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PATH (ADVANCED)CODE: 01
PACE APTITUDE & TALENT HUNT
(QUESTION PAPER & SOLUTION)
1.
Ans.
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20
minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that
of the cars is:
(A) 2:3
(B) 3:2
(C) 3:4
(D) 4:3
(C)
Let the speed of the train be x km/hr and that of the car be y km/hr.
120 480
1 4 1

 8    ..... i 
x
y
x y 15
200 400 25 1 2 1

   
..... ii 
And,
x
y
3
x y 24
Then,
Solving (i) and (ii), we get: x = 60 and y = 80.
2.
Ans.
3.
Ans.
4.
Ans.
5.
Ans.
Sachin Tendulkar, The God of the cricket, scored 6000 runs in a certain number of innings. In the
next five innings he was out of form and hence could make only a total of 90 runs, as a result of
which his average fell by 2 runs. How many innings did he play in all?
(A) 105
(B) 95
(C) 115
(D) 104
(A)
6000  90
6000
2
 x  100
x 5
x
So total innings = 105
If aliens from planet of apes are in the habit of replacing every 2 by 3 and then finding the value of
anything, what according to them is the value of 32 ?
(A) 9
(B) 5
(C) 27
(D) 37
(D)
32 
 33  27  37
What is 20 % of 50 % of 75 % of 70 ?
(A) 5.25
(B) 6.75
(A)
1 1 3
. . . 70   5.25
5 2 4
(C) 7.25
(D) 5.5
Ben Johnson and Carl Lewis run at speeds of 10 m/s and 11 m/s respectively on a circular track.
They start simultaneously from the same point on the 420 m long track and run in opposite
directions. After they start, how many times would they have met each other when Ben Johnson has
run a distance of 1 km?
(A) 6
(B) 5
(C) 4
(D) 7
(B)
VB  VC  21
1000 m
time 
 100 s
10 m s
420 m
1  Round 
 205
21m s
So in 100 s = 5 – Round
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PATH (ADVANCED)CODE: 01
6.
Ans.
7.
Ans.
8.
The smallest of
3
4, 4 5, 4 6, 3 8 is
(A) 3 8
(B) 4 5
(B)
Rice power by 12
44, 53, 63,84 so smallest is 4 5
If sin A :cos A  4 : 7 , then the value of
(A) 3 14
(D)
4
7  3
1
7

6
4
7   2
7
 
(B) 3 2
9.
Ans.
10.
Ans.
3
4
7 sin A  3cos A
is
7 sin A  2 cos A
(C) 1 3
(D)
4
6
(D) 1 6
The difference between an exterior angle of  n  1 sided regular polygon and an exterior angle of
 n  2  sided regular polygon is
Ans.
(C)
6 , then the value of ' n ' is
(A) 13
(B) 14
(C) 12
(A)
  2     2  
0
    n  1        n  2    6  n  13
  

 
(D) 15
The incomes of A, B, C are in the ratio of 12 : 9 : 7 and their spendings are in the ratio is 15 : 9 : 8. If
A saves 25% of his income. What is the ratio of the savings of A, B and C?
(A) 15 : 18 : 11
(B) 11 : 18 : 15
(C) 11 : 15 : 18
(D) None
(A)
Let
A
B
C
Income
12 x
9x
7x
Spending
15 y
9y
8y
3
Given 15y  9x  y  x
5
27  
29 

