AMCAT ( Aspiring Minds) previous questions with solutions-5

2. Cp of 4 calculators and 2 pencil is is 6200 what is the cost of ten calculators and five pencils

Sol: 4C+2P=6200-->2C+P=3100

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

3.12 men can complete work in 6 dayswhereas 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

4. 27^18/14 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

5. Probability of getting sum of odd numbers in two throw of a dice?

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)

total cases=18

therefore probability=18/36=>1/2

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

6. if log(p+q)(p-q)= -1 then find the value of: log(p+q)(P2-q2) (this p square - q square)

Sol: log(p+q)(p-q)=-1

log(p^2-q^2)= log(1/10)

adding both side log(p+q)

then,

log(p^2-q^2)+ log(p+q) = log(p+q)+ log(1/10)

log (p+q)(p^2-q^2)= log ((p+q)/10)

7. If peacock = 11526312316 then feather=?

Sol: 21526201959

alphabet order reverse and straight

8. 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)

72/5 + 72/25 := 14 + 2 := 16 zeros.

9. 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 2s in 77! is: = 38 + 19 + 9 + 4 + 2 + 1 = 73

So number of 2^4 will be: = 73%4=18

Number of 3s in 77! is:= 25 + 8 + 2 = 35

So number of 3^2 will be: = 17

number of 5s will be:= 15 + 3 = 18

Hence the number of 720s that we can obtain by analyzng d minimm value 17...

10. 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: 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

Sol:It is always true that

log(base 10)(a three digit no)= gives a characteristics value 2".and to find the no of digits we use formula "characteristics value+1".So 264 is a 3 digit no and its characteristics value for log(base 10) 264 is 2 and after decimal there is xyz anything so according to formula it is 2 + 1 = 3...

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

Sol: log y raised to the power of 3;

y^3 = 1369y

y^2 = 1369

y =37.

Sol: log y raised to the power of 3;

y^3 = 1369y

y^2 = 1369

y =37.

13. What is the remainder when 17^23 is divided by 16?

Option 1 : 0 Option 2 : 1 Option 3 : 2 Option 4 : 3

Sol: 17^23 /16

(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

14. antilog10^100

Sol: antilogx=10^x(for base 10) so ans is 10^(10^100)

Sol: antilogx=10^x(for base 10) so ans is 10^(10^100)

15. 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:

Option 1 : 14 times Option 2 : 15 times Option 3 : 13 times Option 4 : 11 times

Sol: given each interval time is 6s.7s,8s,9s.

Now all together toll is nothing but lcm of all individuals i.e., lcm of6,7,8,9=504s

convert 2 hours in secondsi.e., 2hrs=2*60*60=7200s

now no.of times=total time in seconds/together lcm

i.e,7200/504=14

Option 1 : 14 times Option 2 : 15 times Option 3 : 13 times Option 4 : 11 times

Sol: given each interval time is 6s.7s,8s,9s.

Now all together toll is nothing but lcm of all individuals i.e., lcm of6,7,8,9=504s

convert 2 hours in secondsi.e., 2hrs=2*60*60=7200s

now no.of times=total time in seconds/together lcm

i.e,7200/504=14

16. 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

Sol: 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

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

Sol: 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

18. 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?

Option 1 : 1043 Option 2 : 1073 Option 3 : 1103 Option 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

divided by 5 leaves a remainder of 3?

Option 1 : 1043 Option 2 : 1073 Option 3 : 1103 Option 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

19. A, B, C started a business with their investments in the ratio 1:3:5. After 4 months, A invested the same 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.

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.

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

Sol: 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)

Sol: 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)

21. If (a^4 - 2a^2b^2 + b^4)^x-1 = (a-b)^2x (a+b)^-2, then x equals to:

Sol: Taking log on both sides,

=>(x-1)log(a2-b2)^2=log[(a-b)^2x(a+b)^-2]

=>2(x-1)log(a2-b2)=2xlog(a-b)-2log(a+b)

Cancelling 2 on both sides,and expanding log(a2-b2) into log [(a+b)(a-b)]=>log(a+b)+log(a-b),

=>(x-1)[log(a+b)+log(a-b)]=xlog(a-b)-log(a+b)

=>xlog(a+b)-log(a-b)-log(a+b)+xlog(a-b)=xlog(a-b)-log(a+b)

=>xlog(a+b)=log(a-b)

x=log(a-b)/log(a+b).

