ACE OF PACE (ADVANCED) CODE 01
ACE OF PACE
1.
Sol.
The ages of two friends is in the ratio 6:5. The sum of their ages is 66.After how many years will the
ages be in the ratio 8:7?
(A) 12 Years
(B) 15 Years
(C) 9 Years
(D) 18 Years
(A)
x1 6
  x1  6k and x2  5k
x2 5
Also, x1  x2  66
k 6
 current age  36 & 30

2.
Sol.
3.
Sol.
36  x 8
  x  12 years
30  x 7
An athlete decides to run the same distance in 1/4th less time that she usually took. By how much
percent will she have to increase her average speed?
(A) 45 %
(B) 33.33 %
(C) 23.33 %
(D) 43 %
(B)
3t
4
S  t  S '  S '  S
4
3
The expression (1- tan A + sec A) (1 –cot A+ cosec A) has value
(A) –1
(B) 0
(C) 1
(D) 2
(D)
1  tan A + sec A 1  cot A  cosec A 
2
 cos A  sin A  1  sin A  cos A  1  1   cos A  sin A 

2


cos A
sin A
cos A  sin A



4.
Sol.
What percent decrease in salaries would exactly cancel out the 20% increase?
2
1
(A) 16
(B) 18
(C) 20
(D) 33
3
3
(A)
x
1.2  1    1
 100 
x
1
x
1

1



100 1.2 100 6
100 50
x

6
3
40
5.
36120   36  x  then x 
Sol.
(A) 44
(D)
(B) 42
(C) 62
(D) 64
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ACE OF PACE (ADVANCED) CODE 01

36120  36  362

40
Hence x  64
6.
Sol.
Three numbers are in ratio 2:3:4 and their LCM is 240.Their H.C.F. is
(A) 40
(B) 30
(C) 20
(D)10
(C)
Numbers are 2 x, 3 x, 4 x  Lcm  12 x  240
x  20
 40, 60, 80, HCF = 20
7.
Sol.
How many two-digit numbers are divisible 3?
(A) 25
(B) 27
(C) 30
(C)
99
Required no. 
 3  30
3
8.
If tan  
Sol.
9.
Sol.
a
cos   sin 
then
is
b
cos   sin 
ba
ba
a
(A)
(B)
(C)
ba
ba
b
(A)
cos   sin  1  tan  b  a
tan   a / b 


cos   sin  1  tan  b  a
cos 2 250  cos 2 650
Find the value of
sin 2 590  sin 2 310
(A) 0
(B) 1
(B)
(C) 2
(D) 33
(D)
b
a
(D) 3
cos2 25o  cos 2 65o cos2 65o  sin 2 65

1
sin 2 59o  sin 2 31o sin 2 59o  sin 2 59o
10.
Sol.
Three quantities A, B, C are such that AB  KC , where K is a constant. When A is kept constant,
B varies directly as C ; when B is kept constant, A varies directly C and when C is kept constant,
A varies inversely as B . Initially, A was at 5 and A : B; C was 1: 3 : 5 . Find the value of A when
B equals 9 at constant C .
(A) 8
(B) 8.33
(C) 9
(D) 9.5
(B)
AB
 constant
C
5 15 A  9


25
25

A  8.33
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ACE OF PACE (ADVANCED) CODE 01
11.
If
a
b
c


, then each fraction is equal to
bc ca ab
(A)  a  b  c 
2
(B) 1 2
(C) 1 4
(D) 0
Sol.
(B)
a
b
c
a bc
1




b  c c  a a  b 2a  b  c 2
12.
If 30 men working 7 hours a day can do a piece of work in 18 days, in how many days will 21 men
working 8 hours a day do the same work?
(A) 24 days
(B) 22.5 days
(C) 30 days
(D) 45 days
(B)
Total man-hours  30  7 18
.... 1
Sol.
Let no. of days be ‘x’ then total man-hours is
 x  21 8
.....  2 
1   2 

