ssc mains (math) mock test - 9 (solution)notes.sscexamtutor.com/maths/solutions/ssc mains (maths)...
TRANSCRIPT
Centres at:
==================================================================================
1
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
SSC MAINS (MATH) MOCK TEST - 9 (SOLUTION)
1.(D) Let x = 8 2 8 2 8......... � � � , then
x = 8 2� x
squaring both sides, we get,
x2 = 8 + 2x
� x2 – 2x – 8 = 0
� (x – 4) (x + 2) = 0
� x = 4
2.(A)
2 2
2 2
3.07 .0193
.307 .00193
�
�
2 2
2 2
3.07 .0193
3.07 .0193
10 10
��� � � �� � � ��� � � �� �
2 2
2 2
3.07 .0193100
3.07 .0193
��� � � �� � �
= 100
3.(B) Men to be arranged = (6000 – 71) = 5929Number of men arranged in each row
= 2 5929 77�
4.(B) 7 5, 5 3, 9 7, 11 9� � � �
On rationalizing each term
=2 2 2 2
, , ,7 5 5 3 9 7 11 9� � � �
Smallest denominator = 5 3�
So largest value = 5 3�5.(C) Let the price of 1 book = ` x
and price of 1 pencil = ` y
According to the question:
4x + 3y = 8x + y
2y = 4x
y = 2x
Price of 4 books and 3 pencils
= 4x + 3y
= 4x + 6x = 10x
and if 6 books and 2 pencils = 6x + 2y
= 6x + 4x = 10x
Both are equal
6.(B)
So 47 is wrong
7.(C) 1000 is not a perfect square so we need
to make a perfect square.
we need 24 more plants
8.(A) 2557 –1 � 71–1 = 6 25 = 4×6+1����
9.(A) Let the five consecutive integers be x,
x + 1, x + 2, x + 3, x + 4, then
x+x+1+x+2+x+3+x+4 = a
� 5x + 10 = a..................(i)
Also, 5x + 35 = b..................(ii)
b – a 5 35 – 5 –10 1
100 100 4
�� �
x x
10.(A) I no. × II no. = L.C.M. × H.C.F.
(x2 +2x –3) × P = (x3–7x+6) × (x +3)
P =
3
2
( –7 + 6) ( +3)
2 – 3�
x x x
x x
P =
2
2
( +3)( –3 2)( +3)
2 – 3�
x x x + x
x x
=
2( +3) ( –2) ( –1)
( +3) ( –1)
x x x
x x
= (x +3) (x –2)
= x2 +x –6
11. (A)91 2 34 1 4 4 4� � �
= 914 1 4 16 64� � �
= 914 85
= 5 × 17 × 491
Divisible by 17.
12. (B) 1 + 2 + 3 +..........99 + 100 + 99+...3 + 2 + 1
(1 + 2 + 3+ ........100) + (99 + 98+....3+2+1)
By using the formula of sum of first 'n'
Natural Number = 1
2
�n n
We get, 100 100 1 99 99 1
2 2
� ��
= 5050 + 4950 = 10000
13.(A) Given
x = 0.037
Centres at:
==================================================================================
2
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
x 1\3 = 1/3
0.037
= (37/999)1/3
= (1/27)1/3
= 1/3
= 0.3
14.(B) Average age of the couple is 25 years
The sum = 2 × 25
= 50 years
After 3 years, sum = 50 + 2 × 3
= 56 years
Age of baby = 2 years
Then average (X) = � x
n
= 56 2 1
58 /3 193 3
�� � years
15.(C) Let minors be x
Consumption by adults = 8 × 15 = 120
Total Consumption = (x + 8) × 10.8Average consumption by minors
= 8 10.8 120
6 7� �
� � �x
xx
16. (C) Sum of 8 numbers = 20 × 8 = 160
1 115 2 21 3 4 7 160
2 3
� � � �� � � �� � � � � � � � �� � � �� �
x x x
31+ 64 + 3x + 11 = 160
3x = 160 – 106 � x = 54/3 � x = 18
8th Number = x + 7
= 18 + 7 = 25
17.(C) Total brass = 33 kg
copper =33 7
21 kg11
��
33 4zinc = 12 kg
11
��
copper 21 21 7 : 6
zinc 12 6 18� � �
�18.(A)
����1 1
I : II : 10 :18 5 : 918 10
� � �
Clearly, for 5l of I-Mixture required amount
of II mixture = 9l.
