# Aspiring Minds – AMCAT Previous Years Question Paper with Answers set 5

Ques1. Statement:
The chairman of the car company announced in the meeting that all trials of its first product, the new car model ‘M’,are over and company plans to launch its car in the marked after six months.
Courses of action:
I. The network of dealers is to be finalised and all legal, financial and other matters in this connection will have to be

finalised shortly.
II. The company will have to make plans for products other than the car.
III. The Material, managerial and other resources will have to be in fine tune to maintain production schedule.
Op 1: I and III only
Op 2: Only I
Op 3: All the three
Op 4: Only II
Op 5: None of these

Correct Op : 1

A program is compiled by Tarun on his machine. Whether it will run on a different computer will depend upon:

Option 1 : Operating system on the computer
Option 2 : Hardware configuration of the computer
Option 3 : Both operating system and hardware configuration
Option 4 : The language of the program

correct option 3) Both OS and Hardware configuration

platform = OS + Hardware cofiguration

so to run on different computer

it has to provide same platform

or

virtual machine that can provide same environment(e.g JVM in case of java)

so either of the above case we are providing platform(directly or indirectly)

the language of the program is even dependent upon the 2 factors

1)operating system of the computer.
2)Hardware configuration (that mainly include processor) of the computer

que3:If logx (0.1) = -1/3, then the value of x is:

Op 1: 10
Op 2: 100
Op 3: 1000
Op 4: 1/1000

ans:op 3(1000)
loga x=m => a^m=x

x^(-1/3)=0.1
x=(0.1)^-3
x=1/(0.1)^3
=1/0.001=1000

If log8 x + log8 (1/6) = 1/3 then the value of x is: Op 1: 12 Op 2: 16 Op 3: 18
log8(x*1/6) = 1/3 log88
x/6=2
x=12
solve (144^(-3/2))^(-1/6)?????
(12)^2(3/2*1/6)
12^(1/2)
3.46
n(n^2-1) is always divisible by:_______
a)5 b)6 c)7 d)8
verify with options by taking some n value
log(base 4)2 * log(base 4) 32
1)let log(base 4)2=n;
then 4^n=2;
n=1/2;
same for log(base 4)32=m;
m=5/2;
m*n=5/4
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log (base x) 0.8 = 16..X=?
log (base x) 0.8 = 16..X=?
0.8= 4/5
log (base x)4/5 = 2^4
x^4/5 = 2^4
x =(2^4)^5/4
x = 2^5 =32
R.S. AGGARWAL L0G(EX-23, Q11, PAGE-490)

If LCM and HCF of two numbers are equal and their product is 3136 find the number.?

a.44
b.46
c.54
d.56
for que 3, It is by property that multiplication of two numbers is equal to the multiplication of their HCF and LCM so the answer is sq root of 3136 i.e, 56

log(base 27(power 1/3)) X = 1..Find X?
log(base( 3^3))^1/3)x=1
log(base3)x=1
x=3
If x increases linearly, how will a to the power -x
behave (a>1)?
a. Increase linearly
b. Decrease linearly
c. Increase exponentially
d. Decrease exponentially
Decrease linearly..
Which is more-successive discount of 40% of 30 % OR flat 70% ?
70% is more successive
Case1:If the price is 100rs, 40% of 100rs is 60rs and 30% of 60rs is 48rs.
70% of 100rs is 30rs
If log(base p)25p=2.Find the value of P?
25p=p^2
p=25
In demand paging scheme, let p be the probability of a page fault such that 0<=p<=1 and memory access time=ma,then What is the effective access time?
a. (1-p)*ma+p*page fault time
b. p*ma+p*page fault time
c. (1-p)*ma+(1-p)*page fault time
d. p*ma+(1-p)*page fault time

there is a direct formula in operating system to find the effective access time.
EAT=P*(PAGE FAULT)+(1-P)*MEMORY ACCESS TIME
=P*PAGE FAULT+(1-P)*Ma.so 1st one is correct option.

