qas inotes

8/19/2019 Qas Inotes

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Smart Instructor Note | BPO

QAS-1

Home Work Assignment Time: 20 minutes

Direction: Choose the correct alternative

1. The Difference between largest and smallest of the following

13 5 3 16 7, , , ,16 8 4 29 12

is

a!

1

4

b!

1131

1392

c!

121

464

d!

3

4

e!

768

1392

2. 334.41 " 47.26 " 1.25 " 5 " #.66 $%

a! 411.24 b! 396.15 c! 388.58 d! 376.85 e! &one of these

3.

1 1 3 2

%2 4 4 3+ + + =

a!

12

5

b!

1

16

c!

12

16

d!

12

6

e! &one of these

4. 11.#93 ' #.##1 $%

a! #.##11#93 b! #.#11#93 c! #.11#93 d! 1.1#93 e! &one of these

5.

13 3 111 3 1 %

32 4 14+ + =

a!

832

224

b!

853

224

c!

812

224

d!

2115

224

e! &one of these

6. (f

9) $ * " 32

5

and ) $ 39.7, what is +ale of *%

a! 2 b! 2.3 c! 4.# d! 3.2 e! &one of these

7.

2345"1983"1#28! 47 %÷ =

a! 1## b! 95 c! 12# d! 115 e! 12#

8. (f

11.644

#.6#84=

then

1

1.644=

a! #.8 b! #.9 c! #.7 d! 1.1 e! &one of these

9.

( )2

4.36 66.5 - 371 %× =

a! 9#2 b! 893 c! 871 d! 888 e! 88#

1#. (f

31.36 5.6= then

#.##36 #.1539 #.#625 %+ − + =

/mart0re dcation 0+t td. 1.# 0age 1



a! 5.6 b! #.##56 c! #.56 d! 56# e! 56

11.

3.5 4 2.6 5 3

7.8 5 5.5 4 %

× − ×=

× − ×

a! 34 b! 28 c! 17 d! 51 e! &one of these

12.

2 2 5 11 1 of %

9 5 21 3

÷ + = ÷

a! 3 b!

23

3

c! 4 d! 12 e! &one of these

13. #.85 of 4#5! " 2.25 of 55#! $%

a! 15.7175 b! 115.715# c! 13.8275 d! 14.7134 e! &one of these

14. % of 398 " 31 of 993 $ 4#3.35


15.

#.9 of 65# #.4 of 75#! 368.25 %÷ + =

a! 38#.15 b! 37#.2# c! 369.25 d! 375.2# e! &one of these

16. :hich of the following nmber is e;actl< di+isible b< 88%

a! 575412# b! 559#464 c! 345654 d! 121121 e! 234422

17. nmber is when di+ided b< 437 lea+es remainder as 19. :hen the same nmber is di+ided b< 123 the

remainder is 5#. (f the =otient in second case is 39 more than 3 times the =otient in first case, thenthe nmber is

a! 34#15 b! 31#46 c! 25648 d! 56418 e! 53462

18. nmber when sccessi+el< di+ided b< 2, 3 and 4 lea+es the remainders 1, 2 and 3 resecti+el<. Thenmber is


19. (f 546>342 is di+isible b< 9, then what smallest digit is there in lace of >%

a! 1 b! 2 c! 3 d! 4 e! 5

2#. :hich one of the following will di+ide

34 35 36 374 4 4 4+ + +

comletel<%a! 5 b! 7 c! 9 d! 1# e! 11




Solution to Home Work Assignments

1. Correct Option: (c)

13 5 3 16 7, , , ,

16 8 4 29 12

is e=i+alent to

1131 87# 1#44 768 812, , , ,1392 1392 1392 1392 1392

1131 768 363 121

1392 1392 1392 464∴ − = =


% $ 334.41 " 47.26 " 1.25 " 5 " #.66 $ 388.58

3. Correct Option: (d)

1 1 3 2 13 1% 2

2 4 4 3 6 6= + + + = =

4. Correct Option: ()

11.#93 ' #.##1 $ #.#11#93

5. Correct Option: (e)

13 3 111 3 1

32 4 14+ + ( )

13 3 111 3 1

32 4 14

= + + + + + ÷

91 168 176 211

5 6224 224

+ + = + = ÷

.


)rom the gi+en e=ation

( )5 ) - 32*

9=

( )5 39.7 - 3 24.28 4

9= = ≈


( )2345 1983 1#28 47

5356 47 114 115

+ + ÷

= ÷ ≈ ≈

8. Correct Option: (!)

11.644

#.6#841

#.6#841.644

1#.6#84 #.78 .8

1.644

=

⇒ =

⇒ = = ≈


( ) ( )2

% 4.36 66.5 - 371

19 66.5- 371

1264 -371 893

= ×

≈ ×

≈ ≈

1#. Correct Option: (c)

?i+en

31.36 5.6=




#.##36 #.1539 #.#625

#.##36 #.1539 #.25

#.##36 #.#961

#.##36 #.31 #.56

+ − +

= + − +

= +

= + =


et the missing nmber be x

3.5 4 2.6 5 3

7.8 5 5.5 4 x

× − ×=

× − ×

14 13 3

39 22 x

−⇒ =−

1 3

17 x ⇒ =

∴ x $ 51


11 7 5 1% of

9 5 21 3

= ÷ + ÷

11 1 1 11 3 1 11 1 124

9 3 3 9 1 3 3 3 3

= ÷ + = × + = + = = ÷ ÷


( ) ( )

( )

#.85 4.#5 2.25 5.5

3.4425 12.375 15.8175

= × + ×

= + =


4#3.35 3#7.83% 24

3.98

−= =


( )

( )

( )

#.9 of 65# #.4 of 75# 368.25

#.9 6.5 #.4 7.5 368.25

5.85 3 368#25 1.95 368.25 37#.2

÷ +

= × ÷ × +

= ÷ + = + =


(f the nmber is di+isible b< 8 and 11,

then it is di+isible b< 88.

@ere, onl< 559#464 is di+isible b< 8 and 11.


et the nmber be x and =otient be y when di+ided b< 437.

:hen the nmber is di+ided b< 123 =otient will be 3y "39

x $ 437y " 19 and x $ 1233y " 39! " 5#

437y " 19 $ 1233y " 39! " 5#

437y " 19 $ 369y " 4797 " 5#

437y - 369y $ 4847 - 19

68y $ 4828 and y $ 71

x $ 437 ' 71 " 19 $ 31#46


et x be the nmber.Then x $ 2 p " 1, where is the =otient. ...1

nd p $ 3r " 2, where r is the =otient. ...2

r $ 4s " 3, where s is the =otient. ...3




/ol+ing e=ation 1, 2, 3, we get x $ 24s " 23

:hen s $ 1, we get x $ 47 which is in otion c!


5 " 4 " 6 " x " 3 " 4 " 2 $ 24 " x is di+isible b< 9.

:hen x $ 3.27 is di+isible b< 9..

2#. Correct Option: (d)434 " 435 " 436 " 437

$ 4334 " 16 " 64 " 256! $ 433 ' 34#.

34# is di+isible b< 1#.

QAS-2


1. The @*) of two nmbers is 9 and their *A is 22338. (f one nmber is 3#6, then the other nmber is

a! 567 b! 2#8 c! 1#4 d! 657 e! &one of these

2. )ind the smallest nmber of for digits which when di+ided b< 8, 9, 12 and 15 lea+es remainder of 3.

a! 1#8# b! 1#83 c! 11#4 d! 1#28 e! &one of these

3. )or different digital wrist watches maBe a bee after e+er< 15 mintes, 3# mintes, 45 mintes and 9#mintes resecti+el<. ll the watches beeed together at 1# 0A. The< will again bee together at

a! 11.15 0A b! 12.15 0A c! 11.3# 0A d! 12.45 0A e! &one of these

4. Three nmbers are in the ratio of 3 4 5 and their *A is 24## find the sm of the nmbers%

a! 6# b! 48# c! 4# d! 24## e! &one of these

5. The least mltile of 9, which lea+es a remainder of 3, when di+ided b< 7, 15, 21 and 35 is


6. The *A of two rime nmbers a and b a C b! is 341. The +ale of 3b - a - 2 is

a! 2 b! 11 c! 1# d! # e! &one of these

7. The sm of two nmbers is 132 and their *A is 36#. )ind the nmbers%

a! 6#, 72 b! 5#, 84 c! 48, 72 d! 36, 6# e! &one of these

8. )ind the +ale of

!b c a

x −

× !c a b

x −

×

!a b c

x −

a! 1 b! # c!

abc x

d!

a b c x + +

e! &one

9.

3 22 3 5 6

328 9 1#

a a a a

aa a

× × ×

× × $%

a! 5 b! 3 c! # d! 1 e! &one

1#.

3 2 12 2 2 16## x x x + + −+ + =, )ind the +ale of x .

a! 8 b! 9 c! 1# d! 7 e! &one




11. )ind the +ale of x in the e;ression

33# 6 x + =

a! 36 b! 6 c! 72 d! 216 e! &one

12. )ind the +ale of

352

3 12 9 3 !

−

× +

a! 1 b! 3 c! # d! 9 e! &one of these

13. /imlif< the e;ression

( ) ( ) ( )3 137 3 9 x x x x x ×

a! x b!

2 x

c!

3 x

d!

1

2 x

e! &one of these

14. (f

3 9 27a b c = =and

1 1 1 25

3 2 4 24+ + =a b c

.)ind a.

a! 2 b! 3 c! 4 d!

2

3

e! &one

15. The +ale of

−2 23.47! 2.53!

#.94

a! 8 b! 3.47 c! 2.53 d! 6 e! 5

16.

2.47 2.47 2.47 1.53 1.53 1.53 3 2.47 2.47 1.53 3 2.47 1.53 1.53

2.47 2.47 2 2.47 1.53 1.53 1.53

× × − × × − × × × + × × ×× − × × + ×

is e=al to

a! 4 b! #.94 c! #.84 d! 3 e! &one of these

17. (f

3

7

a

b=

, find the +ale of

2 2

2 2.

a b

a b

+−

a!

3

7b!

29

2#−

c!

−5

2

d!

1#

3

e! &one of these

18. (f

3.671 3.6712.6

3.671

b b

b

× − × =+

, find the +ale of b%

a! #.6271 b! 1.71 c! 2.#71 d! 1.#71 e! &one of these

19. )ind the +ale of

( ) 2

2.6 2.6 1.4 1.4 2 2.6 1.4

2.6 2.6 2.6 1.4 1.4 1.4 3 2.6 2.6 1.4 3 2.6 1.4 1.4

× + × − × ×

× × − × × − × × × + × × ×

a! 1.2 b! 4 c! .5 d! .2 e! &one of these

2#. )ind the +ale of

( ) ( )

( ) ( ) ( )

3 3

2 2

3.2 2.3 3.2 2.3

3.2 2.3 3.2 2.3 3.2 3.2 2.3 2.3

+ + −

+ + − − × − ×




a! 6.4 b! 4.6 c! 5.5 d! .9 e! &one of these

Solution to Home Work





*A ' @*) $ 0rodct of the nmbers

et the other nmber be x

22338 ' 9 $ 3#6 ' x

x $ 657


*A of 8, 9, 12, 15 $ 36#./mallest nmber of 4 digits, which is a mltile of 36# is, 36# ' 3 $ 1#8#

/o, the re=ired nmber $ 1#8# " 3 $ 1#83.


