1) In how many ways 16 identical things can be distributed among 4 persons if each person gets at least three things?
2) How would the answer differ if the things were different?
B) There are 5 balls and 3 boxes.Each box can hold all 5 balls and no box is to remain empty. Find the number of ways if
1) boxes and balls are all different
2) balls identical, boxes different
3) balls different, boxes identical
4) balls as well as boxes identical
C) Between two junction stations A and B, there are 12 intermediate stations. The number of ways in which a train can be made to stop at 4 of these so that no two of the halting stations are consecutive is _______
D) The greatest possible no of intersections of 8 straight lines and 4 circles is
a) 32 b) 64 c) 76 d) 104
E) The no of times digit 3 will be written when listing the integers from 1 to 1000 is
a) 269 b) 300 c) 271 d) 302
F) In a certain test, there are n questions. In this test, 2^(n-k) students gave wrong answers to atleast k questions, where k=1,2,3,...n.
If the total no of wrong answers is 2047, then n is equal to
a) 10 b) 11 c)12 d)13
6.N represents a series in which all the terms are
consecutiveintegers and the sum of all the terms of N is 100. If the number of N is greater than one, find the difference b/w the MAX n MIN possible numbers of terms of N.
a)20 b)30 c)45 d)125 e)195
7.How many natural numbers have their cube ending with 24.
a)20 b)18 c)16 d)22 e)11
How many integers between 6000 and 7000 have their thousands digits equal the sum of the other three digits?
A. 20 B. 22 C. 24 D. 26 E. 28
A. 20 B. 22 C. 24 D. 26 E. 28
If for real numbers a, b and c, a/b > 1 and a/c < -1, then which of the following must be correct?
A. a + b - c > 0 B. a > b C. (a - c)(b - c) > 0 D. a - b + c > 0 E. abc > 0
A. a + b - c > 0 B. a > b C. (a - c)(b - c) > 0 D. a - b + c > 0 E. abc > 0
For natural numbers a, b and c, 0 < a < b and the expression x(x- a)(x- b) - 17 is divisible by (x - c). What is the value of a + b + c ?
A. 14 B. 17 C. 21 D. 24 E. 27
A. 14 B. 17 C. 21 D. 24 E. 27
Two consecutive natural numbers have exactly four divisors and also have the same sum of divisors. What is the number of divisors of their product?
A. 6 B. 12 C. 16 D. 20 E. 24
A. 6 B. 12 C. 16 D. 20 E. 24
Let N denote the number obtained by listing all the two-digit numbers from 10 to 99 in order, i.e. N = 10111213 : : : 99. What is the remainder when N is divided by 99?
A. 0 B. 10 C. 45 D. 54 E. 90
A. 0 B. 10 C. 45 D. 54 E. 90
Let S(n) = n when n is a single natural number and S(n) = sum of the digits of n, when n > 10. Let N denote the smallest positive integer such that N + S(N) + S(S(N)) = 99. What is S(N)?
A. 9 B. 10 C. 12 D. 15 E. 18
A. 9 B. 10 C. 12 D. 15 E. 18
There are Red, Blue and Green balls in a bag. To ensure that you pick out at least one Red ball from the bag without seeing the color of the ball, you need to pick out 26 balls. To ensure the same for Blue and Green balls, you need to pick out 17 and 20 balls, respectively. How many balls you need to pick out to ensure that you have at least two balls of each color?
1. if N denotes the sum of squares of first m natural nos.. for how many values of m<55; N is multiple of 4
2. remiander when 2^2+22^2+222^2+.........+(22....49 twos)^2 is divided by 9
3. N = 202 x 20002 x 200000002 x 2..(15 zeros)..2 x 2 ...(31 zeros) ...2 sum of the digits in this multiplication is
4. product on n consecutive integers is x.
(x+1)(x+2)(x+3)..........(x+n-1) = 1000; then n = ?
5. find 28383rd term of the series 12345678910111213........
