The document is a collection of answers and solutions for a mock test consisting of various mathematical problems. It covers algebra, geometry, trigonometry, and basic arithmetic operations, providing detailed calculations and explanations for each solution. The problems included involve finding distances, solving equations, calculating areas, and working with ratios, among others.

1. P-1
Mains Mock Test – 14 [Answer with Solution]
1. (A)  OM  ABAB
 AM = MB =
1
2
AB
=
1
2
× 10
BA
C D
M
O
5cm
13cm
13cm
12cm N
AB = 10 cm
CD = 24 cm
= 5 cm
[perpendicular from centre to anychord bisects
the chord]
Similary, CN = ND =
1
2
CD= 12cm
In AMO,
OM2
= OA2
– AM2
= 132
– 52
= 169 – 25 = 144
OM = 12 cm
In CNO,
ON2
= OC2
– CN2
= 132
– 122
ON = 5 cm
 The distance between the two parallel
chords = MN = MO + ON
= 12 + 5 = 17 cm
2. (B) (x – 1) (x + 2) (x – 3) (x + 4)
 (x2
+ x – 2) (x2
+ x – 12)
 x4
+ x3
– 2x2
+ x3
+ x2
– 2x – 12x2
– 12x + 24
 x4
+ 2x3
– 13x3
– 14x + 24
So, cofficient of x3
= 2
3. (A) Put x=8
y= 1
Condition satisfies
 x4
+ y4
= 84
+ 14
= 4097
4. (C) x = 3 – 5
Squaring both sides
x = 8 – 2 15
x – 8 = – 2 15
Again square both side
x2
+ 64 – 16x = 60
x2
– 16x = – 4
x2
– 16x + 16 = – 4 + 16 (adding '16' on both sides)
x2
– 16x + 16 = 12
5. (B) We know,
sin30 =
1
2
, cos60 =
1
2
So, replacing values
sin (+–) = sin30
So, +– = 30 ...(i)
Similarly,
+– = 60 ...(ii)
+–= 45 ...(iii)
Adding (i), (ii) and (iii)
++=135 ...(iv)
Adding (i) and (iv)
2 (+)=165
6. (B) x2
=
5 2 6
5 2 6



(5 2 6)(5 2 6)
25 24
 

(after rationalizing)
 x2
= 2
(5 2 6)
To find  x2
(x – 10)2
Replacing values of x
 (5 2 6) 2
(5 2 6 10) 
 (5 2 6) 2
(2 6 5)
 (a + b) (b – a)
 (b2
– a2
) = (25 – 24) = 1
7. (B)
a
b
–
b
a
= 3
a
b
+
b
a
=
2
a b
4
b a
 
   
 
a
b
+
b
a
= 13
After cubing both sides
So,
3
3
a
b
+
3
3
b
a
= ( 13 )3
– 3 13
 13 13 – 3 13 = 10 13

2. P-2
8. (C) For equation, ax2
+ bx + c = 0
Product of roots =
c
a
So, product of roots for
x2
– 2x + 4 
4
1
From equation (c)
x2
– 4x + 8
Product of roots =
8
1
So, option (C) is satisfied.
9. (A)
40
41
9
As it is right angle 
In Radius =
9 40 41
2
 

8
2
= 4
Circumradius =
41
2
= 20.5
So, sum = 20.5 + 4 = 24.5
10. (D) Let x = 1
The equation will be satisfied
x –
1
x
= 1 – 1 = zero
11. (C) D C
BA
x
y
180
180 


C exterior = 
C interior = 180 – 
A interior = 180 –
Weknowin a quadrilateral sum ofall angles remain
360º.
So, x + y+ 180 –+ 180 –= 360
x + y = +
12. (A) Let
2 2
2 2
2a
b c
x 

= 2,
2 2
2 2
2b
c a
x 

= 2 ,
2 2
2 2
2c
a b
x 

= 2
So,
2 2
2 2
2a
b c
x 

= 2
x2
= 2b2
+ 2c2
+ 2a2
13. (D) Usual
7
Present
6Speed
Time 6 7
Difference oftime = 1 unit = 15 min.
So, usual time to be taken 
15× 6 =90min.
 1hr. 30min.
14. (B) A B
5 8
7 5
2 days 3 days
So, A (2 days) = B (3 days)
A
B
=
3
2
So, total work = 5 × 3 + 8 × 2 = 31
So, time to be taken =
31
5
= 6
1
5
15. (A) 60
20 – 1530
– 43
2
As the pipes are opened alternatively for in
first 3 minutes = 3 + 2 – 4 = 1 unit is filled
165 min.
55×3
55 unit
per
Now, in next twominutes 5 litres will be filled and
total timewill be 165 + 2 = 167 min.
16. (D) Downstream = 5 + 1 = 6
Upstream = 5 – 1 = 4
24
6 4
64
6 – 4 = 2 unit = 2 hr.
So, distance = 24 km.

