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Exam Name: PP-CP-T-2014-A004

Date of completion: 23-Sep-2013 00:44

Duration: 80 mins

Exam Result

Number Of Questions Correct/(Unattempted + Wrong) Ratio: 27 : 3

Marks Correct/(Unattempted + Wrong) Ratio: 27.0 : 3.0

Percentage Score: 87 %

Total Time Taken Including Stoppage and Resume Time = 430 seconds

Negative Marking Factor 0.33

Attempted Wrong Weightage of Questions: 3.0

Score Report

Overall Score

Correct

Wrong90%

10%

Section wise report

Sub Section wise report

Wrong

Correct

Wrong

Correct

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Question wise report

Sub Section wise report

No. DescriptionResult

AllMarks Time Spent Solution

1

(QID=9734) :

Anand, Babu and Chiyaan ran a race, with Anand finishing 36 meters ahead of Babu and 54

meters ahead of Chiyaan. Babu finishes 24 meters ahead of runner Chiyaan. Each runner travels

the entire distance at a constant speed. What is the length of the race?

132m

140m

144m

152m

Correct 1.0 17s

Ans: 144m

Let the distance of

race be d. Let the

speed of the

runners be a,b,c

Let Anand take t

seconds to cover

d.

at = d,

bt = d-36

ct = d-54

d = at = bt+36 =

ct+54.

Time taken by

Babu to cover

distance d = d/b =

t + 36/b

c(t + 36/b) = d-24

d = ct + 36c/b + 24

so overall we have

d = at = bt+36 =

ct+54 = ct + 36c/b

+ 24

Using ct + 54 = ct

+ 36c/b + 24,

b=6c/5.

Using bt+36 =

ct+54 and b=6c/5.,

ct=90.

As on the top we

have ct=d-54, 90 =

d-54,

Solving d=144m.

2

(QID=9729) :

If f(x)=9x-15 what is f (x) (the inverse function)?-1

x/9 - 15

x/15 - 9

(x+15) / 9

None of the above

Correct 1.0 3s

Ans: (x+15)/9

The function f(x)

does the following.

1. Multiplies x by

9.

2. Subtracts 15

The inverse is to

undo using the

steps in the

reverse order.

1. Add 15

2. Divide by 9.

Hence answer is

(x+15)/9

Wrong

Correct

3

(QID=9728) :

What is the remainder when 6 1̂7 + 17 6̂ + 169 2̂35 is divided by 7?

0

1

3

5

Correct 1.0 3s

Ans: 1

6 1̂7 mod 7=(7-

1) 1̂7 mod 7=

(-1) 1̂7 mod 7=-1

(As 17 is odd

finally -1).

17 6̂ mod 7=(3) 6̂

mod 7= 729 mod 7

= 1 (3 6̂ - 729)

169 2̂35 is 1.

The remainder

when 6 1̂7+17 6̂ is

divided by 7 is 1-

1+1=1

4

(QID=9685) :

What is the largest positive integer that divides all the three numbers 23400, 272304, 205248

without leaving a remainder?

48

24

96

72

Correct 1.0 4s

Ans: 24

23400 =

2*3*3*13*2*5*2*5

272304 =

2*2*2*2*3*3* 1891

205248 =

2*2*2*2*2*2*3*1069

Common factors

are = 2*2*2*3

Hence the required

number = 8*3 = 24

5

(QID=9724) :

22 passengers are to travel by a double decker bus which can accommodate 12 in the upper deck

and 10 in the lower deck. If 5 refuse to sit in the upper deck and 7 refuse to sit in the lower deck,

in how many ways can the passengers be distributed?

140 ways

252 ways

360 ways

440 ways

Correct 1.0 5s

Ans: 252 ways

The 5 who refuse

to sit in upper deck

must be in lower.

The 7 who refuse

to sit in lower deck

must be in upper.

Hence these 12

people have been

utilized and now

upper deck can

accommodate 12-7

= 5 people and

lower can

accommodate 10-5

= 5 people.

The remaining 22-

12 = 10 people can

be distributed in

10C5 * 5C5 = 252

ways

6

(QID=9721) :

A owes B Rs. 250. He agrees to pay B over a number of consecutive days starting on a Monday,

paying single note of Rs 50 or Rs 100 on each day. In how many different ways can A repay B.