Sol: Taking log on both sides,

=>(x-1)log(a2-b2)^2=log[(a-b)^2x(a+b)^-2]

=>2(x-1)log(a2-b2)=2xlog(a-b)-2log(a+b)

Cancelling 2 on both sides,and expanding log(a2-b2) into log [(a+b)(a-b)]=>log(a+b)+log(a-b),

=>(x-1)[log(a+b)+log(a-b)]=xlog(a-b)-log(a+b)

=>xlog(a+b)-log(a-b)-log(a+b)+xlog(a-b)=xlog(a-b)-log(a+b)

=>xlog(a+b)=log(a-b)

x=log(a-b)/log(a+b).

22. In an examination, 70% of students passed in physics, 65% in chemistry, 27% failed in both subjects. The percentage of students who passed is: Op 1: 66% Op 2: 62% Op 3: 69% Op 4: None of these

Sol: pass % in physics n(P)=70%

pass %in chemistry n(c)=65%

fail % in both=27%

pass %in any one or both=n(pUc)=73%

n(p intersection c)=n(p)+n(C)-n(pUc)=70+65-73=62%

Sol: pass % in physics n(P)=70%

pass %in chemistry n(c)=65%

fail % in both=27%

pass %in any one or both=n(pUc)=73%

n(p intersection c)=n(p)+n(C)-n(pUc)=70+65-73=62%

23. 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

Sol: 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.

Sol: 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.

24.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? Op 1: Rs. 9.50 Op 2: Rs. 8.50 Op 3: Rs. 10.50 Op 4: Rs. 12.00

Sol: 30 Kg->8.50*30=255

20kg -> 9*20 =180

so

50Kg=435; 1kg=435/50=8.7.

So CP of 1kg=8.7

now 20% profit it means -> (20/100)*8.7=1.74.

so the final ans is 8.7+1.74="10.44"

Sol: 30 Kg->8.50*30=255

20kg -> 9*20 =180

so

50Kg=435; 1kg=435/50=8.7.

So CP of 1kg=8.7

now 20% profit it means -> (20/100)*8.7=1.74.

so the final ans is 8.7+1.74="10.44"

25. 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

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

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

26. 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 ?

Op 1: 212 Op 2: 266 Op 3: 272 Op 4: 312

Sol: Here total question 10 will be answered so there are 3 ways .i know A has 6 nd B has 7 so

A. B

4. 6

5. 5

6. 4

so total no of ways is 6c4*7c6+6c5*7c5+6c6*7c4=266 so its the ans.

Op 1: 212 Op 2: 266 Op 3: 272 Op 4: 312

Sol: Here total question 10 will be answered so there are 3 ways .i know A has 6 nd B has 7 so

A. B

4. 6

5. 5

6. 4

so total no of ways is 6c4*7c6+6c5*7c5+6c6*7c4=266 so its the ans.

27.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: 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

Sol: 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

28. synonym of OBTRUSIVE

Sol: conspicuous, obvious,unmistacable

Sol: conspicuous, obvious,unmistacable

29. HCF of 3.68 & 5.35 what is 1/LCM:

Sol: 3.68=368/100=92/25

5.35=535/100=107/20

Now, hcf will be : hcf of numerator divided by lcm of denominator i.e

hcf of 92 and 107 = 1

lcm of 20 and 25 = 100

so, 1/100= 0.01

Sol: 3.68=368/100=92/25

5.35=535/100=107/20

Now, hcf will be : hcf of numerator divided by lcm of denominator i.e

hcf of 92 and 107 = 1

lcm of 20 and 25 = 100

so, 1/100= 0.01

30. 3*(4^4+4^3+4^2+4+1) = ?

Sol: 3*4^2(4^2+4+1)+15

48^(21)+15

1023

Sol: 3*4^2(4^2+4+1)+15

48^(21)+15

1023

31. log10 (2)=.6096

log10 (3)=.4709

then log10 (12) = ?

Sol: 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

32. log (base25) 625 - log (base31) 961 + log (base29) 841 = ?

Sol: log25^2 base 25 - log31^2 base 31 + log 29^2 base 29

2-2+2 = 2

log10 (3)=.4709

then log10 (12) = ?

Sol: 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

32. log (base25) 625 - log (base31) 961 + log (base29) 841 = ?

Sol: log25^2 base 25 - log31^2 base 31 + log 29^2 base 29

2-2+2 = 2

33. 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)

34. NATION - 1412091514 THEN

REMOTE- ?

Sol: Ans: 1851315205

A=1 B=2 ...................Z=26

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

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

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

34. NATION - 1412091514 THEN

REMOTE- ?

