A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?

A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?
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A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?
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A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?

Description : A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?

Last Answer : A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page ... `(3)/(50)` D. `(7)/(125)`

A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

Description : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

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In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Description : A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Last Answer : (d) \(rac{1}{4}\)The box contains 100 balls numbered from 1 to 100. Therefore, there are 50 even and 50 odd numbered balls. The sum of the three numbers drawn will be odd, if all three are odd or one is even and 2 are odd. ∴ Required probability = P(odd) × P(odd) × P(odd) + P(even) × P(odd) × P(odd)

A box contains 15bolts of which 5 are defective. If 5 bolts are selected at random from the box, what is the probability that at least one of them is defective? a) 91/143 b) 101/143 c) 111/143 d) 121/143 e) 131/143

Description : A box contains 15bolts of which 5 are defective. If 5 bolts are selected at random from the box, what is the probability that at least one of them is defective? a) 91/143 b) 101/143 c) 111/143 d) 121/143 e) 131/143

Last Answer : Probability that at least one is defective = 1 – probability that none is defective Probability that none is defective = 10C5 / 15C5 = 12/143 Required Probability = 1-12/143 = 131/143 Answer: e)

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Description : If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Last Answer : (c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three ... C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

Nine playing cards are numbered 2 to 10. A card is selected at random. What is the probability that the card will be an odd number? a. 1/9 b. 2/9 c. 4/9 d. 3/7

Description : Nine playing cards are numbered 2 to 10. A card is selected at random. What is the probability that the card will be an odd number? a. 1/9 b. 2/9 c. 4/9 d. 3/7

Last Answer : c. 4/9

An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. a. 3/25 b. 2/25 c. 1/25 d. 13/50

Description : A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. a. 3/25 b. 2/25 c. 1/25 d. 13/50

Last Answer : b. 2/25

What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '7'? e) 13/45 f) 14/45 g) 23/90 h) 19/45

Description : What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '7'? e) 13/45 f) 14/45 g) 23/90 h) 19/45

Last Answer : Answer: a

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Description : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability that the selected pair will consist of one perfume a

Description : Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability that the selected pair will consist of one perfume a

Last Answer : Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability ... of perfumes and does the shop can contain ?

What is the probability that a leap year selected at random will have 33 Sunday?

Description : What is the probability that a leap year selected at random will have 33 Sunday?

Last Answer : Need answer

In a hostel, 40% of the students play cricket, 20% play chess and 10% both. If a student is selected at random, then the probability that he plays cricket or chess is: a) 1/2 b) 3/5 c) 1/4 d) 4/7

Description : In a hostel, 40% of the students play cricket, 20% play chess and 10% both. If a student is selected at random, then the probability that he plays cricket or chess is: a) 1/2 b) 3/5 c) 1/4 d) 4/7

Last Answer : Answer: A) Given that, 40% play cricket; that is, P(C) = 40/100=4/10 20% play chess; that is, P(c) = 20/100 =2/10 And, 10% play both cricket and chess; that is, P(C And c) = 10/100 = 1/10 Now, we have ... P(C) + P(c) - P(C And c) = 4/20+2/10-1/10 =5/10=1/2 Hence, the required probability 1/2

In a batch, 40% of the students offered Maths, 30% offered science and 15% offered both. If a student is selected at random, what is the probability that they has offered science or maths? A) 0.55 B) 0.65 C) 0.45 D) 0.75

Description : In a batch, 40% of the students offered Maths, 30% offered science and 15% offered both. If a student is selected at random, what is the probability that they has offered science or maths? A) 0.55 B) 0.65 C) 0.45 D) 0.75

Last Answer : Answer: A) P(M) = 0.40 P(S) =0.30 and P(M∩S) = 0.15 P(M∪S) = P(M) + P(S) - P(M∩S) = 0.55

In a batch, there are 22 boys and 18 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: a) 3754/8854 b) 4158/9880 c) 8514/9880 d) 2078/4920

Description : In a batch, there are 22 boys and 18 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: a) 3754/8854 b) 4158/9880 c) 8514/9880 d) 2078/4920

