#x 2 − 10 + 2 = 0
n=#
:
=# = − 10 5=2
7 x 4 ( −% )
11
−7x 4%11
x1 + x 2
=−
x1 + x 2
=−
x1 + x 2
=2
( −10) #
12
( x + 4% ) .
=#
−1= 4
=
n
10
.
17. Find the 8th ter of the sequene Ans:
4
7 x8 % 4 ( 4 )
4
−3,4,
# n+2 ... 3 2n − 3
10 13
Solution:
4
( 4) = 126'20
84 13. "hat is the su of the oeffiients of the expansion of
when n
=1
when n
=2
when n
=3
when n
=8
1+ 2 2 ( 1)
Ans: 0 Solution:
= 1 $ut
su$trat
( −1)
= −3
=4 =
#
6−3 3 8 + 2 10 16 − 3
20
u of the oeffiients
−3
2+2 4−3 3+2
.
u$stitute x
2
82 ;er involvin! x 8 (0 x8 )2%*2 ;er involvin! x 8 180 x8 %2
( n − )
7 x8 ( 4% )
10 x 8 ( −2% )
;er involvin! x 8
= r −1
( 2x − 1)
= −136#
Solution:
7 xn−r +1% r −1 7 x n − %
20
7
1#
=−
16. Find the ter involvin! x 8 in the expansion of ( x − 2% )
Solution:
12
=−
Ans: 180 !8 "2
Ans: 126720
7
n
7
411 ( n − ) 1# ( 14 ) ( 13) ( 12) 11 7=− 4 ( 3 ) ( 2) ( 1) 11
12. 5opute the nuerial oeffiient of the # th ter of the expansion of
= 11
=
13
18. "hat follows lo!iall% in these series of nu$ers 2,3,4,#,(,1'<.. Ans: 33
20
2 ( 1) − 1 − ( −1) 20 u of the oeffiients
Solution: 2 3
= ( 1) 20 − ( −1) 20 =0
1
# ( 1' 2
4
x
8 16
= 1' + 16 x = 33 x
14. "hat is the su of the oeffiients in the expansion of 8 ( x + % − 9) .
19. Find the value of x if 2 + 4 + 6 + 8 + ...x = sin ! 2 + 4 + 6 + ....2n = n ( n + 1)
Ans: 1
Ans: 20
Solution: u$stitute x 1 % 1 and 9 1 u of the oeffiients ( 1 + 1 − 1) u of the oeffiients ( 1)
= 110
Solution: 20 2 + 4 + 6 + ...2n = n ( n + 1)
8
= n ( n + 1) 110 = n ( n + 1)
8
n2 + n − 110 = 0
( n − 10) ( n + 11) = 0
u of the oeffiients 1
n = 10
= 2n x = 20 x
15. Find the oeffiient of the expansion of )x%* 1# ontainin! the ter x 4%11 Ans: 1365
20. olve for x in x + 3x + #x + 'x + .....4(x Ans: 1 Solution:
Solution:
Page 2 of 13
Mathematics Enhancers
the
= 62#.
followin!
equation.
a1 = x
12
d = 2x
6
= a1 + ( n − 1) d 4(x = x + ( n − 1) ( 2x ) 4( = 1 + 2n − 2 #0 = 2n n = 2#
25. Find x if ' is the fourth proportional to 36 and 28, and x. Ans: 9
n
Solution: 36
2a + ( n − 1) d 2 2# 2 ( x ) + ( 24) ( 2x ) 62# = 2 2# ( #0) x 62# = 2 62# = ( 2#) ( 2#) x
28
=1
x
( x − 4) ( x + 3 ) + 4, when f)x* is divided
%$21. >iven f)x*
( x − 7 ) , the reainder is ?. Find the value of ?.
x
' 36 ( ')
x
=
x
=(
28
26. ;he fore required to strethed a sprin! is proportional to the elon!ation. If 24 @ strethes a sprin! 3 , find the fore required to streth the sprin! 2 . Ans: 16
24 = 7 ( 3)
Solution: f ( x)
= ( x − 4) ( x + 3 ) + 4 f ( ? ) = ( ? − 4 ) ( ? + 3 ) + 4 ( reainder ) ? = ( ? − 4) ( ? + 3) + 4
=8 F = 7x = 8 ( 2) F = 16@ 7
= ? 2 − ? − 12 + 4 ? 2 − 2? − 8 = 0 ( ? − 4) ( ? + 2) = 0 ?=4 ? = −2 ?
