Amcat 3-DOWNLOAD ENABLED

USEFUL FOR ENGINEERING STUDENTS PREPARING FOR APTITUDE

1. AMCAT 3

2. Decimal Number
•In which of the system, decimal number 184 is
equal to 1234?
•184= 1*x^3 + 2*x^2 + 3*x + 4
•X
3
+ 2x
2
+ 3x –180 = 0.
•Has no integral solution.
•Hence answer choose none of these.

3. Numbers
•The citizens of planet nigietare 6 fingered and have
thus developed their decimal system in base 6.
•A certain street in nigietcontains 1000 (in base 6)
buildings numbered 1 to 1000.
•How many 3s are used in numbering these buildings?
a) 256 b) 54 c) 192 d) 108
•(N-1)*(BASE)^(N-2)
N=NUMBER OF LAST DIGIT=4(1000 contains 4 digits)
base=6
•(4-1)*6
2
= 3*36 = 108.

4. Numbers
•The citizens of planet nigietare 8 fingered and
have thus developed their decimal system in base
8. A certain street in nigietcontains 1000 (in base
8) buildings numbered 1 to 1000. How many 3s
are used in numbering these buildings? a) 54 b)
64 c) 265 d) 192
•(N-1)*(BASE)^(N-2)
•N=NUMBER OF LAST DIGIT=4(1000 contains 4
digits)
•base=8
•(4-1)*8^(4-2) =192

5. NUMBER
•If a and b are odd numbers, then which of the
following is even ?
•A. a + b
•B. a + b + 1
•C. ab
•D. ab+ 2
•E. None of these
•
•always add two odd number it gives even number

6. Divisiblity
•12 divides, ab313ab (in decimal notation, where a,b
are digits>0, the smallest value of a+bis a)7 b)6 c)2 d) 4
•If a number is divisible by 12 then it should be divisible
by 4&3
•for divisible by 4 last [2 digit]no's should be divisible by
4
•Also sum of digits should be divisible by 3.
•Hence 2(a+b)+1 should be 3k and ab= 4m.
•That is a + b = (3k-1)/2 . from the given choices only
first choice is possible k = 5 and smallest should be 7.

7. Divisibility
•If a number 774958A96B is to be divisible by 8 and 9,
the values of A and B, respectively, will be:
•Using the divisibility rules,
•For 8, the last three digits have to divisible by 8,
therefore the number 774958A96B is divisible by 8 if
96B is divisible by 8.
•96B is divisible by 8 if it is 960 or 968 thus B is either 0
or 8.
•For 9, the sum of the number has to be divisible by 9,
therefore (55 + A + B) is divisible by 9 if (A + B) is 8.
•Now either of A or B could be 8, but the other has to
be zero.

8. Divisibility
•311311311311311311 is divisible by:
a)3 and 11. b)11 but not 3. c)3 but not 11.
d)none of the above.
•a.3 and 11
•the sum of digits is 30 which is divisible by 3
•the differences of odd place digits -even place
is divisible by 11

9. LCM
•Find the greatest number of five digits, which is
exactly divisible by 7, 10, 15, 21 and 28.
•The number should be exactly divisible by 15 (3,
5), 21 (3, 7), 28 (4, 7).
•Hence, it is enough to check the divisibility for 3,
4, 5 and 7.
•Lcm of 3,4,5 and 7 is 420.
•10
5
/420 = 238 is quotient
•238* 420 = 99960 is the only number which
satisfies the given condition.

10. HCF
•The ratio of two numbers is 3 : 4 and their H.C.F.
is 4.
•Their L.C.M. is: A. 12 B. 16 C. 24 D. 48
•Let the numbers be 3x and 4x.
•Then, their H.C.F. = x.
•So, x = 4(given)
•the numbers are 12 and 16.
•LCM of 12,16 is 48.
•So, ansis OPT (D)

11. Fractions

12. Fractions
•If the numerator of a fraction is increased by 25%
and denominator decreased by 20%, the new
value is 5/4. What is the original value? a) 3/5 b)
4/5 c) 7/8 d) 3/7
•let numerator is x nddenominator is y.
•if we increase x by 25% then it will be 125x/100
and if u decrease y by 20% then value will be
80y/100.
•new value is 125x/80y=5/4 .or x/y=4/5.
•hence b)4/5