Sol: Ans: 1851315205

A=1 B=2 ...................Z=26

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

Sol: 4*5*5*5*5=2500

Sol: 4*5*5*5*5=2500

36. 400 have how many factors

a.8

b.10

c.12

d.24

Sol: 15 factors....

400=2^4*5*2

no of factor=5*3=15

a.8

b.10

c.12

d.24

Sol: 15 factors....

400=2^4*5*2

no of factor=5*3=15

37. 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: prob=at least 1 ball yellow

=1-no ball yellow

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

=1-(1/462)

=0.997

Sol: prob=at least 1 ball yellow

=1-no ball yellow

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

=1-(1/462)

=0.997

38. if north-west is east ,north-east is south then what is east

Sol: south west

Sol: south west

39. 10,14,23,39,64,?

Sol: the difference btwn two numbers is the square numbers in series orderi.e.,

14-10=2^2

23-14=3^239-23=4^2...like that the number will be....100

Sol: the difference btwn two numbers is the square numbers in series orderi.e.,

14-10=2^2

23-14=3^239-23=4^2...like that the number will be....100

40.Please Remember All The Words Becoz 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

Sol: let their ages 6 years back were x and 2x

(x+12)/(2x+12)=3/4

x=6

present ages ratio are 2:3

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

Sol: let their ages 6 years back were x and 2x

(x+12)/(2x+12)=3/4

x=6

present ages ratio are 2:3

42. 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:

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

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

x=36

op 2:36

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

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

x=36

op 2:36

43. |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

Sol: Op 3: 0.5

4.5+4>0

&(0.5)^2

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

Sol: r=at2

s=2at

therefore t=s/2a

r=as^2/4a^2

s^2=4ar

Sol: r=at2

s=2at

therefore t=s/2a

r=as^2/4a^2

s^2=4ar

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

Sol: Lets add 2 two both sides of the equation

x^4+1/x^4 + 2 = 47+2

(x^2 + 1/x^2)^2 = 49 (This is because (x^2 +1/x^2)^2 = x^4+1/x^4+ 2)

x^2 + 1/x^2 = (49)^1/2

x^2 + 1/x^2 = 7

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

so x+1/x=3;

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

3^3=x^3+3*3+1/x^3

so x^3+1/x^3=18

Sol: Lets add 2 two both sides of the equation

x^4+1/x^4 + 2 = 47+2

(x^2 + 1/x^2)^2 = 49 (This is because (x^2 +1/x^2)^2 = x^4+1/x^4+ 2)

x^2 + 1/x^2 = (49)^1/2

x^2 + 1/x^2 = 7

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

so x+1/x=3;

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

3^3=x^3+3*3+1/x^3

so x^3+1/x^3=18

46. If a, b, c are roots of the equation 1x3-4x2+6.5x + 3.5 = 0, then what is the value of a2 + b2 + c2? Op 1: 1 Op 2: 64 Op 3: 169 Op 4: 3 Op 5:

Sol: let l,m,n be roots of 1x3-4x2+6.5x+3.5=0, den l+m+n=-b/a, lmn=-d/a, lm+mn+ln=c/a,

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

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

Sol: let l,m,n be roots of 1x3-4x2+6.5x+3.5=0, den l+m+n=-b/a, lmn=-d/a, lm+mn+ln=c/a,

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

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

47. 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 Op 5:

Sol: First we find no.of terms=9

commn diff d is=0.33,a=0.33

sum=n/2[2a+(n-1)d]=14.850 ans......??

Sol: First we find no.of terms=9

commn diff d is=0.33,a=0.33

sum=n/2[2a+(n-1)d]=14.850 ans......??

48. 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: Answer should be " can not determine "

Reason being 11 coins are picked randomly and placed in a box, and now the coin is picked from those 11..

So how can we know that between those 11 coins how may are 1Rs coins and how many 50 paisa... so "can not determine"

PLEASE REFER ONCE!!!

Sol: Answer should be " can not determine "

Reason being 11 coins are picked randomly and placed in a box, and now the coin is picked from those 11..

So how can we know that between those 11 coins how may are 1Rs coins and how many 50 paisa... so "can not determine"

PLEASE REFER ONCE!!!