27. ;he volue of a heisphere varies diretl% as the u$e of itAs radius. ;he volue of the heisphere with 2.#4 . radius is 20.'# 3. "hat is the volue of a sphere with3.2# . radius of the sae 7ind of aterial Ans: Solution:
22. "hih of the followin! is a fator of 3x 3 2x2 32 Ans: !2
= 7r 3
B
20.'# = 7 ( 2.#4)
Solution:
= 3x3 + 2x 2 − 32 when x = 2
7
= 3 ( 2) 3 + 2 ( 2) 2 − 32 f ( 2) = 0
B
f ( 2)
= 1.266 ( 3.2#) 3
= 43.463 ( heisphere) B = 2 ( 43.46) B
#ote: If the reainder is 9ero, the nu$er we assue is a fator. ;herefore x2 is a fator of 3x 32x2 32
= 86.(23 ( sphere)
B 23. "rite a u$i equation whose roots are )1,2,4* Ans: !3 $ 5!2 % 2! % 8 & 0 Solution: ( x + 1) ( x − 2) ( x − 4 )
28. If x varies diretl% as % and inversel% as 9, and x14, when %' and 92, find x when %16 and 9 4. Ans: 16
=0
Solution:
( x + 1) ( x 2 − 6x + 8 ) = 0 2
3
= 1.266 B = 7r 3
f ( x)
x
=7
x
=
%
9 ' 14 = 7 2 7=4
2
− 6x + 8x + x − 6x + 8 = 0 x3 − #x 2 + 2x + 8 = 0 x
=
Solution: F = 7x
Ans: 4
3
6
x 36 x= 12 x=3
an
=
=
4 ( 16)
4 x = 16
24. ;he ean proportion $etween 12 and x is equal to 6. Find the value of x. Ans: 3 Solution:
29. tan7 is filled with 2 pipes. ;he first pipe an fill the tan7 1 in 10 hours. ut after it has $een opened for 3 hours, 3 the seond pipe is opened and the tan7 is filled up in 4 hours ore. Cow lon! would it ta7e the seond pipe alone to fill the tan7 ;he two pipes have different diaeters. Ans: 15 Solution:
Page 3 of 13
Mathematics Enhancers
'hen,
3 1 1 + 1 + 1 4 = 1 3 ÷ 10 ÷ 10 x ÷ 10
+
3 ( 10) x
4 10
80 ( 40 ) x
4
+ =1 x
= 1# hours
30. and wor7in! to!ether an finish paintin! a house in six da%s. wor7in! alone, an finish it in five da%s less than . Cow lon! will it ta7e eah of the to finish the wor7 alone Ans: 10,15
= 80
34. ;went% )20* en an finish the Do$ in 30 da%s. ;went% five )2#* en were hired at the start and 10 quit after 20 da%s. Cow an% da%s will it ta7e to finish the Do$ Ans: 27 Solution: @o. of anda%s20 ( 30) = 2# ( 20) + ( 2# − 10) x x
Solution: 1 + 1 = 1 ÷ 6
+ #0 ( x ) = '2 ( 100)
= 6.6' sa% ' da%s
;otal nu$er of da%s to finish the Do$ = 20 + '
= −#
= 2' da%s
1 + 1 6 = 1 −# ÷ 6 ( + − # ) = ( −# )
35. >iven two nu$ers suh that the differene of twie the first and the seond nu$er is 12. If the su of the first and the seond nu$er is 36, find the nu$er. Ans: 16,20
2 − 1' + 30
=0 ( − 1#) ( − 2) = 0 = 1# =2 =se = 1# = 1# − # = 10
Solution: x = 1st no.
