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SSC MAINS (MATHS) MOCK TEST–6 (SOLUTIONS)
1.(A)
. .
. . .
0 13 0 07
0 25 0 075 0 2
243 243
7 49 343
=
5 0.13 5 0.07
0.25 2 0.075 3 0.2(3 ) (3 )
(7) (7 ) (7 )
=
0.65 0.35
0.25 0.150 .063 3
7 7 7
=
(0.65 0.35)
(0.25 0.150 0.6)3
7
1
13 3
77
2.(B)
3 32 216 16
3
2 232
1(4 )
16 =
322
32 2
14
(4 )
33
144
16464
4096 164
409764
3.(C) No. of digits required= [{(9–1)+1}×1 + {(50–10)+1}×2]= 9 × 1 + 41× 2 = 9+82 = 91
4.(D) Digit in the unit's place of(251)98+(21)59–(106)100 + (705)35–164 + 259= 1 + 1 – 6 + 5 – 4 + 9= 6
5.(D) The required number must also be divis-ible by (232+1) and among the options given,(296+1) is divisible by (232+1) 296+1 = 296+196
= (232)3 + (132)3 , which is divisible by 232+1[when n is odd, (an+bn) is always divisibleby (a+b) ]
6.(B) Given that,H.C.F. of the two numbers = 27So, Let the numbers are 27 x and 27 y wherex and y are co-prime nos. i.e. prime to eachother.Now, A.T.Q27x + 27y = 216or, 27 (x + y) = 216 x + y = 8So, possible pairs of x and y are (1, 7) & (3, 5)So, The possible pairs of two numbers willbe (27, 189) & (81, 135) possible pairs ofThe possible no. of pairs is 2.
7.(B)221 428[(1931) ]
1932 =
even0dd(1931)
1932
= even1931 ;Re mainder 1
1932
even(a – 1) ;R 1a
8.(B) Term difference = 12–4 = 8Value difference = 70–14=56Value difference per term (i.e. commondifference) = 56/8 = 7So, first term = 4th term– 3 × common dif-ference = 14 – 3 × 7 = - 7
9.(B) Required height at the 1st bounce 3324
Required height at the 2nd
bounce
23324
Required height at the 3 rd
bounce
33324
= 27 132 13 m64 2
10.(A) Remaining no. of total balls after 1st ball ischosen = (12 + 6)-1 = 17 ballsAlso,Remaining no. of black balls after 1st ball(which is black) is chosen = 12–1
= 11 black ballsSo, The probability that the second ball isalso black = 11/17
11.(A) Let x be the initial no. of people in thecompany.So, ATQ,
= xx
35 5 32 345
or, 35x + 160 = 34x + 170 x = 10
12.(*) Initial bowling average = 12.4After improving bowling average by 0.2,new bowling average = 12.4 – 0.2 = 12.2Now, let x be the number of wickets takenbefore the last matchSo, A.T.Q,
= . x .x
12 4 26 12 2
4or 12.4x + 26 = 12.2x + 48.80.2x = 22.8
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22.8 1140.2
x
No. of wickets taken before the last match= 114
13.(A) Average speed during the entire journey
=Total distance
Total time
3584km 3584km2days 8hours 56hours
= 64 km/hourNow, Average speed during the remainingpart (last 8 hr.) of journey
=3584 – (1440 1608)
8 km/hr.
= 3584 30488 = 536/8
=67 km/hrSo, required difference = (67 – 64) km/hr
= 3 km/hr 3 km/hr. more
14.(A) Age Age height
9 yr. 9 3 4 ft.
(9+7) yr.=16 yr 16 4 4 4 ft3
= 15 ft3
15.(A) Let the required ratio is x : ySo, A.T.Q,
=
192 x 150 y 120 194.40x y 100
192 x 150yor, 162
x y
or, 192 x + 150 y = 162 x + 162 yor, (192 – 162)x = (162 – 150)yor, 30x = 12yor, 5x = 2y x : y = 2 : 5
16.(B) weight of lead per kg in the new alloy
3 3 1 kg(5 4 2) 2 24 8
17.(B) Required average rate of interest perannum
=
1 1 1 110 9 1 – 12 %2 3 2 3
= (5 + 3 + 2) % = 10%
18.(C) Let the income of Sanjay two yrs. ago= Rs. x
Saving of Sanjay two yrs ago= 20% of Rs. x = Rs. x/5 Expenditure of Sanjay two yrs. ago.
