Math Arithmetic Problems With Solutions :
Here you can see some arithmetic problems. We have prepared these questions, inorder to help the students who are trying to get better score in competitive exams.
To find question 1 to 7, please visit "Arithmetic Problems With Solutions for Competitive Exams"
Question 8 :
Which of the following is a multiple of 10 ?
Solution :
Multiples of 10 ends with with 0.
Options B ends with 0, So 10,030 is a multiple of 10.
Question 9 :
Which of the following is a multiple of 5 and 2 ?
(A) 2, 203 (B) 2, 342 (C) 1, 005 (D) 7, 790 (E) 9, 821
Solution :
Multiple of 2 ends with one of the following numbers 0, 2, 4, 6 or 8.
Multiple of 5 ends with 5 or 0.
From the given numbers 7790 satisfies the above conditions. So 7790 is a multiple of 5 and 2.
Question 10 :
Which of the following is a multiple of both 3 and 10 ?
(A) 103 (B) 130 (C) 210 (D) 310 (E) 460
Solution :
In order to find which of the following number is a multiple of both 3 and 10, let us use divisibility test.
If the sum of the digits is a multiple of 3, then the given number is divisible by 3.
If the number ends with 0, then it is divisible by 10.
From the given options , option C satisfies both the conditions.
2 + 1 + 0 = 21 (divisible by 3 and ends with 0)
Hence 210 is divisible by both 3 and 10.
Question 11 :
Which of the following is a multiple of 2, 3 and 5 ?
(A) 165 (B) 235 (C) 350 (D) 420 (E) 532
Solution :
165 ==> 1 + 6 + 5 = 12 (divisible by 3), it is not even.
235 ==> 2 + 3 + 5 = 10 (not divisible by 3)
350 ==> 3 + 5 + 0 = 8 (not divisible by 3)
420 ==> 4 + 2 + 0 = 6(divisible by 3, ends with 0)
Hence 420 is divisible by 2, 3 and 5.
Question 12 :
Which of the following is an even multiple of both 3 and 5 ?
(A) 135 (B) 155 (C) 250 (D) 350 (E) 390
Solution :
390 ==> 3 + 9 + 0 = 12 (divisible by 3)
It ends with 0. So it is divisible by 5.
Question 13 :
Professor Jones bought a large carton of books. she gave 3 books to each student in her class and there were no books left over. Which of the following could be the number of books she distributed ?
(A) 133 (B) 143 (C) 525 (D) 271 (E) 332
Solution :
If the required number of books is the multiple of 3, we will not have any book left over.
525 ==> 5 + 2 + 5 = 12 (divisible by 3)
Hence 525 books could be distributed.
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