(Two ways are said to be different if at least one day, a note of a different denomination is given)

8

12

14

None of the above

Correct 1.0 111s

Ans: 8

Let us consider the

possible count of

Rs.50 notes being

paid to pay Rs.250

and the pattern of

payment across

the days.

5 fifty rupee notes

= 5C5 = 1

3 fifty rupee notes

and 1 hundred

rupee note = 4C1 =

4

1 fifty rupee note

and 2 hundred

rupee notes = 3C1

= 3

Hence total ways =

1+4+3 = 8

7

(QID=9736) :

If x 2̂ - 16 > 0, then which of the following must be true?

-4 < x < 4

4 < x

-4 > x > 4

Correct 1.0 7s

Ans: -4 > x > 4

x 2̂ - 16 > 0

x 2̂ > 16

So if x is positive it

should be greater

than 4.

If is negative it

should be less

than -4.

-4 > x < 4

than -4.

Hence -4 > x > 4

is true.

8

(QID=9727) :

In a number system with base b, 15*22 = 440, the value of b is

2

4

5

7

Correct 1.0 7s

Ans: 5

(b + 5)(2b+2) =

4b 2̂ + 4b + 0,

2b 2̂ + 12b + 10 =

4b 2̂ + 4b + 0,

2b 2̂ -8b -10 = 0,

b 2̂ -4b -5 = 0,

Solving b=5 or

b=-1,

Among the given

options, 5 is the

answer.

9

(QID=9719) :

How many positive integers less than 450 can be formed using 1, 2, 5 and 6 for it's digits, with

each digit being used only once?

96

60

42

28

Correct 1.0 8s

Ans: 28

Num of possible 1

digit numbers = 4

Num of possible 2

digit numbers =

4*3 = 12

For 3 digit

numbers, to meet

the condition the

number should be

less than 450, 5

and 6 cannot be in

the hundredth

position.

Num of possible 3

digit numbers =

2*3*2 = 12

Total +ve integers

meeting the

condition =

4+12+12 = 28

10

(QID=9722) :

What is the remainder when the number

123456789101112131415161718192021222324252627282930313233343536373839404142434481

is divided by 45?

16

21

36

41

Correct 1.0 6s

Ans: 36

45 can be written

as 9*5.

The sum of all the

digits is 44*45/2 +

81 = 22*45 + 63 =

1071 which is

divisible by 9.

Hence the entire

number is divisible

by 9.

As 1 is in the last

digit, the last digit

in the quotient

when divided by 9

will be 9 (As 9*9

gives unit digit as

1).

So when divided by

5, the remainder

will be 4.

Because we used

9 already to divide

the number, the

remainder 4 should

be multiplied by 9.

Hence required

answer is 36.

Ans: 4

Let the

committees/groups

be a,b,c

a n b n c = 2

a n b = b n c = c n

a = 3

So total members

in committee a =

Number of

members who are

not common with

11

(QID=9741) :

An organization has three committees. Only two persons are members of all three committees,

but every pair of committees has three members in common. What is the LEAST possible number

of the members on any one committee?

2

3

4

5

Correct 1.0 6s

any of the

committee +

number of

members common

with committee b

and c + number of

members common

with committee b

ALONE + number

of members

common with

committee c

ALONE

For the least

possible number,

Number of

members who are

not common with

any of the

committee = 0.

Hence least

possible number =

0 + 2 + (3-2) + (3-

2) = 4

Note: We have 3-2

because though

there are 3

members common

for a group, 2 are

common across all

groups.

12

(QID=9740) :

A man starts his work on a monday. He works for 10 days and takes rest on every 11th day, that

is every 11th day as holiday. The 1233 rd holiday will fall on

Tuesday

Wednesday

Thursday

Friday

Correct 1.0 7s

Ans: Thursday

From day 1 to day

11 is considered

one cycle. So for

1233 rd holiday it

is 1233*11 =

13563. Excluding

the starting

Monday it is 13562

days.

13562 has 3 odd

days.

So Monday+ 3

days = Thursday.

13

(QID=9720) :

Meena wanted to buy 3 kgs of apples. The vendor kept the 3 kg weight on the right side and

weighed 10 apples for that. She doubted on the correctness of the balance and placed 3 kg weight

on the left side and she could weigh 15 apples for 3 kgs. If the balance was accurate how many

apples she would have got for 6 kilograms?