=
4–5 5xx x
Two years later now,
income of Sanjay = 120 6Rs. x Rs. x100 5
and saving of Sanjay = xRs.5
Expenditure of Sanjay =
6 xRs. x – x5 5
So,% increase in the expenditure
4x – x5 100%
4 x5
x 100% 25%
4x19.(B) Price is reduced by 20%
Consumption can be increased by
20 100%100 – 20
= 25% 25% of initial consumption = 500 gm Initial consumption (ie. 100%) = 2000gm
= 2 kg Original price of the Sugar per kg
= Rs. 36/2kg= Rs. 18/kg
20.(B) Let the maximum marks = xCase (i) Pass marks = 32% of x + 16Case (ii) Pass marks = 36% of x – 10from Case (i) & Case (ii), we get,32% of x + 16 = 36% of x – 10or, 4% of x = 26
or, 4 x 26
100
x =
26 100 650
4So,
Pass% =
1632% 100 %650
= 6 632% 2 % 34 %
13 13
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21.(D ) Re. 1 50pcoins coins
Ratio of respective values = 13 : 11Ratio of value of 1 coin of each = 2 : 1
So, Ratio of no. of coins =132
:111
= 13 : 22 total no. of coins = 210
So, No. of Re.1 coins =
13 21013 22
13 21035
= 78 coins22.(A) By Alligation method
12
35
49
12
49
118
35
12
110==
Required ratio of mixture = 1 1:
18 10= 10 : 18= 5 : 9
A.T.Q, Amount of the former mixture= 3 litre
So, required of the later mixute
935
litre
255
litre
23.(D) Let the original number of boys and girlsbe 5x and 3x respectively and that of newboys and girls be 5y and 7y respectively.5x + 3x + 5y + 7y = 1200 2x + 3y = 300 ......... (i)
and 5 5 73 7 5
x yx y
25x + 25y = 21x + 49y 4x = 24y x = 6y ......... (i)From equation (i),4x + 6y = 600 5x = 600 x = 120 Original no. of students= 8x = 960
24.(A) Ratio of first and second class fares = 3:1
and Ratio of no. of passengers = 1 : 50 Ratio of total amount from 1st & 2nd classpassengers= 3 × 1 : 1 × 50 = 3 : 50So, Amount collected from 2nd class
passengers = 50 132552
=Rs. 1250
25.(C) ATQ,Ratio of money received by each(Son : Daughter : Nephew)= 5x : 4x : xSo, Ratio of amount to5 Sons : 4 daughters : 2 nephews= 25x : 16x : 2x25x : 16x : 2x = Rs. 8600or, 43x = Rs. 8600x = Rs. 200Required money to each daughter= 4 × 200 = Rs. 800
26.(C)2 2A 40 B 205 7
9 1017
C x (let)
5 7( – 40), ( – 20)2 2
A x B x
and, 17 ( –10)9
C x
5 7 17( – 40) ( – 20) ( –10)2 2 9
x x x
= 600x = 100
A’s share = 5. (100 – 40)2
Rs = Rs. 150
27.(D) Extra interest received in 4 years if therate of interest is increased by 1%
= (4×1)% of 1200= 4% of 1200= Rs. 48
Total amount received in 4 years if the rateof interest is increased by 1%
= 1632 + 48= Rs. 1680
28.(C) Reparied gain = 12 6 – 4 % 50004
of
12 2 % 50004
of
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92 50004 100
= Rs. 22529.(A) By method of alligation
-6% 14%
-4%
{14-(-4)}%= (14+4)%=18%
{-4-(-6)}%= (-4+6)%=2%
Ratio of Amount = 18:2 = 9:1 Quantity sold at 14% profit
= 1 50 5
10kg kg
Quantity sold at 6% loss
9 50 4510
kg kg
30.(C) n = 2 years, r = 10% C.I. = 525
. . 1 –1100
nrC I P
210525 1 –1100
P 121 –100100
P
525 10021
P = Rs. 2500
Now, ATQ, n1= 4 years, r1 = 5%
So, 1 1. .100
P r nS I 2500 5 4
100
= Rs. 500
31.(D) Let Rs. x be the sum borrowedSo, Rs. x after two years will become
Rs. x 25 441x1 Rs.
