1. Find the odd no: 2,3,12,36,86,167

Sol: given series. 2,3,12,36,86,167

now, 3-2=1 => 1^2=1; 12-3=9 => 3^2=9;

now, 3-2=1 => 1^2=1; 12-3=9 => 3^2=9;

36-12=24 => here if we change 36 into 37 then 37-12=25 =>5^2=25

86-36=50 => Similary 86-37=49 =>7^2=49

167-86=81 =>9^2=81

so answer will be 36.

86-36=50 => Similary 86-37=49 =>7^2=49

167-86=81 =>9^2=81

so answer will be 36.

2. 8,12,24,60,… what is the next number?

Sol: 12-8=4; 24-12=12; 60-24=36…

the difference like this 4*3=12,12*3=36 then next 36*3=108.so the answer is 108+60=168.

the difference like this 4*3=12,12*3=36 then next 36*3=108.so the answer is 108+60=168.

3. 3,7,10,11,12,17,? find out next number?

Sol: sum of alternates -1 equal to next alternate number.

3+10-1=12; 7+11-1=17; so the next number will be 10+12-1=21

3+10-1=12; 7+11-1=17; so the next number will be 10+12-1=21

4. Cost price of 4 calculators and 2 pencil is 6200. What is the cost of ten calculators and five pencils.

Sol: C+2P=6200–>2C+P=3100

SO 10C+5P=5(2C+P)=5(3100)=15,500 RS

SO 10C+5P=5(2C+P)=5(3100)=15,500 RS

5. 12 men can complete work in 6 days whereas 10 men and 21 women take 3 days to finish the same work .in how many days can 12 women alone complete.

Sol: 10 men’s, 1 day work=10/(12*6)=5/36

If 21 women’s, 1 day work= 21/W ,then

3[(5/36)+(21/W)]= 1 ,On solving, W=108

So, 12 women can complete the work in 108/12=9 days

If 21 women’s, 1 day work= 21/W ,then

3[(5/36)+(21/W)]= 1 ,On solving, W=108

So, 12 women can complete the work in 108/12=9 days

6. 2718/14 find the remainder value?

Sol: Any number of the form (a*x-1)^n /a the remainder will be +1 if the power n is even.

and the remainder will be -1 or (a-1) if the power is odd. According to this the remainder will be 1

7. What is the probability of getting a odd sum when two dice are thrown.

Sol: odd numbers 3,5,7,9,11(between 2(min sum)-12(max sum))

cases:- 3-(1,2),(2,1) 5-(3,2),(2,3),(4,1),(1,4) 7-(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)

9-(3,6),(4,5),(5,4),(6,3) 11-(5,6),(6,5)

cases:- 3-(1,2),(2,1) 5-(3,2),(2,3),(4,1),(1,4) 7-(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)

9-(3,6),(4,5),(5,4),(6,3) 11-(5,6),(6,5)

total cases=18 therefore probability=18/36=>1/2

8. If log(p+q)(p-q)= -1; then find the value of: log(p+q)(P2-q2)

Sol: log(p+q)(p2-q2)=log(p+q)2(p-q)=log(p+q)2 + log(p-q) = 2 log(p+q) + log(p-q) = log(p+q)+log(p+q)+log(p-q)

log(p+q)-1= log(p+q)-log 10= log (p+q)/10;

9. If peacock = 11526312316 then feather=

Sol: First letter is decoded on the position of letter in reverse order

and 2nd letter is decoded on the position of letter in alphabetic order in peacock

p=11 e=5 a=26 c=3 o=12 c=3 k=16

so, feather

and 2nd letter is decoded on the position of letter in alphabetic order in peacock

p=11 e=5 a=26 c=3 o=12 c=3 k=16

so, feather

f=21 e=5 a=26 t=20 h=19 e=5 r=9

Ans: 21526201959

10. Find the number of consecutive zeros at the end of 72!

Sol: By using formula round(n/5)+round(n/25)+……….round(n/5^n)

round(72/5)+round(72/25)=14+2=16

11. Find the maximum value of n such that 77! is perfectly divisible by 720^n

Sol: 720 = (2^4) x (3^2) x (5)

Number of twos in 77! is: 38 + 19 + 9 + 4 + 2 + 1 = 73

So number of 2^4 will be: = 18

Similarly, Number of threes in 77! is:

25 + 8 + 2 = 35

So number of 3^2 will be: = 17

and number of fives will be:

= 15 + 3 = 18

Hence the number of 720s that we can obtain is 17.

Number of twos in 77! is: 38 + 19 + 9 + 4 + 2 + 1 = 73

So number of 2^4 will be: = 18

Similarly, Number of threes in 77! is:

25 + 8 + 2 = 35

So number of 3^2 will be: = 17

and number of fives will be:

= 15 + 3 = 18

Hence the number of 720s that we can obtain is 17.