���� For 3 l of I–mixture required amount of
II mixture = 9 27 2
3 55 5 5� � � l
19.(A) Students Failed in Hindi = 100% – 80% = 20%
Students Failed in mathematics
= 100%–75% = 25%
Students Failed in both subject = 18%
Students Passed in both subject
= 100–(25+20–18)
= 73%
Let total students be x
73438
100
��
x� x = 600
Total students is 60020.(D) Let C.P = x
100 1001
100 100
� �� � �
r rx
100 100
100 r 100 r
��
� �x
= 2 2
10000
100 r�
x = 2
10000
10000 r�21. (C) 100 × 68/100 = 68 Kg.
Dry Fruits = 32 Kg + 20 kg = 52 kg22.(D) Let, the total number of voters be x
���� Number of votes cast in the election = 92
100x
Number of votes obtained by winner = 48
100x
���� Number of votes obtained by the defeated
Candidate = (92 – 48)
100x
= 44
100x
From question,
48 441100
100 100�
x x–
� 4x = 110000
���� x = 27500
���� Total no. of voters = 27,500
23.(C) According to the question
Let the numbers are a and b
(a – b): (a + b) : ab = 1 :7 : 24
Numbers are a = 8 b = 6
so product = 8 × 6 =48
24.(A) Let the Length of candles be L
Centres at:
==================================================================================
3
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
The rate of burn of first candle = L
4 per hour
The rate of burn of second candle = L
3per hour
Let after x hour the ratio be 2:1
L LL – 2 L –
4 3
� �� �� � ��
x x
L 1 – 2 L 1 –4 3
� � � �� � � ��� � � �� �
x x
4 – 3 –2
4 3
� �� �� � ��
x x
x = 2
25
hours = 2 hours 24 min.
25. (A) Cost of raw material = 4x
Cost of labour = 3x
Cost of miscellaneous = 2x
Total cost = 4x + 3x +2x
= 9x
Amount = 4 110 3 108 2 95
100 100 100
� � �� �
x x x
= 9.54x
Percentage rise = 9.54 – 9100
9�
x x
x = 6%
26.(A) Let the number be x
(3–x ) (7–x ) = (5–x ) (6–x )
x = 9
27. (B)2 1 2 5 7 3
a : b : ,b : c : ,d : c :9 3 7 14 10 5
� � �
a : b = 2 : 3 b : c = 4 : 5
d : c = 7 : 6 c : d = 6 : 7
then,
a : b : c : d = 48 : 72 : 90 : 105
= 16 : 24 : 30 : 35
28.(D) Given,
Total earning of
A+B+C = 76000.................(i)
percentage of their saving are
30%, 25% & 20% respectively
Let, savings of A, B & C
be 4x, 5x & 6x respectively Now, 30% of A = 4x
or,A
30 4100
� � x
or,40
A3
� x ..................(i)
also, 25% of B = 5x
or,B
25 5100
� � x
or, B = 20x.......................(iii) also, 20% of C = 6x
or,C
20 6100
� � x
or, C = 30x...........................(iv)
on using (ii), (iii) & (iv) in (i), we get
40 +20 30 76,000
3
xx + x =
or, x = 1200
����40 40
A = 1200 160003 3
� � �x
B = 20x = 20 × 1200 = 24000
C = 30x = 30×1200 = 36000
���� (A + B) – C = (40000 – 36000) = ` 400029.(C) Let money be P.