So req. 12x  9x  :  9x  x  :  7x  x 
5  
5 

 15 :18 :11
A box measures 30 cm × 24 cm × 18 cm. The longest rod that can be placed in it has the length
(Take 2  1.414 )
(A) 16 cm
(B) 30 cm
(C) 42.42 cm
(D) 42.52 cm
(C)
2  b 2  h 2  42.42 cm
11.
Ans.
If the equation 9x2 + 6kx + 4 = 0 has equal roots, then the roots are both equal to
2
3
(A) 
(B) 
(C) 0
(D)  3
3
2
(A)
D  0  36k 2  4.9.4  k   2
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12.
Ans.
13.
Ans.
14.
1 
1 
 1  1 
Find the value of 1   1    1    1 
 ...... .
 2   4   16   256 
1
(A) 1
(B) 2
(C)
3
(B)
1
4
In a coded language TAKE = 1790, PLOT = 5321 then code for PLATE will be
(A) 52701
(B) 53071
(C) 35710
(D) 53710
(D)
Mark A, B, C....  with 1, 2, 3, …, 26 respectively and select the coding
Find the missing number
36
49
26
9
81
64
21
25
Ans.
(D)
25
25
64
16
(A) 24
(D)
(B) 25
?
144
36
(C) 23
(D) 31
25
64
?
144
= 5 + 8 + 6 + 12 = 31
36
15.
The mirror image of the point A  a, b  (where a  0, b  0 ) in the x-axis is
Ans.
(A)  a, b 
(B)
(B)  a,  b 
(C)  a,  b 
(D) None of these
A (a, b)
x
I (a, – b)
16.
Ans.
If xy  0 , then the point  x, y  may lie in
(A) quadrants I and III
(B) quadrants II and IV
(C) quadrants I and II
(D) quadrants III and IV
(A)
xy > 0  both are positive or negative i.e. Ist or IIIrd quadrant.
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17.
A ring of radius 'R' is kept on top of another ring of radius '10R' & touching it at point A as shown
R
A
10R
Ans.
18.
If Smaller ring starts from A and only rolls on top of bigger ring. How many revolutions does it have
to make to reach back at point A
(A) 10
(B) 20
(C) 11
(D) None
(C)
Centre of smaller ring has to travel 2 (11R ) distance. In one rotation it travels 2 R
If tan 86º = m then
2
(A) m 2  1
m 1
Ans.
cot 4º  cot86º
is
m  tan 4º
m2 1
(B)
1 m2
(C)
1 m2
1 m2
2
(D) m 2  1 .
m 1
(A)
tan 860  m  cot 40
1
cot 4  cot 86
m2  1
m


So
1 m2  1
m  tan 40
m
m
0
19.
Ans.
20.
Ans.
21.
Ans.
0
m
The sum of the speeds of a boat in still water and the speed of the current is 10 kmph. If the boat
takes 40% of the time to travel (same distance) downstream when compared to that upstream, then
find the difference of the speeds of the boat when travelling upstream and down stream.
(A) 3 kmph
(B) 6 kmph
(C) 4 kmph
(D) 5 kmph
(B)
x + y = 10
…(2)
d
2 d 
 
… (1)
  3x  7y
xy 5 xy
(1) & (2) x = 7, y = 3
So,  x  y    x  y   6
An athlete decides to run the same distance in 1/4th less time that she usually took. By howmuch
percent will she have to increase her average speed?
(A) 45 %
(B) 33.33 %
(C) 23.33 %
(D) 43 %
(B)
t
v

vt   v  x  .  t    x 
3
 4
33.33%
Find the units digit of the expression 111.121.133.144.155.166 .
(A) 4
(B) 3
(C) 7
(D)
It contain 2  5  xxx...0
So ans. 0
(D) 0
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22.
Ans.
The area of a rectangle gets reduced by 80 square units, if its length is reduced by 5 units and breadth
is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the
area is increased by 50 sq. units. Find the length and breadth of the rectangle
(A) 30, 40
(B) 35, 35
(C) 40, 30
(D) 45, 25
(C)
xy = (x – 5) (y + 2) + 80 = (x + 10) (y – 5) – 50
2x – 5y + 70 = 0 & –5x + 10y – 100 = 0
x = 40, y = 30
23.
Ans.
24.
Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6
respectively.
(A) 60
(B) 65
(C) 78
(D) 63
(D)
445
R4
63
572
R 5
63
699
R6
63
A wheel makes 20 revolutions per hour. The radians it turns through 25 minutes is
c
(A) 50 
7
Ans.
Ans.
3
c
(C) 1 50 
7
c
(D) 50 
3
(D)
60 min 
 20 round
25 min 

