ARITHMETIC PROBLEMS WITH SOLUTIONS FOR COMPETITIVE EXAMS
About "Arithmetic Problems With Solutions for Competitive Exams"
Arithmetic Problems With Solutions for Competitive Exams :
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.
Arithmetic Problems With Solutions for Competitive Exams
Question 1 :
Which of the following is not even ?
(A) 330 (B) 436 (C) 752 (D) 861 (E) 974
Solution :
If the given number is even, then it must ends with one of the following numbers 0, 2, 4, 6 or 8.
In the above numbers 861 ends with 1, which is not even. Hence 861 is not even number.
Question 2 :
What is the least prime number greater than 50 ?
(A) 51 (B) 53 (C) 55 (D) 57 (E) 59
Solution :
A number which is divisible by 1 and itself is known as prime numbers.
51 is divisible by 1, 51, 3, 7. So it not prime.
53 is divisible by 1 and 53. There is no divisor other than 1 and 53. Hence it is prime.
Question 3 :
Which of the following is a multiple of 2 ?
(A) 271 (B) 357 (C) 463 (D) 599 (E) 756
Solution :
Multiple of 2 is also even number. Since 756 is even number, it is multiple by 2.
Question 4 :
(15 ⋅ 7 ⋅ 3) / (9 ⋅ 5 ⋅ 2) =
(A) 2/7 (B) 3/5 (C) 3 1/2 (D) 7 (E) 7 1/2
Solution :
= (15 ⋅ 7 ⋅ 3) / (9 ⋅ 5 ⋅ 2)
= 7/2
By converting the improper fraction as mixed fraction, we get 3 1/2.
Question 5 :
What is the lest common multiple of 18 and 24 ?
(A) 6 (B) 54 (C) 72 (D) 96 (E) 432
Solution :
To find L.C.M of 18 and 24, let us use "ladder method"
LCM of 18 and 24 = 3 ⋅ 2 ⋅ 3 ⋅ 4
= 72
Question 6 :
Which of the following is a multiple of 3 ?
(A) 115 (B) 370 (C) 465 (D) 589 (E) 890
Solution :
In order to find which of the following number is a multiple of 3, let us use divisibility test.
If the sum of the digits is a multiple of 3, then the given number is divisible by 3.
115 = 1 + 1 + 5 = 7 (not divisible by 3)
370 = 3 + 7 + 0 = 10 (not divisible by 3)
465 = 4 + 6 + 5 = 15 (divisible by 3)
Hence 465 is a multiple of 3.
Question 7 :
-6 (3 - 4 ⋅ 3) =
(A) -66 (B) -54 (C) -12 (D) 18 (E) 54
Solution :
According to order of operation, first we have to consider the parenthesis. (BODMAS)
= -6 (3 - 4 ⋅ 3) (performing multiplication)
= -6(3 - 12)
= -6 (-9)
= 54
Hence the answer is 54.
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OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
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Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
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