15?1792 is divisible by 9 only when ?=
1.1
2.4.
3.3.
4.2.
sum of digits should be a multiple of 9
1+5+1+7+9+2=25
so we have to add 2 for make it divisible by 9

2^x+y=2/2^3/2.
2^x-y=2.
2^(x+y)=2/2^(3/2)
2^(x+y)=2^(1-3/2)
x+y=-1/2……………………i
then
2^(x-y)=2^1
x-y = 1……………………..ii

Solve eq i and ii

x=1/4
y=-3/4
8,8,6,2,___.
1.-4.
2.4
3.2
4.0.
Ans :- -4

Solution:-

8-0 = 8
8-2 = 6
6-4 = 2
2-6 = -4

79,64,26,15,6,__.
79= 7*9 = 63+1= 64
64 = 6*4 = 24+2= 26
26= 2*6 = 12+3 = 15
15 = 1*5 = 5+4 = 9
so the ans is 9

GULMOHAR=TFONLSZI….. then PIPAL =??
GULMOHAR = TFONLSZI

G+T = 7+20 = 27
U+F = 21+6 = 27
…..
A+Z = 1+26 = 27
R+I = 18+9 = 27

So,

27-P = 27-16 = 11 = K
27-I = 27-9 = 18 = R
27-P = 27-16 = 11 = K
27-A = 27-1 = 26 = Z
27-L = 27-12 = 15 = O

PIPAL = KRKZO
6,9,__,24,39.
1).18.
2)15.
3).10.
4)12.
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addition of the previous two numbers
9+6=15
15+24=39
4.28 and -3.28 are two numbers on a real number line. If 1is added to both the numbers ,then which of
the following is true?
1.Distance between the two numbers is 2 units more than the distance between 4.28 and -3.28
2.Distance between the two numbers is 2 units less than the distance between 4.28 and -3.28
3.Distance between the two numbers is equal to than the distance between 4.28 and -3.28
4.None
option 3
distance b/n 4.28 and -3.28 = 7.56
distance b/n 5.28 and -2.28 = 7.56
So distance btwn two num is equal
if four dice are thrown simultaneously what is the probability of getting double twice?
since there are four dice hence probabality of getting double twice i.e(2,2) is 6 times.
dice1 dice2 dice3 dice4
yes yes no no
yes no yes no
yes no no yes
no yes yes no
no yes no yes
no no yes yes

total outcomes is 6*6*6*6=6^4=1296
hence,probability=6/(6*6*6*6)=1/216

8,12,24,60,… what is the next number?
8 * 3 = 24 – 12 = 12
12*3 = 36 – 12 = 24
24 * 3 = 72 -12 = 60
60 * 3 = 180 – 12 = 168
multiple by 3 subtract 12 will get answr

3,7,10,11,12,17,? find out next number?
3+10 = 13-1 =12
7+11 = 18-1 =17
10+12= 22-1 =21
sequence would be 3 7 10 11 12 17 21
if peacock = 11526312316 then feather=?
21526201959
alphabet order reverse and straight
Find the number of consecutive zeros at the end of 72!
By using formula round(n/5)+round(n/25)+……….round(n/5^n)
72/5 + 72/25 := 14 + 2 := 16 zeros.
Ans:16
Some persons can do a piece of work in 12 days. Two times the number of such persons will do half of that work in:

let no of persons can do x work
1 day work=1/12
now x/2=1/6
6x=2
x=1/3

ans:3 days

log y 1369y=3 then what is the value of y?

y^3=1369y
y^2=1369
y=37
The students are in the ratio 2:3:5.if 20 students are increased in each batch the ratio changes to 4:3:7
The total number of students in the three batches before the increase was

the ratio is 2:3:5
let students 2x,3x,5x.
after 20 student are increased in each batch
now student will be 2x+20,3x+20,5x+20
now ratio is 4:3:7
now student will be 4x,3x,7x
2x+20=4x
x=10
students before increases : 20,30,50
sum of money doubles itself in 9 years, in how many years it will become 8 times itself?
given, the sum of money doubles itself in 9years
initially let the sum be X
after 9 years it becomes 2X(now this becomes the initial for next 9y)
after 18 years it becomes 4X(now this becomes the initial for next 9y)
after 27 years it becomes 8X.
so it takes 27 years to get 8times of the initial amount invested.