(i m getting the answer as 3)
6. what is the remainder when (1!)^3 + (2!)^3+ (3!)^3 +...........+ (1152!)^3 is divided by 1152
7. remainder when 9+9^2+9^3+.........9^(2n+1) is divided by 6.
answers to these r 1. 12; 2-6; 3- 160; 4- 20; 5-9; 6- 225; 7-3
The group of numbers starting from 1 and including 1, 2, 3, 4, 5, and so on. Zero, negative numbers, and decimals are not included this group.
EXAMPLE
1. If n is an odd natural number, what is the highest number that always divides n(n2 – 1)?
Answer: n∙(n2 – 1) = (n – 1)∙n∙(n + 1), which is a product of three consecutive numbers. Since n is odd, the numbers (n – 1) and (n + 1) are both even. One of these numbers will be a multiple of 2 and the other a multiple of 4 as they are two consecutive even numbers. Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three numbers will be a multiple of three. Hence, the product of three numbers will be a multiple of 8 ´ 3 = 24.
Hence, the highest number that always divides n∙(n2 – 1) is 24.
2. For every natural number n, the highest number that n∙(n2 – 1)∙(5n + 2) is always divisible by is
(a) 6 (b) 24 (c) 36 (d) 48
Answer:
Case 1: If n is odd, n∙(n2 – 1) is divisible by 24 as proved in the earlier question.
Case 2: If n is even, both (n – 1) and (n + 1) are odd. Since product of three consecutive natural numbers is always a multiple of 3 and n is even, the product n∙(n2 – 1) is divisible by 6. Since n is even 5n is even. If n is a multiple of 2, 5n is a multiple of 2 and hence 5n + 2 is a multiple of 4. If n is a multiple of 4, 5n + 2 is a multiple of 2. Hence, the product n∙(5n + 2) is a multiple of 8.
Hence, the product n∙(n2 – 1)∙(5n + 2) is a multiple of 24.
Hence, [b]
Rule: The product of n consecutive natural numbers is divisible by n!, where n! = 1 × 2 × 3 × 4 × 5…. × n
EXAMPLE
3. Prove that (2n)! is divisible by (n!)2.
Answer: (2n)! = 1·2·3·4· … ·(n – 1)·n·(n + 1)· …·2n
= (n)!·(n + 1)·(n + 2)· …·2n.
Since (n + 1)·(n + 2)· …·2n is a product of n consecutive numbers, it is divisible by n!. Hence, the product (n)!·(n + 1)·(n + 2)· …·2n is divisible by n!·n! = (n!)2.
Whole Numbers:
All Natural Numbers plus the number 0 are called as Whole Numbers.
Integers:
All Whole Numbers and their negatives are included in this group.
Rational Numbers:
Any number that can be expressed as a ratio of two integers is called a rational number.
This group contains decimal that either do not exist (as in 6 which is 6/1), or terminate (as in 3.4 which is 34/10), or repeat with a pattern (as in 2.333... which is 7/3).
Irrational Numbers:
Any number that can not be expressed as the ratio of two integers is called an irrational number (imaginary or complex numbers are not included in irrational numbers).
These numbers have decimals that never terminate and never repeat with a pattern.
Examples include pi, e, and √2. 2 + √3, 5 - √2 etc. are also irrational quantities called Surds.
EXAMPLE
Real Numbers:
This group is made up of all the Rational and Irrational Numbers. The ordinary number line encountered when studying algebra holds real numbers.
Imaginary Numbers:
These numbers are formed by the imaginary number i (i = √-1). Any real number times i is an imaginary number.
Examples include i, 3i, −9.3i, and (pi)i. Now i2 = −1, i3 = i2 × i = −i, i4 = 1.
EXAMPLE
Complex Numbers:
A Complex Numbers is a combination of a real number and an imaginary number in the form a + bi. a is called the real part and b is called the imaginary part.
Examples include 3 + 6i, 8 + (−5)i, (often written as 8 - 5i).
Prime Numbers:
All the numbers that have only two divisors, 1 and the number itself, are called prime numbers. Hence, a prime number can only be written as the product of 1 and itself. The numbers 2, 3, 5, 7, 11…37, etc. are prime numbers.