12. 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:

Sol: 3 days

13. If log102 = 0.3010, what is the number of digits in 264

Sol: 264= 64 log2=64*0.3010=19.264

Characteristic(integral part of log) is one less than no. of digits. here characteristic=19; So no of digits=19+1.

Characteristic(integral part of log) is one less than no. of digits. here characteristic=19; So no of digits=19+1.

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

Sol: y3=1369y y2=1369 y=37

15. 4.28 and -3.28 are two numbers on a real number line. If 1 is added to both the numbers ,then which of

the following is true?

a.Distance between the two numbers is 2 units more than the distance between 4.28 and -3.28

b.Distance between the two numbers is 2 units less than the distance between 4.28 and -3.28

c.Distance between the two numbers is equal to than the distance between 4.28 and -3.28

d.None

Ans:c

distance b/w 4.28 and -3.28 = 7.56

distance b/w 5.28 and -2.28 = 7.56

distance b/w 5.28 and -2.28 = 7.56

16. What is the greatest 4-digit perfect square, which is exactly divisible by 3, 5, 7 and 9?

a. 9999 b. 9684 c. 9801 d. Cannot be determined e. 11025

17. What is the full form of OSI ?

Ans: Open system interconnection

18. What is the remainder when 1723 is divided by 16?

a. 1 b.0 c.2 d.3

Sol: Apply bionomial theorem. 17 can be written as (16+1)

similarly 1723= (16+1)23

expand above equation using bionomial theorem then we get 24 terms, in that 23 terms contain 16x as one term for x>0; 24th term will be 24c24 124 =1 when you divide 1 with 16 you get 1 as remainder.

19. antilog10100

Sol: All you need to know is that AntiLog(X) = 10x.

so ans is 10^(10100)

20. Four bells begin to toll together and then each one at intervals of 6 s, 7 s, 8 s and 9 s respectively. The

number of times they will toll together in the next 2 hr is:

Sol: lcm of 6,7,8,9=504 sec; in 2 hrs =3600*2 sec

so no times they will ring =3600*2/504==14 times

21. The students are in the ratio 2:3:5.if 20 students are increased in each batch the ratio changes to 4:5:7

The total number of students in the three batches before the increase was

Sol: Let number of students be x

2x+20 : 3x+20 : 5x+20 = 4:5:7 –> x = 10

Initially the number of the students would be 20,30 and 50 –> 100 ans

2x+20 : 3x+20 : 5x+20 = 4:5:7 –> x = 10

Initially the number of the students would be 20,30 and 50 –> 100 ans

22. sum of money doubles itself in 9 years, in how many years it will become 8 times itself?

Sol: 27 years; 9yrs = 2(sum); 18yrs = 4(sum); 27yrs = 8(sum)

23. What is the smallest four-digit number which when divided by 6, leaves a remainder of 5 and when

divided by 5 leaves a remainder of 3?

1. 1043 2. 1073 3. 1103 4.None of these

Sol: ans is none of these because :

let us assume the smallest 4 digit number be 1000 if we divide it with 6 we get remainder 4 so to get a rem of 5 add 1 to it => 1001.

Then the general form of a number is 1001+6k for every positive integer value of k it always yields rem 5 when divided by 6

then by trail and error if we take k=2 then number is 1013

which when divided by 5 gives a rem of 3

so the right ans is 1013 which is none of these from options

let us assume the smallest 4 digit number be 1000 if we divide it with 6 we get remainder 4 so to get a rem of 5 add 1 to it => 1001.

Then the general form of a number is 1001+6k for every positive integer value of k it always yields rem 5 when divided by 6

then by trail and error if we take k=2 then number is 1013

which when divided by 5 gives a rem of 3

so the right ans is 1013 which is none of these from options

24. A, B, C started a business with their investments in the ratio 1:3:5. After 4 months, A invested the double amount as before and B as well as C withdrew half of their investments. The ratio of their profits at the end of the year is:

Sol: Let their initial investments be x, 3x and 5x respectively. Then,

A : B : C = (x * 4 + 2x * 8) : (3x * 4 + 3x/2 * 8 ) : (5x * 4 + 5x/2 * 8)

= 20x : 24x : 40x = 5 : 6 : 10.

A : B : C = (x * 4 + 2x * 8) : (3x * 4 + 3x/2 * 8 ) : (5x * 4 + 5x/2 * 8)

= 20x : 24x : 40x = 5 : 6 : 10.

25. There are 10 yes or no questions. How many ways can these be answered?

Sol: for 1 question 2 possibilities

for 2nd question 2 possibilities

for 3rd question 2 possibilities

.

.

..

.

210=1024

for 2nd question 2 possibilities

for 3rd question 2 possibilities

.

.

..

.

210=1024

26. In an examination, 70% of students passed in physics, 65% in chemistry, 27% failed in both subjects. The percentage of students who passed is:

1. 66% 2. 62% 3. 69% 4. None of these

Sol: let’s total student be 100

Passed in atleast one subject= 100-27=73.