(nter+al after which all for watches bee together $ *A of 15, 3#, 45, 9# $ 9# Aintes.

/o, the watches will again bee together 9# mintes after 1# 0A, i.e. at 11.3# 0A.


et the nmbers be 3 x , 4 x , and 5 x

Then their *A $ 3 ' 4 ' 5 ' x $ 6# x

/o, 6# x $ 24## or x $ 4#

&mbers are 12#, 16#,2##. /m $ 48#


*A of 7, 15, 21 and 35 is 1#5.

et re=ired nmber be 1#5m " 3, which is mltile of 9.

east +ale of m for which 1#5m " 3! is di+isible b< 9 is 4.

/o, the re=ired nmber is 1#5 ' 4 " 3 $ 423.


@*) of two rime nmbers is 1.

et the nmbers be m and n, Then mn $ 341

&ow, since the nmbers are rime ossible air is 31, 11!

/ince a and b are rime and

a C b, /o a $ 31 and b $ 11.

⇒3b - a - 2 $ 3 ' 11 - 31 - 2 $ #.

7. Correct Option: (!)The @*) of the sm and *A of the two nmbers is also the @*) of the nmbers.

@*) of 132 and 36# $ 12

&ow, let the nmbers be 12 x and 12y

/o, 12 x " 12y $ 132

12 x " y ! $ 132

x + y = 11

gain, *A of 12 x and 12y $ 12 xy

. 12 x < $ 36#

⇒

;y $ 3#

⇒ x $ 5 and y $ 6

Therefore, the re=ired nmbers, 12 x and 12y, are

12 ' 5 $ 6# and 12 ' 6 $ 72


ab bc bc ab ac bc x x x − − −× ×

ab ac bc ab ac bc x − + − + −

⇒

# 1 x =

9. Correct Option: (d)3 2

3 3

2 3 5 2 3

2 3 2 5

a a a a a

a a a a

× × × ×× × ×

4 3

4 3

2 3 51

2 3 5

a a a

a a a

× ×=

× ×

1#. Correct Option: (d)




3

3 2

2 12 2 2 16

22 2 2 2 16

##

##2

x x

x x

x

x + + −

× + × + =

+ + =

12 8 4 16##

2

252 16##

2

x

x

× + + = ÷

× = ÷

22 16##

25

x x =

7

2 128

2 2

7

x

x

x

=

==


33# x +

$ 6

( )2

233# 6 x + =

/=aring both the sides!

⇒ 33# " 36 x =

⇒ 3 36 3# x = −

⇒ 3 6 x =

TaBe cbe on both sides

⇒ ( ) ( )3 33 6 x = ⇒

x $ 216


bo+e e;ression becomes

12

3 2

523 2 3 3

+−

÷ ÷× + ÷ ÷

( )3 3

2 2

3 3#2 2

3 2 3 3 3 3 3

3 3 3 1

− −

−

= × + ×

⇒×⇒ =

⇒


( ) ( ) ( )3 1337 9 x x x x x ×

3 13 9 137

2 3 2 x x + +

× ×

r

277

2 92 7 x x ×× ×1 3 1 3

22 2 2 2 x x x x +

= × = =


23 9 or 3 3a b a b= =

2a b∴ =




gain

3 27a c =

33 3a c =

∴ a $ 3c ⇒

1 1 1 25

3 2 4 24a b c + + =

1 1 1 25

3 244

3

aa a+ + =

×

1 1 3 25

3 4 24a a a+ + =

4 12 9 25

12 24a

+ +=

⇒25 25

12 24a=

⇒ a $

25 24

12 25

××

∴ a $ 2


/bstitte 3.47 $ a and 2.53 $ b

?i+en e;ression becomes

2 2a ba b

a b

−= +

−

⇒

3.47 " 2.53! $ 6




( )

( )

3

2 !

a ba b

a b

−= −

−

2.47 1.53! #.94⇒ − =

17. Correct Option: ()2

2 2 2

22 2

2

2 2

2 2

1

?i+en e;.

1

3 91 1 17 49

93 11 1 497

9 49 5858 49 58 2949 49

9 49 4# 49 4# 4# 2#

49 49

a

a b b

aa b

b

a

b

a

b

++= =

−−

+ + + ÷ ÷

= = = −− − ÷ ÷

+−

= = = × = − = −− −

Sm!rt trick TaBe a $ 3 and b $ 7


/bstitte 3.671 $ a

.@./. becomes

2 2

a b

a ba b

−= −

+

,3.671 2.6b− =

⇒b $ 1.#71







( ) ( )

( )

22 2 4

3 3 2 2 3

2

3 3

a b ab a ba b

a b a b ab a b

+ − −= = −

− − + −

2.6 E 1.4 1.2⇒ =

Fsing

2 2 22a ab b a b− + = ( − )

and

3 3 2 2 33 3a b a b b a a b− − + = ( − )

2#. Correct Option: (!)

/bstitte 3.2 " 2.3 $ a and 3.2 - 2.3 $ b


( ) ( )

( ) ( ) ( )

3 3

2 2 !

a ba b

a b ab

+= ( +

+ −

⇒

3.2 " 2.3 " 3.2 - 2.3! $ 6.4

Fsing

3 3 2 2a b a b a ab b+ = ( + ) × ( − + )

QAS-"


1. The a+erage age of 1## emlo<ees in an office is 31 <ears, where

th2

5

÷

o# emplo$ees !re %omen!nd the ratio of the a+erage ages of women to men is 5 7. The a+erage age of women is

a! 18 <ears b! 25 <ears c! 35 <ears d! 3# <ears e! &one of these

2& T'e !er!ge income o# A * !nd C is +s 1,000 per mont' !nd !er!ge income o# * C !nd is+s 1.000 per mont'& /# t'e !er!ge income o# A is t%ice t'!t o# t'en t'e !er!ge s!l!r$ o# *!nd C is:

(!) +s 1"00 () +s 21000 (c) +s .000 (d) +s 12000 (e) one o# t'ese

"& T'ere is t%ice t'e numer o# %'eelers !s t'ere !re " %'eelers !nd tot!l numer o# 2 %'eelersis s!me !s tot!l numer o# %'eelers& T'e !er!ge numer o# %'eels per e'icle is:

(!) " () (c) (d) .

(e) C!n3t e determined

& T'e !er!ge o# 2. 24 n " !nd " lies et%een 2 !nd "& /# n is !l%!$s !n integer !nd gre!tert'!n t'e !er!ge o# t'e gien integers t'en t'e !lue o# n lies et%een:

(!) "" 5 n 5 6 () " 5 n 5 " (c) "" 5 n 5 2 (d) "0 5 n 5 0 (e) one o# t'ese

& T'e !er!ge !ge o# Adul 'is %i#e !nd 2 c'ildren is 22 $e!rs Adul3s %i#e is 7ust $e!rs $ounger to 'im !nd s'e %!s 2 %'en 'er d!ug'ter %!s orn& Adul %!s "2 %'en 'is son %!s orn& T'e!er!ge !ge o# Adul !nd 'is d!ug'ter is:

(!) 2 $e!rs () 22 $ears c! 26 <ears d! 24 <ears e! 2# <ears




6. (f 3# of a nmber is sbtracted from a second nmber, the second nmber redced to its threeEforth.:hat is the ratio of the first nmber to the second nmber%

a! 6 5 b! 2 3 c! 5 6 d! 3 1# e! &one of these

7. Difference between 2 nmbers is 85. 8# of smaller is 4 less than 4# of the larger the sm ofnmbers is

a! 325 b! 435 c! 235 d! 545 e! 425

8. (f the length of a rectangle increases b< 6# then b< what shold the breadth be diminished so thatthe area of the new rectangle is half of old area%

a! 68

3

4

b! 86

3

4

c! 68

1

4

d! 86

1

4

e! &one of these

9. The rice of a motorc<cle is Gs 57###. (t was insred to 8# of its rices. (n an accident, the motorc<clewas damaged comletel< and the insrance coman< aid 9# of the insrance. :hat was thedifference between the rice of the motorc<cle and the amont recei+ed%

a! Gs 1596# b! Gs 41#4# c! Gs 98#4# d! Gs 64125 e! &one of these

1#. Tara boght 32 ales for Gs 48 and then sold them at the rate of Gs 36 for 16 ales. :hat is her rofitercentage%

a! 75 b! 72 c! 5# d! 65 e! &one of these

11. :hen the selling rice of a watch is Gs 13#, the loss ercentage is 25. :hat is the loss or gainercentage if selling rice is increased to Gs 182%

a! 5 loss b! 2# rofit c! 5 rofit d! 2# loss e! &one of these

12. bsinessman sells goods at a rofit of 1#. @e ses a false balance that weighs 2# less. @is totalercentage gain is

a! 37.5 b! 3# c! 2# d! 47.5 e! &one of these

13. to< seller cheats to the e;tent of 2# while b<ing as well as selling, b< sing false weights. @is total

gain is a! 244 b! 44 c! 144 d! 21 e! &one of these

14. dealer sold an article at a loss of

12 .

2

(f he had sold it for Gs 1## more, he wold ha+e gained 7.

To gain

112 ,

2 in how mch rees shold he sell it%

a! Gs 1184.21 b! Gs 1#52.63 c! Gs 1### d! Gs 2### e! &one of these

15. dishonest milBman mi;es 12 water in re milB and again mi;es 15 more water in the re+iosmi;tre. :hat is the rofit ercentage of milBman if he sells it at *0%

a! 15 b! 12 c! 28.8 d! 27 e! 18#

16. mrtham marBs the rices of sweets b< 4# and he sold onl< 3# of those at this rices. @e sells4# of the rest at 2# discont and rest 25 discont. :hat is the net rofit made b< him%

a! 31.36 b! 42 c! 44.1 d! 17.46 e! &one of these

17. )ind the difference between *( and /( on Gs 17## in+ested at 16 annall< at both simle andcomond interest for 2 <ears.

a! Gs1#.88 b! Gs 43.52 c! Gs 282.88 d! Gs 544 e! &one of these

18. lent Gs 6### to H for 2 <ears and Gs 8### to * for 4 <ears on simle interest at the same rate ofinterest and recei+ed Gs 22## in all from both of them as interest. The rate of interest er annm is

a! 5 b! 6 c! 8 d! 2# e! &one of these




19. :hat will be the ratio of simle interest earned b< certain amont at the same rate of interest for 8 <earsand that for 12 <ears%

a! 3 2 b! 1 3 c! 2 3 d! Data inade=ate e! &one of these

2#. sm of mone< amonts to Gs 98## after 5 <ears and Gs 12##5 after 8 <ears at the same rate of

simle interest. The rate of interest er annm is


Solution to Home Work





+erage of 1## emlo<ee $ 31I Total sm $ 31##

&o. of women $

2 1##4#

5

× =

⇒ &mber of men $ 6#

Gatio of a+erage ages of women and men $ 5 7

et +erage age of women be $ 5 x

@ence, +erage age of man wold be $ 7 x

/m of all ages wold be

⇒

( ) ( )5 4# 7 6# 31## x x × + × =

⇒ ; $ 5

+erage age of women $

5 5×$ 25 <ears


?i+en

( ) H *

3

+ +

$ 18###

⇒

H *+ +$ 54### J1!

