +2

Solution : x-2=√19-2x  by squaring both (x-2)2 =19-2x ie x2-4x+4=19-2x
ie x2-2x-15 =0 ie (x-5)(x+3) =0 ie x=5 or -3 by substituting x=5

202. One root of the equation ax2+bx+c =0 is five times of the other.Show 5b2=36ac.

Solution : Let the roots are a,5a in the equation x2+b/ax+c/a =0 ie the root
a+5a=-b/a ie 6a =-b/a  another (a)(5a)=c/a ie 5a2 =c/a in a=-b/6a by substituting in 5a2=c/a ie 5(b/6a)2 =c/a ie 5(b2/36a2)=c/a ie 5b2a=36a2c
ie 5b2=36ac

203.The roots of the equation 2x2+6x+3=0 are α,β Find the value of 1)α+β 2)αβ 3)3α+3β

Solution : 2x2+6x+3 =0 ie x2+6x/2+3/2 =0 here the roots α+β =-6/2=-3
and  αβ =3/2  then 3(α+β)=3*-3 =-9

204. The equation x2-2px+q =0 has roots a,a+2 prove p2=q+1

Solution : a+a+2 = 2p ie a+1 =p and a(a+1) =q but p2 =(a+1)2 =a2+2a+1 ie (a+1)2=q+1 =P2

205. (2x-1) is a factor of the polynomial f(x)=2x3-5x2-kx+3 =0 find the value of k and other factors ?

Solution : f(x) = 2x3-5x2-kx+3 =0 if (2x-1) is a factor then 2x=1 x=1/2
By substituting in the equation 2(1/2)3-5(1/2)2-k(1/2)+3=0 ie ¼-5/4-k/2+3=0 ie 1-5-2k+12=0 ie -2k=-8 ie k=4 . then 2x3-5x2-4x+3 divided by the factor 2x-1 we get the quotient as x2-2x-3 by factorizing (x-3)(x+1) ie the factors (2x-1)(x-3)(x+1)

206. Let f(x)=2x3+mx2+nx+2 where m,n are constants and x-1 and x+2 are factors for that find m and n ?

Solution : In the equation one factor (x-1) =0 x=1 and another x+2 =0 x=-2
Substituting the value in f(x) ie 2+m+n+2=0 ie m+n=-4 eqn 1 and -16+4m-2n+2=0 ie 2m-n =7 eqn 2 by factorizing m=1 and n=-5

207. Find the value of a/a-b+b/b-a ?

Solution : Since b/b-a = b(-1)/b-a(-1) = -b/a-b then a/a-b-b/a-b =a-b/a-b =1

208. Find the value a+b/a2-b2+a-b/a3-b3-(a2+b2/a3-b3) ?

Solution : a+b/a2-b2= a+b/(a+b)(a-b) =1/a-b and a-b/a3-b3 =1/a2+ab+b2
Ie (a2+ab+b2)+(a-b)-a2-b2/a3-b3 =ab+a-b/a3-b3

209.  Solve 1/x2+3x+2 + 2x/x2+4x+3

Solution : 1/(x+2)(x+1) + 2x/(x+3)(x+1)ie x+3+2x(x+2)/(x+1)(x+2)(x+3)
= x+3+2x2+4x =2x2+5x+3 ie (2x-1)(x+3) ie 2x-1/(x+1)(x+2)

210. Solve 1/(x+5)(x+2) + 1/(x+8)(x+5)

Solution : LCM (x+5)(x+2)(x+8) ie x+8+x+2 / (x+5)(x+2)(x+8) ie 2(x+5)
= 2(x+5)/(x+5)(x+2)(x+8) = 2/(x+2)(x+8)

211. Solve x/x-2a  + x/x+2a +  2x2/x2+4a2

Solution : x/x-2a + x/x+2a = x(x+2a)+x(x-2a)/x2-4a2  ie x2+2ax+x2-2ax/x2-4a2 ie 2x2/x2-4a2 then add 2x2/x2+4a2 = 4x2/x4-16a4

