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
• X3 + 2x2 + 3x – 180 = 0.
• Has no integral solution.
• Hence answer choose none of these.

3. Numbers
• The citizens of planet nigiet are 6 fingered and have
thus developed their decimal system in base 6.
• A certain street in nigiet contains 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)*62 = 3*36 = 108.

4. Numbers
• The citizens of planet nigiet are 8 fingered and
have thus developed their decimal system in base
8. A certain street in nigiet contains 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+b is 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.
• 105/420 = 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, ans is 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 nd denominator 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 hrs in forward journey,
during the return journey
•
increased speed 12km/hr takes
the times 5 hrs.
• What is the distance traveled
A.) 210 kms B.) 30 kms C.) 60
kms D.) 90 kms
• let the speed in the forward
journey be x km/hr, also
•
given time taken in fwd journey
= 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,the journey 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 hr men sees two digit number be(1hr):10x+y
• now after an hr he 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,speed is 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 hrs in forward journey, during the
return journey increased speed 12km/hr takes the times 5 hrs.What is the
distance traveled?
A.) 210 kms B.) 30 kms C.) 60 kms D.) 90 kms
• let the speed in the forward journey be x km/hr, also
given time taken in fwd journey = 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 cheque of amount 35 dollars.
Shop Keeper exchanged the cheque with his neighbor and gave change to
John. After 2 days, it is known that cheque is 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 cheque and 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.s 1000 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.s are 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? 2300 or 3200
Method 1:
2300 = 23*100 = 8100 and 3200 = 32*100 = 9100.
Hence 3200 is greater.
Method 2:
Using log,
300log2 = 300*0.3010 = 90.3 approx and
200log3 = 200*0.4771 = 95.42 approx

33. POWER
•
•
•
•
•
•
•
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find out last two digits of 29573661 + 30813643.
A) 42
B) 38
C) 98
D) 22
29573661= 29573660*2957
so last two digits be 57.
30813643=30813640*30813.
811= 81; 812 = 6561; 813 = 6561*81 = 531441.
so last two digit will be 41
sum 57+41=98.so 98 ans.