Note: 1 is not a prime number.
EXAMPLE
- If x2 – y2 = 101, find the value of x2 + y2, given that x and y are natural numbers.
Hence, x + y = 101 and x – y = 1. -> x = 51, y = 50.
-> x2 + y2 = 512 + 502 = 5101.
- What numbers have exactly three divisors?
Odd and Even Numbers:
All the numbers divisible by 2 are called even numbers whereas all the numbers not divisible by 2 are called odd numbers. 2, 4, 6, 8… etc. are even numbers and 1, 3, 5, 7.. etc. are odd numbers.
Here goes todays set:--
70
Marks: --/1
Choose one answer.
| a. 2 < x | ||
| b. 1 < x < 1 and x not equal to 0 | ||
| c. 1 < x < 2 | ||
| d. 0 < x and x less than or equal to 5/3 |
71
Marks: --/1
What is the sum of the two digits?
Choose one answer.
| a. 5 | ||
| b. 6 | ||
| c. 7 | ||
| d. 8 |
72
Marks: --/2
If 336 - 1 = 1A009463A296999120, where A is a single digit whole number, then the value of A is
Choose one answer.
| a. 5 | ||
| b. 2 | ||
| c. 1 | ||
| d. 8 |
73
Marks: --/1
The product P of three positive integers is 9 times their sum, and one of the integers is the sum of the other two. The sum of all possible values of P is
Choose one answer.
| a. 702 | ||
| b. 336 | ||
| c. 1404 | ||
| d. 540 |
74
Marks: --/1
What is the value of N such that N × [N] = 27, where [N] represents the greatest integer less than or equal to N?
Answer:
75
Marks: --/1
Let x and y be positive integers such that x is prime and y is composite. Then,
(CAT 2003)
(CAT 2003)
Choose one answer.
| a. xy cannot be an even integer. | ||
| b. (x + y)/x cannot be an even integer. | ||
| c. None of the other statements is true. | ||
| d. y – x cannot be an even integer. |
76
Marks: --/1
If a, a + 2, and a + 4 are prime numbers, then the number of possible solutions for a is
(CAT 2003)
(CAT 2003)
Choose one answer.
| a. more than 3 | ||
| b. 2 | ||
| c. 1 | ||
| d. 3 |
77
Marks: --/2
Five consecutive integers are chosen. Let S denote their sum and let P denote their product. If P/S is divisible by 100, what can be the smallest value of S?
Choose one answer.
| a. 125 | ||
| b. 115 | ||
| c. 625 | ||
| d. 615 |
78
Marks: --/1
(CAT 2006)
The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduate and above, to that of employees who are not, can possibly be
The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduate and above, to that of employees who are not, can possibly be
Choose one answer.
| a. 87:100 | ||
| b. 97:84 | ||
| c. 85:98 | ||
| d. 101:88 | ||
| e. 110:111 |
79
Marks: --/1
Which one of the following numbers is a perfect square?
Choose one answer.
| a. 35! x 36! | ||
| b. 34! x 37! | ||
| c. 36! x 37! | ||
| d. 37! x 38! |
80
Marks: --/1
For how many prime numbers p, is p4 + 15p2 – 1 also a prime number?
Choose one answer.
| a. 3 | ||
| b. 0 | ||
| c. 2 | ||
| d. 1 |
81
Marks: --/1
The remainder, when (1523 + 2323) is divided by 19, is
(CAT 2004)
(CAT 2004)
Choose one answer.
| a. 0 | ||
| b. 4 | ||
| c. 18 | ||
| d. 15 |
12 Marks: --/1The single digits a and b a re neither both nine nor both zero. The repeating decimal 0.abababab... <!--[if !vml]--><!--[endif]--> is expressed as a fraction in lowest terms. How many different denominators are possible?
Choose one answer. <TD text ">d. 4
The squares of the natural numbers are written in a straight line 149162536… to form a 200-digits number. What is the 100th digit from the left?
Choose one answer.
