Reasoning Aptitude
ARITHMETIC REASONING MCQs
Arithmetical Reasoning
Total Questions : 278
| Page 22 of 28 pages
Answer: Option C. -> 20
Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.
Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.
Answer: Option B. -> 15
Let son's age be x years. Then, father's age = (3x) years.
Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.
So, 3x - 5 = 4 (x - 5) ⇔ 3x - 5 = 4x - 20 ⇔ x = 15.
Let son's age be x years. Then, father's age = (3x) years.
Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.
So, 3x - 5 = 4 (x - 5) ⇔ 3x - 5 = 4x - 20 ⇔ x = 15.
Answer: Option D. -> 5/9
Answer: Option D. -> 100 days
Answer: Option C. -> 42
Let number of girls = x and number of boys = 3x.
Then, 3x + x = 4x = total number of students.
Thus, to find exact value of x, the total number of students must be divisible by 4.
Let number of girls = x and number of boys = 3x.
Then, 3x + x = 4x = total number of students.
Thus, to find exact value of x, the total number of students must be divisible by 4.
Answer: Option C. -> 10
Answer: Option C. -> 9
'All but nine died' means 'All except nine died' i.e. 9 sheep remained alive.
'All but nine died' means 'All except nine died' i.e. 9 sheep remained alive.
Answer: Option D. -> None of these
Let R, G and B represent the number of balls in red, green and blue boxes respectively.
Then, .
R + G + B = 108 ...(i),
G + R = 2B ...(ii)
B = 2R ...(iii)
From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.
Putting G = 3R and B = 2R in (i), we get:
R + 3R + 2R = 108 ⇔ 6R = 108 ⇔ R = 18.
Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.
Let R, G and B represent the number of balls in red, green and blue boxes respectively.
Then, .
R + G + B = 108 ...(i),
G + R = 2B ...(ii)
B = 2R ...(iii)
From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.
Putting G = 3R and B = 2R in (i), we get:
R + 3R + 2R = 108 ⇔ 6R = 108 ⇔ R = 18.
Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.
Answer: Option D. -> 20
Total runs scored = (36 x 5) = 180.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.
Total runs scored = (36 x 5) = 180.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.
Answer: Option B. -> 16
Suppose the boy got x sums right and 2x sums wrong.
Then, x + 2x = 48 ⇔ 3x = 48 ⇔ x = 16.
Suppose the boy got x sums right and 2x sums wrong.
Then, x + 2x = 48 ⇔ 3x = 48 ⇔ x = 16.
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