212. Solve x2-4x+3 /x2-6x+5  X  x2-7x+6/x2-5x+6

Solution : (x-3)(x-1)/(x-5)(x-1) X (x-5)(x-2)/(x-3)(x-2) ie (x-3)(x-2)/(x-3)(x-2) =1

213. Solve 2x2-7x+3 / 2x2+7x-4  X  3x2+11x-4/3x2+8x-3  X  2x2+x-15/2x2-11x+15 ?

Solution : Factorize each 2x2-7x+3= (2x-1)(x-3)
2x2-7x-4 = (2x-1)(x+4) and 3x2+11x-4 = (3x-1)(x+4) and other factorized
3x2+8x-3=(3x-1)(x+3) and 2x2+x-15 = (2x-5)(x+3) and (2x-5)(x-3)
Ie (2x-1)(x-3)/(2x-1)(x+4) X (3x-1)(x+4)/(3x-1)(x+3) X (2x-5)(x+3)/(2x-5)(x-3) = x-3/x+4 X x+4/x+3 X x+3/x-3 =1

214. Solve 5(x+1)2+7(x+3)2 =12(x+2)2

Solution : 5(x2+2x+1)+7(x2_6x+9)=12(x2+4x+4)
=5x2+10x+5+7x2+42x+63=12x2+48x+48 ie 2x2+52x+68=12x2+48x+48
=4x=-20hence  x=-5

215. Solve x-6/5 +x-4/3 =8-  (x-2)/7

Solution : LCM  for LHS  is 15 then 3x-18+5x-20 /15 = 8x-38 /15 again LCM for both LHS and RHS is 105 ie 56x-266 =840- 15x-30 ie 71x =1136 ie x=16

216. A motorboat, whose speed in 15 km/hr.in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr.) is:

Solution: Let speed of the stream be x then 30/15+x +30/15-x=9/2 ie 9x2=2025-1800 ie x2=25 x =5

217. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?

Solution  : Total time taken =x/5+1+x/5-1=1 ie 10x=24 ie x=24km

218. A boat covers a certain distance downstream in 1 hour, while it comes back in 1.30 hour. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?

Solution :Let x is the speed of boat then  x+3=(x-3)3/2=2x+6=3x-9 ie x=15

219. A man can row 7.5 Kmph in still water and he finds that it takes him twice as long to row up as to row down the river. Find the rate of stream.

Solution : (7.5+x) =2(7.5-x) ie 3x =7.5 then x =2.5

220. A boat takes 19 hours for travelling downstream from point A to point B and coming back to point C midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B?

Solution : Let x be the distance X/18+X/10*2 =19  ie 38x=19*360=180

221. The current of a stream runs at the rate of 2 km per hr. A motor boat goes 10 km upstream and back again to the starting point in 55 min. Find the speed of the motor boat in still water?

Solution : 10/(x +2)+10/(x-2) =55/60 ie 11/12 11x2-240x-44 ie x=22

222.Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance travelled by train was ?

Solution: Let the total trip be x km.Then 2x/5=1200,x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.

223.The sum of 5 consecutive odd number is 1185 what are the numbers?

Solution :Let 2x+1 is the smallest number 2x+1,2x+3,2x+5,2x+7,2x+9 =1185 Ie 10x+25 =1185 , 10x =1160 x =116 Then the numbers are 233,235,237,239,241

224.Divide 28 into two such parts that the difference between their square is equal to 112.

Solution : Let x is the greatest part so the other part is 28-x and x2-(28-x)2 =112 (x+28-x)(x-28+x) =112 ie 28(2x-28)=112 ie x-14 =2 ie x=16. Hence the parts are 16,12

225. Father is aged three times more than his son Kavin. After 8 years, he would be two and a half times of Kavin's age. After further 8 years, how many times would he be of Kavin's age?