H * D

3

+ + ÷

$ 16###

⇒

H * D+ +$ 48### J2!

lso 2D $ then

1! ⇒

2D H * 54###+ + = J3!

/ol+ing e=ation 2! and 3!

⇒ D $ 6###⇒ 2! ⇒ H " * $ 42###

⇒

H *Gs 21###

2

+ = ÷


2wheelers

3wheelers

4wheelers

No. ofvehicles

2 x x 2 x

No. ofwheels

4 x 3 x 8 x

+erage $

( )4 3 8

2 2 !

x x x

x x x

+ ++ +

$ 3


+erage of 26, 29, 35 and 43 is 33.25

?i+en

26 29 35 4325 35

5

125 133 175 8 42

n

n n

+ + + +< <

⇒ < + < ⇒ − < <

Ht gi+en n C 33.25

/o, 33 K n K 42.


Total of ages of 4 $ 422! $ 88 LearsAbdul Wife Daughter Son

28 24 0




32 28 4 0

/m of ages at time of sonMs birth $ 64

/ince sm of ages at resent $ 88

(ncrease in sm of 4 members ages $ 88 - 64 $ 24

(ncrease in each of their ages

24

4=

$ 6 Lears

!resent ages

+erage age of bdl and his daghter

( )36 8

2

+=

$ 22 Lears


et the nmber be and , then

3 3# 33# of of

4 1## 4

x y

y x y y x y − = ⇒ − × = ×

3 34 1#

y y x ⇒ − = 1 3 1# 5 64 1# 12 x y x y ⇒ = ⇒ = =


et the nmbers be x , x " 85.

?i+en

4# 85! 8# 4# 34## 8#4 4

1## 1## 1##

34## 4## 4# 75

85 16# sm 16# 75 235

x x x x

x x

x

+ + −− = ⇒ =

⇒ − = ⇒ =+ = ⇒ = + =

8. Correct Option :( !)

et length and breadth of rectangle be a and b resecti+el<

et new breadth be c

?i+en &ew length $

16# 8

1## 5

a a=

, gi+en new area

2

ab=

8 5

5 2 16

5 11Diminished breadth

16 16

275 3Diminished 68

4 4

ac ab bc

b bb

= ⇒ =

= − =

= =


mont recei+ed $ 9# of 8#

f

9# 8#57### of 57### 41#4#

1## 1##= × =

Difference in the rice of motorc<cle and the amont recei+ed $ 57### - 41#4# $ 1596#


/ose, nmber of ales boght

*A of 32 and 16 32 *0 of 32 ales Gs 48

36/0 of 32 ales Gs 32 Gs72

16

72 480rofit 1## 5#

48

= = ⇒ =

= × =

−= × =





1## 52#*0 13#

75 3

52#182

30rofit on new /0 1## 552#

3

= × =

−= × =


et *0 be Gs x and let the bsinessman b<s y Bg of goods.

Total *0 $ Gs xy I

11 /0 Gs

1#

xy ∴ =

/ince he weighs 2# less, therefore actal weight of y Bg of goods $

4

5

÷

Bg.

4 11 /0 of Bg of goods Gs

5 1#

11 /0 of 1 Bg of goods Gs

8

y xy

x

∴ =

∴ =

3 0rofit in 1 Bg Gs

8

30rofit 1## 37.5

8

x

x

x

∴ =

= × =


Sm!rt Trick:

?ain $

2;2;

1##

+ ÷ ÷

0tting ; $ 2#, :e get, ?ain $ 44


1##*0 1##

9

7 2.5 9.5,

.5

112.5 1##/0 1## Gs 1184.21

1##

9.5 of *0 Gs 1

9.5

##

⇒ = ×

⇒ = × × =

+ = =


et s consider =antit< of milB as 1## l itres

:ater $ 12 $ 12 litres

gain 15 of 112 litres $ 16.28 litres

Total milB $ 1## " 12 " 16.28 $ 128.28 litres

0rofit $ 28.28 litresI 0rofit $

28.281##

1##×

$ 28.8


et there be 1## acBs of sweets with *0 Gs1## each.

?i+en A0 $

14# 1##

1##

×

$ Gs 14#




3#/01 1## 14# Gs 42##

1##

4# 8#/02 7# 14# Gs 3136

1## 1##

75/01 42 14# Gs 441#

1##

= × × =

= × × × =

= × × =

Total /0 $ Gs 42## " 3136 " 441#! $ Gs 11746

Total *0 $ 1##'1## $ Gs1####

0rofit $ Gs 1746I 0rofit ercentage $ 17.46

17 Correct Option: ()

2

17## 2 16/( Gs 544

1##

16*( 17## 1 17## Gs 587.52

1##

× ×= =

= + − = ÷

Difference /( *( 587.52 544 8 43.52d = − = − =

Sm!rt Trick:

1## 1##0 Difference r r = × ×


et the rate be .a.

6### 2 8### 4Then, 22##

1## 1##

22##12# 32# 22## 5

44#

r

r r

r r r

× × × × + = ÷ ÷

⇒ + = ⇒ = =


et the rincial be 0 and rate of interest be G.

0 G 881##

Ge=ired ratio 2 30 G 12 12

1##

× × ÷ ∴ = = =

× × ÷


/( for 3 <ears $ Gs 12##5 N 98##! $ Gs 22#5

/( for 5 <ears $ Gs

22#55

3×

$ Gs 3675

∴ 0rincial $ Gs 98## N 3675! $ Gs 6125

@ence rate $

1## 3675

6125 5

× ÷

× $ 12

QAS-


1. The ratio of incomes of and A is 3 2. The sm of their e;enditre is Gs 8### and the amont ofsa+ing of A is e=al to amont of e;enditre of . :hat is sm of their sa+ings%

a! Gs 12### b! Gs 2#### c! Gs 8### d! Gs 4###

e! *anMt be determined

2. (f x " y ! y " z ! z " x ! $ 8 11 9, x " y " z ! $ 21, then O x " y ! x P $ %a! 7 8 b! 7 5 c! 8 3 d! 7 6 e! 6 5




3. Gatio between two nmbers is 9 1#. (f each of them is increased b< 3, the ratio becomes 19 21, thedifference between their s=ares is

a! 6 b! 19 c! 114 d! 684 e! 36

4. :hat nmber shold be sbtracted from each of the nmbers 16, 29, 21, 39 so that the reslting

nmbers are roortional%a! 1 b! 2 c! 4 d! 5 e! 3

5. The salaries of , H and * are in the ratio 3 4 5. (f the< get 2#, 3# and 4# increment in theirsalaries resecti+el<. )ind the new ratio of their salaries.

a! 19 52 7# b! 18 29 35 c! 18 26 35 d! 18 26 36 e! &one of these

6. The ratio of the nmber of bo<s and girls in a school is 5 2. (f 6# of bo<s and 25 of girls recei+edscholarshi, find the ercentage of stdents those who ha+e not recei+ed the scholarshi.

a! 5 b! 25 c! 5# d! 4# e! &one of these

7. container contains a mi;tre of two li=ids and H in the ratio 8 5. 26 litres of mi;tre is drawn off

and relaced with li=id H. Then the ratio of and H becomes 4 9. @ow mch litres of li=id wascontained b< the container initiall<%

a! 32 b! 2# c! 24 d! 15 e! &one of these

8. (n a icnic team, the ratio of nmber of males to the nmber of females is 3 2 and ratio of nmber ofchildren to the nmber of females is 4 5. (f the nmber of children is 56, then the difference betweennmber of males and females is

a! 7# b! 1#5 c! 175 d! 8# e! &one of these

9. (n what ratio sgar costing Gs 27 er Bg be mi;ed with sgar costing Gs 32 so that the mi;tre worth Gs3#.

a! 2 3 b! 3 2 c! 2 5 d! 5 2 e! 3 5

1#. (n what ratio mst water be mi;ed with milB to gain 25 b< selling the mi;tre at cost rice%

a! 1 5 b! 4 5 c! 1 4 d! 4 1 e! &one of these

11. The cost of ariet< 1 rice is Gs 6# er Bg and ariet< 2 rice is Gs 1## er Bg. (f both +arieties are mi;edin the ratio 5 3, then the rice er Bg of mi;ed +ariet< of rice is

a! Gs 7# b! Gs 75 c! Gs 85 d! Gs 65 e! &one of these

12. (n what ratio mst a grocer mi; two +arieties of tea worth Gs 29.1# er Bg and Gs 62.1# er Bg so thatb< selling the mi;tre at Gs 52.13 er Bg, he ma< gain 3#%


13. Two +essels and H contain sirit and water mi;ed in the ratio 7 6 and 7 3 resecti+el<. )ind the ratioin which these mi;tre be mi;ed to obtain a new mi;tre in +essel * containing sirit and water in theratio 8 5.

a! 11 13# b! 11 1# c! 1# 11 d! 1 11 e! &one of these

14. flasB fll of whisB< contains 64 alcohol. art of this whisB< is relaced b< another containing 29alcohol and now the ercentage of alcohol was fond to be 36. )ind the =antit< of whisB< relaced.

a!

4

5

b!

1

5

c!

1

4d!

4

1e! &one of these

15. 0rofit shared b< and H at the end of a <ear are in the ratio of 4 3. (f HMs eriod of in+estment is twice

that of . :hat is the ratio of in+estments of to H% a! 3 4 b! 4 3 c! 3 8 d! 8 3 e! 2 3




16. (n a bsiness in+ested Gs 6### more than H, HMs caital remains for

17

2

months while Ms caitalremains for 1 month more than HMs. (f total rofit is Gs 52#. t of which gets Gs 16# more than H.)ind Ms caital.

a! Gs 15### b! Gs 9### c! Gs 24### d! Gs 6### e! &one of these

17. 0arl in+ested Gs 9#### in a bsiness. fter few months, nita Qoined her with Gs 27###. t the end ofthe <ear, the total rofit was di+ided in the ratio 8 1. )ind after how man< months nita Qoined her.