P 12 4 P 15 5– 1890
100 100
� � � ��
27P 1890 1001890 P P 7000
100 27
�� � � � � `
30.(A) Let initial amount = ` x
7 2 2 10 2 4 12 21430
3 100 5 100 15 100
� � � � � �� � � � �
�x x x
14 4 81430
300 50 125� � �
x x x
7 2 81430
150 25 125
x x x� � �
35 60 481430
750
� ��
x x x
143x = 1430 × 750
1430 750
143
��x = ` 7500
31.(C) CP of Horse = x
C.P S.P
100 25 100 32 115 60
100 100 100
� �� �� �� � � � �� �� x x
x = 375
32.(C) Let cost of 100 Articles is ` 100
( 1Article 1)� �`
Centres at:
==================================================================================
4
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
25%Profit S.Pif 100 articles
125�������������
`
� Diff = 5 unit = 100
1 unit = 20
33.(B) Let Marked Price = x
(100 –10)108
100�
x� x = ` 120
34.(A) Let C. P. = ` x
S.P. =108
100
�x
Again C. P. = 80
100
�x
S.P. = 80 140
100 100�
x =
112
100
x
����112 108
– 640100 100
�x x
x = ` 16000
35.(A) According to the Question
36.(D) According to the question
He should purchase = 400
320 50%�
= 5 shirts
37.(A) S. P. = 720 115
100
� = ` 828
Marked price = ` x
we get,
90
100
�x = ` 828
x = ` 920
38.(A) Let work done by a man in a day be x
and work done by a woman be y
���� From question,
4x + 6y = 1/8...................(i)
3x + 7y = 1/10...................(ii)
on solving (i) & (ii), we get
11
400�x
1
400�y
required ratio = 11 1
400 400� �
x
y = 11 : 1
39.(B) Given,
1/B = 2/A
or, A = 2B...........(i)
also, given 1/C = 3/A
or, A = 3C .....................(ii)
also, given 1 1 1 1
A B C 2� � � ................(iii)
on using (i) in (iii), we get
1 1 3 1
2B B 3C 2� � �
or,1 1 3 1
2B B 2B 2� � � [using (i) & (ii)]
or,1 2 3 1
2B 2
� ��
or, 2B = 12
or, B = 6 days
40.(A)2 21 1
1 2
M DM D
W W
��
2M 80100 16
1/7 6/7
���
2 2
100 16 6M M 120
80
� �� � �
Required labourers = 120 – 100 = 20
41.(A) Given,
A + B +C = 14,400............................(i)
Let savings of A, B & C are 8x, 9x & 20x
respectively.
Also given percentage of expenditure of A,
B & C are 80%, 85% & 75% respectively.
���� Percentage of savings of A, B & C be 20%,
15% & 25% respectively.
Now, 20% of A = 8x
or,20 A
8100
�� x
or , A = 40x..........................(ii)
Again, 15% of B = 9x
or,15 B
9100
�� x
or, B = 60x...........................(iii)
again, 25% of C = 20x
or,25 C
20100
�� x
or, C = 80x .............................(iv)
on using (ii), (iii) & (iv) in (i), we get
40x + 60x + 80x = 14,400
or, x = 80
���� A = 40x = 40 × 80 = 3,200
B = 60x = 60 × 80 = 4,800
C = 80x = 80 × 80 = 6,400
Centres at:
==================================================================================
5
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
42.(A) Speed = 15 km per hour
= 5 25
15 m /s.18 6
� �
Water flow out in one second
= .2×.15×25/6 m3
Volume of tank = 150× 100 × 3 m3
Time taken = 150 100 3 6
.2 .15 25
� � �
� �
= 100 hours.
43.(D) Given,
1/A = 1/8.....................(i)
also, given 1/A – 1/B = 1/20
or, 1/B = 1/A – 1/20
or, 1/B = 1/8 – 1/20
or, 1/B = 3/40
���� B takes 40/3 min. to empty the tank
also, given rate of water flowing out = 6k l
���� Capacity of tank = 40
3 × 6k l = 80 k l
44.(C) Speed = 350 60
21 1000
�� km / hr
Total time taken = 84
13 621
� �
4 + 78 min = 5 hours + 18 min.