= 2nd no. 2x − % = 12 x + % = 36 3x = 48 x = 16 % = 20 %
31. 5rew no.1 an finish the installation of an antenna tower in 200 anhours while 5rew @o.2 an finish the sae Do$ in 300 anhours. Cow lon! will it ta7e $oth rews to finish the sae Do$, wor7in! to!ether Ans: 120
36. ;he produt of Solution: 1
200
+
= ÷ 300 1
nu$er. Ans: 100
1 x
( 300 + 200) 1 = 200 ( 300) x x
1 1 and of a nu$er is #00. "hat is the 4 #
Solution: 1 x 1 x = #00 4 ÷ # ÷
= 120 an − hours
x2
32. n experiened statistial ler7 su$itted the followin! statistis to his ana!er on the avera!e rate of prodution of transistori9ed radios in an asse$l% line, 1.# wor7ers produed 3 radios in 2 hours. Cow an% wor7ers are eplo%ed in the asse$l% line wor7in! 40 hrs eah per wee7 if wee7l% prodution is 480 radios Ans: 12
= #00 ( 20) x = 100 37. ;he square of the nu$er inreased $% 16 is the sae as 10 ties the nu$er. Find the nu$ers. Ans: 2,8 Solution: x 2 + 16 = 10x
Solution: @o. of anhours to produed 3 radios =
1.# ( 2)
x 2 − 10x + 16 = 0
3
( x − 2) ( x − 8 ) = 0
@o. of anhours to produe 480 radios =
x ( 40) 480
% proportion40x 1.# ( 2)
38. ;he sides of the ri!ht trian!le are 8, 1# and 1' units. If eah side is dou$led, how an% square units will the area of the new trian!le. Ans: 240
=
480 3 x = 12 wor 7er s 33. ertain Do$ an $e done $% '2 en in 100 da%s. ;here were 80 en at the start of the proDet $ut after 40 da%s, 30 of the had to $e transferred to another proDet. Cow lon! will it ta7e the reainin! wor7fore to oplete the Do$ Ans: 80 Solution: &et x no. of da%s it will ta7e for the reainin! wor7fore to oplete the Do$
Page 4 of 13
=2 x=8 x
Solution: 16 ( 30)
=
= 240 sq..
2
39. 5opute the edian of the followin! set of nu$ers 4,#,',10,14,22,2#,30. Ans: 12
Mathematics Enhancers
Solution: ;he iddle nu$er is 10 and 14 ;herefore the edian is the avera!e of 10 and 14 12
45. o$erto is 2# %ears %oun!er than his father, Cowever his father will $e twie his a!e in 10 %ears. Find the a!es of o$erto and his father. Ans: 15,40
40. ;o !et an in a ourse a studentAs avera!e ust $e at least (0)and at ost 100*. If a partiular studentAs !rade so far are 84,88 and (3. "hat sores ust that student !et on the last test to earn an if all tests ourt equall% Ans: 95 Solution: 84 + 88 + (3 + x 4 x
x + 10 = 2x − 30
= 40 father x − 2# = 1# ( o$erto) x
= (0
= (#
41. students has test sores of '#, 83 and '8. ;he final tests ounts half the total !rade. "hat ust $e the iniu )inte!er* sore on the final so that the avera!e will $e 80 Ans: 82
'# + 83 + '8 vera!e = 3 vera!e = '8.6'
3 G + = 2# 3 4 8G + ( = 300 G+
= 3# (G + ( = 31# 8G + ( = 300 G = 1# ( a!e of Garia)
1
+ ( '8.6') = 80 2 2 x = 81.33 sa% 82 42. "hih of the followin! is a prie nu$er Ans: 487
47. ;en )10* %rs fro now the su of the a!es of and is equal to #0. ix )6* %rs a!o, the differene of their a!es is equal to 6. Cow old is and Ans: A&18, +&12
Solution: (actors: 43' = 1( and 23
Solution: Hast)6 %rs. !o* Hresent)10 %rs. &ater* 6 10 6 10
483 = 3 and 161 41' = 3 and 13( 48' = 1 and 48' #ote: prie nu$er is a positive inte!er that has exatl% two fators, the nu$er itself and 1.
+ 10 + + 10 = #0
u
+ = 30 v
43. ;he $oat travels downstrea in 2/3 the tie as it does !oin! upstrea, if the veloit% of the river urrent is 8 7ph, deterine the veloit% of the $oat in the still water. Ans: 40 )*h Solution: E B+8
=
2
= 3#
2
= final sore
x
46. ;he su of the a!es of Garia and nna is 3#. "hen Garia was two thirds her present a!e and nna was 3/4 of her present a!e, the su of their a!es was 2#. Cow old is Garia now Ans: 15 Solution: G+
Solution:
x
Solution: x a!e of o$ertoAs father x + 2# a!e of o$erto x 10 a!e of o$ertoAs father 10 %ears fro now ( x + 10 ) = 2 ( x − 2# + 10 )
( − 6) − ( − 6 ) = 6 − = 6 + = 30
2 = 36 = 18 = 30 − 18 = 12
E
3 ( B − 8)
2 ( B + 8)
= 3 ( B − 8) 2B + 16 = 3B − 24 B = 40 7ph 44. ;he veloit% of an airplane in still air is 12# 7ph. ;he veloit% of the wind due east is 2# 7ph. If the plane travels east and returns $a7 to its $ase a!ain after 4 hours. t what distane does the plane travel due east Ans: 240 )m.
Solution: 20
Solution: x
+
x
12# + 2# 12# − 2# x x + =4 1#0 100 x = 240 7.