15 apples

20 apples

25 apples

None of the above

Correct 1.0 155s

Ans: 25 apples.

Let the additional

weight on the left

side be x.

Let the average

weight of an apple

be a.

x+10a = 3kgs

x+3 = 15a

Solving, a=6/25

kgs.

Hence for 6 kgs

she should have

got 6/1/6/25 = 25

apples.

14

(QID=9733) :

If the price of petrol increases by 30% and Babu intends to spend only an additional 10% on

petrol, by how much percentage should he reduce the consumption?

11.62%

12.64%

14%

15.38%

Correct 1.0 5s

Ans: 15.38%

Let the current

price of unit

quantity of petrol

be x. New

price=1.3x.

Let reduced

quantity be y.

y*1.3x = 1.1x,

y=11/13

Reduction = 1 -

11/13 = 2/13 =

15.38%

15

(QID=9739) :

Diana bought two varities of rice costing 50 Rs per kg and 80 Rs per kg and mixed them in some

ratio. Then she sold that mixture at 72 Rs per kg making a profit of 20%. What was the ratio of the

mixture?

2:1

3:1

2:5

3:5

Correct 1.0 4s

Ans: 2:1

Cost of mixture be

x.

x*1.2 = 72, x =

720/12 = 60

Using alligation

rule,

Qty of 50kg

rice/Qty of 80kg

rice = (80 - 60)/(60-

50)

= 2:1

16

(QID=9737) :

A 3*3 grid is coloured using red, green and blue colors such that if we rotate the grid about it's

center in the plane by 180 degrees, the grid looks the same. The number of ways to color the grid

this way is

6

15

32

None of the above

Correct 1.0 15s

Ans: None of the

above (As answer

is 243)

Based on the

condition the

center can contain

any of the three

colors and need

not have matching

cell.

The 2 pairs of grid

cells which are

positioned

diagonally must

have same color.

The 2 pairs of grid

cells in the edge in

the middle should

match it's color

with the cell in the

directly opposite

side.

As there are 3

colors to be used,

number of ways =

3*3*3*3*3 = 243

17

(QID=9723) :

There are 5 boys and 6 girls in two separate queues. The two queues are merged so that the

positions of the boys (in relation to each other) and those of the girls (in relation to each other)

remain unchanged. In how many ways can this be done?

250

462

684

748

Correct 1.0 6s

Ans: 462

Now when the

queues are merged

there are 11 slots.

In these slots the

girls can be

arranged in a

fashion without

disturbing their

order. Hence num

of ways to arrange

girls = 11C6 = 462

The boys can be

arranged in the

remaining five slots

in only one way as

their order should

be preserved.

Hence total num of

ways = 462 * 1 =

462.

Note: If you

consider arranging

boys first, then

also 11C5 * 1 =

11C6 = 462

Ans: 53

Let the number of

chocs be x.

The price of one

choc will be

between Rs.1

18

(QID=9738) :

A shop used to sell chocolates at Rs.2 each but there were no sales at that price. When it

reduced the price, all the chocolates sold out enabling the shopkeeper to realize Rs 68.90 from

the sales. If the new price was not less than half of the original price quoted, how many

chocolates were sold?

47

53

73

87

Wrong 1.0 6s

between Rs.1

inclusive to Rs.2

which will be y

So xy == 68.90 =

6890 paise.

Now writing 6890

as multiple of

prime factors,

6890 = 2*5*13*53

As the price of a

choc should be

between 100 paise

to 200 paise, the

possible product

combo is either

2,5,13 or 53*2

So the price of a

chocolate and the

num of chocolate

can be either

(Rs.1.30 and 53

chocolates) or

(Rs.1.06 and 65

chocolates)

Among the options

53 is the answer

19

(QID=9669) :

6 positive numbers are taken at random and multiplied together. Then what is the probability that

the product ends in an odd digit other than 5?

2/5

9/10

4/5

None of the above

Correct 1.0 3s

Ans: None of the

above (As answer

is 64/15625)

Let us find the

probability that the

product ends with

odd digit.

