🧠 Arithmetic - PYQ

Correct Answer: (d)

P(E) = Probability of passing in English = 0.6 1

Correct Answer: (d)

P(E) = Probability of passing in English = 0.6 1

Correct Answer: (c)

There can be 3 cases :

Correct Answer: (d)

There can be two cases : Hence, total no. of ways = 9C × 1 = 84 3 Case I : When 1 row contains 3 balls and rest two contains

Correct Answer: (c)

For finding the silver coin, only drawer 1 and 3 remains So probability (P) = Total number of samples (150) in consideration, because the open drawer in any case When we consider the given statements cannot be the drawer that have only gold coins. Now the probability of next coin being a gold coin = 1/2 = 0.5 3+7+10+15+28+33 96 (1) P(X£15)= = = 0.64

Correct Answer: (c)

For finding the silver coin, only drawer 1 and 3 remains So probability (P) = Total number of samples (150) in consideration, because the open drawer in any case When we consider the given statements cannot be the drawer that have only gold coins. Now the probability of next coin being a gold coin = 1/2 = 0.5 3+7+10+15+28+33 96 (1) P(X£15)= = = 0.64

Correct Answer: (b)

No. of different sums of money = any 1 coin at a time + = 5C – 3C = 10 – 3 = 7 any 2 coins + any 3 coins it all 4 coins 3 1

Correct Answer: (c)

Suppose any particular student is always selected. Here, either case I or case II is possible. Hence, total no. of Now, remaining 2 students are to be selected out of the ways = 81 + 18 = 99 remaining 5 students.

Correct Answer: (b)

3 balls can be placed be in any of the 12 squares in This row, containing 1 ball can be selected in 3C ways and 12C ways. 1 Tota 3 l number of arrangements = 12C = 220 number of ways of arrangement in this row = 3C 1 . In other 3 two rows, containing 2 balls each, number of ways of Now, assume that balls lie along the some line. arrangement in each = 3C . There can be 3 cases : 2 Thus, number of ways for case II = 3C × 3C × 3C × 3C Case I: When balls lie along the straight horizontal line. 1 1 2

Correct Answer: (c)

Let E be the event of selecting the three numbers such Case V : Number of ways for (3, 2, 2, 1) = =12 2! that their product is odd and S be the sample space. As, either of these five cases are possible, For the product to be odd, 3 numbers choosen must be odd. Hence total number of ways = 4 + 12 + 1 + 6 + 12 = 35 \ n (E) = 5C 3

Correct Answer: (d)

x y z 7 1 0,1 ï 1 1 4 2 ï 3 0,1,2 ïï 1 4 1 ý21 0,1,2,3 4 1 1 4 ï 0,1,2,3,4 ï 1 2 3 5 ï 6 0,1,2,3,4,5ïþ 1 3 2 3 1 2 8 1 0 ü 3 2 1 ï 2 0,1 ï 2 2 2 3 0,1,2 ï 2 3 1 ï 4 0,1,2,3 ý28 2 1 3 ï 5 0,1,2,3,4 ï Total Number of triplets = 10. 6 0,1,2,3,4,5 ï 7 ï

Correct Answer: (a)

Black White Number of way

Correct Answer: (a)

Required number of ways = = 120 3! 2 (adjacent or opposite faces are 4 2 white)

Correct Answer: (a)

Number of diagonals in an octagon =8C – 8 2 8´7 5 1 1 = – 8 = 20 6 0 1 2 Hence, total number of ways

Correct Answer: (c)

HundredCorresponding Corresponding Digit Ten’s Digit Unit’s Digit = 1 + 1 + 2 + 2 + 2 + 1 + 1 = 10 0 ü

Correct Answer: (b)

Total number of balls 15 + 20 = 35 Total number of red balls numbered 3 `50 `10 `1 Numberofways æ100-20-40ö 40 2 0 7 1 = 15 ×ç ÷ = 15× = 6 è 100 ø 100 1 (0–5) (7–57) 6 Total number of black ball numbered 1 or 2 0 (0–10) (7–107) 11 æ100-30ö 70 Total number of ways = 1 + 6 + 11 = 18. = 20 × ç ÷ = 20 × = 14 è 100 ø 100