(as work is same)
x  21 8  30  7 18
45
x
 22.5 days
2
13.
Sol.
273  272  271 is the same as
(A) 269
(B) 270
(C) 271
(D) 272
(C)
273  272  271  271  2 2  2  1  271
14.
Sol.
On Ashok Marg three consecutive traffic lights change after 36, 42 and 72 seconds respectively. If
the lights are first switched on at 9.00 A.M. sharp, at what time will they change simultaneously?
(A) 9:08:04
(B) 9:08:24
(C) 9:08:44
(D) None of these
(B)
They will charge again together after an integral multiple of LCM of their individual tune period
 LCM of (36, 42, 72)
 504sec .
 8 min and 24 sec.
15.
Find the units digit of the expression 111.121.133.144.155.166 .
Sol.
(A) 4
(B) 3
(D)
Let U  n  denote units digit of n

(C) 7

(D) 0
           
 U 111 121 133 144 155 166  U 111  U 121  U 133  U 14 4  U 155  U 16 6
 U 1 2  7  4  5  6 
0
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ACE OF PACE (ADVANCED) CODE 01
16.
In the above figure, AD is the bisector of  BAC, AB  6 cm, AC  5cm and BD  3 cm . Find DC .
Sol.
(A) 11.3cm
(B)
(B) 2.5cm
(C) 3.5cm
(D) 4cm
A
6
B
5
3
C
D
AD is the angle bisector
AB BD
Hence

AC CD
5
5
CD   3   2.5
6
2
17.
In the given figure, ABC and ACD are right angle triangles with distinct integral sides and
AB  xcm, BC  ycm, CD  z cm and x. y  z and x, y and z has minimum integral value. Find the
area of ABCD
Sol.
(A) 36cm2
(A)
(B) 64cm2
A
(D) 25cm2
D
x
B
(C) 24cm2
z
y
C
As x, y , z are integers. Hence minimum value of x, y and z satisfying the given conditions are
x  3; y  4; z  12

AC  5
1
1
Hence Area ABCD   3  4   5 12
2
2
 36
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ACE OF PACE (ADVANCED) CODE 01
18.
Sol.
Three persons Amar, Akbar and Anthony inverted different amounts in a fixed deposit scheme for
one year at the rate of 12% per annum and earned a total interest of Rs 3,240 at the end of the year. If
the amount invested by Akbar is Rs. 5000 more than the amount invested by Amar and the amount
invested by Anthony is Rs. 2000 more than the amount invested by Akbar, what is the amount
invested by Akbar?
(A) Rs. 12,000
(B) Rs. 10,000
(C) Rs. 7000
(D) Rs. 5000
(B)
Akbar
Amar
Anthony
x  5000
x
x  7000
Total principle  3x  12000
 3x  12000  12 1  3240
S  I 
100

x  5000
19.
Divide a solid right circular cylinder in 8 parts identical to each other such that these parts are not cylinders.
Find the surface area of each part. (Given that Curved surface area of original cylinder is 2πrh and Non curved
surface area of original cylinder is 2πr2)
(A) (πr2 + πrh )/4
(B) (πr2 + πrh )/4 + rh
(C) (πr2 + πrh )/4 + 2rh
(D) None of these
Sol.
(C)
S1
S4
S5
S3
S2
Surface of S1 and S 2 each
r 2
8
Surface area of S3 & S4

 rh
2rh
8
Hence the total surface area
 S1  S2  S3  S 4  S5
Curved surface S5 
  r 2   rh 
 r 2 2 rh


 2  rh  
2
4
8
4



r 2 2rh

 2rh
4
8
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ACE OF PACE (ADVANCED) CODE 01
20.
Sol.
I was born on x/y/198 z (i.e. on the xth day of the yth month of the year 198 z, where z is the unit’s
digit). It is known that y is a prime number & z = 2y. At a particular year in 21st century, which is a
perfect square, I will be exactly ‘x  y’ years old. Find my date of birth in DD/MM/YYYY format
(A) 13/5/1987
(B) 13/3/1986
(C) 23/08/1984
(D) 03/03/1986
(B)
y can be 2, 3 only as z  2 y and z is a digit
hence 0  z  9  either z  4 or 6
year 2025 is a perfect square and age in that year = 45
45  15  3  Hence x = 15
y3
z 6
1
and tan  = 1, then  lies in which quadrant
21.
If sin   
Sol.
(A) First
(B) Second
(C)
sin   0 ; and tan   0
22.
Sol.
(C) Third
(D) Fourth
If sin 1 + sin 2 + sin 3 = 3, then cos 1 + cos 2 + cos 3 =
(A) 3
(B) 2
(C) 1
(D)
sin 1  sin 2  sin 3  3  sin 1  sin 2  sin 3  1