Last Answer : Answer: B) Let , S - sample space E - event of selecting 1 girl and 2 boys. Then, n(S) = Number ways of selecting 3 students out of 40 = 40C3 = 9880 n(E) = 18C1 *22C2 = 18*231  = 4158 P(E) = n(E)/n(s) = 4158/9880 

Which among the following print commands should be selected to print first 5 pages of document: a) Page Setup b) Print All c) From ___to ____ d) None of The Above

Description : Which among the following print commands should be selected to print first 5 pages of document: a) Page Setup b) Print All c) From ___to ____ d) None of The Above

Last Answer : c) From ___to ____

__________ is responsible for hamper the search rankings. a. Connecting to your own website from any random website. b. Utilizing the same colors of texts as that of your background pages. c. Integrating page templates into your page template. d. None of the above

Description : __________ is responsible for hamper the search rankings. a. Connecting to your own website from any random website. b. Utilizing the same colors of texts as that of your background pages. c. Integrating page templates into your page template. d. None of the above

Last Answer : b. Utilizing the same colors of texts as that of your background pages.

A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Description : A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Last Answer : (b) \(rac{7}{99}\)There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(rac{|\underline{7}}{|\underline3|\underline4}\) = \(rac{7 imes6 imes5}{3 ... ) = 35∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Description : A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Last Answer : (d) None of theseThe month having 3 days less than 31 days has 28 days, i.e, it is the month of February. P(Choosing February) = \(rac{1}{12}\).

A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen? a) 16/243 b) 16/283 c) 14/243 d) 23/729

Description : A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen? a) 16/243 b) 16/283 c) 14/243 d) 23/729

Last Answer : a) 16/243

A box contains 2 black, 3 orange and 4 pink ribbons. If two ribbons are drawn at random. What is the probability that both are orange? 1) 5/12 2) 1/13 3) 1/14 4) 1/12 5) 1/24

Description : A box contains 2 black, 3 orange and 4 pink ribbons. If two ribbons are drawn at random. What is the probability that both are orange? 1) 5/12 2) 1/13 3) 1/14 4) 1/12 5) 1/24

Last Answer : 4) 1/12

A cartoon contains 15 torch lights out of which 3 are defective. Two torch light are chosen at random from this cartoon. The probability that at least one of these is defective is. A) 13/35 B) 14/35 C) 11/35 D) 17/35

Description : A cartoon contains 15 torch lights out of which 3 are defective. Two torch light are chosen at random from this cartoon. The probability that at least one of these is defective is. A) 13/35 B) 14/35 C) 11/35 D) 17/35

Last Answer : Answer: A) P (none is defective) = 12c2/15c2 = (12*11/2*1)/(15*14/2*1) = 66/105=22/35 P (at least one is defective) = (1 – 22/35) =13/35

A Package contains 12 pack of variety1 drink, 6 pack of variety2 drink and 8pack of variety3 drink. Three packsof them are drawn at random, what is the probability that the three are not of the same variety? a) 37/325 b) 288/325 c) 188/325 d) None of these

Description : A Package contains 12 pack of variety1 drink, 6 pack of variety2 drink and 8pack of variety3 drink. Three packsof them are drawn at random, what is the probability that the three are not of the same variety? a) 37/325 b) 288/325 c) 188/325 d) None of these

Last Answer : Answer: B) Total number of drink pack= 12+6+8= 26. Let S be the sample space. Then, n(S) = number of ways of taking 3 drink pack out of 26. Therefore, n(S) = 26C3 = 2600 Let Ebe the ... 296/2600=37/325 Then, the probability of taking 3 pack are not of the same variety = 1 - 37/325= 288/325

There are 2 vessels. 1st vessel contains 5white and 5 blue thread roll. 2nd vessel contains 4 white and 6 black thread roll. One roll is taken at random from first vessel and put to second vessel without noticing its color. Now a roll is chosen at random from 2nd vessel. What is the probability of the second roll being a white colored roll? A) 11/13 B) 9/11 C) 13/11 D) 5/12