73= 70+65-x(passed both subjects)

Passed in atleast one subject= 100-27=73.

73= 70+65-x(passed both subjects)

x=62.

27. 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:

1. Rs. 86.80 2. Rs. 86.10 3. Rs. 88.65 4. Rs. 81.60

Sol: S.I for 2 years is 80;

Then S.I for one year is 40. C.I for 2 years = S.I for 2 years + S.I for 40

=80+(40*4*1)/100=80+1.60=81.6

28. Prabodh bought 30 kg of rice at the rate of Rs. 8.50 per kg and 20 kg of rice at the rate of Rs. 9.00 per kg. He mixed the two. At what price (App.) per kg should he sell the mixture in order to get 20% profit?

1. Rs. 9.50 2. Rs. 8.50 3. Rs. 10.50 4. Rs. 12.00

Sol: 30*8.5+20*9=435

20% of 435 is 87

total= 435+87=522

522=50*x; x=10.44= 10.5(approx)

29. 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:

1. 4 hours 2. 19/4 hours 3. 9/2 hours 4. 17/4 hours 5. None of these

Sol: 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

2W=465 minutes(& hours 45 minutes)

2R=?

2(W+R)=375*2=750

2R=750-465=285=19/4

30. In an examination 10 questions are to be answered choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can 10 questions be answered ?

1. 212 2. 266 3. 272 4. 312 5. Correct Op

Sol: 266 = 6c4*7c6+6c5*7c5+6c6*7c4

31. 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.

Sol: sqrt(8^2+6^2)=10

32. synonym of OBTRUSIVE

Sol: conspicuous

33. hcf of 3.68 & 5.35

Sol: prime factorization of 368 is 24*23

prime factorization of 535 is 5*107

hcf is 1. ans is 0.01

34. 3*(44+43+42+4+1) = ?

Sol: 3*(45-1)/4-1 ( applying sum of n terms in G.P )

=45-1= 1023

35. log10 2=.6096 log10 3=.4709 then log10 12 = ?

Sol: log10 12= log10 (22*3)= log10 22+ log10 3= 2 log10 2+ log10 3=2*0.6096+0.4709= 1.6901

36. log xy – log |x| = ?

Sol: log xy – log |x| = log x + log y – log |x| = log y

since x > 0 for log to be defined.

hence |x| = x

since x > 0 for log to be defined.

hence |x| = x

37. log 25625 – log 31961 + log 29841 =?

Sol: 2-2+2=2

38. P3M : N4J :: R3P : ??

Sol: because compare P3M with R3P then N4J=P4M

39. 2,35,104,209,?

Sol: 35-2=33;104-35=69;209-104=105;;

69-33=36;105-69=36;;

so 36+105=141==>>141+209=350(ans)

69-33=36;105-69=36;;

so 36+105=141==>>141+209=350(ans)

40. NATION – 1412091514 THEN REMOTE- ?

Sol: Ans: 1851315205

A=1 B=2 ……………….Z=26

A=1 B=2 ……………….Z=26

41. 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

Sol: Data not 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.

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.

42. how many 5 digit nos are possible from 2,7,0,8,4 if the first digit is not zero

Sol: if there is no repetition then its equal to=4*4*3*2*1=96

if repetitions are allowed = 4*5*5*5*5=2500

if repetitions are allowed = 4*5*5*5*5=2500

43. 400 have how many factors?

Sol: 15 factors….

400=24*52

no of factor=5*3=1

400=24*52

no of factor=5*3=1

44. 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.

Sol: probability=at least 1 ball yellow

=1-no ball yellow

=1-{(6C0*3C3*2C2)/11C5}

=1-(1/462)

=0.997

=1-no ball yellow

=1-{(6C0*3C3*2C2)/11C5}

=1-(1/462)

=0.997

45. if north-west is east ,north-east is south then what is east?

Sol: south-west

46.How many two digit numbers have exactly 5 factors?

Sol: for a two digit number to have 5 factors it must be a squared number and it must not be a square of prime number. the two numbers are,

4^2 = 16 = 1,2,4,8,16

9^2 = 81 = 1,3,9,27,81

4^2 = 16 = 1,2,4,8,16

9^2 = 81 = 1,3,9,27,81

47.How many four digit numbers have exactly 5 factors?

Sol: for any no. greater than 100 to have 5 factors it must be the 4th power of the prime number.

e.g, 5^4 = 625 = 1,5,25,125,625

7^4 = 2401 = 1,7,49,343,2401

11^4 = 14641 = 1,11,121,1331,14641

so the only 4 digit number having 5 factors is 2401

and two digit number having 5 factors are 16 and 81

e.g, 5^4 = 625 = 1,5,25,125,625

7^4 = 2401 = 1,7,49,343,2401

11^4 = 14641 = 1,11,121,1331,14641

so the only 4 digit number having 5 factors is 2401

and two digit number having 5 factors are 16 and 81

48. 15?1792 is divisible by 9 only when ?=

1.1 2.4. 3.3. 4.2.