100 400
Present value of Rs. 441x 882 Rs.800
x
Present value of Rs. 882 = Rs. 800Now, Amount of Rs. x after one year
5 211 .100 20
xx Rs
Present value of = Rs. 2120
x after one year
= Rs. x
Present value of Rs. 882 due after oneyear
20 882 .84021
x Rsx
Required sum= Rs. (800 + 840) = Rs. 1640
32.(A) Let Rs. x be the marked price of the shirt.ATQ, Difference of discounts = 2% 2% of x = 15
2 15
100x
15 100 .750
2x Rs
33.(C) Let the first CP of the commodity be Rs.100 First SP = Rs. 110Second CP = Rs. 90.
Gain = 26 %3
=50 %3
Second SP
50100 % .903
of Rs
350 903 100
= Rs.105
Difference of SPs = Rs. (110 – 105) = Rs. 5 If the difference is Rs.5, then CP = Rs. 100So, If the difference be Rs. 2, then
100 2 .405
CP Rs
34.(C) For the first trader,Let the CP of the article = Rs. 100 SP = Rs. 120Now, For the second trader,SP of the article = Rs. 120& Gain = 20%Let the CP be Rs. x.
120 – 100 20
120x
6120 – 20 245
x
120 – 24 .96x Rs Gain = Rs. 24Now when difference of gains = Rs. 4,then SP = Rs. 120So, When the difference = Rs. 85,
then SP = 120 85
4 = Rs. 2550
35.(A) Let the C.P. = Rs. 100,When sold at 3/4 th of S.P. the loss is 4%. S.P. in this case = Rs. 96 = 3/4 times Actual selling Price.
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Actual selling price = (96)×4/3 = Rs. 128If Gaurav sells at the actual S.P. then hemakes a profit of Rs. 28 on a cost price ofRs. 100 i.e. profit = 28%.
36.(A) Let the marked price of the article= Rs. xSingle discount for successive discounts of30% and 20%
30 2030 20 – %100
(50 – 6)% 44% discountNow, A.T.Q.(100 – 44)% 2240of x
56 2240
100x
2240 100 .4000
56x Rs
37.(D) Let the printed price of the book = Rs. x.So, Selling price = 90% of x
9.10
xRs
Now, if the CP of the book = Rs. y. (let)Then, A.T.Q,
112 9100 10
xy
or, 9 100 45
10 112 56
yx
Required ratio = 45 : 5638.(A)Let x= number of months (from starting)
after which C joined the business.So, Ratio of shares of Profit= 30,000×12 : 40,000×8: 50,000× x= 32 : 36 : 5x
C’s share 5
36 32 5x
x
5
68 5x
x
given, 5
68 5x
x=
1500049000 x = 6
C joined the business (i.e. 6-4) =2 monthsafter joining of B
39.(A) Let 1 child’s 1 day’s work =x& 1 adult’s 1 day’s work = y
Then, 112
16x
1192
x
and 18
12y
196
y
Work done in 3 days by 16 adults
1 116 396 2
part
Remaining work 12
part
Now, (6 adults + 4 children)’s 1 day’s work
6 4 196 192 12
i.e. 112
work is done by them is 1 day
So, 12
work will be done by them in
= 1122
days
= 6 days40.(C)Time taken by A to complete the work
4 32
= 6 days
& Time taken by B to complete the work
= 6 5 10
3
days
A and B together will complete the work
in 6 106 10 days =
334
days.
41.(C) Completed road in 80 days by 280 workers
= 72
km= 3.5 km
Remaining road to be completed in 20days = 1.5 kmLet, total x workers are needed to constructthe 1.5 km road in 20 days.