Description : There are 2 vessels. 1st vessel contains 5white and 5 blue thread roll. 2nd vessel contains 4 white and 6 black thread roll. One roll is taken at random from first vessel and put to second vessel without noticing its color ... second roll being a white colored roll? A) 11/13 B) 9/11 C) 13/11 D) 5/12

Last Answer : Answer: B) Case 1: first was a white roll Now it is put in second vessel, so total white rolls in second vessel = 4+1 = 5, and total rolls in second vessel = 10+1 = 11 So probability of white roll ... = 5/11+4/11 = 9/11 (added the cases because we want one of these cases to happen and not both)

A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Description : A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Last Answer : Answer: B) Probability of drawing 1 blue marker pen =8/16 Probability of drawing another blue marker pen = 7/15 Probability of drawing 1 red marker pen = 3/14 Probability of drawing 2 blue marker pens and 1 red marker pen = 8/16*7/15*3/14=1/20

Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink pencil? a)3/ 49 b)3/686 c)3/14 d)3/545

Description : Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink pencil? a)3/ 49 b)3/686 c)3/14 d)3/545

Last Answer : Answer: B) Here, total number of pencils = 14 Probability of drawing 1 black pencil = 2/14 Probability of drawing another black pencil = 2/14 Probability of drawing 1 pink pencil = 3/14 Probability of drawing 2 black pencils and 1 pink pencil = 2/14 * 2/14 * 3/14 = 3/686

A Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the receptacle. The probability that all of them are purple, is: A)3/55 B)7/55 C)1/55 D)9/55

Description : A Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the receptacle. The probability that all of them are purple, is: A)3/55 B)7/55 C)1/55 D)9/55

Last Answer : Answer: C) Let S be the sample space. Then, n(S) = number of ways of drawing 3 balls out of 12 = 12C3 = 220 Let E = event of getting all the 3 purple balls. n(E) = 4C3= 4 P(E) = n(E)/n(S) = 4/220 = 1/55

A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they are of same colour. A) 91/47 B) 47/105 C) 47/95 D) 95/47

Description :  A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they are of same colour. A) 91/47 B) 47/105 C) 47/95 D) 95/47

Last Answer : Answer: C) Let S be the sample space Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190 Let E = event of getting both bulbs of same colour Then, n(E) = no of ways (2 bulbs out of 12) ... 12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94 Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95

A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries: Image Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU? (A) 1024, 3072, 4096, 6144 (B) 1234, 4012, 5000, 6200 (C) 1020, 3012, 6120, 8100 (D) 2021, 4050, 5112, 7100

Description : A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries:  Which of the following list of virtual addresses (in ... 1234, 4012, 5000, 6200 (C) 1020, 3012, 6120, 8100 (D) 2021, 4050, 5112, 7100

Last Answer : Answer: C Explanation: The pages which are not in main memory are: 1020 will not cause page fault (1024-2047) 3012 will not cause page fault (3072-4095) 6120 will not cause page fault (4096-5119) 8100 will not cause page fault (6144-7167)

Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Description : Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Last Answer : The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten's place. So, Total number of 2-digit numbers that can be formed using these 5-digits = 5 5 = ... 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)

A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Description : A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Last Answer : Without repetition, a five -digit number can be formed using the five digits in 5! ways (5 4 3 2 1) Out of these 5! numbers, 4! numbers will be starting with digit 0. (0 (fixed) 4 3 2 1) ∴ Total ... + 6 + 6 + 4 + 4 + 4 = 30∴ Required probability = \(rac{30}{96}\) = \(rac{5}{16}.\)

My author's uncertain yet my title's the same, I contain random text yet order's my aim. Read me one day and see my pages are totally bare. Try again another day and the words will be there. I'm not a book of magic although it may sound, I can predict the future, and inside, your life can be found. Move my eye, I become involved in lactic extraction. But that's just a clue , a minor distraction.What am I? -Riddles

Description : My author's uncertain yet my title's the same, I contain random text yet order's my aim. Read me one day and see my pages are totally bare. Try again another day and the words will be there. ... I become involved in lactic extraction. But that's just a clue , a minor distraction.What am I? -Riddles

Last Answer : A diary.