14 Marks: --/2A faulty odometer of a car always jumps from digit 4 to digit 6, always skipping the digit 5, regardless of the position. For example, after traveling for one kilometer the odometer reading changed from 000149 to 000160. If the odometer showed 000000 when the car was bought and now it shows 001000, how many kilometers has the car traveled?
Answer:
15 Marks: --/1For how many prime numbers p, is p4 + 15p2 – 1 also a prime number?
Choose one answer.
16 Marks: --/2Let [x] = greatest integer less than or equal to x. Let A = [2x], B = 2[x] and
C = [x + 1/2] + [x – 1/2]. Then
I. A, B and C can be equal for some value of x.
II. A, B and C can all take different values for some value of x.
Choose one answer.
17 Marks: --/1Let S = (3 + 32 + 33 + … + 3400) – (7 + 72 + 73 + … + 7201). The last two digits of S are
Choose one answer.
18 Marks: --/1In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?
Choose one answer.
19 Marks: --/1For how many integers S is
square of an integer?
Choose one answer.
20 Marks: --/1Let S = p2 + q2 + r2, where p and q are consecutive positive integers and
r = p × q. Then (S)1/2 is
Choose one answer.
Choose one answer. <TD text ">d. 4
Choose one answer.
14 Marks: --/2A faulty odometer of a car always jumps from digit 4 to digit 6, always skipping the digit 5, regardless of the position. For example, after traveling for one kilometer the odometer reading changed from 000149 to 000160. If the odometer showed 000000 when the car was bought and now it shows 001000, how many kilometers has the car traveled?
Answer:
15 Marks: --/1For how many prime numbers p, is p4 + 15p2 – 1 also a prime number?
Choose one answer.
16 Marks: --/2Let [x] = greatest integer less than or equal to x. Let A = [2x], B = 2[x] and
C = [x + 1/2] + [x – 1/2]. Then
I. A, B and C can be equal for some value of x.
II. A, B and C can all take different values for some value of x.
Choose one answer.
17 Marks: --/1Let S = (3 + 32 + 33 + … + 3400) – (7 + 72 + 73 + … + 7201). The last two digits of S are
Choose one answer.
18 Marks: --/1In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?
Choose one answer.
19 Marks: --/1For how many integers S is
square of an integer?Choose one answer.
20 Marks: --/1Let S = p2 + q2 + r2, where p and q are consecutive positive integers and
r = p × q. Then (S)1/2 is
Choose one answer.
1.Five consecutive integers are chosen. Let S denote their sum and let P denote their product. If P/S is divisible by 100, what can be the smallest value of S?
Choose one answer.
115,125,615,625
2.In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?
choose one ans 4,6,7,5 1. If a,b,c are in GP and m,n,o are in AP then (n-o)loga+(o-m)logb+(m-n)logc is let m-n=X then n-o=X and m-o = (m-n)+ (n-o) = 2X log aX b-2X cX = log ac2X b-2X = log b2X b-2X = log 1 = 0 2. a square whose side is 2m has its corners cut away so as to form an octagon with all sides equal.then the lengthof each side f octagon is if x is the measure of a side of a octagon then
x + x/sqrt(2) + x/sqrt(2) = 2 ==> x = 2/sqrt(2)+1
Not sure where is the vertex B. Q2. Given the sets of consecutive integers { 1 } , {2, 3}, {4, 5, 6}, . . , where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let Sn be the sum of the elements in the nth set. For example, S1 = 1 and S2 = 5. Then S21 equals Last term of S20 = sigma(20) = 210 and S21={211,212... 21 terms } 4641 ) A 3 digit number N is such that it is divisible by 11. Also, N/11 is equal to the sum of squares of three digits of N. N can take ......... value(s)
550 and 803