Correct Answer: (b)

A cube has 6 symmetric faces. Here m = n = 4. So, Number of prallelograms Number of faces of a cube coloured by: = 4 C ×4 C = 6 × 6 = 36. 2 2 5337_DBE Permutation, Combination & Probability F 87

Correct Answer: (c)

Pages till 9 uses 9 digits i.e.; 1 to 9. 0,1,2,3,4,5,6þ Pages from 10 to 99 uses 2 digits each. 9 1 0 ü Number of pages = (99 – 10) + 1 = 90 ï Number of digits = 2 × 90 = 180. 2 0,1 ï Pages from 100 to 999 uses 3 digits each. 3 0,1,2 ï Number of pages = 999 – 100 + 1 = 900. ï Number of digits = 3 × 900 = 2700. 4 0,1,2,3 ï ý36 Pages from 1000 to 9999 uses 4 digits each. 5 0,1,2,3,4 ï Remaining digits = 3089 – 2700 – 180 – 9 = 200. 6 0,1,2,3,4,5 ï 200 ï Remaining number of pages = =50. 7 0,1,2,3,4

Correct Answer: (b)

Let, there are n number of pages in the book. obtained by the digits 1, 2, 3, 4 and 5 will be divisible by 3. All 5 digits numbers obtained by using all the digits 1, n(n+1) Sum of page numbers = 2, 3, 4 and 5 without repetition of digits, will be divisible 2 by 3. \n(n+1)=2´(Sum of page numbers) Hence, no 5 digits prime number can be obtained by Here given, sum of page numbers = 195. the digits 1, 2, 3, 4 and 5.

Correct Answer: (c)

—B —B — 2 (two adjacent or two opposite 1 2 2 4 2 boys can take their seats in 2! ways and 3 girls can take faces are black) the remaining 3 seats in 3 C2 ´2! ways. 2 (around a corner or not around a Hence, 2!´3 C2 ´2! = 12 ways. 3 3 corner are either black or white) 6!

Correct Answer: (d)

We cannot predict the number of dinners for a According to question particular member from the given data. It may be possible Þ 5 ´ 3x + yx = 100 ... (i) that by choosing members from picking lots, one may have Now, clerk has interchanged socks pairs then price is to host a dinner more than one times. increased by 100%

Correct Answer: (c)

Here are five persons, and 5 tasks (15x+yx)´100 So, When T task is fixed for person 3 3xy + 5x = (15x+yx)+ 2 100 Task Þ 3xy + 5x = 30x + 2xy ... (ii) Þ 30x + 2yx = 3xy + 5x T 2 Þ 25x = xy 1 2 3 4 5 y = 25 \ So, number of brown socks = 25 F 86 Topicwise Solved Papers 58. (*)

Correct Answer: (c)

Let x be the number of women. These can be arranged in 3! ways and A, B can be arranged Let y be the number of men. among themselves in 2! ways. Total number of hand shakes = xy = 24 Hence, no. of arrangements such that A and B are always Then, the possible factors of x and y are x = 6 or 4, y = 4 or 6 together = 3! × 2! = 3 × 2 × 2 = 12 Number of hugs =xC + yC

Correct Answer: (d)

Number of ways to select Principal = 2C we know, 1 Number of ways to select Vice Principal = 5C n(C È F) = n(C) + n(F) – n(C Ç F) Total number of ways= 2C + 5C 2 = 60 + 30 –10 = 80 1 2 n(C¢ Ç F¢) = n(C È F)¢ = n(U) – n(C È F) 5´4 = 2+ = 2 + 10 = 12 = 100 – 80 = 20 2´1

Correct Answer: (b)

Let us take books A and B as one i.e., they are always = 4C + 4C + 4C + 4C 1 2 3 4 continuous. = 4 + 6 + 4 + 1 = 15 Now, number of books = 4 – 2 + 1 = 3

Correct Answer: (d)