Sol: 1+5+1+7+9+2=25=2+5=7+?=9 so ?=2

49. 2x+y=2/23/2; 2x-y=2; Find the values of x and y?

Sol: 2x+y=21-3/2; x+y=-1/2;

and 2x-y=21; x-y=1;

and x=1/4 y=-3/4

and 2x-y=21; x-y=1;

and x=1/4 y=-3/4

50. 6,9,__,24,39. 1).18. 2)15. 3).10. 4)12.

Sol: its simply the addition of the previous two numbers

9+6=15; 15+24=39; so the answer is 15

9+6=15; 15+24=39; so the answer is 15

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

Sol: 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

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

52. 79,64,26,15,__

Sol: 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

64 = 6*4 = 24+2= 26

26= 2*6 = 12+3 = 15

15 = 1*5 = 5+4 = 9

so the ans is 9

53. 8,8,6,2,___.

1.-4. 2.4 3.2 4.0.

Sol: 8-8 = 0; 8-6 = 2; 6-2 = 4; 2-(-4) = 6; So Ans is: -4

54. Please remember all the words because amcat always repeat synonym and antonym…

1.conceit 2.conceal 3.preamble 4. Engendered 5.veteran 6..instigate 7.blighting

8.overt 9.verdict 10.avarice 11.set off 12. gruesome 13.vent

these words were in my set and my friend set so prepare the synonym and ant both because when they repeat question they change ant into synonym or vice versa…..

Sol: Synonyms : 1.arrogance 2.hide 3.preface 4.arouse,create,stir 5.experienced,old time

6.influence,provoke 7.ruin,destroy 8.unconcealed 9.judgement 10.greedy 11.compensate, redeem

12.horrible, awful 13.outlet

6.influence,provoke 7.ruin,destroy 8.unconcealed 9.judgement 10.greedy 11.compensate, redeem

12.horrible, awful 13.outlet

Antonyms: 1.humility,shyness 2.disregard,uncover,disclose 3.discusion,meeting 4.calm, destroy, discourge

5.unskilled, inexperenced 6.discourage, negelect 7.Encourage, improve 8.unclear, hidden

9. announcement, commitment 10.generosity, philanthopy 11.disagree, spoil

12.attrective, beautiful 13.closing, clouse

5.unskilled, inexperenced 6.discourage, negelect 7.Encourage, improve 8.unclear, hidden

9. announcement, commitment 10.generosity, philanthopy 11.disagree, spoil

12.attrective, beautiful 13.closing, clouse

55. 6 years back, Rom and Dom had their ages in the ratio 1:2. 6 years from now the ratio of their ages would be 3:4. What is the ratio of their ages today?

Sol: (x-6)/(y-6)=1/2 ….(1)

(x+6)/(y+6)=3/4 ..(2)

by solving these eqn, we get x=12 and y=18 i.e. 2/3

(x+6)/(y+6)=3/4 ..(2)

by solving these eqn, we get x=12 and y=18 i.e. 2/3

56. 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:

57. 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:

Sol: 5u=2t u=2/5t

(33*2t*15*12)/(5*xt*11*9)=2/3

x=36 op 2:36

(33*2t*15*12)/(5*xt*11*9)=2/3

x=36 op 2:36

58. |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 5:

Sol: Op 3: 0.5; 4.5+4>0; &(0.5)^2

59. If r = at2 and s = 2at, the relation among s, r and a is: 1. s2=4ar 2. s=ar 3. s=2ar 4. s2=ar 5. None of these

Sol: s=2at; squaring on both sides; s2=4a2t2=4a* at2=4ar; so option 1 is correct

60. If |x| + |y| = 7, then what is the sum of minimum and maximum values of x + y? 1. 3/2 2. -7 3. 7 4. 0 5.none

Sol: 0; as mod has property; |x|=x;x>0; =-x;x

61. If x4 + 1/x4 = 47, then find the value of x3 + 1/x3 1. 18 2. 27 3. 9 4. 12

Sol: (x2+1/x2)2=x4+1/x4+2

so x2+1/x2=7

(x+1/x) 2=x2+1/x2+2=7+2=9

so x+1/x=3;

now (x+1/x) 3=x3+3*x+3*1/x + 1/x3

33=x3+3*3+1/x3

so x3+1/x3=18

62. If a, b, c are roots of the equation x3-4×2+6.5x + 3.5 = 0, then what is the value of a2 + b2 + c2? a. 1 b. 64 c. 169 d.3 Sol: let l,m,n be roots of ax3+bx2+cx+d=0; then l+m+n=-b/a, lmn=-d/a, lm+mn+ln=c/a,

here a=1, b=-4, c=6.5

a2+b2+c2= (a+b+c)2-2(ab+bc+ca)=3.

here a=1, b=-4, c=6.5

a2+b2+c2= (a+b+c)2-2(ab+bc+ca)=3.