3. P-3
17. (C) Speed =
1000
60
x
×
18
5
km/hr
Time = yhour
Distance = speed × time
 x × 20 × 3 × y
= 60 x y= 4 cm = 60000 xymeter
18. (C)
135
15 12
=
135
27
 5 hours.
19. (A) Volumes of two solid = Volume of cone
4
3

 3 3
1 2r r =
1
3
× r2
h
4
3
(1 + 27) =
1
3
× r2
× 7
Solving we get
r= 4
diameter = 8cm.
20. (D) 20% 5n
6n
put n = 4 condition satisfied
21. (C) Circumference
Cone base
r = Radius of cone
Let r = Radius of sector
Sector Radius will be equal to slant height of cone
also circumference of cone will be equal to cone
circular base
So, 2r ×
90
360
= 2r1
r = 4r1
r1
=
r
4
So, CSA of cone =r1
l =
22
7
×
r
4
× r =
2
r
4

22. (A) Newrate = 4%
Time = 2 times
So, overall rate = 4 + 4 +
4 4
100

 8.16
23. (B) Height of cone = 40
Volume of smaller cone Volume of Large cone
1 : 64
Taking cube root
1 : 4
10 : 40 (Given)
×10 ×10
So, height from base was (40 – 10) = 30 cm.
24. (D) 16
25. (C) 18
26. (B)
27
45
× 100 = 3 × 20 = 60%
27. (A)
28. (A)
29. (A) Using option
30. (B)
Walk
Ride
A B =37min. .....(i)
Walk
Walk
A B =55min .....(ii)
From (ii) ifI walkAtoBonly, it will take
=
55
2
min= 27.5min
So, from(i)
Walk+ Ride = 37min
27.5min + ride= 37min
 byride I will take = 37 – 27.5 min
[to go A to B only]
=9.5 min
Thus, to ride both wayit will take me
= 9.5 × 2= 19min.
31. (A) Boys Girls
2 3
×25% ×30%
50% 90%
Average =
50 90
2 3


= 28%
Percentage of students who are not scholarship
holder is = 100 – 28 = 72%

4. P-4
32. (A) 461
+ 462
+ 463
+ 464
 461
[1 + 4 + 42
+ 43
]
 461
[1+ 4 +16+ 64]
 461
× 85
 [85 is onlydivisible by17 from option]
So, option (A) is correct
33. (C)
a
x
= seccos,
y
b
= secsin,
z
c
= tan
According to question,

2
2
a
x
+
2
2
y
b
–
2
2
z
c
 sec2
. cos2
+ sec2
. sin2
– tan2

 sec2
[sin2
+ cos2
] – tan2

 sec2
– tan2

 1
34. (D) (a2
+ 7a + 12) , a2
+ 8a + 15
 (a2
+ 4a + 3a + 12) , (a2
+ 5a + 3a + 15)
 (a + 4) (a + 3) , (a + 5) (a + 3)
So, HCF = (a + 3)
35. (A) (34
)10
= (81)10
 So, last disit will be 1
36. (A) Milk : Water
A 4 : 3 =7]×10
B 2 : 3 =5]×14
New, C 1 : 1 =2]×35
So, We have
10×4
7
14×2
5
35×1
So, Aand Bbemixed in 7 : 5toform newmixture
in 1 : 1
37. (B) Surface area of box = 88 cm2
Sides are in ratio 1 : 2 : 3
 Sidesbe1x, 2x, 3x
 2(2x2
+ 6x2
+ 3x2
)= 88
11x2
= 44
x=2
 Sides are 2, 4, 6
 Volume = l × b × h
= 2 × 4 × 6 = 48 cm3
38. (A) 2×
1
3
r2
h =
4
3
r3
(given)
h = 2a (radii = a)
39. (B) coment
360
72º
5
 
  
 
40. (A) Steel : Cement : Bricks
36 : 72 : 54
2 : 4 : 3
41. (B) 3% of360º = 108º
On miscellaneous
42. (C) Expenditure on labour
1
4
of total.
43. (C) Steel + Bricks = 90º =
1
4
= 25%
44. (A) Let same side of the triangle is 'a'
So,
It's perimeter
a
a
2a
2a + 2 a
Given that 
2a + 2 a = 2p
a =
2P
2 2
Area 
1 2 2
2 2 2 2 2
 
 
p2

2
4 2 4 2 

1
3 2 2
p2
 (3–2 2) p2
45. (A) Let 3
2 4..... K
So, 3
2 4K = K
Squaring both sides
2(4K)1/3
= K2
Cubing both sides
8 × 4 K = K6
 K5
= 32
 K = 2