First person can shake hand with the other 9 i.e, in 9 No. of ways for Task T = 2 ways. Second person can shake hand with the remaining 8 1 No. of ways for Task T = 1 persons and so on. 2 No. of ways for Task T = 3 \ total no. of hands shaked = 9 + 8 + ....... + 2 + 1 3 No. of ways for Task T = 3 4 = 9(9+1) =45 No. of ways for Task T 5 = 3 Total number of ways for condition II 2 = 3 + 3 + 3 + 2 +1 = 12

Correct Answer: (b)

No. of different sums of money = any 1 coin at a time + = 5C – 3C = 10 – 3 = 7 any 2 coins + any 3 coins it all 4 coins 3 1

Correct Answer: (c)

8 7 7 (i) Number of ways of arranging three colours taken 1 at = ´ = 104 103 1339 3´2!

Correct Answer: (b)

White MarblesRed Marbles i.e. n(F) = 30 10 13 Number of students who play both the games = 10 White MarblesGreen Marbles i.e. n(C Ç F) = 10 10 . 0 5 Now, total number of Marbles = 5 + 10 + 13 = 28 To find : n(C¢Ç F¢) = ?

Correct Answer: (d)

P(E) = Probability of passing in English = 0.6 1

Correct Answer: (c)

Suppose any particular student is always selected. Here, either case I or case II is possible. Hence, total no. of Now, remaining 2 students are to be selected out of the ways = 81 + 18 = 99 remaining 5 students.

Correct Answer: (b)

Take, A and B to be always together as a single entity. Total number of ways for condition I and II = 12 + 12 = 24 Now, total no. of children = 4 – 2 + 1 = 3

Correct Answer: (c)

Suppose three people have been given a, b and c 2! number of items. Now, either of case (I) or case (II) is possible, Then, a × b × c = 30 Hence, total number of ways = 60 + 90 = 150 Now, There can be 5 cases :

Correct Answer: (c)

Total number of arrangements, when any 3 boys are R B B R selected out of 5 = 5C . Now, when groups contains both C 3 R W W R and D, then their selection is fixed and the remaining 1 boy can be selected out of the remaining 3 boys. It can be done There are eight such arrangements, if we start with Red ball. in 3C ways. Similarly, there are 8 arrangements, if we start with black or 1 So, number of groups, when none contains both C and D white ball. = total number of arrangements-number of arrange

Correct Answer: (c)

Since, every question has five options, so no. of Case I : number of ways = 2C × 3C × 3C × 2C = 36 1 1 1 1 choices for each question = 5 Case II : number of ways = 3C × 3C = 9 \ total no. of choices = 5 × 5 × 5 × 5 = 625 2 2 Case III : number of ways = 2C × 2C = 1 Now, no. of choices of all correct answer = 1 2 2 Hence, no of choices for all the four answers not correct Hence, total number of ways = 36 + 9 + 1 = 46 = total no. of choices – no. of choices of all correct answer

Correct Answer: (a)

Contains 1 ball each. Case II - One box contain 1 ball and rest two x Contains 2 balls each. y Case 1 : Number of ways = 5C × 2C × 1C = 20 3 1 1 3! There can be two cases: Now, these 3 boxes can be arranged in among themselves, 2! Case (I) : When x row contains 6 balls: as two of them contains similar number of balls. Then the 2 balls can be arranged in y row in 6P ways = 15 2 3! or the 2 balls can be arranged in any of the 4 two box row in So, number of ways = 20´ =60 2! 4 ways. Case II : Numbe

Correct Answer: (c)

Probability of a particle lying in any particular half = 2

Correct Answer: (c)

2 Women can occupy 2 chairs out of the first four Thus, no of ways, when Ajit and Mukerjee are always together = 5 ! × 2! chairs in 4P 2 ways. 3 men can be arranged in the remaining Hence, no of ways when they are never together 6 chairs in 6P ways. 3 = Total no of ways – no of ways when they are always Hence, total no. of ways = 4P × 6P = 1440 together 2 3

Correct Answer: (d)

Let the total no. of participants be 'n' at the beginning. = 6 ! – (5 ! × 2 !) = 6 × 5 ! – (5 ! × 2 !) = 5 ! (6 – 2) = 480 Players remaining after sometime = n – 3

Correct Answer: (b)

Maximum number of triangles can be formed by 365 365 selecting 3, 4 or 5 points out of 6 at a time. 1 1 Hence, required probability = 1´ = So, maximum no. of triangles = 6C + 6C + 6C which is ( 365 )2 ( 365 )2 clearly more than 18. Now, triangles fo 3 rmed 4 will be 5 minimum

Correct Answer: (c)

8 7 7 (i) Number of ways of arranging three colours taken 1 at = ´ = 104 103 1339 3´2!