For a product's unit

digit to be odd, no

unit digit in the 6

numbers should be

even. Else even *

odd will be even.

For the unit digit to

be an odd digit

other than 5, 5

should not be in

the unit digit.

So possible digits

in unit digit of all

the six numbers

are 1,3,7,9 out of

1,2,3,4,5,6,7,8,9,0

Hence required

probability =

(4/10) 6̂ = (2/5) 6̂ =

64/15625

20

(QID=9735) :

If f(x)=ax+b, then f(f(f(x))) = 27x + 39. Then a+b is

5

6

7

8

Correct 1.0 5s

Ans: 6

f(f(f(x))) = f(f(ax+b))

= f(a 2̂*x + ab + b)

= a 3̂*x + a 2̂*b +

ab + b = 27x + 39

Comparing the

coefficients of x,

a 3̂=27 and hence

a=3.

Also a 2̂*b + ab +

b = 39,

As a=3, b=3 and

a+b=6

(QID=9725) :

Ans: 126654

Thousandth

position. 2,3,5,9

can occupy there.

Sum = 1000 *

(2+3+5+9) * 3! =

21

(QID=9725) :

Find the sum of all the four digit numbers that can be formed using the digits 2,3,5,9. No digit is to

be repeated.

524254

324454

126654

None of the above

Wrong 1.0 10s

(2+3+5+9) * 3! =

1000 * 19 * 6

Hundredth position

sum = 100 * (19) *

6

Tenth position sum

= 10 * (19) * 6

Unit position sum

= 1 *(19) * 6

Hence overall sum

= 19 * 6 * (1000 +

100 + 10 + 1)

= 19 * 6 * 1111 =

126654

22

(QID=9731) :

How many 5 digit numbers are there that do not contain the digits 3 or 6?

25088

32768

28672

None of the above

Correct 1.0 6s

Ans: 28672

Possible digits are

0,1,2,4,5,7,8,9

Num of ways =

7*8*8*8*8 = 28672

(As the digits can

be repeated and 0

cannot be in first

place.)

23

(QID=9730) :

How many 6-digit odd numbers can be formed from the digits 1,2,3,4,5,6,7 so that the digits do

not repeat and the second last digit is even?

1440

1820

2400

None of the above

Correct 1.0 6s

Ans: 1440

Number of ways to

fill last digit = 4

(one among

1,3,5,7), second

last = 3 (one

among 2,4,6)

Reqd num of ways

= 4*3*5*4*3*2 =

1440

24

(QID=9560) :

A man travels by bus for 20 hours and then by train for 5 hours. If the average speed of the bus

was 20kmph and that of the entire journey was 24 kmph, what was the average speed of the train?

30 km/hr

32 km/hr

36 km/hr

40km/hr

Correct 1.0 3s

Ans: 40km/hr

Total distance =

25*24 kms

Distance by bus =

20*20

Distance by train =

25*24 - 20*20

Average speed for

train = (25*24 -

20*20)/5 = 5*24 -

4*20 = 40km/hr

25

(QID=9732) :

There are 6 boxes colored violet, pink, red, yellow, blue and green. If 3 boxes are selected at

random, how many combinations are there for atleast one green box or one red box to be

selected?

12

14

16

18

Wrong 1.0 4s

Ans: 16

Num of ways only

green box is

selected = 1C1 *

4C2 = 6 (PVG,

YVG, BVG, PYG,

PBG, YBG)

Num of ways only

red box is selected

= 1C1 * 4C2 = 6

Num of ways one

green box and one

red box is selected

= 2C2 * 4C1= 4

Total ways = 16.

26

(QID=9726) :

In how many ways can we arrange letters from A to Z, when C and D can have only eight letters

between them?

24! * 34

24! / 8!

26! * 7!

None of the above

Correct 1.0 3s

Ans: 24! * 34

Out of the 26 slots,

two slots are

allocated for C and

D and can be filled

in 2,1 ways (The

first slot can be C

or D and the

second slot the

remaining).

The number of

ways in which C

and D can occur in

different slots = 26-

(8+1) = 17

Hence total num of

ways = 24! * 2 * 17

= 24! * 34

27

(QID=9718) :

Find the last two digits in 4761 2̂5

21

41

81

None of the above

Correct 1.0 3s

Ans: None of the

above (As answer

is 01)

The unit digit will

be 1.