So, 280 80 3.5x 20 1.5
x = 80 1.528020 3.5
x = 480 workers No. of more workers needed
= (480–280) people= 200 workers
42. (B) Ratio of wages of A, B, C= (6×5): (4×6): (9×4) = 30 : 24: 36
= 5 : 4 : 6
A’s Share = 5 1800
15 = Rs. 600
43.(D)Part of tank filled in one hour by inlet pipe
1 1 1
12 15 60 part
So, the inlet pipe can fill the tank in 60 hrs. Inlet pipe fills water at the rate of 5 litresper minute
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Capacity of tank= (60 × 60 × 5) litres = 18000 litres
44.(A) Part filled by 1st pipe from 8 a.m. to 11 a.m.
= 3 1
15 5 part
Part filled by 2nd pipe from 9 a.m. to 11 a.m.
= 2 1
12 6 part
Part filled till 11 a.m.
1 1 6 5 115 6 30 30
part
At 11 a.m. 3rd pipe is also opened to emptyit.Now, time taken by 3 pipes together tocompletely empty the full cistern =
1 1 10 hrs.
1 1 1 1– –4 15 12 10
So, Time required to empty 1130 filled part
= 11 111030 3
hr =233 hrs.
i.e. 3 hours 40 minutesi.e. at 11. 40 a.m.
45.(D)Let the speed of the bus = x km/hrThen to take a lead of 60m, he will have tocover a distance of (60+40)m= 108m, withthe speed of (30–x) km/hr in 20 sec.100m = (30–x) km/hr × 20 sec.
100 = (30 – x) 1000 20
3600
1 (30 – )10 180
x
or , 180= 300–10x 10x = 120 x = 12 km/hr
46. (C)
r
Circumference = 2 r
Time taken for one round= 408 = 5 min.
Now, new radius = 10 r
So, New circumference = 2 ×10r = 20 r
So Required time = 20 r 52 r
minute
= 50 minutes47.(B) In such type of questions, required ratio of
the speeds of the two trains = 9 3
24 = 3 : 2
48.(A) Let the speed of goods train = x km/hrDistance travelled at the speed of 80 km/hr in 4 hours.
= 4 × 80 = 320 km
32010
x = 32 km/hr
49.(B) Let the length of train or platform = x metre.Speed = 90km/hr
= 90 5
18
metre/sec.
= 25 metre/sec. Distance covered in 60 sec.= 25 × 60 = 1500 metresNow, according to question,2x = 1500 x = 750 metre
50.(A) Speed in still water = 12km/hrspeed against the current
4 123
= 9km/hr (80 min= 43 hr)
Speed of current = 12 – 9 = 3km/hrSpeed with the current = 12 + 3 = 15 km/hr
So, required time =45 4515 9
= 8 hr.
51.(A) Let the cost of one saree = Rs. xand the cost one shirt = Rs. yAccording to question
20x + 4y = 1600 x + 2 y = 800 ........... (i)andx + 6 y = 1600 ........... (i)on solving equations (i) and (ii), we getx = 400; y = 200; cost of 12 shirts= 12 × 200 = Rs. 2400
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52.(D)4 21– 1 1
1 (1– )x x Ax x x x
)1(1
11)1)(1(
2
22
xxxx
xxx
= A
)1(1
1)1()1)(1)(1(
2
2
xxxx
xxxx
= A
A = 1
53.(*) 1/3( 2 1)x
3 2 1x
Now, 31 1
2 1
x
1 2 –12 1 2 –1
= 2 1
2 –1
2 –1
Now, 33
1– ( 2 1) – ( 2 –1)xx
= 2 1– 2 1 2 54.(C) Let p(x) = (x+1)7 + (2x + k)3
(x+2) is a factor of p(x) p(–2) = 0 [by factor Theorem] (–2 + 1)7 + (2 × –2 + k)3 = 0 (–1)7 + (k–4)3 = 0 (k – 4)3 =1