25.
c
(B) 250 
20
50
 25 
.25 round =   2  
60
3
 3 
An examination consists of 100 questions. Two marks are awarded for every correct option. One
mark is deducted for every wrong option and half mark is deducted for every question left, then a
person scores 135. Instead, if half mark is deducted for every wrong option and one mark is deducted
for every question left, then the person scores 133. Find the number of questions left unattempted by
the person.
(A) 14
(B) 16
(C) 10
(D) 12
(A)
Let
x  correct
y  wrong
(100 – x –y)  un attempted
 1 
2x    y   100  x  y     135
 2 
2x    y 2   100  x  y  1  133
x + y = 86
So (100 – x – y) = 14
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PATH (ADVANCED)CODE: 01
26.
Ans.
27.
Ans.
A man bought 50 dozen fruits consisting of apples and bananas. A banana is cheaper than an apple.
The number of dozens of apples he bought is equal to the cost per dozen of bananas in rupees and
vice versa. If he has spent a total amount of Rs 1050, find the number of dozens of apples and
bananas he bought respectively.
(A) 12 and 38
(B) 14 and 36
(C) 15 and 35
(D) 28 and 32
(C)
Let x number of dozen are apple with rate y Rs / dozen
& y number of dozen are banana with rate x Rs / dozen
Then, xy  yx  1050  xy  525  15.35
So, (15, 35)
After Yuvraj hit 6 sixes in an over, Geoffery Boycott commented that Yuvraj just made 210 runs in
the over. Harsha Bhogle was shocked and he asked Geoffery which base system was he using? What
must have been Geoffery’s answer?
(A) 9
(B) 2
(C) 5
(D) 4
(D)
2x2 + 1. X + 0.x0 = 36
2x 2  x  36  0
9x 8x
2x (x – 4) + 9 (x – y) = 0
9
x  4,  x  4
2
28.
Ans.
29.
Ans.
30.
Ans.
Mr. Srivastava has five children – Dolly, Polly, Molly, Solly and Lolly – named in the decreasing
order of their ages. The age difference between any two consecutive children is the same (an integral
number of years). If dolly is 14 years old, what are the possible ages of Molly?
I. 13 years
II. 12 years
III 11 Years
(A) only II
(C) only I and II
(A)
Dolly  Molly
= Polly must be an integer
2
So Molly can be 12 yrs
(B) only II and III
(D) All three of these
1
1

If  a    3 , then a 3  3 equals
a
a

10 3
(A)
(B) 3 3
(C) 18
3
(C)
1

a    3
a

Cubing (i) both sides, we get
1
1
1

a3  3  3  a    27  a 3  3  27  9  18
a
a
a

Abscissa of all points on the y-axis is
(A) 0
(B) 1
(A)
Abscissa of all points on y axis is 0.
(C) 1
(D) 7 7
(D) None of these
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31.
Ans.
32.
Ans.
33.
Ans.
If the height of a cone is doubled, with same radius of base then its volume is increased by
(A) 100%
(B) 200%
(C) 300%
(D) 400%
(A)
1
Volume of cone  r 2 h
3
1 2
2
New volume  r  2h   r 2 h
[ height of cone  2h ]
3
3
 2 1 2
 3  3  r h

 % increase in volume  
100
1 2
r h
3
 100 %
In the given figure, ABCD is square of side 14 cm and APD and BPC are semicircles. The area of
the shaded region is (Take   22 7 ) :
(A) 42 cm2
(B) 40 cm2
(C) 43 cm2
(A)
Area of the shaded region
 Area of square ABCD  Area of two equal semicircles
 1 22