5. P-5
46. (C) A B
CD 12
12
12
12
60
Let ABCD is rhombus
A= C= 60º
So ABC is a equilateral triangle
So AC= 12
For BD =
3
2
× 12 × 2 = 12 3
So Ratio between diagonels
12: 12 3
1: 3
or
3 : 3
47. (B)  
 
 
 
 
8 7
15
8 7
15 35
60
10 10
 

       
 
15 15 8 7
15
6 10 15 35
10
  
 (2 × 3)15
× (5 × 3)8
× (5 × 7)7
= 215
× 323
× 515
× 77
= 15+ 23+ 15+ 7= 60
48. (A) Moon : Earth
Diameter 1 : 4
Radius 1 : 4
So, Volume (1)3
: (4)3
Volume= 1 : 64
 VolumeofEarth : Volume ofMoon = 64 : 1
49. (D)
1
x
+
1
y
+
1
z
= 0 ...(i)
x2
+ y2
+ z2
= 25 ...(ii)
From(i)

yz z y
yz
x x
x
 
= 0
 yz + xz + xy= 0 ...(iii)
Now,
(x + y + z)2
= x2
+ y2
+ z2
+ 2 (xy+ yz + zx)
= 25+ 2 (0)
=25
So, x + y+ z = 25
= 5
50. (A) MRP= 65
So, according to question
Price after Ist
discount of 10% =
65×
90
100
= 58.50
So, money discounted in 2nd
discount
 58.50–56.16 = 2.34
So, 2nd
discount =
2.34
58.50
× 100 = 4%
51. (B)
1
2 1
= 2 – 1
2 – 1 + 3 – 2 + 4 – 3 _ _ _ 121 – 120
 121 – 1
 11 – 1 = 10
52. (D) According to question :-
A B C
3 ×
150
100
: 5 ×
160
100
: 7 ×
150
100
45 : 80 : 105
 9 : 16 : 21
53. (C) 12% = 3/2 5
Label Price 25  22
×32 ×32
1000 25×32 704
×25×8 ×25×8
5 4
54. (B) Sekhar total investment during three years
25000× 1 + 35000× 1+ 45000 × 1105000
Rajivtotal amount = 35000 × 2 = 70000
Jatin invested amount = 35000
So, ratio of their amount =
Sekhar Rajiv Jatin
105 : 70 : 35
3 : 2 : 1
So, Rajiv profit =
2
6
× 150000
= ` 50,000

6. P-6
55. (D) Cost of papering the wall will be 4 times of initial
price.
So, 475 × 4 = ` 1900
56. (C) Use alligation
I
20%
st
15% 10%
10%
II
–5%
nd
 3 : 2
So, part of loss =
2
5
× 24  9.6 kg.
57. (A) A B C
6 : 2 : 3
Total = 11
B's share =
2
11
× 6600 = ` 1200.
58. (B) 3×20% = 60%
1
10%
2
25%
×10% 60–10 = 50%
 S. P. =
5
4
× 7

` 8.75
59. (B) Total
A 2: 1  3 ×20 = 40: 20
B 3: 1  4 ×15 = 45: 15
C 3: 2  5 ×12 = 36: 24
 Final ratio= 121: 59
60. (C) 8% =
2
25
, 5% =
1
20
, 2% =
1
50
 Net selling price = 7500 ×
23
25
×
19
20
×
49
50
=6423.9
61. (C) 18×4 (Total work)
Ram Shyam
18
43
25% =
1
4
So, Ram can do the work in =
18 4
3

days
= 24 days
Now, Ram can do two times of the work in =
24 × 2 = 48 days
62. (B) [Note : Prefer the statement of this question in
english]
 A + B will finish in =
1 1
5 8
3 3
 days
=
16 25
3 3
 days
=
4 5
3

days = 6
2
3
days
63. (B) 60%=
3
5
 If he earns ` 5 then ` 3 is deducted per day
 Loss = 5 + 3 = 8
5 4
 20
 8 4
 32
Loss = 40 + 20 – 576 = 224
 No of days how was absent =
224
32
= 7
64. (D) A man can fill 5litre in 3min.
Everymin. he fill 
5
3
litre/min.
Woman fills 
3
4
litre/min.
Boyfills 
2
3
litre/min.
Total theyfill = M + W + B

5
3
+
3
4
+
2
3

20 9 8
12
 

37
12
litre/min.
Time taken to fill 111litre

111
37 /12
 3 × 12= 36 min.