45.(A) Let total coaches be N
decrease in the speed = x
���� N�x
K N�x [K = constant]
4 K 4�K = 2
���� 24 K N�
24 2 N�
N = 24/2
N = 144 coaches
���� Number of coaches that can be exactly
pulled by the engine = 144 – 1 = 143 coaches
46.(A) Time taken by C = 114 min.
Time taken by B = 114/3 = 38 min.
Time taken by A = 38/2 = 19 min.
47.(A) Let correct time = t1���� From question,
1 1
11 540 t 50 t
60 60
� � � �� � � �� � �� � � �� �
1
19 19 t 60
60 60h� � � � = 19 mins
48.(C) Length of train = 120m.
Speed of train Ist = 120
12 m/s10
�
Speed of train IInd = 120
8 m/s15
�
time taken = 120 120
12 8
�
� =
24012
20� seconds
49.(D) Let the speed of the man = x km and the
speed of the current = y km/h
According to the question
D15�
x – y .......(i) (Where D is distance)
D10�
x + y........(ii)
Divide (i) by (ii)
�15 3
10 2� �
x + y
x – y
� 2x + 2y = 3x – 3y
� x = 5y
�5
5 :11
� �x
y
50.(A)1 1 1
3 3 3� �x y z
zyx �� 33131 )(
zyxyxyx ���� )(3)()( 31313131331331
zzxyyx ���� 3131)(3
31)(3 xyzzyx ����
xyzzyx 27)( 3 ����
027)( 3 ���� xyzzyx
51.(C) Given,
or, x4 – 3x3 – 2x2 + 3x +1 = 0
or, x4 – 2x2 +1 – 3x3 + 3x = 0
or, (x2 –1)2 – 3x (x2 – 1) = 0
or, (x2 –1) (x2 – 1 – 3x) = 0
taking, x2 – 3x – 1 = 0
x = 2–(–3) (–3) – 4(1) (–1)
2 1
� �
�
or, x = 3 9 4
2
� �
or, x = 3 13
2
�
52.(D) (1+m2) x2 + 2mcx + c2 – a2 = 0
B = 2mc
A = (1+m2)
C = c2 – a2
Centres at:
==================================================================================
6
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
���� roots are equal ���� D = 0
B2–4AC = 0
(2mc)2 – 4 (1 + m2) (c2 – a2) = 0
4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2 = 0
–c2 + a2 + a2 m2 = 0
c2 = a2 (1 + m2)
53.(D) Given,
l x2 + nx + n = 0.................(i)
& / /� �� p q .................(ii)
Equation (i) � – /���� n l .........(iii)
& /��� n l .................(iv)
Equation (ii) � / /� �� p q .........(v)
& / /� �� q p .................(vi)
���� / / /� �p q q p n l
= ���� ���� �� (using (v), (vi) & (iv))
= 1
� ���� �
� �
= ( ) ( )
.
��� � ��
� �
= .� �
–n / l + n / l
= 0/ .� � = 0
54.(B) 3x2 + 2x +1 = 0
������ – 2/3
���� 1/3
Product of roots = 1 – 1 –
31 1
� �� �
�� ��
sum of roots = 1 – 1 –
21 1
� �� �
�� ��
Required equation is
x2 – (sum of the roots) x + product of roots = 0
x2 – 2x + 3 = 0
55.(B) 4x2 + 4x + 9
(2x)2 + 2.2x +1+ 8 make a perfect
square
(2x + 1)2 + 8
for minimum (2x +1)2 = 0 ���� 0 + 8 = 8
We have minimun value of 4x2 + 4x + 9 is 8
56.(B) Given, x = 1 2 3� � ..........(i)
����1 1
–1 1 2 3 –1�
� �x
= 1
3 2�
= 1 3 – 2
3 2 3 – 2�
�
= 3 – 2
3 – 2
13 – 2
–1�
x
(i) + (ii) �1
1 2 3 3 – 21
� � � � ��
xx
= 1 2 3�57.(A) x 2+ y 2+ z 2+2 = 2(y – x )
(x2+2x+1)+ (y2–2y+1)+z2 =0
(x+1)2 + (y –1)2 +z2 = 0
x = –1, y = 1, z = 0
���� x 3+y 3+z 3 = (–1)3 + (1)3 + 03
= – 1+1+0 = 0
58.(A)1 1 – 1 1 –
1 – 1 – 1 1 –
� � � ��
� � �
x x x x
x x x x
21 1 – 1 1/2 3
3 /2
� �� � �
x
x
59.(C) 5 2 6� �x
5 2 6� �x
= 2 2
3 2 2. 3. 2� �
= 2
3 2�
x = 3 2�
1
x= 3 – 2
13 2 3 – 2� � � �x
x
= 2 3
60.(B)
According to the questionLet the height of the telegraph pole = h
so, in ABC�
Centres at:
==================================================================================
7
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
tan30º20
�h
h = 20 20
3 AB33
� �
and cos 30º = 20
AC
AC = 20 2 40
3 3
��
so height of the pole= AB + AC
= 20 40 60
3 3 3� �
=60 3
20 33
�
61.(B) 14 20sin cos� �� �xEven power means no negative value andmax value will be 1 and never be zero.