Solution : Kavin age is x  now  after 8 years the fathers age 3x+8=5/2(x+8) ie x=24 another 8 years after  son age 24+8+8 =40 fathers age 72+8+8=88 ie 11/5 times

226. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

Solution : Let x be the age of youngest child then the given equation is x+x+3+x+6+x+9+x+12=5x+30 =50 ie x+6 =10 then  x=4

227.  A certain fraction becomes 2 when 7 is added to its numerator and 1 when 2 is subtracted from the denominator. Find the fraction ?

Solution : Let the fraction br x/y then given x+7/y=2 and x/y-2 =1 by solving this equation x+7=2y ie x-2y=-7 and x=y-2 ie x-y =-2 ie y=5 and x=3 then the fraction is 3/5

228. 2 men 7 boys can do in 4 days a work which would be done in 3 days by 4 men and 4 boys . How long would it take one man or one boy do it ?

Solution : Let x be the number of days in which one man would do the work
And y be the numer of days one boy would do the same work.
Given 2/x+7/y =1/4--1 and 4/x+4/y =1/3 –2 eqn1x2 ie 4/x+14/y=1/2 by solving with eqn2 10/y = 1/6 then y=60 and x= 15

229. Two plugs are opened in the bottom of a cistern containing 192 litres of water. After 3 hrs one of them becomes stopped, and the cistern is emptied by the other in 11 hrs , had 6 hrs elapsed before the stoppage, it would have only required 6 hrs more to empty it. How many litres will each plug discharge in one hour?

Solution : Let x.y be the number of litres each plug will discharge in one hour
In the Ist case the first plug remains 3 hrs and the second 3+11 hrs. Hence 3x+14y =192 ---1 and  in II case 6x+12y =192 ---2 by solving both y=12 and x=8 .

230. The sum of the two integers is 5 the leftside integer is added by 1 then the 1/8 of the integers . Find the integers?

Solution : Let be the integers are x, y Given x+y =5 and the second one x+1=1/8(10x+y)  ie 8x+8=10x+y ie -2x-y=-8 by solving both equation x=3 and y=2 then the required integer is 32.

231. Divide the number 39 into four parts such that if the first be increased by 1 ,the second diminished by 2, the third multiplied by 3 the fourth divided by 4 the result will be all equal.

Solution : Let x be the result in all cases. By the condition of the problem
Ist part +1 = x ie the first part is x-1 IInd part -2=x ie IInd part is x+2
IIIrd part X3 = x  ie IIIrd part is x/3 IV th part is 4x then given
x-1+x+2+x/3+4x=39 ie 19x=114 ie x=6 then the required parts are like 5,8,2,24

232. The dimension of a rectangular court is such that if the length were increased by 3 meters and the breath diminished by the same its area diminished by 18sq. m and if its length and breath increased by 3 meters than its area would be increased by 60 sq.m Find the dimension ?

Solution : Let x be the length and y be the width Given (x+3)(x-y)=xy-18
Then (x+3)(y+3) = xy+60 by solving both equation y-x =-3 and x+y=17
Then solving x=10 and y=7

233. Find two numbers such that if the first added to 5 times the second , the sum is 52 and if the second be added to 8 times the first the sum is 65 ?

Solution : Let the numbers are x,y Given x+5y =52 and 8x+y =65 by solving the equation by elimination method -39x =-273 ie x=7 and y =9

234. What numbers are those whose difference is 45 and the quotient of the greater by lesser is 4 ?

Solution : Let the greater is x and the lesser is y Given x-y=45 and x/y=4
Ie x=4y ie x-4y=0 by solving 3y=45 ie y=15 and x=60

235. Two persons ,27KM apart , setting out at the same time ,meet together in 9 hrs , if they walk in the same direction. But in 3hrs if they walk in the opposite direction, Find their rate of walking ?