63. If 13 + 23+ 33+ ……….+ 93 = 2025 ,then the value of (0.11)3 + (0.22)3 + ….+ (0.99)3 is ?

Sol: (11/100)3 +(22/100)3 + ……(99/100)3

take (11/100)3 as common then

(11/100)3[13+23+33+ …..+93]

take (11/100)3 as common then

(11/100)3[13+23+33+ …..+93]

=(11/100)3*2025= 2.695275

64. 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 Op 5:

Sol: 2/3

65. A ,B, 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 attend the class is 7/10,Exactly one out of B and C attends is 4/10 exactly one out of A and C attends is 7/10.if the probability that all the three attend the class is 9/100 then find the probability that all at least one attends the class.

Sol: 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 =7/10B,C

b(1-c)+c(1-b) = 4/10

b+c -2bc =4/10

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 =7/10B,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

Add all 3 u get

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

66. 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 number than amit.

Sol:

67. A, B, C, D and E play the following game. Each person picks one card from the cards numbered 1 through 10. The person who picks the greatest numbered card loses and is out of the game. Now the remaining four return their cards to the pack and draw again, and again the person with the greatest numbered card loses. This process is repeated till only one person is left in the game that is declared the winner. What is the probability that A is the

winner?

Sol: Ans:1/5

Total five persons so anybody can be the winner.

68. a buy clips at 12 for R.s. 60 .How many clips should he sellfor Rs. 60 to earn a profit of 20% ?

Sol: x+0.2x=60;

x=50;

12 clips for 60

so 10 clips for 50.

ans:10

69. An article was sold for Rs. 2770. Had it been sold for Rs. 3000 there would have been an additional gain of 10%. Cost Price of the article is:

Sol: given selling price is = 2770

he said if we sell it for 3000 there would be a 10% more gain

3000-2770=230

because of this Rs.230 he can gain 10% more

from profit percentage formula

230*100/ cost price =10

from the above equation cost price is 2300

he said if we sell it for 3000 there would be a 10% more gain

3000-2770=230

because of this Rs.230 he can gain 10% more

from profit percentage formula

230*100/ cost price =10

from the above equation cost price is 2300

70. The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is:

Sol: Hitting the target at least 3 times means it can be greater than 3 also i.e.3,4,5

in 5 chances hitting target by 3 times is

5c3*(3/4)^3*(1/4)^2 = 10*27/1024= 270/1024

probability of hitting by 4 times is

5c4*(3/4)^4(1/4)^1=5*81/1024= 405/1024

probability of hitting 5 times is

5c5*(3/4)^5 = 243/1024

total is (270+405+243)/1024= 918/1024

=459/512

in 5 chances hitting target by 3 times is

5c3*(3/4)^3*(1/4)^2 = 10*27/1024= 270/1024

probability of hitting by 4 times is

5c4*(3/4)^4(1/4)^1=5*81/1024= 405/1024

probability of hitting 5 times is

5c5*(3/4)^5 = 243/1024

total is (270+405+243)/1024= 918/1024

=459/512

71. A 5-digit number is formed by the digits 1,2,3,4 and 5 without repetition. What is the probability that the number formed is a multiple of 4?

Sol: any number is divisible with 4 iff last two digits should be divisible with 4.

so if last digit is 2 then 12, 32 and 52 can be last two digits. so 3* 3!

if last digit is 4 then only possibility is 24 so 3!

total= 3* 3! + 3!= 4!

so 4!/5!=1/5

72. In how many ways can a number 6084 be written as a product of two different factors ?

Sol: 6084 = 6 * 1014 = 6*6*169 = 22 * 32 * 132

So the pairs will have either 0,1, or 2 powers of each of three prime numbers. But one of these has two identical numbers, and the rest come in pairs of duplicates.

The answer is ((3*3*3)-1)/2 + 1= 13.

So the pairs will have either 0,1, or 2 powers of each of three prime numbers. But one of these has two identical numbers, and the rest come in pairs of duplicates.

The answer is ((3*3*3)-1)/2 + 1= 13.

73. A lady gives dinner party to five guests to be selected from 9 friends .The number of ways of forming the party of 5,given that two of the friends will not attend the party together is

Sol: No of guests to be invited=5

Therefore,

No of ways forming the party= (9-2)c5*2c0+(9-2)c4*2c1 =7c5*1+7c4*2 =91

Therefore,

No of ways forming the party= (9-2)c5*2c0+(9-2)c4*2c1 =7c5*1+7c4*2 =91

74. There are 5 letters and five addressed envelops. the number of ways in which all the letters can be put in wrong envelops is:

Sol: We have N letters and N envelopes. The Letters can be put in the N envelopes in N! ways . We want to count the Number of “Derangements” ( The no. of ways that no letter goes into right envelope ).N!( 1- 1/1! + 1/2! – 1/3! +………………+(-1)n 1/n! ) (this the the formula).