0 1� x
62.(A)
Here cos 10º > cos 80ºso, cos 10º – cos 80º > 0 or positive
63.(D) 1 + s inx + s in 2x .......� = 4 2 3�
�1 1
4 2 31 sin 1 sin
� � �� �x x
�1
4 2 31 sin
� �� x
�1 4 2 3 3
1 sin 14 24 2 3
�� � � � �
�x
�3 2
sin sin sin2 3 3
! !� � �x
so 2
or2 3
! !� �x
64.(A)
Let height of the tower = AB
So, In ABC�
ABtan45º
AC�
so, AB = 60 m. [�tan 45 = 1]
Also, In ADB�
60 1 60tan30º
60 603� � �
� �x x
� 60( 3 1) 60 (1.73 1)� � � � �x
� x = 60 × 0.73 = 43.8 m
Required Speed = 43.8 18
31.55 5
� � km/hr
65.(A)
In AFC�
tan30º 3 h� � �h
xx
.....(i)
In AEC�
tan60º 3160
� ��
h
x
� 3 (160- ) = x h
� 3 (160 – 3 )h = h From (i)
� 4 160 3�h � h = 160 3
3 ft = 340 ft
66.(A)tan3 sin3 .cos
tan cos 3 .sin�
x x x
x x x
=
3
3
(3sin – 4sin )cos
(4cos – 3cos )sin
x x x
x x x
or,
2
2
tan3 (3 – 4 tan )
tan (4cos 3)�
x x
x x –...........(i)
tan3 3 – 4 0 3Max 3
tan 4 – 3 1
�� � �
x
x
(when x =0 º)
tan3 3 – 4 –1 1Min
tan –3 –3 3� � �
x
x
����tan3
tan
x
x vary from
1
3 to 3.
{clearly, tan3x /tan x never lies between –1/3and 0}
67.(B) tan 20º tan 80º cot 50º
= tan 20º tan 80º tan 40º
[���� cot (90º–40º) = tan 40º]
Centres at:
==================================================================================
8
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
= tan [3×20º]
[���� tan tan (60º+) tan (60º–) = tan 3�]
= tan 60º
= 3
68.(C) sin 18º + cos 36º
= 5 –1 5 1
4 4
��
= 5 – 1 5 1 2 5 5
4 4 2
� �� �
69.(B) We have,
2 4 6cos cos cos
7 7 7
! ! !� �
1 2 4 62sin cos 2sin cos 2sin cos
7 7 7 7 7 72sin
7
! ! ! ! ! !� �
!
�� �� � �
1 3 5 3 5sin –sin sin –sin sin –sin
7 7 7 7 72sin
7
! ! ! ! !� � � !
!
�� � � �
1–sin sin
72sin
7
�!� � � !
!� � �
1 1– sin 0 –
7 22sin
7
�!� � � �
! � � �
70.(C)
BL is bisecter of ABC"
so, MBL LBC" � " � x (let)
ML BC�
LBC MLB" � " � x
MLB BML" � "Can't be 90º then triangle does not exist
71.(C)
In ADC�
DAC 180 – 60 – 30 90º" � �
In DAC�
ACtan60º 3
AD� � �
a
x
3�
ax
72.(C) It is true that congruent triangles will be
similar but opposite is not true. Also III
will be true.