7. P-7
65. (A)
C
P
Q
B
D
A
85º
40º
In ADP
DAP = 180º – 85º – 40º = 55º
ABCD is a cyclic quadrilateral
ADC + ABC = 180º
ABC = 180º –85º = 95º
Now, in AQB
AQB= 180º – 55º – 95º
AQB= 30º
 AQB= CQD = 30º
66. (C) No. of diagonals =
n (n 3)
2
 
n = No. of sides
For decagon, n = 10 
10 7
2

= 35
67. (C) Put  = 0
so, a × cos 0 – sin 0 = c
a = c ........ (i)
Also, b cos 0 + a sin 
 
b + 0 = b  2
Now using option putting a = c in third option it will be
2 2 2
a b a   b
68. (B) Here, AOB is equilateral triangle
So, OAB= 60º, OBA= 60º
DAB= 90º, CBA= 90º
As the one side of square is equal to equilateral
triangle. All sides will be mutuallyequal.
In triangle DOA 
OAD = DAB –  OAB
 90 – 60 = 30
As DA is equal to OA
So it is Isosles triangle
 So, angle  ADO = DOA
we will get DOA =
180 30
2

 75º
So, similarly ODA = 75º
Applying same in triangle BOC, OBC = 30º
we will get BCO = 75º , BC = OB
As, C =  D = 90º (square angles)
In triangle DOC
ODC = 90 – 75 = 15º
 OCD = 90 – 75 = 15º
So, DOC = 180 –(15 + 15)
 150º
67. (A)
B
A
c
b
C
x
a
LetABC is right angle triangle
So, C2
= a2
+ b2
.........(i) (pythagorus theorum)
BC = x (length of perpendicular)
x =
a b
c

........... (ii)
putting c =
ab
x
in ..........(i)
we will get
2 2
2
a b
x
= a2
+ b2
Dividing both sides by a2
b2
2
1
x
= 2
1
b
+ 2
1
a
70. (A)
71. (D) Total age offamily3 year before 5 × 17 = 85 yrs.
Their present age  85 + 15 = 100 yrs.
Their present age with baby  17 × 6 = 102 yrs.
Babyage  102 – 100 = 2 yrs.

8. P-8
72. (B)
B
A
C
3
4
D
5
BC2
= 16
BC = 4
CA2
= 9
CA = 3
CD 
3 4
5

=
12
5
73. (A) As, A+ B+ C = 180
A + B = 180 – C ......... (1)
Putting in equation sin (180 c)
2

 sin
c
90
2
 
  
 
 cos
c
2
74. (A) tan + cot = 3
cubing both sides
tan3
+ cot3
+ 3.tan. cot. 3 = 3 3
 tan3
 + cot3
 = zero
75. (A) tan2
 + cot2
 = 7
Adding '2' both sides
tan2
 + cot2
 + 2 = 7 + 2
(tan + cot)2
= (3)2
tan + cot = 3

sin
cos


+
cos
sin


 sec × cosec = ± 3
76. (D) As we know
HCF × LCM = Ist
no. × 2nd
No.
6 × 72 = 24 × ?
18 = 2nd
number.
77. (B) Present population = 1,00,000 , 10% =
1
10
Population of the 2nd year
100000×
11
10
×
11
10
 121000
78. (B) 248 52 144 
 248 52 12 
 248 64
 248 8 = 256
79. (D) Initial radius 1
Final radius  2
 Initial area  12
= 1
Final area  22
= 4
 Increase =
4 1
1

× 100 = 300%
80. (D) A
B C
O
40º
20º
8 5º
75º
BOC = 2 ABC (Circle center property)
BOC = 40º
OA= OC (radius of circumcircle)
OAC = OCA = x
40 + x + x = 180º
2x= 140
OAC = x = 70º Ans.
81. (C) tan +
1
tan
= 2
Squaring both sides
tan2
+ 2
1
tan 
+ 2 = 4
tan2
+ 2
1
tan 
= 2 Ans.