Here N = 5.

So When We put N = 5 in Formula we get 44 ans.

Here N = 5.

So When We put N = 5 in Formula we get 44 ans.

75. A five -digit number divisible by 3 is to be formed using numerals 0,1,2,3,4 and 5 without repetition. The total number of ways this can be done is:

Sol: we have 5 place to arrange this no. and the total should be divisible by 3

no. are :- 0,1,2,3,4,5

take five no.

0+1+2+3+4=10(not divisible by 3)

1+2+3+4+5=15(divisible by 3) possible combination 5!=120

2+3+4+5+0=14(not divisible by 3)

3+4+5+0+1=13(not divisible by 3)

4+5+0+1+2=12(divisible by3)

so these no. we can take, but remember that we can not take 0 at 1st place so

posible combination is 4*4*3*2*1=96

5+0+1+2+3=11(not divisible by 3)

total no.is=120+96=216

no. are :- 0,1,2,3,4,5

take five no.

0+1+2+3+4=10(not divisible by 3)

1+2+3+4+5=15(divisible by 3) possible combination 5!=120

2+3+4+5+0=14(not divisible by 3)

3+4+5+0+1=13(not divisible by 3)

4+5+0+1+2=12(divisible by3)

so these no. we can take, but remember that we can not take 0 at 1st place so

posible combination is 4*4*3*2*1=96

5+0+1+2+3=11(not divisible by 3)

total no.is=120+96=216

76. Mark price of a good is 45 Rs. If seller sells it at 42 Rs as discount price and also want 5 % profit then what will be cost price?

Sol: x+0.05x=42; then x=40

77. In a bag there are 5 white, 8 red, 2 black and 3 blue balls. what is probability that ball picked is red or black?

Sol: Total balls=18; Probability=red/tot + black/tot ; 10/18 =5/9

78. How many 4 digit even no. is possible by 1,2,3,4 if no one is repeated?

Sol= 3! + 3! =12

79. log3 9-log4 256+log5 125=?

Sol: 2-4+3=1;

80. If a=2 & b=1 then log(a+b)(a2-b2)=?

Sol: substitute a= and b=1 in log(a+b)(a2-b2)= log33=1

81. A coin is tossed 3 times by raju. what is probability that raju win all three time?

Sol: 1/2 * 1/2 * 1/2 = 1/8

82. If there are 5 different roads to go into a city then no. of ways to go and back to home?

Sol: 25,if one goes using 1st road,there are 5 roads to come back…..so 5*5, 25 is the ans

83. probability of finding 9 of hearts from deck of 52 cards ?

Sol: there is only 1,9 of heart is present in a deck of 52 cards. so probability of finding 9 of heart=1/52

84. opposite of ‘ instigate’

Sol: instigate means get something started: to cause a process to start.

opposite— stifle,halt

opposite— stifle,halt

85. log √6 1296= ?

Sol: log √6 64= 8;

86. Hemant and Ajay start a two-length swimming race at the same moment but from opposite ends of the pool. They swim in lane and at uniform speed, but Hemant is faster than Ajay.They first pass at a point 18.5 m from the deep end and having completed one length, each one is allowed to rest on the edge for exactly 45 seconds.After setting off on the return length, the swimmers pass for the second time just 10.5 m from the shallow end.How long is the pool?

Sol:

87. A and B start together from the same point on a circular track and walk in the same direction till they both again arrive together at the starting point. A completes one circle in 224 s and B in 364 s. How many times will A have passed B?

Sol: 13 times; LCM of 224,364=2912; so, A does 13 circles while B does 8 in 2912s.

Thus A crosses B 13 times.

Thus A crosses B 13 times.

88. What is opposite meaning of– DISCREET?

Sol: Discreet means tactful: careful to avoid embarrassing or upsetting others.

so opposite — careless

so opposite — careless

89. The North is a North-East, North-East is a East, East is a South-East ….then which direction point a South-East?

Sol: Anticlockwise shift; so south is ans

90. 6:60:: 120:?

Sol: 6+0=6; so 1+2+0=3; so 3 is the answer

91. Log √64 base 8==?

Sol: 1

92. Log2 0.5==?

Sol: Log2(1/2)= -1

93. Which one used as gobal aparator from Fuction

1. opaarator:: 2.oparator;; 3.oparator% 4.oparator !! (two bars)

Sol: :: scope resolution operator

94. Which is invalid?

1.10!6 2.false && True 3. bool(x)=(bool)10 4. flat= 12.67

Sol: flat has no meaning. float will be decimal point

95. meeru has lost her way t to home and was standing 25 meters away from her house in the S-W Diraction. she walks 20m north and reachs Point A . How far and in which direction would she have to walk to reach her House

1.20 meter.east 2. 15 meter , east 3. 15 meter west 4. 20 meter, west

Sol: 15 meter,east

96. 46:64::82:

1.100 2.104 3.48 4. 42

Sol: 100

97. Q. A __________ is a constructor that either has no parameters, or if it has parameters, all the parameters have default values.