 14  14 cm 2  2    7 2  cm 2
2 7

2
2
 196 cm  154 cm  42cm2
(D) 44 cm2
If the circumference of two circles are in the ratio 2 : 3, then the ratio of their area is :
(A) 4 : 9
(B) 2 : 3
(C) 3 : 2
(D) 9 : 4
(A)
C1 : C2  2 : 3
2r1 : 2r2  2 : 3
 r1 : r2  2 : 3
2
2
Now, r : r     4 : 9
3
2
1
34.
Ans.
2
2
The area of the circle is 220 cm2. The area of a square inscribed in it is (Take   22 7 )
(A) 49 cm2
(B) 70 cm2
(C) 140 cm2
(D) 150 cm2
(C)
Area of the circle  220 cm 2
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PATH (ADVANCED)CODE: 01
 r 2  220  r 2  70
 r  70 cm
Diagonal of the square inscribed  2 70 cm
2 70
 140 cm
2
Area of the square  140  140  140 cm 2
Side of the square 
35.
Ans.
36.
If  means  , – means  ;  means + and  means –, then the value of 36 12  4  6  2  3 when
simplified is
1
(A) 2
(B) 18
(C) 42
(D) 6
2
(C)
Using proper signs, we get:
36  12  4  6  2  3
 36  3  3  3
 36  3  9
 42 .
So the answer is (C)
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of
the largest square that can be fitted into the remaining space?
(A) 9
Ans.
(B) 12
1
2
(C) 15
(D) 15
1
2
(C)
We draw a square as shown
We wish to find the area of the square. The area of the larger square is composed of the smaller
square and the four triangles. The triangles have base 3and height1 , so the combined area of the four
3
 6 The area of the smaller square is 9. We add these to see that the area of the large
2
square is 9  6  15
triangles is 4.
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PATH (ADVANCED)CODE: 01
37.
Ans.
38.
Ans.
In ABC, D is a point on side AC such that BD = DC and BCD measures 70o. what is the
degree measure of ADB ?
(A) 100
(B) 120
(C) 135
(D) 140
(D)
BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are
supplementary, ADB  180  40  140
The circumference of the circle with center O is divided into 12 equal arcs, marked the letters A
through L as seen below. What is the number of degrees in the sum of the angles x and y?
(A) 75
(B) 80
(C) 90
(D) 120
(C)
For this problem, it is useful to know that the measure of an inscribed angle is half the measure of its
corresponding central angle. Since each unit arc is
1
of the circle's circumference, each unit central
12
o
 360 
o
o
o
angle measures 
  30 . Then, we know that the inscribed arc of x = 60 so mx  30 ;
 12 
and the inscribed arc of y  120o so my  60o . mx  my  30  60  90
39.
Ans.
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180
and 594?
(A) 110
(B) 165
(C) 330
(D) 625
(C)
To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of 180  32  5  22 .
The prime factorization of 594  33 11 2 .
Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one
number is one but not the other, use it (this is 33, 5, 11, 22). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations
and multiply. 32 × 2 = 18
Thus the answer =
5940
 330
18
Similar Solution
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We start off with a similar approach as the original solution. From the prime factorizations, the GCF
is 18 .
It is a well known fact that gcd  m,n  1cm  m, n   mn . So we have,
Dividing by 18 yields 1cm (180, 594) = 594  10  5940
Therefore,
40.
Ans.
1cm 180,594  5940

 330 .
gcf 180,594 
18
Quadrilateral ABCD is a trapezoid, AD = 15, AB = 50, BC = 20, and the altitude is 12. What is the
area of the trapezoid?
(A) 600
(D)
(B) 650
(C) 700
(D) 750
If you draw altitudes from A and B to CD the trapezoid will be divided into two right triangles and a
rectangle. You can find the values of a and b with the Pythagorean theorem.
a  152  122  81  9
b  202  122  256  16
ABYX is a rectangle so XY  AB  50
CD  a  XY  b  9  50  16  75
The area of the trapezoid is
12.
41.
Ans.
 50  75   6 125  750
 