Solution : Let the rates are x,y In the same direction the rate is x-y but in the opposite direction x+y . Given 27/x-y =9 ie 9x-9y=27 and 27/x+y=3 ie 3x+3y=27 by solving 18x =108 x=6 and y=3

236. Factorize x+4y =14 and 7x-3y =5 ?

Solution : Method of substitution : In eqn 1 x=14-4y by substituting this in eqn 2 ie 7(14-4y)-3y=5 ie 98-28y-3y=5 ie -31y=-93 ie y=3 and x+12=14 ie x=2.

Solution : Method of comparision : In eqn 1 x=14-4y and in eqn 2  7x=5+3y ie x=5+3y/7 ie 14-4y = (5+3y)/7 ie 98-28y =5+3y ie -31y =-93 y=3 and x=2

Solution : Method of elimination : Multiply eqn1 by 7 ie 7x+28y=98 by solving with eqn 2 7x-3y=5 ie 31y =93 ie y=3 and x=2

237.    Find the reminder 3x3+7x2+11x+2 is divided by 3x-1 ?

Solution : The reminder = the value of expression 3x3+7x2+11x+2  when nd 3x=1 x=1/3 is put up hence we get 3(1/3)3+7(1/3)2+11(1/3)+2 =0 ie 3(1/27)+7(1/9)+11/3+2 ie 1/9+7/9+11/3+2 ie 1+7+33+18 = 59

238. Find 6x3+13x2+17x+6 divisible by 2x+1 ?

Solution : The value of expression 6x3+13x2+17x+6 is put up when x=-1/2
6(-1/2)3+13(-1/2)2+17(-1/2)+6 = -6/8+13/4-17/2+6 = -6+26-68+48 =0 Hence it is divisible by 2x+1

239. Simplify (a3+b3)/(a2-b2)   /  (a2-ab+b2)/a-b ?

Solution : a3+b3/a2-b2 x a-b /a2-ab+b2 ie (a+b)(a2-ab+b2)/(a-b)(a+b) = a2-ab+b2/a-b x a-b/a2-ab+b2 =1

240. Simplify x2-x-2/x2+7x+12  /  x2-3x-10/x2+x-12 X x2-4x-5/x2-4x-3

Solution : (x-2)(x+1)/(x+4)(x+3) X (x+4)(x-3)/(x-5)(x+2) = (x-2)(x+1)/x+3)X (x-3)/(x-5)(x+2) x (x-5)(x-1)/(x+3)(x-1) = x+1/x+3

241. Simplify x2-49/x2-25   /   x+7/x+5

Solution : (x+7)(x-7)/x+5)(x-5) X  (x+5)/(x+7) = x-7/x-5

242. Simplify x2-x-30/x2-36   /   x2+3x-10/x2+2x-8   / x+4/2x2+12x

Solution : (x-6)(x+5)/(x+6)(x-6) X (x+4)(x-2)/(x+5)(x-2) X 2x(x+6)/x+4
= x+5/x+6 x x+4/x+5 x 2x(x+6)/x+4 = 2x

243. x-6/8 – (2x-15)/9 +1 = x/15 –(x-12)/6 find x ?

Solution :Multiply by LCM72on LHS 9(x-6)-8(2x-15)+72=9x-54-16x+120+72
= -7x+138/72 on RHS 6x-15(x-12)/90 ie -9x+120 then LCM 360 -35x+690=24x-60x+720 ie -35x+690 =-36x+720 ie x=30

244.  Solve (5x+6)/4 +(64x-35)/15 =(20x+23)/16 +(13x-7)/3

Solution : Rearrange (5x+6)/4 – (13x-7)/3 + (64x-35)/15 – (20x+23)/16 LCM 240 then 300x+360-1040x+560+1024x-560-300x-345/240 ie -16x=15
Then x = -15/16

245. The length of a field is twice its breath , another field which is 50metres longer and 10metres broader, contains 6800sq.m more than the former. Find the size of each ?