13. Factorization
•The prime factorization of integer N is A*A*B*C where A, B
and C are all distinct prime integers. How many factors
does N have? a)12 b)24 c)4 d)6
•n is A*A*B*C =A^2*B*C
no. of factors=(2+1)(1+1)(1+1)
=12
•A HAS POWER OF 2,B HAS POWER OF 1,C HAS POWER 1
SO PRIME FACTORIZATION CAN BE CALCULATED AS:
IF A^P+B^Q+C^R,THEN
PRIME FACTORIZATION IS (P+1)*(Q+1)*(R+1)
SO IN THIS CASE P=2,Q=1,R=1
SO (2+1)(1+1)(1+1)=12 FACTORS

14. ALGEBRA
•A person drives with constant speed
and after some time he sees a
milestone with 2 digits. Then he travels
for 1 hour and sees the same 2 digits in
reverse order. 1 hour later he sees that
the milestone has the same 2 digits
with a 0 between them. What is the
speed of the car (in mph)? (a) 54 (b)
45 (c) 27 (d) 36
•s=speed "Then travels for 1 hour and
sees the same 2 digits in reverse
order."
10x + y + 1s(speed *time) = 10y + x
10x -x + s = 10y -y
9x + s = 9y
•1 hours later he sees that the
milestone has the same 2 digits with a
0 between them."
10y + x + s(again same) = 100x + y
10y -y + s = 100x -x
9y + s = 99x
•Rearrange the above two equations for
elimination
9x -9y + s = 0
-99x+9y + s = 0
----------------adding eliminates y
-90x + 2s = 0
2s = 90x
s = 45x
•x has to be equal to 1, then s = 45 mph
•(If x=2 and s=90 mph, when added to a
two digit milestone, could not be at
another two digit milestone.)

15. SPEED, TIME & DISTANCE
•A train travels a certain distance
taking 7 hrsin forward journey,
during the return journey
increased speed 12km/hrtakes
the times 5 hrs.
•What is the distance traveled
A.) 210 kmsB.) 30 kmsC.) 60
kmsD.) 90 kms
•let the speed in the forward
journey be x km/hr, also
given time taken in fwdjourney
= 7 hrs
•during return journey
speed=(x+12)km/hr
time=5hrs
•now distance travelled in both
the cases is same therefore
•forward journey
distance=return journey
distance
7x=5(x+12)
7x-5x=60
2x=60
x=30km/hr
•therefore distance travelled=
30*7=210km
•or =5*(30+12)=210km

16. SPEED, TIME & DISTANCE

17. SPEED, TIME & DISTANCE
•a train travels a distance of 300km at a constant speed.
•if the speed of the train is increased by 5km/hr,thejourney would
have taken two hours less.
•find the initial speed of the train?
•we know that, s = d / t
•distance = 300 km
•initial speed ( s) , new speed = s + 5
•initial time taken ( t ) = 300 / s and new time taken = 300 / s + 5
•so now, according to the ques:
•300 / (s + 5) + 2 = 300 / s
•2 = 300 / s -300 / (s + 5)
•s = 25 or -30
answer: 25 km / hr..

18. SPEED, TIME & DISTANCE
•A person drives with constant speed and after
some time he sees a milestone with 2 digits.
Then he travels for 1 hour and sees the same
2 digits in reverse order. 1 hour later he sees
that the milestone has the same 2 digits with
a 0 between them. What is the speed of the
car (in mph)?
(a) 54 (b) 45 (c) 27 (d) 36

19. SPEED, TIME & DISTANCE
•s=speed
"Then travels for 1 hour and sees the same 2 digits in reverse order."
10x + y + 1s(speed *time) = 10y + x
10x -x + s = 10y -y
9x + s = 9y
•1 hours later he sees that the milestone has the same 2 digits with a 0 between
them."
10y + x + s(again same) = 100x + y
10y -y + s = 100x -x
9y + s = 99x
•Rearrange the above two equations for elimination
9x -9y + s = 0
-99x+9y + s = 0
----------------adding eliminates y
-90x + 2s = 0
2s = 90x
s = 45x
•x has to equal 1, then s = 45 mph (If x=2 and s=90 mph, when added to
a two digit milestone, could not be at another two digit milestone.)