18. and H started a bsiness with in+estments Gs 1#### and Gs 6### resecti+el<, while * Qoins themafter 2 months with Gs 8###. withdraws from the bsiness after 1 <ear and * withdraws 4 monthsbefore the comletion. (f the rofit at the end of 2 <ears is Gs 51##. )ind who all gets e=al rofits andhow mch.

a! , * Gs 15## b! H, * Gs 15## c! , H Gs 18## d! H, * Gs 18## e! , * Gs 2###

19. , H and * started a bsiness with some in+estments. (f ratio of their time for in+estments is 12 1# 3and that of rofit earned b< them is 4 5 6 and if in+estment of H is Gs 15###, find the total amontin+ested b< other two.

a! Gs 85### b! Gs 75### c! Gs 7#### d! Gs 55### e! &one of these

2#. (n a bsiness , H, * and D are artners Ms in+estment is thrice of *Ms in+estment, HMs in+estment isoneEthird of DMs in+estment and , H in+ests e=all<. The total rofit at the end of the <ear is Gs 16###.)ind the sm of rofits of and D.

a! Gs 9### b! Gs 6### c! Gs 1#### d! Gs 12### e! Gs 7###

Solution to Home Work Assignments


A

(ncome 3 x 2 x

;enditre k 8### - k

/a+ings &ot Bnown k

8### - 2

4###

k k x

x

+ ==

Total income $ 3 x " 2 x $ 2####

Total e;enditre $ 8###

Total sa+ings $ 2#### - 8### $ Gs 12###


x " y ! y " z ! z " x ! $ 8 11 9, x " y " z ! $ 21

⇒ x " y ! $ 8k , y " z ! $ 11k and z " x ! $ 9k ⇒ 2 x " y " z ! $ 28k

⇒ 42 $ 28B

⇒

42 3

28 2k = =

/o,

3312,

2 x y y z + = + =

and

27

2z x + =

⇒

33 921

2 2 x = − =

⇒ x " y ! x $ 8 3


et nmbers be 9 x , 1# x




?i+en

9 3 19

1# 3 21

6

x

x

x

+=

+⇒ =

/o, the nmbers are 54, 6#

Difference between their s=ares $ 684


et the re=ired nmber be x .

2 2

16 21

29 39

624 55 6#9 5#

3

x x

x x

x x x x

x

− −=− −

⇒ − + = − +⇒ =

5.. Correct Option: (c)

et the salaries of , H and * is 3B, 4B and 5B resecti+el<.

Gatio of new salaries after increment is

3 1.2 4 1.3 5 1.4

3.6 5.2 7

18 26 35

k k k

k k k

k k k

× × ×=

=


et the nmber of bo<s and girls be 5B and 2B resecti+el<.

0ercentage of stdents who ha+e not recei+ed the scholarshi is

4# 755 2

1## 1## 1##5 2

2 1.51## 5#

7

k k

k k

k k

k

× + ××

++

= × =

Sm!rt Trick:

et the nmber of stdents be 5## and 2## then the nmber of stdents who get scholarshi will be 2## and 15# resecti+el<which is half of the total nmber of stdents.

7. Correct Option: (!)/ose the container initiall< contains 8; and 5; of mi;tres and H resecti+el<.

antit< of in mi;tre left $

88 26 litres

13 x

− × ÷

antit< of H in mi;tre left $

55 26 litres

13 x

− × ÷

88 26

13 4

5 95 26 26

13

8 16 45 16 9

4

x

x

x x

x

− × ÷ ⇒ =

− × + ÷

−⇒ =+⇒ =

/o, the container contained 32 litres of .


Aale )emale $ 3 2, *hildren )emale $ 4 5

Aale )emale *hildren $ 15 1# 8

Aale $

15 561#5

8

×=

)emale $

1# 567#

8

×=

Difference $ 1#5 - 7# $ 35




9. *orrect tion a!

H< the rle of alligation

*heaer *ostlier

27 32

3#

32 3# 2 3# 27 3− = − =

" N

N "

Gatio of mi;tre $ 2 3


H< the rle of alligation

et *0 of 1 litre of milB be Gee 1

/0 of 1 litre of mi;tre $ Gee 1, ?ain $ 25

Therefore, *0 of 1 litre of mi;tre $

4

5

H< the rle of alligation,

*0 of water *0 of milB

# 1

4

5

1 4

5 5

1 4Gatio 1 4

5 5

= =

" N

" N

N "

N "

11. Correct Option: ()H< the rle of alligation,

ariet< 1 rice ariet< 2 rice

6# 1##

1## 6#

x

x x

− −

" N

" N

N "

N "

?i+en,

1## 5

6# 375

x

x x

−=

−⇒ =



T<e 1 tea T<e 2 tea

29.1# 62.1#

1##52.13

13#

22 11

×

" N

" N

N "

N "




Gatio $ 22 11 $ 2 1



/irit in /irit in H

7 7

13 1#

8

13

11 1

13# 13

" N

" N

N "

N "

@ence the re=ired ratio is

11 1 111#

13# 13

= ÷ ÷



st nd1 flasB strength 2 flasB strength

64 29

36

7 28

" N

" N

N "

N "

Gatio 7 28 1 4

th4

means of flasBE1 has been relaced b< flasBE25

4 Ge=ired =antit< $

5

= =

÷ ÷ ÷

∴


H

(n+estment x y

Time k 2k

@ence, Gatio of rofit $ xk 2ky $ 4 3

⇒ x 2y $ 4 3

⇒ x y $ 8 3


H

(n+estment x " 6### x

Time 17

2

15

2

0rofit 34# 18#

where x is the in+estment b< H!

( )17 6### 15 34# 18#

2 2

x x +=

⇒ 153 x " 918### $ 255 x /o, x $ 9###

Ms caital $ Gs 15###





et nita Qoins after ; months.

Gatio of rofit wold be

( )9#### 12 8

27### 12 1

4# 8

12 1

7

x

x

x

×=

× −

⇒ =−⇒ =


H *

(n+estment 1#### 6### 8###

Time 12 24 18

0rofit ratio $ 1#### ' 2 6### ' 4 8### ' 3 $ 5 6 6

@ence and H will get the same share of rofit.

The rofit amont is

6 51##Gs 18##

17

×= =


Gatio of in+estments

4 5 6 2 3 12

12 1# 3= =

et the total amont in+ested b< all three be Gs ;.

Then,

315###

17 x × = 85### x ⇒ =

Therefore, the amont in+ested b< and * $ Gs 85### - 15###! $ Gs 7####


et *Ms and DMs in+estments be x and y resecti+el<.

Ms in+estment $ 3 x , HMs in+estment

3

y =

Ht

33

y x =

⇒ y $ 9 x

/o, , H, * and DMs in+estments are 3 x , 3 x , x and 9 x resecti+el<.

(n+estment ratio $ 3 3 1 9

@ence rofits ratio $ 3 3 1 9

Ms rofit

3 16###3###

16

×= =

DMs rofit

9 16###9###

16

×= =

/m $ 3### " 9### $ Gs 12###

QAS : Arit'metic


1. De to a technical snag, a train tra+els at twoEthird of its sal seed. @ow mch longer will it taBe to

co+er the distance that it normall< co+ers in one hor% a! 15 mintes b! 3# mintes c! 45 mintes d! 2# mintes e! &one of these




2. /Qit lans to tra+el 18# Bm. @e tra+els first 3# Bm at 45 BmRh, ne;t 6# Bm at 9# BmRh and the remainingdistance at 75 BmRh. *alclate the a+erage seed of the entire Qorne<.

a! 71 BmRh b!

272

19

BmRh c!

171

19

BmRh d! 72 BmRh e! &one of these

3. car starts with a seed of 9# BmRh with its seed increasing b< 2# BmRh in e+er< two hors. (n howman< hors will it co+er a distance of 465 Bm%

a! 2.25 hors b! 45 hors c! 4.5 hors d! 4 hors e! &one of these

4. man can reach a certain lace in 32 hors. (f his seed decreases b<

th1

16

÷

, he tra+els 14 Bm less inthe same time. )ind his original seed.

a! 7 BmRh b! 28 BmRh c! 14 BmRh d! 3# BmRh e! &one of these

5. ssme that the distance tra+elled b< a car on one litre of etrol +aries in+ersel< as the s=are of theseed at which it is dri+en. (f it tra+els at the rate of 25 BmRlitre at a seed of 3# BmRh, at what seed itshold be dri+en to tra+el at the rate of 36 BmR litre%

a! 2#

5

6

BmRh b! 25 BmRh c! 36 BmRh d! 43.2 BmRh e! &one of these

6. Two trains starting at the same time from 2 stations 2## Bm aart and going in oosite direction s,cross each other at a distance of 11# Bm from one of the stations. :hat is the ratio of their seeds%


7. :ith a niform seed a car co+ers a distance in 1# hors. @ad the seed been increased b< 5 BmRh the

same distance cold ha+e been co+ered in 8 hors, what is the distance co+ered% a! 64 Bm b! 8 Bm c! 78# Bm d! 2## Bm e! &one of these

8. thief steals a car at 33# m and dri+es it at 7# BmRh. The theft is disco+ered at 4 m and the ownersets off in another car at 85 BmRh. :hen will he o+ertaBe the thief if the< both are mo+ing in samedirection%

a! 53# 0A b! 6#2 0A c! 62# 0A d! 43# 0A e! &one of these

9. n athlete rns from to H and retrns at 1# BmRh, while another athlete simltaneosl< rns from H to and goes bacB at 15 BmRh. (f the< cross each other for the first time 12 mintes after the< start, the<will ne;t cross each other after

a! 18 mintes b! 3# mintes c! 24 mintes d! 36 mintes e! 42 mintes

1#. train asses two ersons who are walBing in the direction oosite to which the train is mo+ing, at therate of 5 mRsec and 1# mRsec in 6 sec and 5 sec resecti+el<. )ind the length and seed of the train.

a! 13# m, 1# mRsec b! 15# m, 2# mRsec c! 17# m, 3# mRsec d! 19# m, 4# mRsec e! &one of these

11. stdent walBs from his hose at 5 BmRh and reaches his school 1# mintes late. (f his seed had been6 BmRh he wold ha+e reached 15 mintes earl<. The distance of his school from his hose is

a! 2.5 Bm b! 12.5 Bm c! 5.5 Bm d! 3.6 e! &one of these

12. Two trains 131 m and 1#9 m in length resecti+el< are rnning in oosite directions, one at the seedof 5# BmRh and other at the seed of 4# BmRh. (n what time will the< be comletel< clear of each otherfrom the moment the< meet%

a! 14.8 sec b! 9.6 sec c! 2.6 sec d! 1#.6 sec e! &one of these




13. Train train lea+es from Delhi to Ambai, a distance of 16## Bm and at the same time another train Hlea+es from Ambai to Delhi. These trains meet at Hhoal at a distance of 6## Bm from Delhi. :hat isthe ratio of seeds of train to train H.


14. man standing on a railwa< latform obser+es that a train going in one direction taBes 4 seconds to

ass him. nother train of same length going in the oosite direction taBes 5 seconds to ass him. Thetime taBen in seconds! b< the two trains to cross each other will be

a!

4#

9

seconds b!

3#

11

seconds c!

31

9

seconds d!