73.(C)
Given OA = 3 cm
and OM AC#
Let AOM" ��
In AOM�
AMsin
OA��
AC 3 1sin
20A 2 3 2�� � �
�
sin sin30º��
30º��
74.(C)
We know that tangents from an external
point on the circumference of the circle
will be equal.
AP = AS
BP = BQ
CR = CQ
DR = DS
� AP+BP+CR+DR = AS+BQ+CQ+DS
� AB+CD = AD + BC
75. (A) According to the questions sides are 5xand 12x
=1
5 12 = 2702� x × x
x2 = 9x = 3
= Hypotenuse = 13x = 13 × 3 = 39 cms.
Centres at:
==================================================================================
9
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
76.(A)
Radius of circle APBQ = 5 4 9
2 2
��
Area of circle APBQ = �R2 =
29 81
sq. cm.2 4
!! � �� � � �
77.(C) Area of square > Area of rectangle
78.(A)
ABC 50º" �DAB ABC" � " [alternate angle]
�DAB = 50ºIn �ADB
A D B 180º" � " � " �50º ADB 65º 180º�" � �
ADB 180º –115º" �ADB 65º" �
79.(A)
� PAQ 68º" �
���� PAO 68 /2 34º" � �
In APD
APD PAD ADP 180º" � " � " �
APD 34º 90º 180º" � � �
APD 56º" �
APD APQ 56º" � " �
APQ 56º" �
80.(A)
Area of the rhombus = 21
1d d
2� �
= 1
55 48 13202� � � cm2
Area of the rhombus = base × height= DC×AE
� so DC × AE = 1320
2 255 48
P 13202 2
� � � �� � � �� � �� � � �� �
5329P 1320
4� �
1320P 36.16
36.5� �
so, 36. < P < 37 cm
81.(B)
Area of circlular segment = 23
360º
�� �!r
= 260
3 (4)360º
� �!
= 8! cm2
Area of the 2 23
ABC (8) 16 3 cm4
� � � �
���� Area of the portion included between
them = (16 3 – 8 )! cm2
82.(B) Let the radius of the circle be rand side of the square be a.
so, 2 r 4! � a � given
2
!�
ra
Area of the square =
2r
2
� �!� �� ��
= 2.46r2......(i)
Area of the circle = 2r!= 3.14r2.......(ii)
so (ii) > (i) option (B) is correct
10
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
83.(B) GivenAC2 = AB × CB
x2 = 2 (2–x)
� x2 +2x–4 = 0
� x = 2 4 16
–1 52 1
� � �� �
�
� BC = 2– ( –1 5� )
= 3 – 5 [leaving (3 5)� because
it is greater than 2]
84.(C) By the property of triangle
85.(C) According to the question
86.(D)
Let height of the pillar = h metres.
In DAB, tan 30º = /3
50
h
1
3 503�
�
h� h = 50 3 metres
87.(D) Given,
l + b + h = 19............(i)
& diagonal of cuboid = 11
or, 2 2 2 11� � �l b h
or, l 2+ b 2+ h 2 = 121........(ii)
we know that surface area of cuboid
= 2 (l b + b h + h l )
Now, (l + b + h )2 = l 2+ b 2+ h 2 + 2(l b + b h + h l )
� 2(l b + b h + h l ) = (l+b+h)2 – (l 2+ b 2+ h 2)
= (19)2 – (11)2
= 361 – 121
= 240 cm2
���� surface area of cuboid = 240 cm2
88.(C) Let height be h m.