k – 4 = 3 1 1
5k 55.(C) (y – z) + (z – x) + (x – y) = 0
[a3 + b3 + c3 = 3abc if a + b + c =0] (y – z)3+ (z – x)3 + (x – y)3= 3(y – z) (z – x)(x –y)
56.(D) 1 1 1 11 1 1 1
1 2 3x x x x
1 2 3 41 2 3
x x x xx x x x
4xx
57.(B) 5 12 13x x x
The above statement is true only for 2x
x = 22 = 4
58.(C) 3( 3 2) ,a
3( 3 – 2)b
a.b = 3
( 3 2)( 3 – 2)
3 33 – 2 (1) 1
= 1 1( 1) ( 1)a b
= 1 1
1 1a b
1 11
b aab b a
= 2
1 1a b
a b
22
a ba b
[ ab =1]
= 159.(B) x y za b c k (Say)
zyx kckbka111
,,
2b ac
yk2
= zxk11
)(
2 1 1y x zk k
2 z xy zx
2zxyz x
60.(C) let p(x) = ax3 + 3x2 – 8x + b (x + 2) is a factor of p(x) P(–2) = 0 a (–2)3 + 3(–2)2– 8(–2) + b =0 –8a + 12 + 16 + b = 0 –8a + b + 28 = 0 .............. (i)Again, (x–2) is factor of p(x) P(2) = 0 a(2)3 + 3(2)2 – 8.2 + b = 0 8a + b – 4 = 0 .............. (ii)On adding (1) & (2), we have2b + 24 = 0b = –12On substituting b = -12 in (2)8a –12 – 4 = 0 a = 2
61.(C) sinB = 21
= sin30º
030B Now, 3 cos B - 4cos3B= 3 cos 300 - 4cos3300
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3 3 33 – 42 8
3 3 2 3– 02 2
Another method33cos – 4cosB B
= – cos3B= -cos 3× 300
= - cos900
= 062.(A) tanA =1 A = 450
tan B = 3 = B = 600
Now, cosA. cosB + sinA.sinBcos450.cos600 + sin450.sin600
1 1 1 32 22 2
=
2213
63.(B)
A
P C B300 m
Pole
tan = 125
5
12ABBP
5
300 12AB
BC
------------- (1)
tan B = 43
34
ABBC
------------- (2)
On dividing (1) by (2), We have
5 4 5300 12 3 9
BCBC
9BC = 5BC + 1500
BC = 41500
= 375m
Height of the pole = AB = 34
BC
= 3 3754
= 1125 281.25
4m
64.(A) (sinA+cosecA)2 + (cosA + secA)2
=sin2A + cosec2A + 2sinA.cosecA+ cos2A+ sec2A + 2cosA.secA= sin2A +cos2A + cosec2A + sec2A + 2.1 + 2.1= 1 + 1 + cot2A +1 + tan2A + 4= 7 + cot2A + tan2A
65.(D)
A B300 m
Tower
b m
In ABQ
ABQA
tan
ABb
tan
tanbAB
In ABP
ABPB
tan
tan
tan bPB
PQ =
tan.tan b
= cottanb
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66.(C)
h2
In CBD,
tan = h
BD
BD =
cottan
hh
In CBA,
tan = BACB
= DABDCB
= 2
cot hh
h
(BD= hcot )
tan =
2cot hh
h
1 1cot2 tan
cot +12
= cot
cot – cot = 12
67.(B)
Let the height of the tower be x unitIn CBA
tan CB xBA BA
BA = tan
x = x cot ....................... (i)
In DBA
tan =DBBA
= coth x
x ( BA = x cot )
xcot = tan
h x= ( )h x cot
x(cot –cot ) = hcot
cot
cot cothx
tan
1 1–tan tan
h
=
tantan
tantantan
h
tantan – tan
h
68.(B) 6 6 4 42(sin cos ) – 3(sin cos ) 1q q q q
= 3 3 2 22 2 2 42 sin cos 3 sin cos 1 q q q q
32 2 2 2 2 22 sin cos – 3sin cos (sin cos )q q q q q q
22 2 2 2–3 sin cos 2sin .cos 1q q q q
3 2 22 1 – 3sin .cos .(1)q q -3 [(1)2 – sin2q.cos2q] + 1
= 2 – 6sin2q cos2q – 3 + 6sin2q cos2q + 12 – 3 1 0
69.(C ) 2 3sin sin sin 1 3 2sin sin 1– sin
On squaring both sides,
422 cos)]sin1)([(sin