5#

9

seconds e! &one of these

15. man rows 27 Bm downstream and 18 Bm stream taBing 3 hors each time. :hat is the seed of thecrrent%

a! 3 BmRh b! 4.5 BmRh c! 3 BmRh d! 1.5 BmRh e! &one of these

16. , who is tra+elling at 3.5 BmRh, starts 2.5 hor before H who tra+els at 4.5 BmRh in the same direction as . (n how mch time will H o+ertaBe %

a! 8.75 hors b! 8 hors c! 6 hors d! 7.75 hors e! &one of these

17. monBe< tries to ascend a greased ole 14 metres high. @e ascends 2 metres in first minte and slisdown 1 metre in the alternate minte. (f he contines to ascend in this fashion, how long does he taBe toreach the to%

a! 26th min b! 28th min c! 25th min d! 3#th min e! &one of these

18. fter tra+elling 42# Bm that is 87.5 of the distance, 0radee was stoed de to traffic Qam. The Qamwas cleared in 9 mintes. @e then increased his seed b< 2# BmRh and reached his destination 6 minearlier than he had anticiated. :hat was the initial seed of the car%

a! 8# BmRh b! 75 BmRh c! 72 BmRh d! 6# BmRh e! &one of these

19. The dri+er of a car sees a bs 2# m ahead of him. fter 1# seconds, the bs is 3# m behind. (f theseed of the car is 3# BmRh, what is the seed of the bs%

a! 12 BmRh b! 12 mRsec c! 5 mRsec d! 5 BmRh e! &one of these

2#. Two trains - a slower train 49# m long and a faster train 21# m long - were tra+elling on arallel tracBstowards each other. The dri+er of the faster train noticed the dri+er of the slower train when it was 7## mawa<I 28 sec later the dri+er assed each other. )ind the relati+e seeds of each train, if we Bnow thatthe slower train taBes 35 sec longer to ass the signal than the faster train.

a! 36 BmRh S 54 BmRh b! 36 BmRh S 45 BmRh

c! 3# BmRh S 5# BmRh d! 1# BmRh S 15 BmRh

e! 36 BmRh S 6# BmRh

Solution Home Work Assignments


s the train is tra+elling at

rd2

3

÷

of its sal seed then,

it will taBe

3

2

of its original time

Therefore,

3

2

of 6# mintes $ 9# mintes

Therefore, it taBes 3# mintes more than its normal time.





Total distance $ 18# Bm

Total time $

3# 6# 9# 2 2 6 38 hors

45 9# 75 3 3 5 15+ + = + + =

+erage seed $

Total Distance 18# 17138Total time 19

15

= = =

BmRh


Distance co+ered in first 2 hors $ 9# ' 2 $ 18# Bm

Distance co+ered in ne;t 2 hors $ 9# " 2#! ' 2 $ 22# Bm

Gemaining distance $ 465 - 22# - 18# $ 65 Bm

/eed in fifth hor $ 11# " 2# $ 13# BmRh

@ence, time taBen in co+ering 65 Bm

65 1hor #.5 hor

13# 2= = =

Total time taBen $ 2 " 2 " #.5 $ 4.5 hors


et seed of the man be x BmRh, then

1 15&ew /eed $ -

16 16

15&ow, 32 - 32 14

16

32 E 3# 14

7 BmRh

x x x

x x

x x

x

=

× =

⇒ =⇒ =


2

2

ccording to the =estion, we ha+e

1 1 125 25 5 .

3# 3# 3#

5 :here is a constant.!...........1!3#

et the re=ired seed be BmRh

1 1, 36 6

6 :here is a cons

k k

x

Hence x x

k k

x

µ ⇒ µ ⇒ µ ÷ ÷

⇒ =

µ ⇒ µ ÷ ÷

⇒ = tant.!............2!

5 3#)rom e=ations 1 and 2, we ha+eI6

25 BmRh

k

k

x

x

=

⇒ =


(n same time the< co+er 11# Bm and 2## N11#! $ 9# Bm resecti+el<.

)or the same time, seed and distance are directl< roortional.

/o, ratio of their seed $ 11# 9# $ 11 9


et the distance be x Bm.

Then difference in seed $

;

8 1#

x −




;5

8 1#

5 45

4#

5 4# 2## Bm

x

x x

x

⇒ − =

−⇒ =

⇒ = × =


/ose the thief is o+ertaBen x hors after 3 3# 0A

Then, distance co+ered b< the thief in x hors

$ distance co+ered b< the owner in

1 hors

2 x

− ÷

17# 85

2

14 2 172 1!

17 17 6# hors mintes

6 6

17# mintes 2 hors 5# mintes

x x

x x

x

⇒ = − ÷

⇒ × = −

×⇒ = =

= =

@ence, he will o+ertaBe the thief at 6 2# 0A


et the total distance co+ered b< them when the< first meet is d Bm

:hen the< meet for the second time, the total distance co+ered b< them is 3d .

@ence, the< will meet after 12 ' 3 $ 36 mintes

1#. Correct Option: ()

et the seed of the train be x mRsec.

ength of the train $ x " 5!6 $ x " 1#!5

/o, x $ 2# mRsec and length of the train $ 2# " 5! ' 6 $ 15# m


et the distance $ d Bm

Difference in time is of 25 minte for the two cases

/o,

1# 15! -

5 6 6#

d d +=

hors

5

3# 12

12.5 Bm

d

d

⇒ =

=


s the two trains are mo+ing in oosite directions their relati+e seed $ 5# " 4#! BmRh $ 9# BmRh $ 25 mRsec.

/o, the re=ired time $

Total length

Gelati+e seed

131 1#99.6 seconds

25

+= =


Gatio of seeds $ Gatio of distances co+ered

( )

/eed 6## 6## 3

/eed H 16## 6## 1### 5= = =

−


Gatio of times $ 4 5

@ence ratio of seeds $ 5 4

4sec

5 x =

∴

2#

x =

et the total time taBen b< trains to cross each other be t




t $

Total length

/m of the seeds

$

( )

( )

2

5 4 9 x x x

+=

+

( )2 4#

2# seconds9 9

= =


/eed downstream

279 BmR h

3= =

/eed stream

186 BmR h

3= =

et the seed of the man and crrent be v and u resecti+el<.

@ence

9

6

v u

v u

+ =− =

n sol+ing we get u = 1.5 BmRh


taBes a lead of 3.5 BmRh! ' 2.5 h! $ 8.75 Bm o+er H

Gelati+e seed of H with resect to

$ 4.5 BmRh N3.5 BmRh! $ 1 BmRh

∴ Time taBen b< H to o+ertaBe $ 8.75 hors.


(n e+er< 2 min he is able to ascend 2 - 1 $ 1 metre. This wa< he ascends to 12 metres becase when he reaches at the to, hedoes not sli down. Ths, to 12 metres he taBes 12 ' 2 $ 24 mintes. and for the last two metres he taBes 1 min. Therefore, hetaBes

24 " 1 $ 25 mintes to reach the to.

(n

th25

mintes he reaches the to.

18 Correct Option: (d)

(t is gi+en that

7

8

of the distance

$ 42# Bm or total distance $ 42# '

8

7

$ 48# Bm

et x BmRh be the initial seed of the car.

48# 42# 9 6# 6&ow,

6# 2# 6# x x x

÷

− + + =+

n sol+ing,

x $ 6# BmRh


Total relati+e distance tra+elled b< the car

$ 2# " 3#! m $ 5# m

The relati+e seed $ 5# mR1# sec $ 5 mRsec

The relati+e seed is also e=al to 3# - v ! BmRh

$ 3# - v ! '

5

18

mRsec, where v is the seed of the bs.

Therefore,

( )5

3# - 518

v × =

r, v $ 12 BmRh

2#. Correct Option: (!)

et the seed of the slower train be x mRsec and faster train be y mRsec.




(n 28 seconds the slower train co+ered 28 x metres and faster train 28y metres.

Therefore,

28 28 7## x y + =

The slower train asses the signal lights in

49#

x

seconds.

nd the faster train in

21# secy

Therefore,

49# 21# - 35

x y =

/ol+ing the 2 e=ations, we get

x $ 36 BmRh and y $ 54 BmRh

QAS .: Arit'metic.


1. can do a Qob in 15 da<s and H can comlete the same Qob in 3# da<s. Hoth started the Qob and H leftafter 5 da<s. Then in how man< da<s comleted the remaining Qob%

a! 7.5 da<s b! 12.5 da<s c! 1# da<s d! 22.5 da<s e! &one of these

2. is for times as good as worBman as H and therefore is able to finish a Qob in 9# da<s less than H.)ind how man< da<s the< will taBe to comlete the worB if the< worB together.

a! 3# da<s b! 4# da<s c! 24 da<s d! 12# da<s e! &one of these

3. can do a iece of worB in 4 hors. H and * together can do it in 3 hors, while and * together cando it in 2 hors. @ow long will H alone taBe to do it%

a! 12 hors b! 2.4 hors c! 1.3 hors d! 24 hors e! &one of these

4. alone can do a iece of worB in 8 da<s and H alone can do it in1# da<s. and H ndertooB to do it forGs 4###. :ith the hel of *, the< comleted the worB in 4 da<s. @ow mch is to be aid to *%

a! Gs 4## b! Gs 16## c! Gs 4### d! Gs 8## e! &one of these

5. 1# trained ersons can dig a ond in 12 da<s. (f trained ersons worB twice as fast as the nonEtrained,then 4 trained ersons and 4 nonEtrained ersons can dig the same ond in

a! 1# da<s b! 2# da<s c! 15 da<s d! 5 da<s e! &one of these

6. )ill ie is fi+e times faster than fill ie H and it taBes 4# mintes less than fill ie H. :hen will thecistern be fll, if both the ies are oened together%

a!

5#3

mintes b!

253

mintes c! 25 mintes d! 5# mintes e! &one of these

7. does 8# of a worB in 2# da<s. @e then calls in H and the< together finish the remaining worB in 3da<s. @ow long H alone wold taBe to do the whole worB%

a! 13.6 da<s b! 15# da<s c! 75 da<s d! 37.5 da<s e! &one of these

8. 5 men or 6 women can finish the worB in 18 da<s. @ow long will 1# men and 6 women taBe to finish thesame worB%

a! 4 da<s b! 6 da<s c! 9 da<s d! 8 da<s e! &one of these




9. Two ies and H can fill a cistern in 5# mintes and 65 mintes resecti+el<. 0ie is oened for 4#mintes and then it is closed and ie H is oened. )ind the time taBen b< ie H to comletel< fill thecistern.


1#. Two ies and H fill the cistern in 8 mintes and 1# mintes, resecti+el<. (f the< are oened on

alternate minte and if ie is oened first, find the time in which the tanB gets filled.

a!

48

5

mintes b!

38

5

mintes c!

28

5

mintes d!