Correct Answer: (d)

For each combination, let us name the females and males remaining 4 seats in four ways and the 2nd man occupy any Female (3) Male (5) of the remaining 3 seats in three ways. A B Hence, no of ways = 1 × 4 × 3 = 12 C D (ii) E F Leaving the reserved seat, 1st man can occupy any of the G 4 seats in four ways. 2nd man can occupy any of the H remaining 3 seats in three ways. Lady can occupy any of Since A can’t go with B, it will make team with four males in the remaining 2 seats in two ways. four way

Correct Answer: (d)

For (10 + A + B) = 12 stations, no of tickets required, i.e., zero, when the points will overlap on the same line and when 4 new stations are added for one way journey all the points are along the same vertical line. = 12 × 4 = 48

Correct Answer: (b)

There can be two cases : ah ch eh gf (i) Lady occupies the reserved seat Now, team ad can play only with: cb, cg, ch, eb, eh, (ii) Lady does not occupy the reserved seat. gb, gf, i.e. 7 5337_DBE Permutation, Combination & Probability F 83 The same will apply with all teams. 4! So no. of total match = 12 × 7 = 84 Case IV : Number of ways for (3, 3, 1, 1) = = 6 2!´2! Since every match includes 2 teams, so the No. of matches = 84/2 = 42 4!

Correct Answer: (d)

Total number of ways in which 9 balls occupy any of

Correct Answer: (a)

On the first attempt four coins are overturned. No of ways of arranging the items = 5 × 4 × 3 × 2 × 1=120 Now, six coins are left. Now, in this arrangement there is one such arrangement, In the next turn, four more are overturned. Now only two which is the correct answer. would be left. We take one more from the left over two coins \ Maximum number of examinees = no. of ways of and any three from the previously turned ones. Finally, the arrangement of items = 120 – 1 = 119 leftover coin and the

Correct Answer: (c)

For finding the silver coin, only drawer 1 and 3 remains So probability (P) = Total number of samples (150) in consideration, because the open drawer in any case When we consider the given statements cannot be the drawer that have only gold coins. Now the probability of next coin being a gold coin = 1/2 = 0.5 3+7+10+15+28+33 96 (1) P(X£15)= = = 0.64

Correct Answer: (d)

Since, 8 Green balls + 7 White balls = 15 balls Now, option (c) is the graph of y = cos t. 7 White balls + 5 Red balls = 12 balls

Correct Answer: (d)

Probability that the light is not green 7 and probability for the jack in second draw = 103 time for which light is not green ( 5+30 ) 35 7 = = = = Since both the events are independent events. time taken for the entire cycle 60 60 12 Hence the probability that both of them are jacks.

Correct Answer: (d)

P(E) = Probability of passing in English = 0.6 1

Correct Answer: (a)

As DTQS is an isosceles triangle. a time = 3P 1 = 2! = 3 \ ÐTSQ = ÐTQS = 47° (ii) Number of ways of arranging three colours taken 2 at a Now, in equilateral triangle DQRS, 3! time = 3P = =6 ÐRQS = ÐRSQ = ÐQRS = 60° 2 1! Now, ÐRQS = ÐRQT + ÐTQS = 60° (iii) Number of ways of arranging three colours taken 3 ÐRQT = Y = 60° – 47° = 13° at a time = 3P = 6 3 F 82 Topicwise Solved Papers Hence, Maximum no. of codes = No. of ways of arranging (i) Lady these flags = 3 + 6 + 6 = 15 Fixing one seat for the

Correct Answer: (c)

Let the radius of the circle be unity 150 150 Equation of the circle, x2 + y2 = 1 28+33+24+11 96 (2) P(13