MATH ARITHMETIC PROBLEMS WITH SOLUTIONS

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.

Divisibility rule of 3 :

If the sum of the digits is a multiple of 3, then the given number is divisible by 3. 

Divisibility rule of 10 :

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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• Arithmetic Quiz Questions

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