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1. If a number is divisible by 102 then it is also divisible by which of the following numbers ?
A) 2 B) 3 C) 17 D) All
Solution: 102/2= 51,51/3=17 :. 102= 2x3x17
:. 102 is divisible by 2, 3& 17.
Ans: (D)
2. Which of the following is divisible by 2 and 7?
A) 365 B) 461 C) 524 D) 385 E) 350
Ans: (E)
3. If a certain integer x is divided by 4, the remainder is 3, then all of the following could be an integer except---
A) x/11 B) x/23 C) x/8 D) None
Solution: Divisible by 4 are 4,8,12,16, 20, 24. The integer could be 7,11, 15, 19, 23, 27.
:. x/11= 11/11= 1 integer
x/23= 23/23= 1 integer
But x is not divisible by 8.
:. x/8 is not an integer.
4. A store raised the price of an item by exactly 10 percent. which of the following could not be the resulting price of the item in dollar?
A) 5.50 B) 7.60 C) 11.00 D) 12.10 E) 75.90
Solution: Let, x be the original price. Hence price after 10% rise is x+(10/100)x = 1.1x.
Hence the price must be a multiple of 1.1. We see that 7.6 is not divisible by 1.1. So, This cannot be the price.
Ans: (B)
5. The 180 students in a group are to be seated in rows, so that there are an equal number of students in each row. Each of the following could be the number of rows except-
A) 4 B) 20 C) 30 D) 40 E) 90
Solution: Since each row has to have equal number of students, the number of rows must divide 180 without ay remainders. Hence, 40 cannot be the number. Hence, 40 cannot be the number of rows,
180/40= 4(20/40) i.e 20 is the reminder.
6. The number 10 30 divisible by all of the following except :
A) 250 B) 125 C) 32 D) 16 E) 6
Solution: to find if 10 30 is divisible , we will first express it as 2 30 x 5 30
250 = 2x 5 3 ; So, ( 2 30 x 5 30 )/(2x 5 3) = 2 29 x 5 27
:. 10 30 is divisible by 250
125 = 5 3 so it can divide 10 30 ( 2 30 x 5 30 )
32 = 2 5 " " " 10 30 ( 2 30 x 5 30 )
16 = 2 4 " " " 10 30 ( 2 30 x 5 30 )
6=2x3; Since there is no power of '3' in 10 30 , it cannot be divided by 6= (2x3)
Ans: (E)
7. What is the total number of integers between 100 & 200 that are divisible by 3 but not by 5 or 7?
A) 23 B) 22 C) 21 D) 20 E) None
Ans: The first number greater than 100 & divisible by 3 is 102 . And the last number smaller than 200 & divisible by 3 is 198. He3nce the total number of integers between 100 & 200 which are divisible by 3 is ((198-102)/3)+1 = 96/3+1=33
From these 33 numbers we will have to omit those that are divisible by 5 or 7. So, we try to find out numbers which are divisible by 3 & 5. i.e. 15 and by 3&7 i.e. 21.
1 st number after 100, divisible by 15 is 105. And the last number before 200 divisible by 15 is 195.
:. Number of integers within this range which are divisible by 15
= ((195-105)/15)+ 1= (90/15)+1 = 7
Similarly, number of integers within 100 & 200 which are divisible by 21
=((189-105)/21)+1 = 5
Please note that the number 105 is divisible by both 15 & 21. So,it has been included twice in the count. To make correction, we will add a 1 to the result.
Desired number = 33-7-5+1
=33-12+1=33-11
= 22
Ans: (B)
8. The 356 members of metro-chamber of commerce will vote to choose a president. With 5 candidates seeking office, what is the least number of votes a successful candidate could receive and yet have more votes than any other candidate?
A) 71 B) 72 C) 89 D) 178 E) 179
Solved: For the successful candidate to get the minimum amount to win, the others should get the maximum possible. So, first we evenly distribute the votes among all the candidates.
356/5= 71(1/5). So, 1 vote remains.
The winner must have got this extra vote.