Solution : Let the length is 2x then the breath is x then given (2x+50)(x+10) = 2x2+6800 ie 2x2+70x+500 =2x2+6800 ie 70x =6300 then x =90 then the length of field 180 metres and breath 90

246. A certain fraction becomes 2 when 7 is added to its numerator, and 1 when 2 is subtracted from the denominator. What is the fraction?

Solution : Let x/y represent the fraction. Then x+7/y =2 and x/y-2 =1 ie x+7=2y ie x-2y =-7 and x =y-2 ie x-y = -2 by solving   -y =-5 ie y=5 and x=3 then the fraction be 3/5

247.Factorize x4+2x3+3x2+2x+1

Solution : An algebraical expression in which co-efficients of the terms equidistant from the beginning and end are same , is called reciprocal or recurring expression  . If the given expression is in reciprocal then rearrange the terms with equal co-efficient .

The above expression is reciprocal hence rearrange as x4+1+2x3+2x+3x2
={ (x2+1)2-2x2} +2x(x2+1) +3x2 = (x2+1)2+2x(x2+1)x-2x2+3x2 ie (x2+1)2+2x(x2+1)x+x2 ie (x2+1+x)2 ie (x2+x+1)2

248. Factorize a4-5a3-12a2-5a+1

Solution : Rearrage a4+1-5a3-5a-12a2 ie (a2+1)2-2a2-5a(a2+1)-12a2 ie (a2+1)2-5a(a2+1)-14a2 put a2+1 =x then x2-5ax-14a2 by factorizing this x2-7ax+2ax-14a2 ie (x-7a)(x+2a) Hence (a2+1-7a)(a2+1+2a)=(a+1)2(a2-7a+1)

If any expression containing integral powers of x the sum of the co-efficient is zero then x-1 is a factor of that expression.

If any expression containing integral powers of x the sum of the co-efficient of odd terms is equal with the even terms then x+1 is a factor of that expression.

249. Factorize x3-x2-x+1

Solution : the sum of the co-efficient 1-1-1+1 =0 hence x-1 is a factor.
= x2(x-1)-1(x-1) = (x-1)(x-1)(x+1)

250. Factorize 8x3+16x-9

Solution : let 2x =a then a3+8x-9 the sum of co-efficent 1+8-9=0 hence a-1 is a factor .  then (a-1)(a2+a+9) ie (2x-1)(4x2+2x+9)

251. Factorize 6a2+7ab+2b2+11ac+7bc+3c2

Solution : if a=0 then 2b2+7bc+3c2 ie (2b+c)(b+3c)
If b=0 then 6a2+11ac+3c2 ie (3a+c)(2a+3c) and if c=0 6a2+7ab+2b2 ie (3a+2b)(2a+b) ie (3a+2b+c)(2a+b+3c)

Alternate method: Arrange the expression in descending powers of any letter say a ie 6a2+a(7b+11c)+2b2+7bc+3c2 ie 6a2+a(7b+11c)+(2b+c)(b+3c) then split 6X(2b+c)(b+3c) into 7b+11c 6a2+2(2b+c)a+3(b+3c)a+(2b+c)(b+3c) ie (3a+2b+c)(2a+b+3c)

252. Factorize 2a2-2bc+6b2+ac-7ab ?

Solution : Here terms 1,3,5 in second degree in a,b and 2 and 4 th term in first degree in those letters. Then the expression 2a2-7ab+6b2-2bc+ac ie 2a2-4ab-3ab+6b2-2bc+ac ie 2a(a-2b)-3b(a-2b)+c(a-2b) ie (2a-3b-c)(a-2b)

253. Find the reminder x6-19x5+69x4-151x3+229x2+166x+26 when x=15?

Solution : By reminder theorem 156-19*155+69*154-151*153+229*152+166*15+26 =0 ie 11390625-19*759375+69*509625-151*3375+229*225+166*15+26 =11390625-14428125+3493125 - 50925 +51525+2490+26 =41

II . x=15 means x-15 =0 arrange the expression in terms of x-15 as below.
=x5(x-15)-4x4(x-15)+9x3(x-15)-16x2(x-15)-11x(x-15)+(x-15)+41 hence reminder 41.