There are 10 yes or no questions. How many ways can these be answered?
each question can be answered either yes or no, so every question will have 2 possibilities
therefore, for 10 questions, 20 possibilities(2+2+2+….10(times)

If the simple interest on a sum at 4% per annum for 2 years is Rs. 80, then the compound interest on the same sum for the same period is: Op 1: Rs. 86.80 Op 2: Rs. 86.10 Op 3: Rs. 88.65 Op 4: Rs. 81.60
Amount from simple interest is 1000.then calculate the amount from this formula
a=p(1+r/100)^n.
so,1000(1+4/100)^2 =1081.6.
so, interest is 81.6.
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Mohan walks a certain distance and rides back in 6 hours and 15 minutes. If he walks both ways he takes 7 hours and 45 minutes. If Mohan rides both ways the time which he will take will be: Op 1: 4 hours Op 2: 19/4 hours Op 3: 9/2 hours Op 4: 17/4 hours Op 5: None of these
W+R=375 minutes(6 hours 15 minutes)
2W=465 minutes(& hours 45 minutes)
2R=?
2(W+R)=375*2=750
2R=750-465=285=19/4
A boy move 6 m in west then he turn towards south and move 20 m then turn towards east and move 12 m again move toward north and move 12 m . How much dist he is away from his starting point.
it forms a right angle triange so
hypotenuse square= side_1 square + side_2 square
one side it is 6 n the other it is 8 so
6^2 + 8^2=100 so ans 10

synonym of OBTRUSIVE
conspicuous, obvious,unmistakable
hcf of 3.68 & 5.35
hfs fro 368 & 535 is 1.
so the ans is 0.01

3*(4^4+4^3+4^2+4+1) = ?
3*4^2(4^2+4+1)+15
48^(21)+15
1023
log10 (2)=.6096
log10 (3)=.4709
then log10 (12) = ?

log10(12)=log10(6*2)=log10(6)+log10(2)
log10(3*2)+log10(2)
log10(3)+log10(2)+log10(2)
.4709+.6096*2=1.6901
ln xy – ln |x| = ?
ln xy – ln |x| = ln x + ln y – ln |x| = ln y
since x > 0 for log to be defined.
hence |x| = x

P3M : N4J : R3P : ?
P4M
becauz compare P3M with R3P then N4J=P4M

product of two no u and v is 42.
conclusion
1.u is less than v
2. u is even.
options
1. conclusion 1 and 2 both are necessary
2.only one is sufficient.
3.only two is sufficient
4. data not sufficient
Data in sufficient why because….
2*21
3*14
6*7 are the possibilities.
u is even means u can be 2,14,6 and u less than v again two possibilities are there…. 2*21 and 6*7… so data in sufficient.

400 have how many factors
a.8
b.10
c.12
d.24
15
400=2^4*5^2
no of factors=(4+1)*(2+1)=5*3=15
A box contain 6 yellow,3 red and 2 green ball 5 ball is randomly selected what is the probability that at least one ball is yellow.
prob=at least 1 ball yellow
=1-no ball yellow
=1-{(6C0*3C3*2C2)/11C5}
=1-(1/462)
=0.997
10,14,23,39,64,?
14-10=4 or 2^2
23-14=9 or 3^2
39-23=16 or 4^2
64-39=25 or 5^2
so next will be 6^2=36
that means 64+36=100

Ques. A locomotive engine, without any wagons attached to it, can go at a speed of 40 km/hr. Its speed is diminished by a quantity that varies proportionally as the square root of the number of wagons attached. With 16 wagons, its speed is 28 km/hr. The Op 1: 99 Op 2: 100 Op 3: 101 Op 4: 120 Op 5:
option1>99
let x be the quantity that varies proportionally with sqrt of no. of wagons.
so,x=K*w , where k=constant,w=no of wagons.
so the speed decreases by 16 with 16 wagons.
so, 16=K*sqrt(16)
K=4