A. default constructor B. copy constructor C. Both A and B D. None of these

98. if 15 oxen or 20 cow can eat the grass opf the field in 80 days. then in how many days will 6 oxen and 2 cows eat the same grass. 1.40 2.60 3. 100 4. 160

Sol: 15 oxen take 80 day so, 6 oxen take 15*80/6 =200 day

20 oxen also take 80 day so, 2 cow take 20*80/2 =800 day

together work will be 800*200/(800+200) =160 days

20 oxen also take 80 day so, 2 cow take 20*80/2 =800 day

together work will be 800*200/(800+200) =160 days

99. 754:310::976:??

1.565 2.654 3.643 4.854

Sol: There should be 532;

but a/c to options, 754-310=444;

976-643=333;

so 643 may be an answer

but a/c to options, 754-310=444;

976-643=333;

so 643 may be an answer

100. Which is more-successive discount of 40% of 30 % OR flat 70% ?

Sol: flat 70%;

because on Rs. 100 ,

successive discount of 40% of 30 %=100*0.6*0.7=42;

flat 100*0.3=30,

so discount of 70% is more

because on Rs. 100 ,

successive discount of 40% of 30 %=100*0.6*0.7=42;

flat 100*0.3=30,

so discount of 70% is more

101. If log(base p)25p=2.Find the value of P?

Sol: p2=25*p so p=25;

102. 49 pumps can empty a reservor 6 1/2 days, working 8 hours a day. if 196 pumps are used for 5 hours a day , then the same work will completed in. 1. 2.6 days 2. 3 days 3. 2.5 days 4. 2 day

Sol: 49*13/2*8=196*x*5

x=2.6 days

x=2.6 days

103. If 7 spider make 7 webs in 7 days then 1 spider will make 1 web in how many days? 1.1 2.7 3. 7/2 4.49

Sol: 7 Days

104. Ravi brought 300 liter of milk at Rs 19 per liter. he added 200 liter ofd water to it and sold 400 of milk at rs 20 per liter.to the rest, he added 10 liter more water to it and then sold it for rs. 15 per liter.if he used miniral water that cast iRs 10 per liter.then the money earned by ravi is: 1. 4000 2.4500 3.1800 4.1850

Sol: total cost price will be.. (300*19 + 210 *10 )as 300 lt of was purachsed @ rs 19 and 210 lts of water is added @ rs 10. sp will be , (400*20 + 110 *15 )

so CP = 7800 & SP = 9650

profit = 9650 -7800 = 1850

so CP = 7800 & SP = 9650

profit = 9650 -7800 = 1850

105. 2x+y=2*(2)1/2 and 2x-y= 21/2 , the value of x is. 1.1 2.2 3.3 4.4 5.none of these

Sol: as compare the powers of both eq’s

such that x+y= 3/2; x-y=1/2; on solving x=1

such that x+y= 3/2; x-y=1/2; on solving x=1

106. whats the value 1%(modulus) 160/130; 1.160/130 2.1/130 3.1/160 4. 130/160

Sol: 1 %160/130

ie % having higher priority compare to /

so it is calculated as (1%160)/130

so 1/130

ie % having higher priority compare to /

so it is calculated as (1%160)/130

so 1/130

107. FIND THE SERIES: 2,12,36,80,150,__

Sol: 1^2 + 1^3 = 1+1=2

2^2 + 2^3 = 4+8=12

3^2 + 3^3 = 9+27=36

4^2 + 4^3 = 16+64=80

5^2 + 5^3 = 25+125=150

so the next number will be

6^2 + 6^3 = 36+216=252

2^2 + 2^3 = 4+8=12

3^2 + 3^3 = 9+27=36

4^2 + 4^3 = 16+64=80

5^2 + 5^3 = 25+125=150

so the next number will be

6^2 + 6^3 = 36+216=252

108. if RESULT is coded as SFTVMU then EXAM is coded as_____

Sol: answer is FWBN

adding one to every alphabet to get the next one.

adding one to every alphabet to get the next one.

109. revati brought a machine of 4,50,000 and sold it to raghu at profit . Raghu sold the machine to danush at loss of 10% for 4 95 000 . wt profit got revati ?

Sol: raghu’s cp is revati’s sp.

raghu’s cp= (100/90)*4,95,000= 550000.

profit %= (100000/450000)*100= 22.22.

answer is 4. 22.22%

110. what is prabablity to getting at least one of tail . when two coins are tossed simultaineouly?

Sol:3/4

111. (789101112131415x)/8 leaves a reminder of of 0 .wts th value of x?