2 2 2 4(1– cos )(2 – cos ) cos
]coscos44)[cos1( 422 = 4cos2 4 2 44 – 4cos cos – 4cos 4cos
6 4– cos cos 6 4 2– cos 4cos – 8cos 4 0
6 4 2cos – 4cos 8cos 4
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BC2 = BO2 + CO2 pythograms Theorms )CD2 = CO2 + DO2
AD2 = AO2 + OD2
Now, AB2 + CD2 = AO2 + BO2 + CO2 + DO2
AD2 + BC2 = AO2 + OD2 + BO2 + CO2
Hence AB2 + CD2 = AD2 + BC2
73.(A)
A
B D C
E
E is the mid point of AD BE is the median. ar ( BED) = ar ( ABE)
= 1 ar ( ABD)2
.............. (1)
[A medians divides each into two parts ofequal areas]Similarly, we can write
ar ( CED) = ar (AEC)= 12
ar ( ACD)
On adding (1) & (2)
ar ( BEC) = 12
ar ( ABD)+12
ar ( ACD)
=12
ar ( ABC)
74.(C) Let one of the two adjacent angles be of x0,
other adjacent angle = 23 x0,
Now, 0 0 02 1803
x x
[Adjacent angles of a gm aresupplementary]
021 1803
x
x = 0 03180 1085
Smallest angle = 0 02 2 108 723 3
x
75.(B) BCE CBD BDC (Exterior angle of a is equal to the sum ofopp. interior angles. )650 = 280 + BDC
65 – 28 = BDC
370 = BDCAlso, ABDC & BD works as a transversal.
70.(C ) tan aqb
sincos
q aq b
2
2sincos
a q ab q b
Applying C & D, we have2 2
2 2sin cossin – cos –
a q b q a ba q b q a b
71.(B) D C
A B
O
P
Q
ABCD is a gm whose diagonal BD=18cm.Let both the diagonals bisect at 'O' DO = OB = 9 cm.
DO and BO are medians of ADC & ABCAlso P & Q are centroids of ADC & ABC
1PO BO3
& 1QO DO3
[centroid of a divides each median in theratio of 2 : 1]
1PO 93
& 1QO 93
= 3 cm & = 3 cm PQ = PO + QO = 3 + 3 = 6cm
72.(B)
ABCD is a quad. where diagonals AC & BDintersect each other at O. Also AC BDAB2 = AO2 + BO2 (By using
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BDC = ABD (Alternate interior angles)370 = ABD
76.(D) Let the side of the square be x cmLenght of the rectangle = (x + 5) cmits breadh = (x – 3) cmATQx2 = (x + 5) (x – 3)x2 = x2– 3x + 5x –152x = 15
x = 15 7.52
cm
Perimeter of the rectangle = 2 (l + b)= 2 [(7.5 + 5)+ (7.5 – 3)]= 2 [12.5 + 4.5]= 2 × 17 = 34 cm.
77.(D) EF || DC (Given)EGF CGD (by AA Similarity)
EG EFGC DC
510 18
EF
EF = 10518
= 9 cm
78.(C) ||C F A B BCF ABC
(alternate interior angles) = 850 (Given) BCE = BCF + ECF
= 850 + 200
= 1050
BAD BCF
(Angles in the alternate segment)= 1050
79.(*) coordinates of the mid-point of (4,2) & (6,4)
4 6 2 4,2 2
= (5, 3)Equation of the required straight line is
–3 – 3– 3 ( – 5)–5 – 5
y x
6– 3 ( – 5)10
y x
6x – 30 = 10y – 30 6x – 10y = 0 3x – 5y = 0
80.(A) The equation the circle is(x +1) (x + 2) + (y –1) (y + 3) = 0 x2 + 3x + 2 + y2 + 2y – 3 = 0 x2 + y2 + 3x + 2y –1=0
On comparing with the standard eqn. ofcircle.x2 + y2 +2gx + 2fy + c = 0
g = 32
, f = 1, c = –1
rad. of the circle = 2 2 –g f c
= 2
23 1 – (–1)2
= 9 1724 2
Area of the circle = 2r
=
2172
174
sq. unit.