23.
2
1 , 2 , 3 are odd multiples of
(D) 0

2
4
then sin  is
3
4
4
(A)  but not
5
5
4
4
(C) but not 
5
5
If tan  
4
5
(B)  or
4
5
(D) none of these
Sol.
(B)
tan   0   lies in second or 4th quadrant
4
4
 sin  
or
5
5
24.
The coordinates of the points where the lines 3x – y = 5, 6x – y = 10 meet the y–axis
(A) (0, – 5), (0, – 10)
(B) (– 5, 0), (– 10, 0)
(C) (–5, 0), (0, – 10)
(D) (0, – 5), (0, 10)
(A)
Line 3 x  y  5 meets y-axis at  O1  5
Sol.
And line 6 x  5  10 meets y-axis at  O1  10 
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ACE OF PACE (ADVANCED) CODE 01
25.
Sol.
If the length and breadth of a room are increased by 1 m each, its area would increase by 31 m2. If
the length is increased by 1m and breadth is decreased by 1m, the area would decrease by 9 m2. Find
the area of the floor of the room, in m2
(A) 200
(B) 209
(C) 250
(D) 199
(B)
L = length of room, B = breath of room
Then ( L  1)  B  1  LB  31  L  B  30
and
 L  1 B  1  LB  9   L  B  8

B  11;
Area  LB  1119
 209
26.
Sol.
27.
Sol.
Sridevi purchased cakes of two varieties of soap, Lux and Dove – spending a total Rs. 360. If each
Lux costs Rs 30 and each Dove costs Rs 40, then in how many different combinations could she
have purchased the cakes?
(A) 3
(B) 4
(C) 5
(D) 2
(D)
L  no. of Lux soap cakes purchased
D  no of Dove soap cakes purchased

30 L  40 D  360 
   L, D    8, 3 4, 6 

3L  4 D  36 
Venu has as many sisters as he has brothers. If Karuna, Venu’s sister has thrice as many brothers as
she has sisters, then Venu has how many sisters?
(A) 1
(B) 2
(C) 3
(D) 4
(A)
m  no. of male children
f  no. of female children
Then,  m  1  f
3  f  1  m

28
Sol.
L  19
..... 1
.....  2 
3 f  3  1  f  2 f  4  f  2,
m3
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)
L = Length of rectangle
B = Breadth of rectangle

 L  5  B  2   LB  80  2 L  5B  70
 L  10  B  5   LB  50  10 B  5L  100

5B  2 L  70
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ACE OF PACE (ADVANCED) CODE 01
2 B  L  20
B  30, L  40

29.
Sol.
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)
Let no. of x then
445  x  k1  4
572  x  k2  5
k1 , k2 , k3 are integers
699  x  k3  6

x  k1  441
x  k2  567
x  k3  693
Hence x is the H.C.F. of 441, 567, 693
 63
30.
In the triangle ABC, ABC or B = 90°. AB : BD : DC = 3 : 1 : 3. If AC = 20 cm, then what is the
length of AD (in cm)?
A
B
Sol.
(A) 5 2
(D)
(B) 6 3
C
D
(C) 4 5
(D) 4 10
A
20
3x
B