18

5

mintes e! 8 mintes

11& , H and * can do a iece of worB in 14 da<s, 12 da<s and 3# da<s resecti+el<. The< started doing theworB together bt H lea+es off 3 da<s before the comletion of worB and * lea+es off 2 da<s before thecomletion of worB. )ind the total nmber of da<s to comlete the worB.


12. water tanB can be filled b< two tas and H indi+idall< in 9# mintes and 18# mintes, resecti+el<.To fill the tanB, both tas are oened for 2# mintes and then ta H is trned off. )ind the total time

taBen to fill the tanB. a! 6# mintes b! 12# mintes c! 3# mintes d! 8# mintes e! 14# mintes

13. *istern has a leaB in the bottom. :ithot leaB, it gets filled in 16# mintes and with leaB, it taBes 4#mintes more. (f the cistern is fll of water, how long in mintes! the leaB will taBe to emt< it, in casethe water leaBs at doble the rate after half the cistern becomes emt<%

a! 4## mintes b! 2## mintes c! 6## mintes d! 5## mintes e! &one of these

14. , H and * finished a iece of worB and get a total amont of Gs 625. and H together comletedth

23

25

÷

of the worB and H and * together comlete

th8

25

÷

of the worB. @ow mch shold be aid% a! Gs 325 b! Gs 425 c! Gs 5# d! Gs 15# e! &one of these

15. 9 bo<s finish oneEthird worB in 1# da<s. The nmber of additional bo<s re=ired for finishing theremaining worB in 2 more da<s will be

a! 81 b! 78 c! 55 d! 3# e! 54

16. certain Qob was assigned to a gro of women to do in 2# da<s. Ht, 12 women did not trn for the Qob and the remaining women did the Qob in 32 da<s. The original nmber of women in the gro was

a! 32 b! 34 c! 36 d! 4# e! 42

17. 3# girls can do a worB in 16 da<s and 16 bo<s can comlete the same worB in 15 da<s. :hat is the ratio

between the caacit< of a bo< to that of a girl% a! 2 1 b! 1 2 c! 1 3 d! 3 1 e! &one of these

18. , H and * can do a Qob together in 6 da<s, is half as good as H and twice as good as *. alone willtaBe how man< da<s for comlete worB.

a! 2# da<s b! 24 da<s c! 21 da<s d! 3# da<s e! 15 da<s

19. Two fill tas can searatel< fill the tanB when a waste ta is closed in 2# mintes and 3# mintes,resecti+el<. :hen the waste ta is oened, the< together fill it in 4# mintes. @ow long will waste iewill taBe to emt< the tanB when both the fill tas are closed%

a! 17.1 mintes b! 14.3 mintes c! 15.7 mintes d! 8.6 mintes e! &one of these




2#. cistern can be filled b< two ies and H indi+idall< in 3# mintes and 4# mintes resecti+el<. )or

certain time, both ies are oened bt de to their being cloggedI onl<

th3

4

÷

of fll =antit< of water

flows throgh and

rd

23 ÷

throgh later. The obstrctions are now remo+ed and now cistern taBes 15mintes more to get fll. )ind the time before the fll flow of water began.

a! 3 mintes b! 6 mintes c! 2 mintes d! 18 mintes e! &one of these






:orB done b< S H $

1 1 15

15 3# 2

× + = ÷

:orB left

1

2=

Da<s re=ired b< $

115 7.5

2× =


Gatio of times taBen b< and H $ 1 4.

The time difference is 4 - 1! $ 3 da<s while H taBes 4 da<s and taBes 1 da<.

(f difference of time is 3 da<s, H taBes 4 da<s.

(f difference of time is 9# da<s, H taBes

49# 12# da<s

3× =

/o, taBes 3# da<s to do the worB.

:orB done b< in 1 da<

1

3#=

:orB done b< H in 1 da<

1

12#=

Ths, :orB done b< and H together in 1 da<

1 1 4 1 1

3# 12# 12# 24

+= + = =

@ence, Time taBen b< and H to comlete the worB together

124 da<s

1

24

= =


1:orB done b< in 1 hor 6

1:orB done b< H and * together in 1 hor

4

1:orB done b< and * together in 1 hor

3

1 1 5:orB done b< , H and * together in 1 hor

6 4 12

5 1 1:orB done b< H in 1 hor

12 3 12

=

=

=

∴ = + =

∴ = − =

Ths, H alone can comlete the worB in 12 hors.


1 1 1 1 9 1

:orB done b< * in 1 da< 4 8 1# 4 4# 4#

1 1 1Gatio of their wages

8 1# 4#

1 *s share for 4 da<s 4 4### Gs 4##

4#

= − + = − = ÷

=

∴ = × × =


1# trained ersons can maBe a ond in 12 da<s

Therefore, one trained ersonMs one da< worB $

1

12#

Therefore, one nonEtrained ersonMs $

1

24#




&ow, one da<Ms worB $

4 4 12

12# 24# 24#+ =

Ge=ired nmber of da<s $

24#2# da<s

12=


#i$e

?i+en that

4#5

x x − =

, so x = 5# mintes

@ence time taBen b< H $ 5# mintes and b< $ 1# mintes

/o, time to fill cistern $

1 25 mintes

1 1 3

1# 5#

=+

7& Correct Option: (d)

( )

5:hole worB can be done b< in 2# da<s $ 25 da<s

4

1&ow, 2# worB i.e. worB is done b< and H in 3 da<s.

5

:orB will be done b< and H in 5 3 da<s $ 15 da<s

1 s 1 da<s worB and H!s 1 da

25

× ÷

×

= +1

<s worB $15

1 1 2Hs 1 da<s worB

15 25 75

/o, H alone wold do the worB in 37.5 da<s.

∴ = − =


/ince 5 men $ 6 women

∴ 1# men $ 12 women⇒

1# men " 6 women $ 12 women " 6 women $ 18 women

/ince, 6 women taBe 18 da<s

∴ 18 women will taBe

618 6 da<s

18

× = ÷

9& Correct Option: (c)

et the time taBen b< ie H to fill the cistern is x mintes. Then,

1 14# 1

5# 65

13

x

x

× + × = ÷ ÷

⇒ =

1#& Correct Option: (!)

(n 2 mintes, ortion of tanB filled

1 1 9

8 1# 4#= + =

(n 8 mintes, ortion of tanB filled wold be

36

4#=

10ortion of tanB remained

1#=

&ow time taBen b< $

8 4 mintes1# 5

=




@ence, total time taB4 4

8 8 mintes5 5

en = + =


et the time taBen to finish the worB be t . Then, adding eachMs share of worB done

3 2

114 12 3#

3# 35 3! 14 2!1 7 da<s

42#

t t t

t t t t

− −+ + =

+ − + −⇒ = ⇒ =

12& Correct Option: (d)

(n 2# mintes, art of the tanB filled

$

1 1 12#

9# 18# 3

× + = ÷

Gemaining art $

1 21

3 3− =

fter 2# mintes, time taBen b< ta to fill the tanB $

29# 6# mintes

3× =

Therefore, total time taBen $ 2# " 6#! mintes $ 8# mintes


et the leaB emties the tanB in‘t’ mintes.

1 1 1

16# 2##t − =

@ence, t $ 8## mintes

&ow time to emt< the cistern 1 2t t = +

1 1 4## mintes1 28## 2

t t × = ⇒ =

lso,

1 12## mintes2 2

4## 2t t × = ⇒ =

Therefore, total time taBen to emt< the cistern $ 6## mintes


23 H!s share 625 Gs 575

25

8H *!s share 625 Gs 2##

25

lso H *!s share Gs 625 s share 625 2## Gs 425

+ = × =

+ = × =

+ + == − =


9 bo<s finish

1

3

worB in 1# da<s

x bo<s finish

2

3

worB in 2 da<s

3=ating worB 9 1# 3 2

2

9# bo<s

x

x

⇒ × × = × ×

⇒ =

dditional bo<s needed $ 9# - 9 $ 81





x women taBe 2# da<s

x - 12! women taBe 32 da<s

Therefore,

( ) ( )2# 32 12 x x = −

2# 32 38412 384 32

x x x x

⇒ = −⇒ = ⇒ =

17& Correct Option: (!)

&mber of da<s re=ired b< a girl to comlete the worB

3# 16 48# da<s

1 The worB done b< a girl in 1 da<

48#

&mber of da<s re=ired b< a bo< to comlete the worB

16 15 24# da<s

The worB done b< a bo<

= × =

∴ =

= × =

∴1

in 1 da<24#

1 1Ths, the re=ired ratio is 2 124# 48#

=

=

18& Correct Option: (c)

. fficienc< ratio of , H S * is 2 4 1

" H " *!Ms one da< worB $

1

6

Ms one da< worB $

2 1 1

7 6 21× =

*omlete worB will be done b< in 21 da<s.

19& Correct Option: (!)

To fill the tanB $

1 1 5 12# 3# 6# 12

+ = =

:orB done b< waste ie in 1 minte $ worB done b< all the three tas N worB done b< two fill tas

$

1 5 7

4# 6# 12#− = −

N+e means to emt< the tanB!

:aste ie will emt< the tanB in 17.1 mintes.

2#& Correct Option: (!)

0art of cistern filled b< tanB when being clogged

$

( )3 1

er minte4 3#

×

0art of cistern filled b< tanB H when being clogged

$

( )2 1

er minte3 4#

×

et time before the fll flow of water began be t .

3 1 2 1 1 115 1

4 3# 3 4# 3# 4#

5 715 1

12# 12#

3 mintes

t

t

t

× × + × + × + = ÷ ÷

× + × = ÷ ÷

=

QAS 6: 8eometr$





1. (f m nits are added to the radis of a circle, then b< how man< nits does the circmference of newcircle will increase%

a!

1

2

mπ

b!

4 mπ c!

3 mπ d!

πm

e!

2 mπ

2. :hat is the area of the largest triangle that can be fitted into a rectangle of length / nits and width Tnits%

a!

/ T

2

−

b!

/ T

2

+

c!

/ T

4

×

d!

/

2

T×

e!

/ T

8

×

3. The diagonals of a rhombs are 64 cm and 48 cm, find the area of the rhombs%

a! 1536

2cm

b! 3#72

2cm

c! 768

2cm

d! 16##

2cm

e! &one of these

4. The area of a s=are field is

216## Bm .

@ow long will a lad< taBe to cross the field diagonall< at a seed

of

4 2

BmRh%

a!

1# hors.

b!

1# 2 hors

c!

1# 3 hors

d!

2# hors

e!

15 hors

5. circle of ma;imm ossible siUe is ct from a s=are sheet of board. /bse=entl<, a s=are ofma;imm ossible siUe is ct from the resltant circle. Then again same rocedre is reeated with newthe s=are. Then area of newl< formed s=are is

a! 25 of the area of the original s=areb! 5# of the area of the original s=are.

c! 35 of the area of the original s=are

d! 15 of the area of the original s=are.

e! 12.5 of the area of the original s=are

6. ach interior angle of a reglar ol<gon is 1## degrees greater than each e;terior angle. @ow man<sides are there in the ol<gon%

a! 9 b! 19 c! 14 d! 13 e! 1#

7. :hat is the s=are of the circmradis of a triangle whose sides are 9, 4# and 41%

a! 2#.25 b! 4## c! 42#.25 d! 16## e! 81

8. )ind the area of the sector co+ered b< the hor hand after it has mo+ed throgh 5 hors and the lengthof the hor hand is 14 cm.

a!