9. P-9
82. (B)
30º
BA
180 m tan 30º =
1
3
AB= 180 3 m.
83. (C) Let x be the height of tree
60 : x : : 18 : 24
60
x
=
18
24
x =
60 24
18

= 80 m Ans.
84. (B) Numbers are 20, 30
(This can be checked through option)
85. (A)
86. (A) By using option
2
2
4
9
36
36


=
52
117
=
4
9
.
87. (C) A C
4 years (ago) 1 2
A B
Present 3 1
4 year laterA= 31 (Given)
So, A's present age = 31 – 4 = 27
Hence, C = (HereA= 27 – 4 = 23)
C= 2× 23 = 46
88. (C) Let total number of boys = 100
10% of boys =
1
10
× 100 = 10
1
4
th ofgirl = 10
Total girl = 40
Hence, Boys : Girls
100: 40
5: 2
89. (D)
cot 30º cot 75º
tan15º tan 60º


Now, tan 15º = cot (90º – 15º) = cot 75º
tan 60º = cot (90º – 60º) = cot 30º
cot 30º cot 75º
(cot 30º cot 75º)

 
= – 1
90. (A) Total cost of Radio = 225 + 15 = ` 240
Selling price of Radio = ` 300
Profit % =
300 240
240

=
60
240
× 100 = 25%
91. (B) 400 6400
1 16
×7% ×9%
7 144

144 7
17

=
15
8
17
%
92. (B) S.P M.R.P
25% discount =
1
4
= 3 4) × 3
2
16
3
% discount =
1
6
= 5 6) × 2
Making M.R.P same we get S.P = 9 and 10
9 10
1R = 600
Hence, he bought it at = 9 × 600 = ` 7200
93. (A) SP of1 article = ` 1.25
 S.P. of10 articles = 10 × 1.25 = ` 12.5
and CP = ` 8
 Profit % =
12.5 8
8

× 100 = 56
1
4
%
94. (B)
Let A be side of  .
In radius =
a
2 3
Circum radius =
a
3
Area of Incircle :Area ofCircumcircle

2
a
4 3
: 
2
a
3
1: 4

10. P-10
95. (B) Replace angle 20º and 70º
by45º each (BCZ sum of 20º and 70º is 90º)
1
1
1
2

+
1
1 2

2
3
+
1
3
= 1
96. (A)
1
2
Weight 8
Weight 17

We know, Density =
Mass
Volume
or
Volume =
Mass
Density
1
2
V 64 8
V 289 17
  
 
 
3 3
1
3
3
2
4
r 83
4 17r
3




1
2
r 8
r 17

97. (C) 2 –3x –4x2
To find its greatest value we find vertex.
3
8

will have max. value.
98. (D) 4 cosec2
 + 9 sin2

Least value = 2 ab
 2 4 9
2 × 2× 3 = 12
99. (D) Line parallel toy-axis for x = k
100. (A) This type of questions can be solved easilly, if we
assume values.
For example :
 If we put a = 0 in both questions and options
a and c both will give – 45 value.
 If we put a = 1, again two option will collide
But if we put a = 2, in question
(a2
+ 2a)2
+ 12 (a2
+2a) – 45
 (4 + 4)2
+ 12(4 + 4) – 45
 64 + 96 – 45
 160–45 = 115
Now putting, a = 2 in first option
(1)(5) (4+ 4 + 15)
23 × 5 = 115 (satisfied)
So, onlya will 115. So it will be answer.
Answer Key
1. (A) 2. (B) 3. (A) 4. (C) 5. (B)
6. (B) 7. (B) 8. (C) 9. (A) 10. (D)
11. (C) 12. (A) 13. (D) 14. (B) 15. (A)
16. (D) 17. (C) 18. (C) 19. (A) 20. (D)
21. (C) 22. (A) 23. (B) 24. (D) 25. (C)
26. (B) 27. (A) 28. (A) 29. (A) 30. (B)
31. (A) 32. (A) 33. (C) 34. (D) 35. (A)
36. (A) 37. (B) 38. (A) 39. (B) 40. (A)
41. (B) 42. (C) 43. (C) 44. (A) 45. (A)
46. (C) 47. (B) 48. (A) 49. (D) 50. (A)
51. (B) 52. (D) 53. (C) 54. (B) 55. (D)
56. (C) 57. (A) 58. (B) 59. (B) 60. (C)
61. (C) 62. (B) 63. (B) 64. (D) 65. (A)
66. (C) 67. (C) 68. (B) 67. (A) 70. (A)
71. (D) 72. (B) 73. (A) 74. (A) 75. (A)
76. (D) 77. (B) 78. (B) 79. (D) 80. (D)
81. (C) 82. (B) 83. (C) 84. (B) 85. (A)
86. (A) 87. (C) 88. (C) 89. (D) 90. (A)
91. (B) 92. (B) 93. (A) 94. (B) 95. (B)
96. (A) 97. (C) 98. (D) 99. (D) 100. (A)