let XY be a two digit number. The problem resolves to the following two equation10X+Y = (X)2 + (X+Y)2 + Y2 (because XY * 11 = XY0 + XY) = 2X2+2Y2+2XY = 2(X2+Y2+XY) (put Y=0,2,4,6,8 because multiple of 2) for Y=0 we can solve the above equation. we get XY=50 and N=550 10X+Y = (X+1)2 + (X+Y)2 + Y2 (because XY * 11 = XY0 + XY) = 2X2+2Y2+2XY+2X+1 (put Y=1,3,5,7 because Y whould be odd). Solving this we need to get 803 but since X+Y is 0 I am not able to solve this. can you help us on this 2) Let N=1000!. N is divided successively by distinct numbers of the form X^x. Where x is a prime number. What is the maximum number of divisions that can be performed, before getting a non-zero remainder? 11 is the answer. 2,3,5,7,11,13,17,19,23,29,31. (total 11 terms). The question is same as the number of prime numbers for which its square is less than 1000 There are 100 children numbered 1 to 100. There are also boxes numbered 1 to 100. Each box contains 1000 toffees. The first child takes 1 toffee from each box numbered 1,2,3,....100. The second child takes 2 toffees from each box numbered 2,4,6,....100, the third child takes 3 toffees from boxes numbered 3,6,9,...99 and so on till the 100th child. Which among the following is / are false? 1. The box left with minimum number of toffees is box numbered 64. 2. The number of boxes from where exactly 2 children took toffees is 25. 3. The number of boxes from where the number of toffees were taken as the number on the box is 2. 1. 72 is the box which has minimum number of toffees. so 1 is false. The number with largest number of factors is 72 and not 64. 2. Total number of prime numbers between 1 to 100 is 25. 3. Obvious only 2 satisfies the condition. (b) 1 1> if a,b,c are in GP and m,n,o are in AP then (n-o)loga+(o-m)logb+(m-n)logc is? Ans: ologasoln :- we have given b2 = ac as a,b,c are in GP and 2n = m + o as m,n,o are in AP Now (n-o)loga + (o-m)logb + (m-n)logc can be written as (logan + logbo + logcm) - (logao + logbm + logcn) => log(an-o * bo-m * cm-n) =>log{an-o * (a*c)o-m/2 * cm-n} using b = (a*c)1/2 upon solving you will get log(ao * c0) => logao + logc0 => logao => ologa2> a square whose side is 2m has its corners cut away so as to form an octagon with all sides equal.then the lengthof each side f octagon is ? Ans: 2/sqrt(2) + 1 3>Given the sets of consecutive integers { 1 } , {2, 3}, {4, 5, 6}, . . , where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let Sn be the sum of the elements in the nth set. For example, S1 = 1 and S2 = 5. Then S21 equals Ans: 4641 soln :- You can easily see that the sum of the number of the element there are in a set is the last element of the set. For eg: set 1 contain one element i.e {1} set 2 contains two elements i.e {1,2} set 3 contains three elements i.e {1,2,3} set 4 contains four elements i.e {1,2,3,4} and so on S5 = 1 + 2 + 3 + 4 + 5 = 15 means 15 will be the last element of 5th set and 15 - 5 + 1 = 11 will be its first element i.e 5th set contains {11,12,13,14,15} similarly 21st set contains 21 element and 1 + 2 + 3 + 4 +......+ 21 will be its last element i.e 21 * 11 = 231 and 231 - 21 + 1 = 211 is the first element so sum is S21 = 211 + 212 + 213 +.......+ 231 = 21 * 221 = 4641 4>The probability that Nike will get Cricket team contract is 2/3 and the probability that Nike will get Hockey team contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability of getting both the contract? Ans : 19/45 soln : - we have given P(C) = 2/3 P(H) = 5/9 and P(C or H) = 4/5 We have a formula P(C and H) = P(C) + P(H) - P(C or H) = 2/3 + 5/9 - 4/5 = 19/45 5> A 3 digit number N is such that it is divisible by 11. Also, N/11 is equal to the sum of squares of three digits of N. N can take ......... value(s) For this question i am not getting the approach. I can only think of N = abc = 100a + 10b + c and as N is divisible by 11 so a + c = b and given is N/11 = a2 + b2 + c2 If any one of solve this problem, then do post your solution too. |
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