Sol: when last 3 digit number is divisible by 8, then the number is divisible by 8

ans== 2

ans== 2

112. in out of 52 cards, 4 cards to be are selected and one card of it should be spade and one card card should be heart. in How many ways can these card selected

Sol: There are 13 spades and 13 heart cards in pack of 52

we need to select 4 cards and one is from spades and one is from hearts and remaining 2 are form remaining cards

ie 13c1 * 13c1 * 50c2

13^2*50c2

we need to select 4 cards and one is from spades and one is from hearts and remaining 2 are form remaining cards

ie 13c1 * 13c1 * 50c2

13^2*50c2

113. In the election, the wining candidate won by 15% of votes. if a total 5000 votes were cast of which is 86% where eligible. then how many votes the wining candidate get?

Sol: 86% votes are eligible in 5000

so the number of votes are 5000*86/100=4300

now if loss candidate get x votes then winning one gets 15%more than that of x

ie the total votes is equal to the winning and loss candidate votes

x + (x*115/100)= 4300

from it x = 2000

now we need to calculate for winning candidate

ie x*115/100 = 2000*115/100 = 2300

so the number of votes are 5000*86/100=4300

now if loss candidate get x votes then winning one gets 15%more than that of x

ie the total votes is equal to the winning and loss candidate votes

x + (x*115/100)= 4300

from it x = 2000

now we need to calculate for winning candidate

ie x*115/100 = 2000*115/100 = 2300

114. 47,322 bulb are to be packed in several boxes. Each box should contains equal Numbers of bulbs and no bulb should be unpacked number of boxed used can be:

1. 12 2. 11 3. 8 4. 14

Sol: Only 11 can divide the given number.

115. How many 4 digit number can we made from 1 2 3 4 5 6 and 7 with none of digits being repeated?

Sol: 7*6*5*4=840

116. what is the value of (10101)2 in decimal form

Sol: 21

117. Wts is price of a pair of sandles is decreased by 10% the number of pair sold incresed by 20%. wt is nxt effect on sells?

1.8% decreases 2.10% decreases 3.10% increases 4. 8% increases

Sol: let price of sandles is x= 100rs.

price is decreased by 10%; x=90rs;

now it is increased by 20%; x=x + 20%(90)

x=90+18; x=108; means increases 8%;

price is decreased by 10%; x=90rs;

now it is increased by 20%; x=x + 20%(90)

x=90+18; x=108; means increases 8%;

118. What is the value of log7(1/49)

Sol: -2

119. 18.3454545nis equalent to;

1 1009/55 2.1009/99 3.342/990 4. 345/99

Sol: option 1.

120. price of salt is increased by 25% in order not to increase the expenditure a lady must reduce her consumption by:

Sol: let’s say lady need 1 salt packet per month and cost is 100 rupees

now salt packet rate is increased by 25% so new price will be 125.

but lady should not increase the expenditure, then how much she can buy with 100 rupees= 100/125

rate of decrease is given by ((1- (100/125))1)*100 = 20

121. How money factors does 400 have?

Sol: 15

122. What is square root of 54 05 625?

Sol: use square root division method…. taking groups of 5 40 56 25

ans is :2325

ans is :2325

123. Not a keyword in exception handling?

1. try 2. rethrow 3.catch 4.access

Sol:access

124. log42+log432 is equals to

Sol: 4

125. product of any two odd numbers is:

a.always odd b.always even c.sometimes odd and same times even d.divisible by 6

Sol: (a)

126. A 8-bit signed integer has the following range:

a. 0 to 255 b. -128 to 127 c. -255 to 254 d. 0 to 509

Sol: (b)… -2(n-1) to 2(n-1)-1

127. Conceal

**a. Hide**b. Seal c. Ceiling d. Horrifying128. Vinod took his meals after he . . . .

a.

**Had completed his work**b. Had been completing his work c. Was completing his workd. Had been completed his work e. Had got completed his work

129. SOLICITUDE

a. insouciance b. ingenuity c. propriety d. austerity

130. what is next number in the below series?

8,8,15,23,38,?

Sol: Ans:54

8 8 15 23 38

first diff=0 7 8 15

second diff=7 1 7 1

15+1=16 16+38=54

8 8 15 23 38

first diff=0 7 8 15

second diff=7 1 7 1

15+1=16 16+38=54

131. Wt is largest and smallest value no divisible by 6,15,21 and 24

a.9235,420 b.9980,840 c.9240,840 d.9999,999

Sol: take lcm of the numbers 6,15,21,24

ie we will get smallest number 3*2*4*5*7=840

now check the answers

b,c are having 840 so take option which is having higher number and divisible by 840

if its divisible it is its largest if not then option c

840*11=9240

ans : C

ie we will get smallest number 3*2*4*5*7=840

now check the answers

b,c are having 840 so take option which is having higher number and divisible by 840

if its divisible it is its largest if not then option c

840*11=9240

ans : C