Page 5 of 13
48. ;wo !allons of 20 salt solution is ixed with 4 !allons of #0 salt solution. Eeterine the perenta!e of salt solution in the new ixture. Ans: 40
=4
2
Mathematics Enhancers
50 4
! 6&
2 ( 0.2)
+ 4 ( 0.#) = 6 ( x )
Solution: >ross ar!in4# of sales
2.4 = 6x
= 0.40 x = 40 x
Kperatin! expenses 1# of sales @et profit 4#1#
49. ;wo alohol solution onsists of a 40 !allons of 3# alohol and other solution ontainin! #0 alohol. If the two solutions are o$ined to!ether, the% will have a ixture of 40 alohol. Cow an% !allons of solutions ontainin! #0 alohol Ans: 20 Solution: 35
50
40
x
40 40+x
@et profit 30 of sales ;ax 40 Hrofit 0.60 of 30 of sales Hrofit 18 of sales 53. "hat is the su of the pro!ression 4,(,14,1(< up to the 20th ter Ans: 1030 Solution:
+ x ( 0.#0) = ( 40 + x ) ( 0.40 ) 14 + 0.#0x = 16 + 0.40x 0.10x = 2 x = 20 !allons 40 ( 0.3#)
a=4 d=(− 4=# n = 20 =
50. If %ou owned a sarisari store in ?uwait, at what prie will %ou ar7 a sall aera for sale the ost H600 in order that %ou a% offer 20 disount on the ar7ed prie and still a7es a profit of 2# on the sellin! prie Ans: -1000 Solution: x = ar7ed prie x − 0.20x
= 600 + 0.2# ( 0.80x ) 0.80x = 600 + 0.20x 0.60x = 600 x = 1000
n
2a + ( n − 1) d
2 20
2 ( 4 ) 2 = 1030 =
+ ( 1( ) ( # )
54. ;he are ( aritheti ean $etween 11 and #1. 5opute the su of the pro!ression. Ans: 341 Solution: =
n
[ a1 + an ] 2 11 = [ 11 + #1] 2 = 341
51. ehanial Jn!ineer $ou!ht 24 $oxes of srews for H2200.00. ;here were three t%pes of srews $ou!ht. rew osts H300 per $ox, srew osts H1#0 and srew 5 osts H#0 per $ox. Cow an% $oxes of srew did he $u% Ans: 2 Solution: xno. of $oxes of srew %no. of $oxes of srew 9 no. of $oxes of srew 5
55. n aritheti pro!ression starts with 1, has ( ters and the iddle ter is 21. Eeterine the su of the first ( ters. Ans: 189
= 24 v 300x + 1#0% + #09 = 2200 6x + 3% + 9 = 44 x + % + 9 = 24 #x + 2% = 20 ;r% % = # #x + 2 ( #) = 20 x=2 2 + # + 9 = 24 9 = 1' u
x+%+9
Solution: 3rd
4th
a 1
ad
a2d
a3d
#th a4d
6th
'th
a#d a6d
8th
(th
a'd a8d
d=# n
2a + ( n − 1) d 2 ( = 2 ( 1) + ( ( − 1) ( # ) 2 = 18( =
300x + 1#0% + #09 = 2200
+ 1#0 ( #) + 1' ( #0) = 2200 2200 = 2200
52. Ealisa% 5orporationAs !ross ar!in is 4# of sales. Kperatin! expenses suh as sales and adinistration are 1# of sales. Ealisa% is in 40 tax $ra7et. "hat perent of sales is their profit after taxes Ans: 18
Page 6 of 13
2nd
Giddle ter a 4d 21 = 1 + 4d
5he7 300 ( 2)
1st
56. If the su is 220 and the first ter is 10, find the oon differene if the last ter is 30. Ans: 2 Solution:
Mathematics Enhancers
=
n 2
Ans: 7
( a1 + an )
220 =
n 2
Solution: x + 2 = 16 − x
( 10 + 30)
2x
n = 11
57. Eeterine the su of the = 2 + # + 8 + 11.....with 100 ters Ans: 15050
61. Eeterine the positive value of x so that x,x 2# and 2x will $e a haroni pro!ression. Ans: 3 followin!
series
Solution: 1
n
2a + ( n − 1) d 2 100 2 ( 2) + ( 100 − 1) ( 3 ) = 2 = 1#0#0
Solution: n=6 a r n−1 = '2(
n
a r# 3r #
= '2( = '2(
= 243 r=3
)
(
a rn − 1 r −1 3 ( 3)
=
6
− 1 3 ( '28) =
3 −1 = 10(2
59. ;he su of the three nu$ers in H is 33, if the su of their squares is 461, find the nu$ers. Ans: 4,11,18 Solution:
Solution: 2x + ' 10x − ' x
=
2x + '
2
( 2x + ' ) = 10x2 − 'x
a = 2nd no.