49. . 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

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 =7/10

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

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

50. 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

Sol: 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

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 =7/10

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

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

50. 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

Sol: 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

51. A, B, C, D and E play the following game. Each person picks one card from 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 Op 1: 3/14 Op 2: 4/17

Op 3: 1/5 Op 4: 5/24

Sol:

Op 3: 1/5 Op 4: 5/24

Sol:

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

a>5

b>8

c>6

d>10

Sol: no.of clips =60/12=5

let x clip should be sell to earn a 20% profit.

in this contrast,

x clip sell for rs 60

1 clip sell for rs:60/x;

proffit=(60/x-5)/5*100=20/100;

x=10(ans)

Op 1: Rs. 2100

Op 2: Rs. 2200

Op 3: Rs. 2300

Op 4: Rs. 2400

Sol: ans : 3

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

a>5

b>8

c>6

d>10

Sol: no.of clips =60/12=5

let x clip should be sell to earn a 20% profit.

in this contrast,

x clip sell for rs 60

1 clip sell for rs:60/x;

proffit=(60/x-5)/5*100=20/100;

x=10(ans)

Op 1: Rs. 2100

Op 2: Rs. 2200

Op 3: Rs. 2300

Op 4: Rs. 2400

Sol: ans : 3

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

54. 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:

Op 1: 291/364

Op 2: 371/464

Op 3: 471/502

Op 4: 459/512

Sol: ans: 4

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

Op 1: 291/364

Op 2: 371/464

Op 3: 471/502

Op 4: 459/512

Sol: ans: 4

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

55. 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?

Op 1: 1/4

Op 2: 1/5

Op 3: 2/5

Op 4: 1/120

Op 5: 4

Sol: last two digits must be,

***12

***24

***32

***52

case1) ***12

rest three digits can be filled in 3*2*1 or 6 ways

similarly,

case2. ***24

6 ways

case3. ***32

6ways

case4. ***52

6 ways

total required ways = 6*4 = 24

total arrangements = 5*4*3*2*1 = 120

probability = 24/120 = 1/5

Op 1: 1/4

Op 2: 1/5

Op 3: 2/5

Op 4: 1/120

Op 5: 4

Sol: last two digits must be,

***12

***24

***32

***52

case1) ***12

rest three digits can be filled in 3*2*1 or 6 ways

similarly,

case2. ***24

6 ways

case3. ***32

6ways

case4. ***52

6 ways

total required ways = 6*4 = 24

total arrangements = 5*4*3*2*1 = 120

probability = 24/120 = 1/5

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

Op 1: 27

Op 2: 26

Op 3: 13

Op 4: 14

Sol: First find the prime factors of 6084

2*3042

2*2*1521

2*2*3*507

2*2*3*3*169

2*2*3*3*13*13

2^2*3^2*13^2

Number of factors = (2+1)*(2+1)*(2+1) = 3*3*3 = 27

Using each of these factors, we can write 6084 as a product of 2 factors.

Half of these will remain same.

Let us take a simple example, 6

number of factors = 4

1,2,3,6

can be written as

1*6

2*3

We cant repeat 6*1 and 3*2 as they are already taken.

Hence only 2 ways.

Another example: take 9.

Number of factors = 3

1,3,9

can be written as

1*9 only.

Hence only 1 way.

So it turnsout ot be number of factors/2 if number of factors is even and

(number of factors-1)/2 if number of factors is odd.

Therefore: 27-1/2 = 26/2 = 13

Hence pick Op 3.

Op 1: 27

Op 2: 26

Op 3: 13

Op 4: 14

Sol: First find the prime factors of 6084

2*3042

2*2*1521

2*2*3*507

2*2*3*3*169

2*2*3*3*13*13

2^2*3^2*13^2

Number of factors = (2+1)*(2+1)*(2+1) = 3*3*3 = 27

Using each of these factors, we can write 6084 as a product of 2 factors.

Half of these will remain same.

Let us take a simple example, 6

number of factors = 4

1,2,3,6

can be written as

1*6

2*3

We cant repeat 6*1 and 3*2 as they are already taken.

Hence only 2 ways.

Another example: take 9.

Number of factors = 3

1,3,9

can be written as

1*9 only.

Hence only 1 way.

So it turnsout ot be number of factors/2 if number of factors is even and

(number of factors-1)/2 if number of factors is odd.

Therefore: 27-1/2 = 26/2 = 13

Hence pick Op 3.

57.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

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

58. 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.

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.

59. 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: Total 5-digit Number formed by 0,1,2,3,4,5 (without repetition) are 5*5*4*3*2 = 600.

Number divisible by 3 means sum is divisible by 3.

so case 1. 5+4+3+2+1 = 15(exclude 0).

possibilities are = 5*4*3*2*1 = 120

case 2. 5+0+4+2+1 = 12(exclude 3).

possibilities are = 4*4*3*2*1 = 96.

so total answer = case1 + case2 = 216

Sol: Total 5-digit Number formed by 0,1,2,3,4,5 (without repetition) are 5*5*4*3*2 = 600.