So, the winner got 71+1=72 votes.
Ans: (A)
A) 2 B) 3 C) 17 D) All
:. 102 is divisible by 2, 3& 17.
Ans: (D)
2. Which of the following is divisible by 2 and 7?
A) 365 B) 461 C) 524 D) 385 E) 350
Ans: (E)
3. If a certain integer x is divided by 4, the remainder is 3, then all of the following could be an integer except---
A) x/11 B) x/23 C) x/8 D) None
Solution: Divisible by 4 are 4,8,12,16, 20, 24. The integer could be 7,11, 15, 19, 23, 27.
:. x/11= 11/11= 1 integer
x/23= 23/23= 1 integer
But x is not divisible by 8.
:. x/8 is not an integer.
4. A store raised the price of an item by exactly 10 percent. which of the following could not be the resulting price of the item in dollar?
A) 5.50 B) 7.60 C) 11.00 D) 12.10 E) 75.90
Solution: Let, x be the original price. Hence price after 10% rise is x+(10/100)x = 1.1x.
Hence the price must be a multiple of 1.1. We see that 7.6 is not divisible by 1.1. So, This cannot be the price.
Ans: (B)
5. The 180 students in a group are to be seated in rows, so that there are an equal number of students in each row. Each of the following could be the number of rows except-
A) 4 B) 20 C) 30 D) 40 E) 90
Solution: Since each row has to have equal number of students, the number of rows must divide 180 without ay remainders. Hence, 40 cannot be the number. Hence, 40 cannot be the number of rows,
180/40= 4(20/40) i.e 20 is the reminder.
6. The number 10 30 divisible by all of the following except :
A) 250 B) 125 C) 32 D) 16 E) 6
Solution: to find if 10 30 is divisible , we will first express it as 2 30 x 5 30
250 = 2x 5 3 ; So, ( 2 30 x 5 30 )/(2x 5 3) = 2 29 x 5 27
:. 10 30 is divisible by 250
125 = 5 3 so it can divide 10 30 ( 2 30 x 5 30 )
32 = 2 5 " " " 10 30 ( 2 30 x 5 30 )
16 = 2 4 " " " 10 30 ( 2 30 x 5 30 )
6=2x3; Since there is no power of '3' in 10 30 , it cannot be divided by 6= (2x3)
Ans: (E)
7. What is the total number of integers between 100 & 200 that are divisible by 3 but not by 5 or 7?
A) 23 B) 22 C) 21 D) 20 E) None
Ans: The first number greater than 100 & divisible by 3 is 102 . And the last number smaller than 200 & divisible by 3 is 198. He3nce the total number of integers between 100 & 200 which are divisible by 3 is ((198-102)/3)+1 = 96/3+1=33
From these 33 numbers we will have to omit those that are divisible by 5 or 7. So, we try to find out numbers which are divisible by 3 & 5. i.e. 15 and by 3&7 i.e. 21.
1 st number after 100, divisible by 15 is 105. And the last number before 200 divisible by 15 is 195.
:. Number of integers within this range which are divisible by 15
= ((195-105)/15)+ 1= (90/15)+1 = 7
Similarly, number of integers within 100 & 200 which are divisible by 21
=((189-105)/21)+1 = 5
Please note that the number 105 is divisible by both 15 & 21. So,it has been included twice in the count. To make correction, we will add a 1 to the result.
Desired number = 33-7-5+1
=33-12+1=33-11
= 22
Ans: (B)
8. The 356 members of metro-chamber of commerce will vote to choose a president. With 5 candidates seeking office, what is the least number of votes a successful candidate could receive and yet have more votes than any other candidate?
A) 71 B) 72 C) 89 D) 178 E) 179
Solved: For the successful candidate to get the minimum amount to win, the others should get the maximum possible. So, first we evenly distribute the votes among all the candidates.
356/5= 71(1/5). So, 1 vote remains.
The winner must have got this extra vote.
So, the winner got 71+1=72 votes.
Ans: (A)
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