6x2
− 3#x − 4( = 0 ( x − ') ( 6x + ' ) = 0 x='
a + d = 3rd no. a − d + a + a + d = 33 a = 11 2 2 2 ( 11 − d) + ( 11) + ( 11+ d) = 461
64. Find the value of x fro the !iven >eoetri Hro!ression 1 2 4 , , ... # x 4#
= 461
= (8
d='
Ans: 15
11 − ' = 4
Solution:
11 = 11 11 + ' = 18 ;he nu$ers are 4,11,18 60. Find the value of x if the followin! fors a haroni 1 1 1 pro!ression. − , , 2 x 16
Page 7 of 13
2
63. ;he nu$er x, 2x ', 10x + ' for a >.H. Find the value of xAns: 7
a − d = 1st no.
2d
a=3
r#
=
2
1 2
62. ;here are 4 !eoetri ean $etween 3 and '2(, Find the su of the >.H. Ans: 10(2
d=4
121 − 22d + d2 + 121+ 121+ 22d + d2
−
− 4x − 1# = 0 ( x − 3) ( 3x + # ) x=3
2a1 + ( n − 1) d 2 n = 2 ( 4) + ( n − 1) 4 2 n = [ 4n + 4] 2 = 2n ( n + 1) =
1
3x 2
58. Find the su of the first n positive ultiples of 4. Ans: 2nn%1/ Solution: a1 = 4
1
− =
2
− # x 2x x − # 2x − 2 ( x 2 − # ) = x 2 − # − 2x x
Solution: =
='
x
= a1 + ( n − 1) d 30 = 10 + ( 11 − 1) d d=2 an
= 14
Mathematics Enhancers
2/ x
=
1/ #
= 1.01+ 1.1 + 1.21 + 1.331 + ....up to the #0th ter Ans: 1163.91
4 / 4# 2/ x
2
2 = 4 1 x ÷ 4# # ÷ x2 x
Solution: a1 = 1
4# ( # ) ( 4)
=
= 1#
Solution: let x one nu$er x 4 other nu$er = 1' 6 = 102
Solution: a1 = aount of reainin! air
an
2
1.21
r −1
11.1#0 − 1
1.1 − 1 = 1163.(1
−%
=x
( + 3 = x x ÷
2
x 2 − 3x − 18
=0 ( x − 6) ( x + 3 ) = 0 x=6 x = −3 70. ;he su of a !eoetri series is as follows = 1.0 + 1.1 + 1.21 + 1.331 + ...up to the #0th ter Ans: 1163.91
2 3
Solution: a1 = 1
n=6
3
= a1 r n−1
r
#
2 2
= ÷ 3 3 an = 0.088)reainin! air an
2% + 3
66. If one third of the air in a tan7 is reoved $% eah stro7e of an air pup, what frational part of the total air is reoved in 6 stro7es Ans: 0.9122
=
1.331
a1 r n − 1
%+3=x
= 23 x + 4 = 2'
3
1.1
=
For .H.
x
=
1.21
Solution: −3 % = x −3 x% = (
= 1( 8 102 + 2x + 4 = 1#2
1
=
=
69. x and % are positive nu$ers. If x,3,% fors a >.H and 3,%,x fors an LH. Find the value of x. Ans: 6,3
+x+x+4
a1 = 1 −
=
1.1
1 r = 1.1
4
65. ;he aritheti ean of 6 nu$ers is 1'. If two nu$ers are added to the pro!ression, the new set of nu$er will have an aritheti ean of 1(. "hat are the two nu$ers is their differene is 4 Ans: 23 an 27
r
r =
=
1.1
=
1
=
1.21 1.1
ir reoved after 6 stro7es -
1.331 1.21
= 1.1
11.1#0 − 1
1.1 − 1 = 1163.(1
after 6 stro7es*
=
= 1 − 0.088 = 0.(122 67. If a stro7e of a vauu pup reoved 1# of the air fro the ontainer, how an% stro7es are required to reoved (#of the air Ans: 19 Solution: a1 = 8# aount of air left after 1st stro7e a2
= 0.8# ( 8#) = '2.2# aount of air left after 2nd stro7es r
=
'2.2#
8# n −1 an = a r
= 0.8#
= # aount of an = a r n −1 an
# = 8# ( 0.8#)
air left after n stro7es
n −1
n−1
= 0.0#882 ( n − 1) In 0.8# = In ( 0.0#882) n = 18.43 sa% 1( stro7es ( 0.8#)
71. Find the total distane traveled $% the tip of a pendulu if the distane of the first swin! is 6 . and the distane of eah sueedin! swin! is 0.(8 of the distane of the previous swin!. Ans: 300 Solution: a = 6 r
= 0.(8
=
1 − r 6 = 1 − 0.(8 = 300 72. uppose a $all re$ounds one half the distane if falls. If it is dropped fro a hei!ht of 40 fet, how far does it travel $efore oin! to stop Ans: 120 eet Solution:
68. ;he su of a !eoetri series is as follows-
Page 8 of 13
a
Mathematics Enhancers
)n − 2*180 = 16#n
a1
=
n = 24
a − r 1 a1 = ( 40) ( 2) 2 a1 = 40
77. Find the lar!est an!le of a trian!le if the su and differene of two an!les are 100 and 20 de!rees, respetivel%.
a1
=
1 − r 40 = 1 1− 2
Ans: 80 erees
= 80
Solution: &et-
= one an!le % = seond an!le x
'otal istance the all has traele 80 40 120 feet
;hen, x + % = 100
73. poliean is pursuin! a thief who is ahead $% '2 of his own leaps. ;he thief ta7es 6 leaps while the poliean is ta7in! # leaps, $ut 4 leaps of the thief are as lon! as 3 leaps of the poliean. Cow an% leaps will eah a7e $efore the thief is au!ht Ans: 648 lea*s 540 lea*s Solution: &etx = additional leaps of the thief # x = nu$er of leaps of the poliean 6 % proportionx + '2 4 = # 3 x 6 x = 648 leaps of the thief # 6
x
= #40 leaps of
x − % = 20
olvin! the two equations siultaneousl%, we !et-
= 60 % = 40 x
;hus, the lar!est an!le is the third an!le-
= 180 − ( 60 + 40) = 80° 78. ;he two orrespondin! sides of two siilar pol%!on is 4-6 If the perieter of the lar!er pol%!on is 48 , what is the perieter if the saller pol%!on Ans: 32 cm. Solution: 4 x = 6 48 4 ( 48) x= 6 x = 32.
the polie an
74. Find the an!le whose suppleent exeeds # ties its opleent. Ans: 67.5o
79. ;he orrespondin! sides of two siilar pol%!on is 4-#. If the area of the saller pol%!on is #6 sq. ., what is the area of the $i!!er pol%!on Ans: 87.5 s.cm. Solution:
Solution: &et-
= an!le 180 − x = sup pleent (0 − x = opleent x
;hen, 180 − x = #)(0 − x* x
= 6'.#°
75. Find the an!le whose suppleent exeeds 6 ties its opleent $% 20o. Ans: 76o
= an!le
180 − x = sup pleent
;hen, )180 − x* − 6)(0 − x* = 20 x
=
#6 x
#6 ( #)
2
x
=
x
= 8'.# sq..
( 4)
2
80. ;he sides of a penta!on easures 6,#,2,8 and 4 . respetivel%. ;he shortest side of a siilar penta!on is 1 eter, find the perieter of this penta!on. Ans: 12.5 m.
1
= '6°
76. Cow an% sides has an equian!ular pol%!on if eah of its interior an!les is 16# de!rees Ans: 24 sies &etn = nu$er of sides
= nu$er of interior an!les Page 9 of 13
( #)
2
atio of orrespondin! saller sides ratio of perieter
(0 − x = opleent
;hen,
2
Solution: Herieter of !iven penta!on = 6+ #+ 2+ 8+ 4 Herieter of !iven penta!on & 25 m.
Solution: &etx
( 4)
=
H
2 2# H = 12.# . )perieter of the saller penta!on* 81. ;he orrespondin! sides of two siilar pol%!ons is 2-4. Eeterine the ratio of the area to the perieter of the $i!!est pol%!on if the sallest pol%!on has a perieter of 24 . and an area of 36 sq. . Ans: 3 Solution:
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2 24 = 4 x x = 48 . ( perieter of $i!!est pol%!on)
( 2)
2
( 4)
2
=
86. n en!ineer plaes his transit alon! the line tan!ent to the irle at point suh that H 200 . Ce loates another point on the irle and finds H 80 . If a third point 5, on the irle lies alon! H, how far fro point will it $e Ans: 420 m.