Number divisible by 3 means sum is divisible by 3.

so case 1. 5+4+3+2+1 = 15(exclude 0).

possibilities are = 5*4*3*2*1 = 120

case 2. 5+0+4+2+1 = 12(exclude 3).

possibilities are = 4*4*3*2*1 = 96.

so total answer = case1 + case2 = 216

60. 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: sp=gain% * cp

42=105% * cp (5% profit)

cp= (42*100)/105 = 40

Sol: sp=gain% * cp

42=105% * cp (5% profit)

cp= (42*100)/105 = 40

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

Sol: unit place can be 2 or 4 only (2 ways) rest 3 places can be filled in 3P3 = 3! = 6 ways total no. = 2*6 = 12

Sol: log3 9 - log4 256 + log5 125

= log3 3^2 -log4 4^4 + log5 5^3

= 2*log3 3 - 4*log4 4 + 3*log5 5

= 2 - 4 + 3 [log x (x) = 1]

= 1

Sol: unit place can be 2 or 4 only (2 ways) rest 3 places can be filled in 3P3 = 3! = 6 ways total no. = 2*6 = 12

Sol: log3 9 - log4 256 + log5 125

= log3 3^2 -log4 4^4 + log5 5^3

= 2*log3 3 - 4*log4 4 + 3*log5 5

= 2 - 4 + 3 [log x (x) = 1]

= 1

63. If a=2 & b=1 then log(a+b)(a^2-b^2)=?

Sol: if(a+b) is base then

log(a+b)(a^2-b^2)

= log(2+1) (2^2-1^2)

= log3 (3)

= 1

Sol: if(a+b) is base then

log(a+b)(a^2-b^2)

= log(2+1) (2^2-1^2)

= log3 (3)

= 1

64. a coin is tossed 3 times by raju.what is probability that raju win all three time.

Sol: every time event is independent

prob. of win 0.5 for one time

for three times 0.5*0.5*0.5 = 0.125

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

a)5

b)10

c)25

d)20

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

Sol: every time event is independent

prob. of win 0.5 for one time

for three times 0.5*0.5*0.5 = 0.125

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

a)5

b)10

c)25

d)20

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

66. Find next no. in sequence

8, 12, 24, 60, ?

Sol: every times difference increases three times.

4,12,36..,108

so the nxt number should be 60+108=168

8, 12, 24, 60, ?

Sol: every times difference increases three times.

4,12,36..,108

so the nxt number should be 60+108=168

67. 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

68. log rootover(6) 1296= ?

Sol: log rootover(6) 1296

= log 6^.5 (6^4)

= 1/.5 * 4 * log 6(6)

= 2*4*1

=8

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

68. log rootover(6) 1296= ?

Sol: log rootover(6) 1296

= log 6^.5 (6^4)

= 1/.5 * 4 * log 6(6)

= 2*4*1

=8

69. 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: LCM of 224,364=2912

so, A does 13 circles while B does 8 in 2912s.

Thus A crosses B 13 times.

Sol: LCM of 224,364=2912

so, A does 13 circles while B does 8 in 2912s.

Thus A crosses B 13 times.

70. 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?

Options:

a.55.5m, b.45m, c.66m, d.49m.

Sol: When they are meeting for the first time, sum of distance travelled by both swimmers is = d, if 'd' is the required length of the pool. Now when they are meeting for the second time (i.e. in their reverse journey), sum of the distance travelled by both swimmers = 3d.

If we exclude the waiting time, then time taken for first meet and second meet is in the ratio 1 : 3. Also distance travelled by each swimmer will be in the ratio 1 : 3 (as both are moving with constant but distinct speeds). So equating the ratio of distance travelled by one of the swimmer with the ratio 1 : 3, we get

18.5/(d + 10.5) = 1/3

or, d + 10.5 = 55.5

or, d = 45m. option b.

Options:

a.55.5m, b.45m, c.66m, d.49m.

Sol: When they are meeting for the first time, sum of distance travelled by both swimmers is = d, if 'd' is the required length of the pool. Now when they are meeting for the second time (i.e. in their reverse journey), sum of the distance travelled by both swimmers = 3d.

If we exclude the waiting time, then time taken for first meet and second meet is in the ratio 1 : 3. Also distance travelled by each swimmer will be in the ratio 1 : 3 (as both are moving with constant but distinct speeds). So equating the ratio of distance travelled by one of the swimmer with the ratio 1 : 3, we get

18.5/(d + 10.5) = 1/3

or, d + 10.5 = 55.5

or, d = 45m. option b.