The tens digit will

be the unit digit in

the product of

6*5(the unit digit in

power and the tens

digit of the number)

= unit digit in 30 =

0

Hence last two

digits will be 01

28

(QID=9682) :

A farmer has a rose garden. Every day he either plucks 7 or 6 or 24 or 23 roses. The rose plants

are intelligent and when the farmer plucks these numbers of roses, the next day 37 or 36 or 9 or

18 new roses bloom in the garden respectively. On Monday, he counts 189 roses in the garden.

He plucks the roses as per his plan on consecutive days and the new roses bloom as per

intelligence of the plants mentioned above. After some days which of the following can be the

number of roses in the garden?

249

232

27

26

Correct 1.0 3s

Ans: 249

Situation where the

number of roses

increases is when

he plucks 7 or 6.

Increase = 37-7 or

36-6= 30.

Situation where the

number of roses

decreases is when

he plucks 24 or 23.

Decrease = 24-9 =

15 or 23-18 =5.

Initial count was

189.

Option 1: 26

To arrive at 26, it

has to be

decremented by

189-26=163 which

cannot happen as

189 + 30x -5y -15z

will not give 3 in

unit digit.

Similarly Option 3

and 4 are not

possible

Option 2: 249

To arrive at 249, it

has to be

incremented by

249 - 189 = 60

which is possible

as 189 + 30*2 =

249

(QID=9575) :

Absentminded professor forgot the day of the week. He met his 2 students Eesha and Usha in the

Ans: Thursday

Esha lies on

Mon,Tue,Wed

Usha lies on

Thu,Fri,Sat

Esha says truth on

Thu,Fri,Sat,Sun

Usha says truth on

Sun,Mon,Tue,Wed

Case 1: Both are

saying truth.

This can happen

only on Sunday.

But saturday

cannot be a lying

day for Esha.

Hence this case is

invalid

29

Absentminded professor forgot the day of the week. He met his 2 students Eesha and Usha in the

class room. Eesha lies on Mondays, Tuesdays, and Wednesdays and tells the truth on the other

days of the week. Usha lies on Thursdays, Fridays, and Saturdays but tells the truth on the other

days of the week. They made the following statements.

Eesha: Yesterday was one of my lying days.

Usha: Yesterday was one of my lying days.

From these two statements, the professor was able to deduce the day of the week. What was the

day?

Monday

Tuesday

Wednesday

Thursday

Correct 1.0 3s

Case 2: Both are

lying.

This cannot

happen at all as

per the info given

Case 3: Esha lies

and Usha is saying

truth.

Usha saying truth

and the previous

day being her day

of telling lie can

happen when

current day is

Sunday.

But on Sunday

Esha will always

say the truth.

Hence this case is

not valid.

Case 4: Usha lies

and Esha is saying

truth

Esha saying truth

and the previous

day being her day

of telling lie can

happen when

current day is

Thursday.

Also on Thursday's

Usha lies. Hence

this case is valid.

Hence answer is

Thursday

30

(QID=9567) :

A mother has 3 babies – Usha, Nisha and Eesha. If Usha is sleeping, Eesha is drinking milk. If

Nisha is not sleeping, Eesha is not drinking. It never happens that both Usha and Nisha are

sleeping. Father concludes that Usha never sleeps. Mother concludes that Nisha never sleeps.

Nurse concludes that Eesha always drinks. Who has made a correct deduction?

Only Father and Mother

Only Father

Only Nurse and Mother

Only Nurse

Correct 1.0 6s

Ans: Only Father

and Mother

Consider the

possible scenarios.

Case 1: Father is

right. Now, Usha

never sleeps.

This implies Eesha

never drinks milk.

Hence nurse is

wrong in this case.

Mom's conclusion

can happen as it

does not contradict

with father's

assumption.

Case 2: Mom is

right. Now, Nisha

never sleeps.

This implies Eesha

never drinks milk.

Hence nurse is

wrong in this case.

Father's conclusion

can happen as it

does not contradict

with mom's

assumption.

Case 3: Nurse is

right. Now, Esha

always drinks.

This implies Usha

is always sleeping.

This also implies

Nishal is always

sleeping.

But as both Usha

and Nisha cannot

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sleep

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