36
= 144 area of
Solution: 200
$i!!est pol%!on
atio of area to perieter = atio of area to perieter
=
144
5 = #00 − 80 5 = 420 . 87. irle havin! a enter at point K has a radius of 20 . trian!le 5 is insri$e in a irle. If the an!le 5 30 o and 5 #0o opute the area of the trian!le. Ans: 301.73 m2 Solution:
Solution:
2
2
=
80
H5 200 H5 = #00
48 3
82. even re!ular hexa!ons, eah with 6 . sides are arran!ed so that the% share the sae sides and the enters of the six hexa!ons are equidistant fro the seventh entral hexa!on. Eeterine the ratio of the total area of the hexa!ons to the total outer perieter enlosin! the hexa!ons. Ans: 6.06
=
2
( 6) in60 ( 6 ) ( ')
5 = 3(.3( .
2
= 6#4.'2 2 H = 6 ( 18) H = 108
atio = atio =
rea of trian!le =
83. hord is 36 . lon! and its id point is 36 . fro the idpoint of the lon!er ar. Find the radius of the irle. Ans: 22.5
Solution: 2θ + 60 = 180 2θ = 120
x
θ = 60 10 + r = 2r r = 10
=( 2r = ( + 36 r = 22.# . x
84. irular pla% area is $ein! laid out in a field. ;he point H, and M have $een found suh that H24 ., HM66 ., and 12 . If ; is to $e another point on the irle, how far fro will it lie Ans: 96 m. Solution: 12x
2 in 100o
88. ;he entral irle has 10 . radius. ix equal saller irles are to $e arran!ed so that the% are externall% tan!ent to the entral irle and eah tan!ent to the adDaent sall irle. "hat should $e the radius in . of eah sall irle Ans: 10
H 6#4.'2
= 18 ( 18)
2 ( 3(.3() in 30o in #0o
rea of trian!le = 301.'3 sq..
108 atio = 6.06
Solution: x ( 36)
2
( 5) = ( 20 ) + ( 20 ) − 2 ( 20 ) ( 20 ) 5os 160o
—
= 42 ( 24)
= 84 ; = 84 + 12 ; = (6 . x
89. ;he distane $etween the enter of the irles whih are utuall% tan!ent to eah other externall% are 10,12 and 14 units, ;he area of the lar!est irle isAns: 64 π
Solution: u r1 + r3
= 10 v r1 + r2 = 12 wr2 + r3 = 14 u
and v
− r2 = −2 r3 + r2 = 14 2r3 = 12 r3 = 6 r2 = 14 − 6 = 8 r1 = 10 − 6 = 4 r3
85. ;wo seants 5 and J have len!ths of 80 . and 100 . respetivel% are drawn fro point outside a irle and intersets the irle at points 5,, E and J. If the an!le $etween the two seants is 2# o, and is equal to 40 . opute the len!th of E. Ans: 32 m.
π ( 8) 2 rea of lar!est irle 64 π rea of lar!est irle
Solution: J = E 5 ( ) ( 5) 40 ( 80)
= E ( J )
= E ( 100)
E = 32 .
Page 10 of 13
90. ;he area of a irular setor is 800 2. If the radius is equal to 16 , opute for the len!th of ar Ans: 100 m. Solution:
Mathematics Enhancers
= θ
θ
=
=
=
94. ;he ratio of the volue to the lateral area of a ri!ht irular one is 2-1. if the altitude is 1# ., what is the ratio of the slant hei!ht to the radius. Ans: 5:2
π2 360o
πθ2 2π θ2
Solution: = πr&
2
B
3 B 2 = 1 B = 2
θ ( 16) 2
800 =
2
θ = 6.2# = θ = 16 ( 6.2# ) = 100 .
2 =
91. setor is $ent to for a one. If the an!le of a setor is 30 de!rees and radius of 6 ., what is the altitude of the one Ans: 5.98 cm.
6 ( 30)
π
r & r &
= 2π
π = 2π
r
1 2
92. Find the volue of a one to $e onstruted fro a setor havin! a diaeter of '2 . and a entral an!le of 1#0o. Ans: 7711.82 Solution: = θ 36 ( 1#0)
= ( 36 ) 2 − ( 1#) 2 h = 32.'3 =
B2
Solution: = πr& 40π = π ( 40) &
Page 11 of 13
=4 =
( 8)
3
h3
( 8)
3
h3 h = #.04 .