71. What is opposite meaning of-- DISCREET?

Sol: imprudent,not wise

Sol: imprudent,not wise

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

Sol:

Sol:

73. 6:60:120:?

a) 210

b) 3

c) 240

d) 280

Sol: 6:60::120:?

It can be either thought of as 60/10=6 so 120/10=12. Or another way can be 60=6+0=6 so 120=1+2+0=3

Ans can be 12 0r 3

74. Log rootover(64) base 8==?

Sol: log8(64)^1/2=log8 (8)=1

a) 210

b) 3

c) 240

d) 280

Sol: 6:60::120:?

It can be either thought of as 60/10=6 so 120/10=12. Or another way can be 60=6+0=6 so 120=1+2+0=3

Ans can be 12 0r 3

74. Log rootover(64) base 8==?

Sol: log8(64)^1/2=log8 (8)=1

75. Log2 0.5==

Sol: Ans:-1

log2 (0.5)

=log2 (1/2)=log2 (2^-1)

=(-1) log2 (2)

= -1

Sol: Ans:-1

log2 (0.5)

=log2 (1/2)=log2 (2^-1)

=(-1) log2 (2)

= -1

76. Which one used as gobal aparator from Fuction

1. opaarator::

2.oparator;;

3.oparator%

4.oparator !! (two bars)

Sol: (::) is global operator or scope operator

1. opaarator::

2.oparator;;

3.oparator%

4.oparator !! (two bars)