2
Three circle of radii 3 cm, 4cm and 5 cm are drawn such that each touches the other two externally.
Then area of the triangle formed by their centers is:
(A) 12 5 cm 2
(B) 6 5 cm 2
(C) 15 cm 2
(D) None
(A)
A
5
3
C
4
B
For  ABC
AB = 7, AC = 8, BC = 0
Apply Heron’s formula
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789
 12
2
area  12 12  7 12  8 12  9   12 5 cm 2
s
42.
Ans.
43.
Ans.
44.
Ans.
45.
Ans.
The total age of A and B is 12 years more than the total age of B & C. Then the age difference
between A & C is
(A) greater than 10 years
(B) less than 8 years
(C) greater than 14 years
(D) less than 11 years
(A)
(a + b) – (a – b) = 12
 a – c = 12
Two numbers a and b are respectively 12.5% and 25% more than a third number c. The percentage
by which b is more than a is
(A) 12.5%
(B) 50%
(C) 11%
(D) none of these
(D)
a = (1.125) C
b = (1.25) C
ba
100
Required percentage = 
100 
%
a
9
The average of 20 results is 30 and that of 30 more results is 20. For all the results taken together, the
average is
(A) 25
(B) 50
(C) 12
(D) 24
(D)
20  30  30  20
Required average =
30  20
= 24
The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and
50% respectively. Find the percentage change in the volume of a cuboid
(A) 77%
(B) 75%
(C) 88%
(D) 98%
(D)
v   bh
v1  1.1 1.2b 1.5 h
 1.98 bh
= 1.98 V
% change = 98%
46.
Ans.
Let a, b be real numbers such that a 2  b 2  4a  2b  5  0 . Then
(A) a 2  b 2  3a  b
(B) a 2  b 2  a  2b
(C) a 2  b2  4
(D) None of these
(A)
a2 + b2 – 4a + 2b + 5 = 0
2
2
  a  2    b  1  0
 a = 2, b = –1
 a2 + b2 = 5 = 3a + b
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47.
In the figure given, what is the measure of ACD ?
A
D
25o
75
o
B
Ans.
48.
C
(A) 75o
(B) 80o
(B)
ABD  1800  750  250  800
ACD  ABD
(C) 90o
(D) 105o
In the figure given below ABCD is a rectangle, ABE is an isosceles triangle and BC  2 2 AE and
the area of triangle ABE is 7 cm2, then the area of ABCD is
A
D
B
C
E
2
Ans.
(A) 35 cm
(B) 42 cm2
(C)
Let AB = x
AE  x 2  BC  4x
Area of  ABE  7
(C) 56cm2
(D) 50cm2
x2
7
 x 2  14
2
Area of ABCD = 4x2 = 56

49.
Ans.
The perimeter of a right angled triangle is 36 cm and the sum of the squares of its sides is 450 cm2.
The area of the triangle is
(A) 42 cm2
(B) 54 cm2
(C) 62 cm2
(D) 100 cm2
(B)
a + b + c = 36
… (1)
2
2
2
c =a +b
… (2)
a2 + b2 + c2 = 450
 2c2  450
 c = 15
 a + b = 21
2
 a  b   a 2  b2 
ab 
2
441  225

 108
2
ab
Area 
 54
2
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50.
Ans.
In a recent survey 40% houses contained two or more people. Of those houses containing only one
person, 25% were having only a male. The percentage of all houses, which contain exactly one
female and no males is
(A) 15%
(B) 40%
(C) 55%
(D) 45%
(D)
Let total no. of houses be x.
No. of houses having one person = 0.6 x
No. of housed having only one female & no male = (0.75) (0.6 x)

= 0.45 x
 required percentage = 45%
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