x
 3x 
D
2
3x
C
 
 4 x 2  202

25 x2  25 16

x4
Hence AD 2  42  122
 160
AD  4 10
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ACE OF PACE (ADVANCED) CODE 01
31.
In the following figure, ABCD is a parallelogram, CB is extended to F and the line joining D and F
intersect AB at E. Then,
F
A
E
B
C
D
(A) AD  BF
AE
Sol.
(B) AD  CF
BE
AE
CD
(C) BF  CF
BE
CD
(D) All of them are true
(D)
F
A
D
E
B
C
 ADE and  BFE are similar
AD BF
Hence
=
AE BE
 BEF and  DCE are similar
BF CF
Hence
=
BE CD
32.
Sol.
Dheeraj has twice as many sisters as he has Brothers. If Deepa, Dheeraj’s sister, has the same
number of brothers as she has sisters, then how many brothers does Deepa have?
(A) 2
(B) 3
(C) 4
(D) Cannot be determined
(B)
m  no. of male child in the family
f  no. of female child in the family
 2  m  1  f and f  1  m
 2m  2  m  1  m  3; f  4
33.
A wheel makes 20 revolutions per hour. The radians it turns through 25 minutes is
(A)
Sol.
50 c
7
(B)
250c
3
(C)
150 c
7
(D)
50 c
3
(D)
In the revolution when terms 2 reduces
2  20  25
 angle turned 
60
50C
3
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
ACE OF PACE (ADVANCED) CODE 01
DIRECTIONS (Q. NO. 34-36) : CODING DECODING.
34.
Sol.
35.
Sol.
36.
Sol.
37.
If MADRAS can be written as ARSARS, how can ARKONAM be written in that code?
(A) ROAAKNM
(B) ROAKANM
(C) ROAKNNM
(D) ROAKNAM
(A)
In a certain code, TELEPHONE is written as ENOHPELET. How is ALIGATOR written in that
code?
(A) ROTAGILA
(B) ROTAGAIL
(C) ROTAGILE
(D) ROTEGILA
(A)
If ‘sky’ is ‘star’, ‘star’ is ‘cloud’, ‘cloud’ is ‘earth’, ‘earth’ is ‘tree’ and ‘tree’ is ‘book’, then where
do the birds fly?
(A) Cloud
(B) Sky
(C) Star
(D) Data inadequate
(C)
Sol.
Pointing to a man, a woman said, “His mother is the only daughter of my mother.” How is the
woman related to the mother?
(A) Mother
(B) Daughter
(C) Sister
(D) Grandmother
(A)
38.
How many squares in the picture?
Sol.
(A) 33
(D)
39.
Sol.
(B) 39
(C) 41
(D) 40
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)
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ACE OF PACE (ADVANCED) CODE 01
40.
Sol.
41.
Sol.
42.
Sol.
43.
Sol.
44.
Sol.
45.
Sol.
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)
If starting from very first ball of the match, each batsman gets bowled out on the very first ball he
faces. Which batsman will be left not out in the end?
(A) Batsman number 11
(B) Batsman number 8
(C) Batsman number 7
(D) None of these
(B)
A boat takes 90 minutes less to travel 36 Kms downstream than to travel the same distance upstream.
If the speed of the boat in still water is 10 Km/h , the speed of the stream is
(A) 2 Km/h
(B) 2.5 Km/h
(C) 3 Km /h
(D) 4 Km/h
(A)
After Yuvraj hit 6 sixes in an over, Geoffery Boycott commented that Yuvraj just made 210 runs in
the over. HarshaBhogle 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)
A teacher throws a question in front of his class and says that he will give a chocolate as a prize to
the student who solves it. He says, “ I have a two-digit number in my mind.” If I square the number,
then the last digit of both the numbers have the same last digit. None of the digits in the original
number is zero. When the digits of the original is written in the reverse order, the square of the new
number obtained has a last digit 6 and is less than 3000. Now find the number of distinct possibilities
for the number. “Rajiv found the exact answer. What was his answer?
(A) 3
(B) 6
(C) 8
(D) 9
(A)
Many years ago when Mr. Waugh was asked as to who among Steve and Mark was elder , he said , “
Two years from now Steve will be twice as old as he was two years ago and in three years from now
Mark will be three times old as he was three years ago”. How old is Mark today, if Mr. Waugh was
asked the question 27 years back and who is older,Mark or Steve?
(A) 31 years ,Steve
(B) 33 years , both are of the same age
(C) 37 years ,Mark
(D) 43 years , can’t say
(B)
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ACE OF PACE (ADVANCED) CODE 01
46.
Sol.
47.
Sol.
48.
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)
(B) only II and III
(D) All three of these
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)
If an ant can travel only in the North or in the East direction, in how many possible routs can an ant
travel from cell P to cell Q?
Q
North
East
P
Sol.
49.
(A) 4
(C)
(B) 8
(C) 6
(D) 5
Sol.
What is the sum of all two digit numbers that give a remainder of 3 when they are divided by 7?
(A) 666
(B) 676
(C) 683
(D) 777
(B)
50.
Evaluate x  12  12  12  .......
Sol.
(A) 6
(C)
(B) 3
(C) 4
(D) 5
CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 12