Posts and Comments

bishoo123 Sayscan you explain the bold one.

Buddy..after naga's post..i hope u got it now.

convalesce SaysWhat will be the last four digits in expansion of 2^999

We need to find 2^999 mod 10000

So, to make them co-primes, we cancel 2^4 from Nr. & Dr.

(we will multiply the remainder thus obtained with 2^4)

So, it becomes 2^995 mod 625

By Totient Theorem(already stated in previous posts)

(625) = 500

So, 2^995 mod 625 = 2^495 mod 625

Let R be the remainder 2^495 is divided by 625.

Now, mod 625 = 001

So, 32* 2^495 = 625k + 1

32R = 625k + 1

32R = 608K + 17K +1

The least integer value possible for k = 15

R = 19*15 + 8 = 293

So, 2^995 mod 625 = 293

So, 2^999 mod 10000 = 293 * 2^4 = 4688

So, Last 4 digits of 2^999 = 4688

- 2 Likes

Total points formed from the intersection : (6c2)c2 = 105

Number of existing lines from these points : 6*5c2 = 60

So, 105-60 = 45

Please explain the part in bold...i am not very good with P&C;:oops:

N is the sum of the squares of three consecutive odd numbers such that all the digits of N are the same. If N is a four-digit number, then the value of N is:

a. 7777 b. 9999 c. 3333 d. 5555 e.2222

Let the nos. be 2k+1, 2k+3 and 2k+5

So,Sum of the squares is

(2k+1)^2 + (2k+3)^2 + (2k + 5)^2

= 12(k)^2 + 36k + 35

= 12(k^2 + 3k + 2) + 11

So, It gives a remainder of 11 when divided by 12

Of the given options,

only (d) satisfies this condition.

- 2 Likes

A sincere request to all fellow puys.

Puy who posts the set is obliged to post the correct solution for the same within a day or two.This is very important for guys like me who are really ignoramus as far as CAT VA is concerned.!!!

Puy who posts the set is obliged to post the correct solution for the same within a day or two.This is very important for guys like me who are really ignoramus as far as CAT VA is concerned.!!!

- 3 Likes

(CAT 2005)

A chemical plant has four tanks (A, B, C, and D), each containing 1000 litres of a chemical. The chemical is being pumped from one tank to another as follows:

From A to B @ 20 litres/minute

From C to A @ 90 litres/minute

From A to D @ 10 litres/minute

From C to D @ 50 litres/minute

From B to C @ 100 litres/minute

From D to B @ 110 litres/minute

Which tank gets emptied first, and how long does it take (in minutes) to get empty after pumping starts?

Choose one answer.

a. A, 16.66

b. C, 20

c. D, 20

d. D, 25

2

(CAT 2005- 2 Marker)

A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and a female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?

Choose one answer.

a. 15

b. 14

c. 12

d. 10

Comparing the scenarios of all tanks

'-' => Outflow

'+" => Inflow

A : -20 + 90 - 10 = +60 litres/minute

B : +20 - 100 + 110 = +30 litres/minute

C : -90 -50 + 100 = -40 litres/minute

D : +10 + 50 - 110 = -50 litres/minute

So,D gets emptied first

Time = 1000/50 = 20 min

So, Option (c)

Let No. of males employed = M

No. of Females employed = F

7<= F <= 12

So,

40M + 50F = 1000

4M + 5F = 100

So, we have only 2 integral solutions for this eqution in the given range of F

(M,F) = (15,

(10,12)

Total Cost = M + F

= 850M + 800F

By simple observation Cost is min. for (M,F) = (10,12)

So, Option (d)

- 5 Likes

15627

Which number is on the opposite side of number 7?

Choose one answer.

a. 12

b. 11

c. 10

d. 6

My answer :option :6

7 - 6

8 - 10

9 - 5

Calculation :

6 + 8 + 5 = 19

7 + 8 + 5 = 20

7 + 10 + 5 = 22

7 + 10 + 9 = 26

The possible sequences are

(1) 4,5,6,7,8,9

(2) 5,6,7,8,9,10

(3) 6,7,8,9,10,11

(4) 7,8,9,10,11,12

Now,

(1) can be rejected as max. possible sum of 3 nos. = 7 + 8 + 9 = 24 (We have 26)

(3) can be rejected as min. possible sum of 3 nos. = 6 + 7 + 8 = 21 (We have 19)

(4) can be rejected as min. possible sum of 3 nos = 7 + 8 + 9 = 24

So, only sequence 2 is possible.

By simple observation,

5+ 6 + 8 = 19

5 + 7 +8 = 20 => (7 is not opp. 5)

5 + 7 + 10 = 22 => ( 7 is not opp. 10)

7 + 9 + 10 = 26

So, 7 is opp. 6

Option (d)

- 6 Likes

15626

Which of the following cannot be the order in which Sarojini types them?

Choose one answer.

a. 24351

b. 13542

c. 45231

d. 23145

My answer : option c

I also go with (c)

- 1 Like

see the prob this way

Pick 1009 distinct numbers from the sequences

{1, 11, 21, ..., 2001}, {2, 12, 22, ..., 2002}, ..., {10, 20, 30, ..., 2010} such that no two numbers in the same sequence are consecutive.

so now take one sequence

{1, 11, 21, ..., 2001}

maximum number of terms u can take is 101 which can be done in one way ...

what if it 100 terms are selected ..in other words in how many ways u can select 100 numbers out of 201 numbers ??

answer is ( 201 C 2 + 201 ) or 202 C 2 { std formula }

now 10 such sequence are there

hope this helps

Sir, please explain the bold part

- 1 Like

Hi folks,

I appreciate you guys for putting up sets for practice, let's add more value to this thread.

Here's the one from my side for practice...

Friday morning, local pediatrician Dr. Johnson N. Johnson had appointments with five infants scheduled at

9:00, 9:30, 10:00, 10:30, and 11:00. Each of the five, including the Ortiz baby, is a different number--1, 2, 3, 4,

or 5--of months old. The following notes are available.

1. Immediately after seeing infant Beth, Dr. Johnson examined the Majors infant, who is 2 months younger than

Beth.

2. Erica isn't the one of the five who is 1 month old.

3. The doctor saw David later in the morning than the 1-month-old.

4. The 9:30 appointment was with the 3-month-old baby.

5. The Luce infant isn't the one who is 5 months of age.

6. Dr. Johnson saw the Nash infant, then examined Alice, who is 2 months older than the Nash baby.

7. The pediatrician's 10:00 appointment was with Chad, who isn't the Majors or the Nash baby.

8. The Prior baby isn't the 1-month-old and wasn't the doctor's 9:00 examinees.

Q.1. How old(in months) is infant Beth?

a. 1 b. 2 c. 3 d. 4 e. 5

Q.2 At what time did infant Erica meet the doctor?

a. 9 b. 9:30 c.10 d. 10:30 e.11

Q.3 Whose baby is infant Chad?

a. Ortiz b. Majors c. Luce d. Nash e. Prior

Q.4 Which of the following is not correct w.r.t infant David

a. He is Nash's baby

b. He is 2 months old

c. His appointment was at 11.00

d. none of these

e. His appointment was at 10:30

Q.5 Who was the last infant to meet the doctor?

a. Alice b. David c. Chad d. Erica e. Beth

come on folks not a tough one....

My take

1) (e)

2) (b)

3) (c)

4) (c)

5) (a)

phuuu..took around half an hour..!!!!:oops:

A very good set i will say..looking forward to more sets of this type from fellow/senior puys.

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