2256 cm

b!

2156.66 cm

c!

23## cm

d!

2266.66 cm

e!

2256.67 cm

9. 12 cm s=ares ct into 3 cm s=ares. )ind the ratio of the erimeter of the larger s=are and sm oferimeters of the smaller s=ares.





1#. /ide of a cbe is 6 cm which is ct into smaller cbes of side 2 cm. Then what will be the ratio of thesrface area of the larger cbe to that of the sm of the srface areas of the smaller cbes.

a! 3 2 b! 2 3 c! 1 1 d! 2 1 e! 1 2

11. (f the srface areas of the three coterminos faces of a cboid are

21#, 8 and 2# cm

resecti+el<. Thenfind ot the ratio between srface area and +olme of the cboid.

a! 76 4# b! 24 37 c! 19 1# d! 2 3 e! &one of these

12. hollow c<lindrical tbe, oen at both the ends is made of iron 5 cm thicB. (f the e;ternal radis ofthe tbe is 15 cm and its length is 21# cm, then find ot the +olme of iron sed in maBing the tbe.

a! 2#24# b! 132### c! 4#56# d! 4### e! &one of these

%3. &e'agon A()D*+ is a regular ,ol-gon of side c$/ thenwhat will be the ratio of the area of rectangle ()*+ tothat of the area of he'agon

1a 2 3 1b 3 2

1c 3 4 1d 4 3

1e None of these

14. The diameter of a roller 1 m long is 14# cm. (f it taBes 1## re+oltions to le+el a la<grond, then findot the cost of le+elling the grond at the rate of Gs 1#.8# er m 2.

a! Gs 3326.4 b! Gs 3### c! Gs 324# d! Gs 1663.2 e! &one of these

15. hall, 14 meters long, 8 meters broad and 6 meters high, has two windows 11# cm ' 1# cm and twodoors 2# cm '11# cm. )ind the cost of aering the walls with aer which costs Gs 1# er s= meter.

a! Gs 2633.34 b! Gs 2533.34 c! Gs 25## d! Gs 2### e! &one of these

16. (f the radis of a shere is dobled, then what will be the ratio of the +olme of original shere to that ofthe new shere%

a! 1 8 b! 8 1 c! 2 1 d! 1 2 e! 4 1

17. )ind the difference between the cr+ed srface area and total srface area of a right circlar cone,ha+ing a radis and height of 3 m S 4 m resecti+el<.

a!

9πb!

64πc!

81πd!

39πe! &one of these

18. well of 14 m inside diameter is dg 21 m dee, the earth taBen ot from it is sread to a distance of 7m all arond it to form an embanBment. )ind the height of the embanBment.

a! 7 m b! 21 m c! 14 m d! 12 m e! 2.1 m

19. (f a cone is ct into two arts b< a horiUontal lane assing throgh the midEoint of its a;is, the ratio ofthe +olme of er art to that of the original cone is

a! 1 4 b! 1 2 c! 1 8 d! 1 6 e! 1 V

2#. The length, breadth and height of a cboid are in the ratio 1 2 3. (f the length, breadth and height ofthe cboid are increased b< 1##, 2## and 1## resecti+el<, then, what will be the increase in the+olme of the cboid%

a! 8 times b! 18 times c! 6 times d! 12 times e! 1# times






et the radis of the circle be r nits.

The circmference of the circle will therefore be

2 nits.r π

if the radis is increased b< m nits, the new radis will be

! ,nits.r m+

Therefore, the new circmference will be

2 ! 2 2r m r mπ + = π + π

r the circmference increases b<

2 mπnits.

2. (d)

The triangle which has its base as the length of the rectangle and its height as the width of the rectangle or the triangle which hasits base as the width of the rectangle and its height as the length of the rectangle will be the largest triangle that can be fitted inthe rectangle.

(f the base of the triangle is / and its height T, then its area will be

/ T

2

×

nits.

/imilarl<, if the base of the triangle is T nits and its height is / nits, then its area will be

/ T2×

nits.

3. (d)

(f diagonals of the rhombs are 64 and 48, then side of the rhombs will be

2 232 24 4# cma = + =

lso, rea of Ghombs $

2164 48 1536 cm

2× × =

4. (!)

et a meters be the diagonal of the s=are field.

Therefore, itMs area $

2 2 16## gi+en!a a⇒ =

4#a⇒ =

,

⇒

Diagonal

2 4# 2a= =

Total tra+elled distance $ length of the diagonal.!

Total time

Distance 4# 21#

/eed 4 2= = =

hors

5. (!)

et the side of the original s=are be' a '

nits. Therefore, the area of the original s=are $

2a

nits.

The diameter of the circle of ma;imm ossible dimension that is ct from the s=are will be

T Ta

nits.

The diagonal of the s=are of ma;imm ossible dimension that can be ct from the circle will be

T Ta

nits.

(f the diagonal of the final s=are is

T Tanits, then its area

2

2

a=

nits.

Therefore, the area of the new s=are will be 5# of the area of the original s=are. The same wa<, when the rocedre is

reeated then the area of new s=are will be

25

of the area of the original s=are.

6. (!)




et e;terior angle be

°

, and then each interior angle will be

1## " °

:e Bnow that in an< reglar ol<gon, the sm of an e;terior and interior angle is alwa<s $ 18#W.

Therefore, " 1## " $ 18#

⇒

$ 4#

:e Bnow,

2! 18#(nterior angle $

2! 18#14# 9

n

n

nn

n

− ×

− ×⇒ = ⇒ =

7. (c)

The gi+en triangle is a right angled triangle. 9, 4# and 41 are 0<thagorean trilet.!

(n a right angled triangle, the circmradis is half the h<otense.⇒

(n the gi+en triangle, the h<otense $ 41

Therefore, the circmradis

41 2#.5

2= =

nits.

/=are of the circmradis( )

2

2#.5 42#.25=

8. (e)

(f the hor hand mo+ed throgh 12 hors, it sbtends 36#W at the centre of the clocB.

i.e. in 12 hors it maBes 36#W.

Therefore, in 5 hors it maBes

536# 15#

12

o o × = ÷

:e Bnow that length of the hor hand

14 cmr =

Therefore, area of sector

2215# 22

156 256.67 cm

36# 36# 7

r θπ = = × × = ÷ ÷

9. (c)

The erimeter of the larger s=are $ 4 ' 12 $ 48 cm&mber of smaller s=ares $ 16

/m of erimeters of the smaller s=ares

$ 16 ' 4 ' 3! $ 16 ' 12 $ 192 cm

Therefore, the re=ired ratio $ 192 48 $ 4 1

1#. (c)

/rface area of original cbe

2 6 6 36 216a= = × =

@ence, there will be 27 sch cbes. /rface area of small cbes

Therefore, the srface area of 27 sch cbes 27 24 648= × =

increase

648 216! 432 1## 2##

216 1## 216

− = = × = ÷×

11. (c)

(f

, ,l b

be the dimensions of the cboid, then olme of the cboid will be

l b × ×

(t is also gi+en that,

1#, 8, 2#l b b l × = × = × =




2

! ! ! 1# 8 2#

! 16##

! 4#

l b b l

l b

l b

⇒ × × × = × ×

⇒ × × =

⇒ × × =

lso, /rface rea is

2 ! 21# 8 2#! 76lb b l = + + = + + =

∴Ge=ired ratio

/rface rea 76 19

olme 4# 1#= = =

12. ()

?i+en, e;ternal radis G! $ 2# cm S height h!

$ 21# cm

∴(nternal radis

! 2# - 5 15 cmr = =

&ow, olme of iron $ ;ternal +olme - (nternal +olme

( ) ( )

( )

2 2 2 2

2 2 3

G !

221521# 132### cm

75

r ! r = π − π = π −

⇒ × − =

13. (!)

(f HX is the altitde of triangleE1 and X) is the altitde of triangleE2.

This reglar he;agon will comrises of si; e=ilateral triangles each of side 6 m. The 6 triangles are nmbered 1 to 6 in the figre

shown abo+e.

@ence,

3HX X) 6 3 3

2= = × =

∴ength of the rectangle, H)

6 3 m=

@ence, the area of the rectangle

H*) $ length ' width $

6 3 6 36 3× =

s=. m

&ow area of he;agon

2 23 3 3 3 3 side! 3 6 36

2 2 2= = × × = ×

∴Ge=ired ratio

rea of rectangle 36 3 2

rea of he;agon 33 336

2

= = =×

14. (!)

@ere

7#cm 7# m, 1 mr = = =

*r+ed srface area $

2 22 2r r π = π

222 #.7# #.7# 1

7×= × × ×

$ 3.#8 m2

rea swet in one re+oltion $3.#8 ' 1## $3#8 m2

*ost of le+elling $ 3#8 ' 1#.8# $ Gs 3326.4




15. (!)

rea of for walls

( ) ( ) 22 2 14 8 6 264 ml b = + = + =×

$

264####

2cm

rea of two windows

22 11# 1# 22## cm= × × =

rea of two doors

22 11# 2# 44## cm= × × =

∴ rea of the wall to be aered $ area of the for wall - area of the two windows -area of the two doors!

2264####- 22##- 44## 26334## cm= = 2263.334 m=

*ost of aering $

263.334 1# Gs 2633.34× =

16. (!)

et YrZ be the radis of original shere, then its +olme,

3

1

4

3v r = π

cbic nit

:hen radis is dobled then +olme of the new shere,

3 3

2

4 42 ! 8

3 3v r r = π = π × ×

&ow re=ired ratio is,

3

11 2

32

413 1 8

4 88

3

r v

v v v

r

= = ⇒ =× ×

π

π

17. (!)

?i+en radis, r $ 3m, height, $ 4 m

∴ /lant height

2 2 2 23 4 5 ml r = + = + =

*r+ed srface area $

3 5 15rl π = π × × = π

cm2

Total srface area $ rea of base " *r+ed srface area $

2 ! 24r rl r r l π + π = π + = π

∴ Difference $ Total srface area - *r+ed srface area $

24 15 9π − π = π

cm2

18. (!)

olme of hollow c<linder $ olme of arth sed in embanBment

( )2 2 27 21 14 7 π × = π − ×

7 7 21721 7

× ×= =×

m

19. (c)

et the radis and height of the original cone be r and resecti+el<

@ence radis and area of the er cone $

2

r

and

2

olme of er art




21 1

3 2 2 8

r = π × = ÷

' riginal +olme

Gatio $ 1 8

2#. (d)

et the length, breadth and height of the cboid be x, 2 x and 3 x

olme of the original cboid,

1

32 3 6 x x x x = × × =

olme of the new cboid,

2 32 6 6 72 x x x x = × × =

∴

(ncrease in olme $

31

32

72;12

6;= =

times

QAS ,: Alger!