if the speed is reduced by 40km/hr, the engine cannot move.
so, 40=4*sqrt(w)
w=100
with 100 wagon engine cannot move,so greatest no of wagons for which engine can move is 99.
Ques. If 33 untrained labourers can do a work in 15 days of 12 hr. each, how many trained labourers can do 50% more work in 11 days of 9 hr each ? (It may be assumed that it takes 2 trained labourers to do the work of 5 untrained labourers) Op 1: 42 Op 2: 36 Op 3: 90 Op 4: 100 Op 5:
36
total work done=15*12*33=5940;
more 50%of work is=5940*150/100=8910;
now,trained worker=8910/9*11=90;
where 5 untrained labouers=2 trained
hence 90/5=18
18*2=36;
Ques. |X – 5| + 4 > 0 and |X2| < 4. Then x can be: Op 1: 4 Op 2: 2 Op 3: 0.5 Op 4: All of these
Op 3: 0.5
4.5+4>0
&(0.5)^2
Ques. If |x| + |y| = 7, then what is the sum of minimum and maximum values of x + y? Op 1: 3/2 Op 2: -7 Op 3: 7 Op 4: 0

op4 :0
as mod has property

|x|=x;x>0
=-x;x
Ques. If 13 + 23+ 33+ ……….+ 93 = 2025 ,then the value of (0.11)3 + (0.22)3 + ….+ (0.99)3 is close to: Op 1: 0.2695 Op 2: 0.3695 Op 3: 2.695 Op 4: 3.695
(11/100)^3 +(22/100)^3 + ……(99/100)^3
take (11/100)^3 as common then
(11/100)^3[1^3+2^3+3^3+ …..9^3]
sum of cubes formula n^2(n+1)^2/4 use it
then
(11/100)^3*[8100/4]
Ques. In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin? Op 1: 4/7 Op 2: 1/2 Op 3: 2/3 Op 4: 5/6
20c1/30c1=2/3
Ques. A, B and C are three students who attend the same tutorial classes. If the probability that on a particular day exactly one out of A and B attends the class is 7/10; exactly one out of B and C attends is 4/10; exactly one out of C and A attends is 7/10. I
Probability(at least one attending ) = 1- Probability(none attending)
Let the Probability of A,B,C attending the class be a,b,c
So not attending will be 1-a,1-b,1-c
Exactly one of A,B
a(1-b) + b(1-a) = 7/10
a+b -2ab =7/10

B,C
b(1-c)+c(1-b) = 4/10
b+c -2bc =4/10

C,A
a(1-c) + c(1-a) =7/10
c+a -2ac=7/10

2(a+b+c) – 2(ab+bc+ca) = 18/10
a+b+c -ab -bc-ca =9/10

P(atleast one ) = 1 – P(none)
1 – [ (1-a)(1-b)(1-c)]
1- [ 1 -a -b-c+ab+bc+ca -abc]
1-[1 – (9/10 + 9/100)]
= 99/100
Ques. A box contains 10 balls numbered 1 through 10. Anuj, Anisha and Amit pick a ball each, one after the other, each time replacing the ball. What is the probability that Anuj picks a ball numbered less than that picked by Anisha, who in turn picks a lesser n Op 1: 3/25 Op 2: 1/6 Op 3: 4/25 Op 4: 81/400
If Amit picks up 10 and Anisha picks up 9 then

(10, 9, 1) (10, 9, 2) …. (10, 9, 8) – 8 ways
(10, 8, 1) (10, 8 2) ….. (10, 8, 7) – 7 ways

8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 8(9)/2 = 36

If Amit picks up 9 then

(9, 8, 1) ….. (9, 8, 7) … = 7 ways
7 + 6 + 5 + 4 + 3 + 2 + 1 = 7(8)/2 = 28

For 8 – (6)(7)/2 = 21

For 7 – (5) (6)/2 = 15

For 6 – (4) (5)/2 = 10

For 5 – (3) (4)/2 = 6

For 4 – (2) (3)/2 = 3

For 3 – (1) (2)/2 = 1

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36

You either add them or apply the expression n(n+1)(n+2)/6 which is the sum of triangular numbers =8(9)(10)/6 = 120

Probability = 120/(10 x 10 x 10) = 3/25