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Number System - Aptitude MCQ Questions and Solutions with Explanations

Number System MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Number System with easy and logical explanations. Number System MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-1 section-2

Each section contains 100 MCQs on Number System .
1
The greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?
Discuss
Answer & Solution
Answer: Option A
Solution:
Required Number can be given by:
HCF of (3026 - 11) and (5053 - 13)
HCF of 3015 and 5040 = 15
To find HCF, we break the given numbers in their prime Factors
3015 = 3 × 3 × 5 × 67
5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7
We take common multiples in these two given numbers to get required HCF
And Common multiples are: 3 × 5
So, Required HCF = 15
2
The difference between three times and seven times of a number comes to 36. What is the number?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let number be X
According to question,
7X - 3X = 36
4X = 36
X = 9
Required Number is 9
3
Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let 'X' be the smallest initial number that results in a win for Amitabh. The sum of the digit of 'X' is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Assume x is the required number.
Step 1 Amitabh 2x Sashi 2x + 50
Step 2 Amitabh 4x + 100. Sashi 4x + 150
Step 3 Amitabh 8x + 300. Sashi 8x + 350
Step 4 Amitabh 16x + 700 Sashi 16x + 750
Now, 16x + 750 > 1000
And, x can be 16
So 7 is the sum
4
Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student can get an A grade in the course if the average of her scores is more than or equal to 90. Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for his last quiz and he realizes that he must score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. how many quizzes did Prof. Suman take?
Discuss
Answer & Solution
Answer: Option D
Solution:
The difference between just getting an A grade and just getting a B grade would be equal to (number of quizzes × 3).
The difference between the required score to manage an A grade and the score achieved to just manage a B grade as given in the problem’s information is equal to 97 – 70 = 27
Thus, number of quizzes × 3 = 27
Hence, number of quizzes is 9
5
A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let he has solved correctly X no. of sums. Therefore incorrect no. of sums = 2X
Now,
X + 2X = 48
3X = 48
X = 16 sums he has done correctly.
6
Three-fifth of a number is equal to 70% of another number. What is the ratio between the first number and second number ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{two}}\,{\text{numbers}}\,{\text{be}}\,X\,{\text{and}}\,Y \cr & {\text{According}}\,{\text{to}}\,{\text{question}}, \cr & X \times {\frac{3}{5}} = 70\% \,of\,Y \cr & \frac{{3X}}{5} = \frac{{70Y}}{{100}} \cr & \frac{X}{Y} = \frac{{ {70 \times 5} }}{{ {100 \times 3} }} = \frac{7}{6} \cr & X:Y = 7:6 \cr} $$
7
A virus known to cause a deadly disease, is also a rapid multiplier. It is known that it can double itself every 30 minutes. If a container is completely full of this virus after 50 hours, when was the container half empty?
Discuss
Answer & Solution
Answer: Option B
Solution:
In every 30 minutes, Virus doubles it self. It means in 49.5 hours, it would be half empty and it will be full in 50 hours.
8
The LCM of two number is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, the other number is:
Discuss
Answer & Solution
Answer: Option C
Solution:
LCM = 45 HCF
LCM + HCF = 1150
45 × HCF + HCF = 1150
46 × HCF = 1150
HCF = 25
Then,
LCM = 45 × 25 = 1125
Now, use the formula,
1st number × second Number = LCM × HCF
125 × second number = 1125 × 25
Second Number = 225
9
The smallest number of four digits which on division by 4, 6, 10 and 15 leaves a remainder 2 in each case is:
Discuss
Answer & Solution
Answer: Option B
Solution:
First of all,we find the LCM of 4, 6, 10 and 15
LCM of 4, 6, 10, 10 = 60
Now, The smallest four digit no. is 1000. We divide it by 60
$$\frac{{1000}}{{60}}$$ It leaves remainder 40
Now, the smallest four digit no which is divisible by 4, 6, 10, 15 is,
1000 + (60 - 40) = 1020
So, required number (as it gives remainder 2 always) would be = 1020 + 2 = 1022
10
One-third of a tower is painted black, $$\frac{5}{{11}}$$th of the remaining part is painted red and the rest is painted white. If the white part measures 60 ft.,the total height of the tower is-
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the tower height will be x ft.
∴ Black part will be = $$\frac{1}{3}x$$
Remaining part = $$x - \frac{1}{3}x = \frac{2}{3}x$$
∴ Red part will be = $$\frac{2}{3}x \times \frac{5}{{11}} = \frac{{10}}{{33}}x$$
White part will be = $$\frac{2}{3}x - \frac{{10}}{{33}}x = \frac{{12x}}{{33}}$$
i.e. $$\frac{{12x}}{{33}}$$ = 60
x = 165 ft.