3
93. ;he lateral area of a ri!ht irular one is 40 $ase radius is 4 . "hat is the slant hei!ht Ans: 10
& = 10
4=
π ( 1#) 2 ( 32.'3)
3 B = ''11.83
2
95. >iven a solid ri!ht irular one havin! a hei!ht of 8 . has a volue equal to 4 ties the volue of the saller one that ould $e ut fro the sae one havin! the sae axis. 5opute the hei!ht of the saller one. Ans: 5.04 cm.
B1
h2
=
=
B2
= 1#
B
6 1# = 6 #
B1
(4.2# = 2πr
πr 2h
h
Solution: B1 = 4B2
π
180 = (4.2# r
=
atio o slant heiht to raius & 5:2
2
1 h = ( 6) − ÷ 2 h = #.(8 . 2
=
rh
& 6
&
180
2
πr =
3 6
rh & 6& = rh
=π
=
πr =
πr 2h
=
Solution: = r θ =
πr 2h
=
π sq.. ;he
96. irular one havin! an altitude of ( . is divided into 2 se!ents havin! the sae vertex. If the saller altitude is 6 ., find the ratio of the volue of sall one to the $i! one. Ans: 0.296 Solution: B1 B2
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=
( 6)
3
( ()
3
= 0.2(6
97. ;he hei!ht of a ri!ht irular one is NhO. If it ontains oil and water at equal depth of h/2, what is the ratio of the volue of the volue of oil to that of the water. Ans: 7
π ( 32) 2 ( #4) 3
= π ( 24) 2 h
h = 32 in.
Solution: B1
( h / 2)
=
B2
3
101. p%raid has a trian!ular $ase whose sides are 10 ., 14 . and 18 . respetivel%. 5opute the volue of the insri$ed one if it has an altitude of 28 . Ans: 323.19
3
h
B 8 B2 = B − B1 ( vol. of oil ) B1 =
B2
Bol. of one = Bol. of 5%l.
=
B2
=
B1 B2
=
( s − a) ( s − $ ) ( s − ) 10 + 14 + 18
= 21 2 − a = 11, − $ = ', −
'
B 8 '/8B
=3
= 2 ( 11) ( ' ) ( 3 ) = 6(.6# sq.. = r 6(.6# = r ( 21) r = 3.32 . πr 2h π ( 3.32) ( 28 ) B= =
B /8
='
B1
=
B 8
= B−
B2
Solution:
98. plane is passed parallel to the $ase of a trian!ular p%raid of altitude of ( . suh that the area of the $ase is ( ties the area of the trian!le of intersetion. Cow far fro the vertex does the plane interset the altitude Ans: 3 m.
3
3
= 323.1( u..
B
Solution: 1
=
2 1
( h)
2
( ()
2
2
=h
(1 81 h=3 .
99. ;he volue of a one havin! an inlined axis at an an!le of 60o with the $ase is equal to 1884.(6 u.., find the len!th of the axis of the one, if the radius at the $ase is 10 . Ans: 20.78 m. Solution: B
=
h 3
π ( 10) 2
1884.(6 =
3
Solution: h
&en!th of axis-
=
h x
18
x
=
x
= 20.'8 . ( axis len!th)
(
100. ;he radius of the $ase of a one of revolution is 32 inhes and its altitude is #4 inhes. "hat is the altitude of a %linder of the sae volue whose diaeter of the $ase is 48 inhes Ans: 32 in.
=
4
πr 3 ( ( ) = (1'.13
103.;he $ases of a ri!ht pris is a hexa!on with one of eah side equal to 6 . If the volue of the ri!ht pris is #00 u. ., find the distane $etween the $ases. Ans: 5.35 cm. Solution: B = h #00 =
6 ( 6) in 60o
h = #.3#
Page 12 of 13
)
B
3 r = 2.(0 .
in 60o
Solution:
h
$ + + $ 3 24 B= 20 + 60 20 ( 60) 3 B = (1'.13 u..
h = 18 in 60o
102. solid !old in the for of a frustu of a p%raid has saller $ase area of 20 2 and a $i!!er $ase area of 60 2 . It has an altitude of 24 . If the !old is elted to for a ( spherial $alls, find the radius of eah $alls. Ans: 2.90 cm.
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2
( 6) h
104.;he $ases of a ri!ht pris is a hexa!on with one of eah side equal to 6 . ;he $ases are 12 . apart. "hat is the volue of the ri!ht pris. Ans: 1122.4 cu. cm. Solution: B = ase x hei!ht
θ=
360 6
ase =
= 60 1 6 6 in 2 ( ) ( )
60 60
ase = (3.#3
= (3.#3 ( 12) B = 1122.4 u. . B
Page 13 of 13
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