81.(B) BDE BCA
(both s are equilateral equi angular) ar ( BDE) : ar ( ABC) = BD2 : BC2
221 BC :BC
2
2 21 BC : BC
4
= 1 : 482.(A) Let the parallel sides of the trapezium be
5x cm & 7x cm
its area = 1 [5 7 ] 142
x x
336 = 12x × 7
336
7 12
x
x = 4smaller of the parallel sides = 5x cm
= 5 × 4 =20 cm.
83.(B) Volume of the wooden block = 5×10×20 cm3
Volume of the required solid wooden cube= 5 × 10 × 20 × x3 cm3
(where x3 is an unknown no.) only ‘8’ is the smallest perfect cube x3 = 8 No. of wooden block = 8
84.(C) Let the original area of the cube= 6x2 sq. unit.
Side of the new cube = 3x unitits area = 6×(3x)2 = 6 × 9x2= 54x 2 sq. unit.
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% increase in area = 2 2
254 – 6 100
6x x
x
=2
248 100 800%6
xx
85. (C) A
B C
I
D
Incentre & circumcentre of an equilateral is same.let ‘a’ unit be the side of ABC
Then AD = 2 2AB – BD2
2 a 3a – a4 2
AI : ID : 2 : 1
AI=23 AD=
23 ×
3 a2
=a3
unit.
ID = 1 1 3 1AD a a3 3 2 2 3
unit.
Now, Area of circumcircle – area of incircle= 44
2 2a a– 443 2 3
222 1 1– a 44
7 3 12
1214
722
a2 = 44
2 44 12 7a 5622 3
Area of the = 2 23 3a 56 14 3cm4 4
86.(A) Let r1 : internal radius r2 : external radius
h : height of the pipe
Volume of the metal = 2 22 1r h – r h
748 = 2 21
22 9 – r 147
21
748 7 81 – r22 14
17 = 81–r1
2
r12 = 81–17 = 64
1r 64 8cm thickness = 9–8 = 1 cm
87.(B) Area of the tank = 2(lb+bh+lh)–lb=2[25×12+12× 6 + 25 × 6]– 25 × 12= 2[300 + 72 + 150] – 300= 2 [522] – 300= 1044 – 300= 744 m2
Cost of plastering = Rs. 744 × 75Rs. 55800
88.(A) h : heightc : curved surface areaV = Volume of the coneC= rl
V = 13 r
2h
3V = r2hNow, qv2–c2h2+3Vh3
= (r2h)2-(rl)2h2 + (r2h)h3
= 2r4h2 + 2r2l2h2+2r2h4
= 2r2h2(r2–l2) + 2r2h4
= 2r2h2×–h2+ 2r2h4
= 2r2h4+ 2r2h4
= 0
89.(B) Area of quad. ABCD = 12
× diagonal ×
(Sum of offsets onthe given diagonal)
= 1 16 (9 7)2
= 8 × 16 = 128 cm2
90.(*) r = 10 cmh = 48 cm
48 cm
10 cm
20 cm
h1
Volume of the water in the conical vessel= Volume of the water in the cylindricalvessel.
13 r
2h =r12h1
13 ×(10)2×48= (20)2×h1
110 10 48h20 20 3
= 4 cm
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91.(A) r = 72
cm
h = 12 cmVolume of the water in pipe in 1 sec. =r2h
= 27
27
722
× 12
= 66 × 7 cm3
Volume of water stored in (3600 seconds)1 hr.= 66× 7 × 3600 cm3
= 1663200 cm3
= 10000001663200
m3
= 1.6632 m3
= 1663.2 l
92.(D) ar ( GBD) = 16 ar( ABC)
6 = 16 ar ( ABC)
ar ( ABC) = 36cm2
93.(B) Area allotted for residential purpose
144 5 2acres.360
Area allotted for road purpose =
=36 15 acres.360 2
Required ratio = 2 : 12
= 4 : 194.(A) % of area allotted for water area and green
zone.