Volume of remaining field = Volume of
soil
[18×15–7.5×6]. h = 7 . 5×6×8
7.5 6 8 .16m 16 cm
270 – 45
� �� � � �h h
89.(C)
Let slant height be h cm
Pyramid height = h/2
In EDF22 2
2 22 4 4
2 4 3 33
� �� �� �� �� � � � � � �� � � �� �
h hh h h
height = 4 1 2
3 2 3� �
Volume of pyramid = 1
3 (Area of base) × height
= 21 3 2
43 4 3
� �� �� �� ��
= 38
3 cm9
90.(C)
In ABC
2 2AC 10.5 14� �
AC = 17.5m
l = 17.5 m
Total surface area = r l + 2 r h
= 22 22
14 17.5 2 14 37 7� � � � � �
= 1034 m2
The cost of painting= 1034 × 2
= ` 2068
91.(A)
Let length be x
breadth be y
(x+14) ( y–6) = xy
xy – 6x +14y – 84 = xy
14y – 6x = 84 ................(i)
(x–14) (y+10) = xy
Centres at:
==================================================================================
11
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
xy + 10x – 14y – 140 = xy
10x – 14y = 140...............(ii)
solving (i) & (ii) we get,
x = 56, y = 30
92.(D) From graph, in standard (viii) the failure
of girls is minimum.
93. (*) Note :Read more in place of less in question
then, the solution is
Average result of the girls =
80 80 40 90 70 70
6
� � � � �=
430
6 = 71.67
clearly from graph in standard (ix) the
results of boys is more than the average
result of the girls.
94. (A) From graph,
The ratio of results of girls and boys in the
standard (vi)
= 80:70
again,
the ratio of results of girls and boys in the
standards (ix)
= 70:80
clearly, in pair (vi), & (ix) of standard the
results of girls and boys are in inverse
proportion.
95.(*) Note : Read less in place of more in question
then, the solution is
Average result of the boys
= 60 70 60 60 80 60
6
� � � � �= 390/6 = 65
Clearly from graph, in standard (vii) the
results of girls is less than the average
result of the boys.
96.(A) No. of students come by bus = 7200 50º
1000360º
��
97.(D) Total angle = 20º + 50º = 70º
required no. = 7200 70º
1400360º
��
98.(B) The ratio of number of students who come byfoot to the students who used two-wheeler
20º 1 1 : 6
120º 6� � �
99.(B) Ratio of students who come by car to that bycycle
= 1200
6 :112200
�
100.(*) Number of students come by (cycle + bus)
= (110º 50º) 160º
7200 7200 3200360º 360º
�� � � �
Number of students come by
Two-wheeler = 120º
7200 2400360º
� �
���� Exceed % = 3200 – 2400100
2400�
= 800 100 1
100 33 %2400 3 3
� � �
Centres at:
==================================================================================
12
����MUKHERJEE NAGAR ����MUNIRKA ����UTTAM NAGAR���� DILSHAD GARDEN ����ROHINI����BADARPUR BORDER
SSC MAINS (MATHS) MOCK TEST - 9 (ANSWER SHEET)
1. (D)2. (A)3. (B)4. (B)5. (C)6. (B)7. (C)8. (A)9. (A)10. (A)11. (A)12. (B)13. (A)14. (B)15. (C)16. (C)17. (C)18. (A)19. (A)20. (D)
21. (C)22. (D)23. (C)24. (A)25. (A)26. (A)27. (B)28. (D)29. (C)30. (A)31. (C)32. (C)33. (B)34. (A)35. (A)36. (D)37. (A)38. (A)39. (B)40. (A)
61. (B)62. (A)63. (D)64. (A)65. (A)66. (A)67. (B)68. (C)69. (B)70. (C)71. (C)72. (C)73. (C)74. (C)75. (A)76. (A)77. (C)78. (A)79. (A)80. (A)
81. (B)82. (B)83. (B)84. (C)85. (C)86. (D)87. (D)88. (C)89. (C)90. (C)91. (A)92. (D)93. (*)94. (A)95. (*)96. (A)97. (D)98. (B)99. (B)100. (*)
41. (A)42. (A)43. (D)44. (C)45. (A)46. (A)47. (A)48. (C)49. (D)50. (A)51. (C)52. (D)53. (D)54. (B)55. (B)56. (B)57. (A)58. (A)59. (C)60. (B)