Sol: (::) is global operator or scope operator

77. Which is invalid

1.10!6

2.false && True

3. bool(x)=(bool)10

4. flat= 12.67

Sol: float has been wrongly denoted as flat

1.10!6

2.false && True

3. bool(x)=(bool)10

4. flat= 12.67

Sol: float has been wrongly denoted as flat

78. 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. Answer

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. Answer

79. Ques. DJ: WQ :: FK :?

Op 1: UR

Op 2: RU

Op 3: PU

Op 4: UP

Sol: Correct Op : 4

Op 1: UR

Op 2: RU

Op 3: PU

Op 4: UP

Sol: Correct Op : 4

80. Ques. 2197:13 :: 3375 : ?

Op 1: 11

Op 2: 17

Op 3: 15

Op 4: 9

Sol: Correct Op : 3

Op 1: 11

Op 2: 17

Op 3: 15

Op 4: 9

Sol: Correct Op : 3

81. Ques. 4, 6, 8, 10, ____

Op 1: 11

Op 2: 12

Op 3: 13

Op 4: 14

Op 1: 11

Op 2: 12

Op 3: 13

Op 4: 14

82. Ques. 32:16::8: ?

Op 1: 6

Op 2: 4

Op 3: 7

Op 4: 3

Sol: Correct Op : 2

Op 1: 6

Op 2: 4

Op 3: 7

Op 4: 3

Sol: Correct Op : 2

83. Ques. 7:11::31: ?

Op 1: 33

Op 2: 37

Op 3: 39

Op 4: 42

Sol: Correct Op : 2

Op 1: 33

Op 2: 37

Op 3: 39

Op 4: 42

Sol: Correct Op : 2

84. Ques. 24:90::56: ?

Op 1: 120

Op 2: 122

Op 3: 118

Op 4: 124

Sol: Correct Op : 2

Op 1: 120

Op 2: 122

Op 3: 118

Op 4: 124

Sol: Correct Op : 2

85. Ques. ACE : 135 :: DFG :?

Op 1: 246

Op 2: 642

Op 3: 467

Op 4: 681

Sol: Correct Op : 3

Op 1: 246

Op 2: 642

Op 3: 467

Op 4: 681

Sol: Correct Op : 3

86. Ques. 5: 124 :: 7 :?

Op 1: 342

Op 2: 343

Op 3: 248

Op 4: 125

Sol: Correct Op : 1

Op 1: 342

Op 2: 343

Op 3: 248

Op 4: 125

Sol: Correct Op : 1

87. Ques. WOLLS: XPMMT :: PILOT : ?

Op 1: QJMPU

Op 2: QJJPU

Op 3: QMMPU

Op 4: QMJPU

Sol: Correct Op : 1

Op 1: QJMPU

Op 2: QJJPU

Op 3: QMMPU

Op 4: QMJPU

Sol: Correct Op : 1

88. Ques. DFO: GIR :: ?

Op 1: EMK : IRP

Op 2: KME : NPH

Op 3: CDO : EGQ

Op 4: MKI : PRE

Sol:correct Op : 2

Op 1: EMK : IRP

Op 2: KME : NPH

Op 3: CDO : EGQ

Op 4: MKI : PRE

Sol:correct Op : 2

89. Ques. OQT : FHK :: DFI :?

Op 1: CEH

Op 2: BEH

Op 3: BFH

Op 4: AFH

Sol: Correct Op : 1

Op 1: CEH

Op 2: BEH

Op 3: BFH

Op 4: AFH

Sol: Correct Op : 1

90. FACE: HACE :: BACE :?

Op 1: DACE

Op 2: CASE

Op 3: NACE

Op 4: LACE

Sol: Correct Op : 1

Op 1: DACE

Op 2: CASE

Op 3: NACE

Op 4: LACE

Sol: Correct Op : 1

91. WINTER : RETNIW ::

Op 1: RMMEUS : SUMMER

Op 2: SPRING : GNIRPS

Op 3: HEAVEN : GOD

Op 4: KNIFE :BLADE

Sol: Correct Op : 2

Op 1: RMMEUS : SUMMER

Op 2: SPRING : GNIRPS

Op 3: HEAVEN : GOD

Op 4: KNIFE :BLADE

Sol: Correct Op : 2

92. MPSV : HKNQ :: PSVY :?

Op 1: CIFL

Op 2: LCIF

Op 3: CFIL

Op 4: ICFL

Sol: Correct Op : 3

Op 1: CIFL

Op 2: LCIF

Op 3: CFIL

Op 4: ICFL

Sol: Correct Op : 3

93. MOQ : TUX :: ACE:?

Op 1: ILH

Op 2: HLI

Op 3: HIL

Op 4: IHL

Sol: Correct Op : 3

Op 1: ILH

Op 2: HLI

Op 3: HIL

Op 4: IHL

Sol: Correct Op : 3

94. QDXM : SFYN ::UIOZ:?

Op 1: PAQM

Op 2: LPWA

Op 3: QNLA

Op 4: WKPA

Sol: Correct Op : 4

Op 1: PAQM

Op 2: LPWA

Op 3: QNLA

Op 4: WKPA

Sol: Correct Op : 4

95. In a code language STAY is written as 9657 SOUND is written as 92348 and DOT IN is written as 826 74.

How would you write SIT STAND in that language ?

Op 1: 967 29348

Op 2: 976 96548

Op 3: 679 92843

Op 4: 796 23984

Sol: Correct Op : 2

How would you write SIT STAND in that language ?

Op 1: 967 29348

Op 2: 976 96548

Op 3: 679 92843

Op 4: 796 23984

Sol: Correct Op : 2

96. If CABLE = 96372 and RISK = 8415, what word is made by 37265 ?

Op 1: TRICK

Op 2: BLEAK

Op 3: BLANK

Op 4: TABLE

Sol: Correct Op : 2

Op 1: TRICK

Op 2: BLEAK

Op 3: BLANK

Op 4: TABLE

Sol: Correct Op : 2

97. Ques. If STUDENT is coded as RUTE DOS, which word would be coded as RDGPKBQ ?

Op 1: SHACKLE

Op 2: SHINGLE

Op 3: SNOBBER

Op 4: SCHOLAR

Sol: Correct Op : 4

Op 1: SHACKLE

Op 2: SHINGLE

Op 3: SNOBBER

Op 4: SCHOLAR

Sol: Correct Op : 4

98. f MACHINE is coded as 19 - 7 - 9 - 14 - 15 - 20 - 11, how will you code DANGER ?

Op 1: 10 - 7 - 20 - 13 - 11 - 24

Op 2: 11 - 7 - 20 - 16 - 11 - 24

Op 3: 13 - 7 - 20 - 9 - 11 - 25

Op 4: 13 - 7 - 20 - 10 - 11 - 25

Sol: Correct Op : 1

Op 1: 10 - 7 - 20 - 13 - 11 - 24

Op 2: 11 - 7 - 20 - 16 - 11 - 24

Op 3: 13 - 7 - 20 - 9 - 11 - 25

Op 4: 13 - 7 - 20 - 10 - 11 - 25

Sol: Correct Op : 1

99. If SHARP is coded as 58034 and PUSH as 4658, then RUSH is coded as

Op 1: 3568

Op 2: 3658

Op 3: 3685

Op 4: 3583

Sol: Correct Op : 2

Op 1: 3568

Op 2: 3658

Op 3: 3685

Op 4: 3583

Sol: Correct Op : 2

100. If CONTRIBUTE is written as ETBUIRNTOC, which letter will be in the sixth place when counted from the

left if POPULARISE is written in that code ?