1. :hat are the +ales of x and y for the e=ations

3 4 4 33 and 1%

x y x y + = − = −

a!

2525,

9 x y = =

b! x $ 25, y $ 25 c!

55,

3 x y = =

x d! x $ 5, y $ 5 e! &one of these

2. /ol+e the gi+en e=ations and choose the correct otion.

4 - 3 3 and 7 - 5 1 x y x y = = −

a!

> x y

b!

≥ x y

c!

< x y

d!

≤ x y

e! x $ y or no sch relation e;ist

3. school boght 12# encils of Gs 4.5 and Gs 6.5. The school sent Gs 68# for the encils. The nmber of encils of Gs 4.5 and Gs 6.5 are resecti+el<

a! 7#, 5# b! 5#, 7# c! 6#, 6# d! 4#, 8# e! &one of these

4. The ath of an enem< tanB and an antiEtanB missile lotted on a grah are

3 2 6 and 2 3 17 x y x y − = + =

The oint at which the enem< tanB can be destro<ed is

a! 2, #! b! 4, 3! c! 4, -3! d! -3, #! e! &one of these

5. )or a show, lower stall ticBet cost

1

4 of the balcon< ticBet. Geser+ation charges are the same for both

the balcon<EticBet as well as the lower stall ticBet. ne reser+ed balcon< ticBet for the show costs Gs

25#.ne reser+ed balcon< ticBet and one reser+ed lower stall ticBet charge cost Gs 32#.The reser+ationcharge is




a! Gs 1# b! Gs 3# c! Gs 2# d! Gs 7# e! &one of these

6. :hat is the +ale of k for which the e=ation

26 11 # x x k + + = has real roots%

a!

121

124

k =

b!

121

24

k =

c!

121

24

k <

d!

121

24

k ≥

e!

121

24

k ≤

7. *onsider

2 25 144 # x x − + = and

2 36 216 #y y + + =.:hich of the otion is tre%

a! x K y b! x C y c! x [ y d! x \ y e! &one of these

8. noo and nil sol+ed a =adratic e=ation. noo made a mistaBe in the coefficient of x and got theroots as 8 and 4 while nil made a mistaBe in the constant term and got the roots as 12 and 6. )ind thecorrect roots of the e=ation.

a! 16, 2 b! N16, N2 c! N12 ,N6 d! 12, 6 e! &one of these

9. (f a and b are the roots of the e=ation

2 - 14 24 #, x x + =

then what is the +ale of

%a b

b a+

a! 38 b!

144

11

c! 256 d!

37

6

e! &one of these

1#. )or a =adratic e=ation, sm of the roots is 2# and their difference is 12. :hat is the e=ation%

a!

2 2# 64 # x x − =+ b!

2 2# 64 # x x + =+

c!

2 - 2# 64 # x x + = d!

2 - 2# 64 # x x − = e! &one of these

11. &e;t term of the following 0 is

192, 243, 300, 363,

JJ

a!

15 2n − b!

413

c!

431

d!

432

e!

469

12. /m of eighth term and fifteenth term of an 0 is 94.(f its 18th term e;ceeds its 7th term b< 44. )ind itsfirst term.

a! 5 b! 6 c! 7 d! 8 e! 9

13. (n an 0 fi+e times its fifth term is twice its 15th term. (f its 7th term is 26. )ind its 13th term

a! 48 b! 38 c! 46 d! 42 e! 44

14. Two cars and H start together in the oosite direction from two cities which are 37# Bm aart. +erage seed of car is 5# BmRh and a+erage seed of car H is 35 BmRh for first hor, 4# BmRh forsecond, 45 BmRh for third hor and so on. fter how man< hors two cars will meet%

a! 3 hor b! 4 hor c! 5 hor d! 6 hor e! 7 hor

15. (f sm of first n terms of an 0 is gi+en b< 7n2 " 5n. (ts nth term is

a!

16 2n +b!

16 2n −c!

14 2n −d!

14 2n +e!

15 2n −

16. )ind the sm of all three digit nmbers which are di+isible b< 13.




a! 37476 b! 37674 c! 37746 d! 34776 e! 34677

17. /m of three nmbers in ?0 is

95

18

and their rodct is

125

27

, their common ratio is%

a!

2 3or

3 2b!

3 or 2 c!

12 or

2d!

3 4 or

4 3e! &one of these

18. 5th term of a ?0 is 4#5, first term is 5. /m of si; terms is

a! 182# b! 91# c! 1315 d! 6#5 e! &one of these

19. /m of some terms of a ?0 is 484. (f second term is 12 and third term is 36, ]second to lastM term is

a! 1#8 b! 324 c! 216 d! 336 e! &one of these

2#. Two series are in ?0 their common ratios are p and " .(f their corresonding terms are mltilied thenthe commonEdifference of the new series we get is

a! #" b!

p"

c! p " " d! p $ " e! &one of these





1. (!)

3 43

4 31

1 1et and H

x y

x y

x y

+ =

− = −

= =

&ow e=ations are

3 " 4H $ 3

4 - 3H $ E1

/ol+ing for % and &

1 3 I H

5 5= =

@ence,

2525I

9 x y = =

2. (!)4 - 3 3 x y =

... 1!

7 - 5 1 x y = −

... 2!

Altil< e=ation 1! b< 7 and e=ation 2! b< 4

28 - 21 21 x y =

... 3!

28 - 2# 4 x y = −

... 4!

/btract e=ations 3! and 4!

- 25 or -25y y = =

-18 x =

3. ()

12# x y + =

...1!

4.5 6.5 68# x y + =

...2!

4.5 4.5 54# x y + =

J3!

( ){ }1 4.5 x

( ) ( )2 3−

2 14# 7# 5#y y x ⇒ = ⇒ = ⇒ =

nswer is 5#, 7#

4. ()

3 2 6 x y − =

J1!

2 3 17 x y + =

...2!

6 4 12 x y − =

...3!

( ){ }1 2×

6 9 51 x y + =

J4! O2!'3!P

3! - 4!

-13y $ -39 I y $ 3

/bstitting the +ale of < in 1!





:e get 3 x $ 12 I x $ 4

x , y ! $ 4, 3!

5. (!)

25# x y + =

J1!

32#4

x x y y + + + =

5 8 128# x y + =

...2!

5 5 125# x y + =

J3! O1! ' 5P

3y $ 3#

y $ 1#

6. (e)

26 11 #

)or real roots

D #211 - 4 6 #

121 - 24 #

24 121

24 121

121

24

x x k

k

k

k

k

k

+ + =

≥⇒ × × ≥⇒ ≥⇒ − ≥ −⇒ ≤

⇒ ≤

7.

)or

2 25 144 # x x − + =

,

/m of the roots $ 25, ositi+e

0rodct of the roots $ 144, ositi+e

/o both the roots are ositi+e. x C #

)or

2 36 216 #y y + + =

/m of the roots $ N36, negati+e

0rodct of the roots $ 216, ositi+e

/o both the roots are negati+e. y K #

*learl< x C y

8. (!)

:hen 8 and 4 are the roots, a " b $ 12 and a b $ 32, where a " b $ 12 is incorrect.

:hen 12 and 6 are the roots, a " b $18 and a b $ 72, where a b $ 72 is incorrect.

*hoosing the correct arts, the correct e=ation is

( ) ( )2

18 32 #I - 16 - 2 # x x x x − + = =.

The roots are 16, 2

9. (d)

/m of the roots

14a b= + =

0rodct of the roots

24ab= =

2 2 2 ! 2 196 2 24 37

24 6

a b a b a b ab

b a ab ab

+ + − − ×+ = = = =

1#. (c)et roots be a and b

2#a b+ =

,

- 12a b =





Fon sol+ing, a $ 16I b $ 4

@ence the e=ation is

2 - 2# 64 # x x + =

/oltion 0!

11. (d)

192 8 3, 243 9 3, 3## 1# 3, 363 11 3= = = =

&e;t term is

12 3 432=

12. (!)

8 15

18 7

94,

44 4

2 21 94 5

a a

a a d

a d a

+ =

− = ⇒ =

+ = ⇒ =

13 (e)

5 15

7

13

5 2

85 4 ! 2 14 !

3

6 26

86 26 3, 8

3

8 12 3 44

a a

a d a d a d

a a d

d d d a

a

=

⇒ + = + ⇒ =

= + =

⇒ + = ⇒ = =

= + × ⇒

14. ()

et two cars will meet after n hors.

(n n hors car will tra+el 5#n km

*ar H will tra+el 35 " 4# " 45 " 5# " J. n terms

( ){ }

{ }

2

5# 35 4# 45 5# ....! 37#

5# 2 35 1 5 37#2

5# 65 5 37#2

165 5 74# #

4

n

nn n

nn n

n n

n

+ + + + + =

⇒ + × + − =

⇒ + + =

⇒ + − ==

15. (c)2

2

1 1

2

2 1 2

1 2

7 5

71 ! 5 1 12

7 2! 5 2 38

12, 26 14

12 1!1414 2

n

n

n

' n n

' a

' a a

a a d

a na n

= +

⇒ = + × = =

= × + × = = +

= = ⇒ =

= + −⇒ = −

16. ()

east three digit nmber di+isible b< 13 is 1#4 and largest three digit nmber di+isible b< 13 is 988. Total three digit nmbersdi+isible b< 13 are

69

988 1#4 1!13

69

692 1#4 68 13!

2

37674

n

n

s

= + −=

⇒ = × + ×

⇒

/oltion ?0!

17. (!)





et first nmber in ?0 be

a

r

and common ratio be r.

Then three nmbers are

, ,

aa ar

r

3

3

2

9518

125

27

5 5,

3 3

5 5 5 95

3 3 3 18

5 1 951

3 18

1 191 6 13 6 #

6

3 2

2 3

a a ar r

aa ar

r

a a

r

r

r r

r r r r

r or

+ + =

× × =

⇒ = = ÷

+ + =

⇒ + + = ÷

+ + = ⇒ − + =

⇒ =

18. Correct Option :( !)

4 4

6

5

4#5, 5 81

3

3 15 182#

3 1

ar a r

r

s

= = ⇒ ==

−= = ÷−

19. (!)

et first term of ?0 be a and common ratio be r . Then2

2

2 3

5

3

4

12, 36

3, 4

4 4 3 4 3 ... 484

1 3 3 3 ... 121

3 1121 3 1 242

3 1

3 243 3

5

4 3 1#8

nn

n

ar ar

r a

n

a

= =⇒ = =

+ × + × + =+ + + + =

−= ⇒ − = ÷−

= ==

= × =

2#. (!)

et first ?0 be

2 3, , , ...a ap ap ap

/econd ?0 be

2 3, , , ...b b" b" b"

(f we mltil< corresonding term of two ?0 we get2 2 3 3, , , ...ab abp" abp " abp "

common difference p"(

qas inotes

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