What is the probability that a randomly selected bit string of length 10 is a palindrome? (A) 1/64 (B) 1/32 (C) 1/8 (D) ¼

Description : What is the probability that a randomly selected bit string of length 10 is a palindrome? (A) 1/64 (B) 1/32 (C) 1/8 (D) ¼

Last Answer : (B) 1/32

If I read 16 pages from a 40 page book what fraction is this?

Description : If I read 16 pages from a 40 page book what fraction is this?

Last Answer : It is 16/40.

If you read a 90-page book in 6 days and if you continue reading at this ratehow long will it take you to read 300 pages?

Description : If you read a 90-page book in 6 days and if you continue reading at this ratehow long will it take you to read 300 pages?

Last Answer : 15 days

Ajay reads at an average rate of 30 pages per hour, while Vikram reads at an average rate of 40 pages per hour. If Ajay starts reading a novel at 4:30 PM and Vikram begins reading an identical copy of the same book at 5:20 PM, at what time will they be reading the same page? a) 7:50 PM b) 8:15 PM c) 8:40 PM d) 9:00 PM e) 9:25 PM

Description : Ajay reads at an average rate of 30 pages per hour, while Vikram reads at an average rate of 40 pages per hour. If Ajay starts reading a novel at 4:30 PM and Vikram begins reading an identical copy of the same book at 5:20 PM, at ... same page? a) 7:50 PM b) 8:15 PM c) 8:40 PM d) 9:00 PM e) 9:25 PM

Last Answer : Ajay reads at average rate of 30 pages per hour ie. 1 page in every two minutes. Therefore, by 5:20 PM, he would have covered 25 pages By 6:20 PM, Ajay would have read 25+30 pages and Vikram would have ... and he will also finish 100 pages Hence, both of them will be on same page at 7:50 Answer: a)

The Shass Pollaks are a community fabled for their feats of memory. With a particular book, they could recall a word from the exact spot where the experimenter stuck a pin into the page before them, and then recall the exact word several pages below into which that pin might have travelled. They take the name Shass from the informal term for the book that they learn so well. What book is this?

Description : The Shass Pollaks are a community fabled for their feats of memory. With a particular book, they could recall a word from the exact spot where the experimenter stuck a pin into the page before them, ... name Shass from the informal term for the book that they learn so well. What book is this?

Last Answer : Talmud. (Also accept Mishna.)

my printer wont print 'selected pages' in word

Description : my printer wont print 'selected pages' in word

Last Answer : Need Answer

What number is chosen at random from 1 to 10. Find the probability of not selecting a multiple of 2 or a multiple of3?

Description : What number is chosen at random from 1 to 10. Find the probability of not selecting a multiple of 2 or a multiple of3?

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Tickets numbered 1 to 37 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 10? A) 11/37 B) 37/11 C) 12/37 D) 37/12

Description : Tickets numbered 1 to 37 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 10? A) 11/37 B) 37/11 C) 12/37 D) 37/12

Last Answer : Answer: A) Here, S = {1, 2, 3, 4, ...., 36,37}. Let E = event of getting a multiple of 4 or 10= {4,8,12,16,20,24,28,32,36,10, 30}. P(E) = n(E)/n(S) = 11/37

Consider a program that consists of 8 pages (from 0 to 7) and we have 4 page frames in the physical memory for the pages. The page reference string is : 1 2 3 2 5 6 3 4 6 3 7 3 1 5 3 6 3 4 2 4 3 4 5 1 The number of page faults in LRU and optimal page replacement algorithms are respectively (without including initial page faults to fill available page frames with pages): (A) 9 and 6 (B) 10 and 7 (C) 9 and 7 (D) 10 and 6

Description : Consider a program that consists of 8 pages (from 0 to 7) and we have 4 page frames in the physical memory for the pages. The page reference string is :  1 2 3 2 5 6 3 4 6 3 7 3 1 5 3 6 3 4 2 4 3 4 5 ... to fill available page frames with pages): (A) 9 and 6 (B) 10 and 7 (C) 9 and 7 (D) 10 and 6

Last Answer : (B) 10 and 7