20. SPEED, TIME & DISTANCE
•let during the 1st hrmen sees two digit number be(1hr):10x+y
•now after an hrhe sees reverse of the digit(2 hr) :10y+x
•again he sees a 3-digit number which is same as previous but with 0 added
so(3hr):100x+0+y
•since,speedis constant then
distance covered in an hr=distance covered in 2 hr
•(10y+x)-(10x+y)=(100x+y)-(10y+x)
-9x+9y=99x-9y
-108x+18y=0
18y=108x
9y=54x
so,x=1 y=6
•ie:16 , 61 AND 106
•so to get the speed when we add 45 to 16 ie:45+16=61
•similarly 61+45=106
so speed is 45mph

21. SPEED, TIME & DISTANCE
•A train travels a certain distance taking 7 hrsin forward journey, during the
return journey increased speed 12km/hrtakes the times 5 hrs.Whatis the
distance traveled?
A.) 210 kmsB.) 30 kmsC.) 60 kmsD.) 90 kms
•let the speed in the forward journey be x km/hr, also
given time taken in fwdjourney = 7 hrs
•during return journey
speed=(x+12)km/hr
time=5hrs
•now distance travelled in both the cases is same
•Therefore forward journey distance=return journey distance
7x=5(x+12)
7x-5x=60
2x=60
x=30km/hr
•therefore distance travelled= 30*7=210km or 5*(30+12)=210km

22. PROFIT / LOSS

23. PROFIT / LOSS

24. PROFIT / LOSS

25. PROFIT / LOSS
•John buys a cycle for 31 dollars and given a chequeof amount 35 dollars.
Shop Keeper exchanged the chequewith his neighbor and gave change to
John. After 2 days, it is known that chequeis bounced. Shop keeper paid
the amount to his neighbour. The cost price of cycle is 19 dollars. What is
the profit/loss for shop keeper? a) loss 23 b) gain 23 c) gain 54 d) Loss 54
•CP of cycle = 19$
•SP of cycle = 31$
•Profit = 31$-19$ = 12$
•Again, shopkeeper gave 35$ to neighbour.
•Loss = 35$
•So, net loss = 35$-12$ = 23$
•Other way there is no profit or loss in exchanging the chequeand change
with neighbour.
•The shop keeper loses the cycle worth 19$ and 4$ change. Hence net loss
= 23$.

26. PROBABILITY
•Mr.Decimal enters a lucky draw that requires him to pick five different
integers from 1 through 37 inclusive.
•He chooses his five numbers in such a way that the sum of their
logarithms to base 10 is an integer.
•It turns out that the winning 5 numbers have the same property, i.e. the
sum of their logarithms to base10 is also an integer.
•What is the probability that Mr.Decimal holds the winning numbers?
•We know log(a)+log(b)+log(c)+log(d)+log(e)=log(a*b*c*d*e).
•From the above question there are only two no.s1000 and 10000 within
range having correct decimal value of 3 & 4.
•Now the factors of 1000 is = 2,2,2,5,5,5
•If the no.sare arranged then we get One sample spaces 2*4*5*25*1.
•Similarly for 10000 we get 2*4*5*10*25 , 5*8*10*25*1 & 4*5*20*25*1.
•totally there are 4 sample spaces and Mr Decimal is choosing one of the
combination than the probability will be equal to 1/4.

27. PERMUTATION
•How many 13 digit numbers are possible by using
the digits 1,2,3,4,5 which are divisible by 4 if
repetition of digits is allowed?
•to be divisible by 4, last two digits must be
divisible by 4. which are 12, 24, 32, 44, 52.
•so 5 combinations are possible for last two digits
•also 5 combinations each for remaining 11 places.
•so the answer is 5^12.

28. Permutation with restriction

29. LOGARITHM

30. LOGARITHM

31. Logarithm

32. POWER
•Which is greater? 2
300
or 3
200
•Method 1:
•2
300
= 2
3*100
= 8
100
and 3
200
= 3
2*100
= 9
100
.
•Hence 3
200
is greater.
•Method 2:
•Using log,
•300log2 = 300*0.3010 = 90.3 approxand
200log3 = 200*0.4771 = 95.42 approx

33. POWER
•find out last two digits of 2957
3661
+ 3081
3643
.
•A) 42B) 38C) 98D) 22
•2957
3661
= 2957
3660
*2957
•so last two digits be 57.
•3081
3643
=3081
3640
*3081
3
.
•81
1
= 81; 81
2
= 6561; 81
3
= 6561*81 = 531441.
•so last two digit will be 41
•sum 57+41=98.so 98 ans.