Op 1: L

Op 2: A

Op 3: I

Op 4: D

Sol: Correct Op : 1

left if POPULARISE is written in that code ?

Op 1: L

Op 2: A

Op 3: I

Op 4: D

Sol: Correct Op : 1

101. From the given choices select the odd man out

Op 1: ADG

Op 2: BEH

Op 3: SUT

Op 4: KNQ

Op 5: CFI

Sol: Correct Op : 3

Op 1: ADG

Op 2: BEH

Op 3: SUT

Op 4: KNQ

Op 5: CFI

Sol: Correct Op : 3

102. Ques. 264 : 275 : 385

Op 1: 145 : 253 : 325

Op 2: 143 : 235 : 246

Op 3: 372 : 563 : 736

Op 4: 233 : 343 : 345

Sol: Correct Op : 3

Op 1: 145 : 253 : 325

Op 2: 143 : 235 : 246

Op 3: 372 : 563 : 736

Op 4: 233 : 343 : 345

Sol: Correct Op : 3

103. Ques. 21:51:15

Op 1: 21:31:51

Op 2: 21:36:41

Op 3: 21:51:61

Op 4: 21:91:35

Sol: Correct Op : 4

Op 1: 21:31:51

Op 2: 21:36:41

Op 3: 21:51:61

Op 4: 21:91:35

Sol: Correct Op : 4

104. 256 : 4086 :: ?

Op 1: 225 : 3365

Op 2: 144 : 3032

Op 3: 132 : 3012

Op 4: 160 : 3600

Sol: Correct Op : 1

Op 1: 225 : 3365

Op 2: 144 : 3032

Op 3: 132 : 3012

Op 4: 160 : 3600

Sol: Correct Op : 1

105. Statement:

The government has decided to pay compensation of Rs. 1 lakh to the family members of those who are killed in

railway accidents.

Assumptions:

I. The government has enough funds to meet the expenses due for compensation.

II. There may be reduction in incidents of railway accidents in near future.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit.

Sol:C orrect Op : 1

The government has decided to pay compensation of Rs. 1 lakh to the family members of those who are killed in

railway accidents.

Assumptions:

I. The government has enough funds to meet the expenses due for compensation.

II. There may be reduction in incidents of railway accidents in near future.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit.

Sol:C orrect Op : 1

106. Ques. Statement:

'Please do not wait for me, I may be late, start taking lunch as soon as the guests arrive.' - a message from a Director

of a Company to his Office managers.

Assumptions:

I. Keeping guests waiting is not desirable.

II.Lunch may not be ready in time.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit.

Sol: Correct Op : 1

'Please do not wait for me, I may be late, start taking lunch as soon as the guests arrive.' - a message from a Director

of a Company to his Office managers.

Assumptions:

I. Keeping guests waiting is not desirable.

II.Lunch may not be ready in time.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit.

Sol: Correct Op : 1

107. Statement:

The government has instructed all the premier institutes offering professional courses to reduce the fees by 50 percent

and increase the number of students.

Assumptions:

I. These institutes may be able to continue providing quality education with less fees and more students.

II. The institutes may continue charging more fees to provide quality education.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit.

Sol: Correct Op : 5

108. Statement:

The railway authority has decided to introduce two additional super-fast trains between Cities 'A' and 'B' during the

vacation time.

Assumptions:

I. All the passengers who desire to travel during vacation time will get a train ticket.

II. All other modes of transport between cities 'A' and 'B' are already overstretched.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit

Sol: Correct Op : 2

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The government has instructed all the premier institutes offering professional courses to reduce the fees by 50 percent

and increase the number of students.

Assumptions:

I. These institutes may be able to continue providing quality education with less fees and more students.

II. The institutes may continue charging more fees to provide quality education.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit.

Sol: Correct Op : 5

108. Statement:

The railway authority has decided to introduce two additional super-fast trains between Cities 'A' and 'B' during the

vacation time.

Assumptions:

I. All the passengers who desire to travel during vacation time will get a train ticket.

II. All other modes of transport between cities 'A' and 'B' are already overstretched.

Op 1: Only Assumption I is implicit.

Op 2: Only Assumption II is implicit.

Op 3: Either Assumption I or II is implicit.

Op 4: Neither Assumption I nor II is implicit.

Op 5: Both Assumptions I and II are implicit

Sol: Correct Op : 2

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1.3,7,10,11,12,17,? find out next number?

Sol: 3,7,10,11,12,17

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

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

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

41. 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?

53. 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:

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

Sol: Correct Op : 2

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