= (108 18) 5
360 1005
= 1.75 100
5
= 35%
95.(D) Land allotted for green zone
= 108 5 1.5acres360
Land allotted for commercial purpose
= 54 5 0.75acres360
difference = 1.5 – 0.75 = 0.75 acres.
3 acres4
96.(C) Total land allotted for residencial &
commercial purpose = 144 54 5
360
198 5 990360 360
acres
= 2. 75 acres = 432 acres
97.(B) Average distribution of loan
87 104 113 120 424 106crores.
4 4
It is clear from the table that the distributionof loan in 2008 is nearest to the averagedistribution.
98.(B) % increse of disbursement of loans by allbanks from 2009 to 2010
120 –113 100113
= 700 226 %113 113
99.(D) Total disbursement of loans by (in cores)
A & B
C & D
45 56 63 71
42 48 50 49
2007 2008 2009 2010
disbursement of loans by A & B is neverequal to the disbursement of loans by C & Din any year.
100.(B) It is clear from the table that the bank Bdistributes more than 30% of the total laonsby all banks in 2010.
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SSC MAINS (MATHS) MOCK TEST–6 (ANSWER KEY)
1. (A)2. (B)3. (C)4. (D)5. (D)6. (B)7. (B)8. (B)9. (B)10. (A)11. (A)12. (*)13. (A)14. (A)15. (A)16. (B)17. (B)18. (C)19. (B)20. (B)21. (D)22. (A)23. (D)24. (A)25. (C)
26. (C)27. (D)28. (C)29. (A)30. (C)31. (D)32. (A)33. (C)34. (C)35. (A)36. (A)37. (D)38. (A)39. (A)40. (C)41. (C)42. (B)43. (D)44. (A)45. (D)46. (C)47. (B)48. (A)49. (B)50. (A)
51. (A)52. (D)53. (*)54. (C)55. (C)56. (D)57. (B)58. (C)59. (B)60. (C)61. (C)62. (A)63. (B)64. (A)65. (D)66. (C)67. (B)68. (B)69. (C)70. (D)71. (B)72. (B)73. (A)74. (C)75. (B)
76. (D)77. (D)78. (C)79. (*)80. (A)81. (B)82. (A)83. (B)84. (C)85. (C)86. (A)87. (B)88. (A)89. (B)90. (*)91. (A)92. (D)93. (B)94. (A)95. (D)96. (C)97. (B)98. (B)99. (D)100. (B)
7. Correct answer is option (3) 12 yrs.Let the ages of the boys are 4x, 5x & 7x
A.T.Q.4 5 7
3x x x
= 16
or, 16x = 48 x = 3 Age of the youngest boy = 4x
= 12 yrs.23. Correct answer is option (1) 10% of rice was spoiled rate of rice must be increased by
10100 10 × 100% = 11.11%
i.e. by 19 th of the intial price
also, 20% profit is required So, Finally the rate of rice must be
150 × (1 9)
9
×120100 = 150×
10 69 5
= ` 200
51. Correct answer is option (3) 1255 2 7x = 3 2x - 7 = 35
or, 2x = 243 + 7
x = 2502 = 125
52. Correct question should be
MOCK TEST PAPER - 3 CORRECTIONS
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32x +1 - 3x = 3x +3 - 32
& according to the above correct question,correct answer is option (4) -1 and 2
87. Correct answer is option (2) 1 : 2 : 3 Vol.(cone) : Vol. (hemisphere) : Vol. (cylinder)
= 213
r r : 323
r : 2r r
= 13 :
23 : 1
= 1 : 2 : 3
MOCK TEST PAPER - 5 ( CORRECTIONS)
14. Accroding to question printed in Englishlanguage answer is option (D) 20yrs.but, Accroding to question printed in Hindilanguage answer is option (A) 27yrs.
20. Correct option is (A) 9800correct solution Suppose the cost price of T.V. = ` xThen, 2 (x - 9400) = (10600 - x )
2x - 18800 = 10600 - x 3x = 29400 x = 980033. In the question,
ratio of C:D should be 56 :
34
and with this correction,answer should be option (C)