254. If any rational and integral expression in x vanishes when –b/a is substituted for x , the expression exactly divisible by ax+b

Show that 3x3-2x2y-13xy2+10y3 is exactly divisble by x-2y and contains x-2y as factor. Putting 2y for x then 24y3-8y3-26y3+10y3 =0

255. If the expression x3+3x2+4x+p contains x+6 as a factor find p ?

Solution : x+6 = x-(-6) the value of expression for x=-6 by substituting
-216+108-24+p =0 ie -240+108+p = 132 ie p=132

256. Divisibility Theorem : I : The expression an-bn is always divisible by a-b if n is any +ve integer odd or even

Theorem II : The expression an-bn is exactly divisible by a+b if n is any even +ve integer but not odd

Theorem III:  The expression an+bn is exactly divisible by a+b if n is any odd +ve integer but not even

Theorem IV: The expression an+bn is never divisible by a-b if n is any odd or even

Eg : Show that 34n – 43n is divisible by 17,if n is any +ve integer ?

Solution : Make it power of n ie (34)n-(43)n ie 81n-64 as per I a-b = 81-64=17 and divisible.

Eg : Show that 26n – 62n is divisible by 100,if n is even +ve integer ?

Solution : Make it power of n ie (64)n-(36)n ie  as per II a+b = 64+36 =100 and is divisible by 100

Eg: Show that (x3+3x2a+3xa2+a3)2m+1+(x3-3x2a+3a2x-a3)2m+1 is divisible by 2x and m is + ve integer?

Solution : {(x+a)3}2m+1 + {(x-a)3}2m+1 3(2m+1) is odd +ve integer as per III a+b is divisable ie x+a+x-a =2x .

257. Show 6x3+13x2+17x+6 is divisible by 2x+1 ?

Solution : 3x2(2x+1)+5x(2x+1)+6(2x+1) here 2x+1 be a factor and no reminder Hence exactly divisible by 2x+1

258. Find 64x6-729y6 divisible by 2x-3y

Solution : (2x6)6 –(3y6)6  = 2x36-3y36  in the form an-bn where n=36 even
As per theorem I 2x-3y is a factor and divisible.

259. (x2+7x+6)n –(2+x)2n where n is a any +ve integer find divisible by 3x+2 ?

Solution : (x2+7x+6)n- (4+4x+x2)n  in the form an-bn with +ve integer then a-b is a factor and divisible hence x2+7x+6-4-4x-x2 =3x+2

260. Find HCF 3x3+5x2+5x+2 and 2x2+5x2+5x+3

Solution : 2x3 ie 6x3+15x2+15x+9  3x3+5x2+5x+2 | 6x3+15x2+15x+9|2
6x3+10x2+10x+4
----------------------
5x2+5x+5
5x2+5x+5 ie x2+x+1        x2+x+1 | 3x3+5x2+5x+2 | 3x
3x3+3x2+3x
-----------------
2x2+2x+2 |2
2x2+2x+2
-----------------
0
-----------------
Hence x2+x+1 is the HCF   in this find LCM
3x3+5x2+5x+2 / x2+x+1  x 6x3+15x2+15x+9 =(3x+2)(6x3+15x2+15x+9)
18x4+45x3+45x2+27x+12x3+30x2+30x+18 = 18x4+57x3+75x2+57x+45

261. Find the value of x+2a/x-2a  + x+2b/x-2b when  x=4ab/a+b

Solution : (x+2a/x-2a -1 ) + (x+2b/x-2b -1)+2 =4a/x-2a + 4b/x-2b +2
=4/(x-2a)(x-2b){(a+b)x-4ab}+2 =0+2 =2 where (a+b)x =4ab