Reasoning Aptitude
ARITHMETIC REASONING MCQs
Arithmetical Reasoning
Total Questions : 278
| Page 25 of 28 pages
Answer: Option C. -> 45 years
Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.
Then, woman's age = (10X + y) years; husband's age = (10y + x) years.
Therefore (10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)
⇔ (9y-9x) = (1/11)(11y + 11x) = (x + y) ⇔ 10x = 8y ⇔ x = (4/5)y
Clearly, y should be a single-digit multiple of 5, which is 5.
So, x = 4, y = 5.
Hence, woman's age = 10x + y = 45 years.
Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.
Then, woman's age = (10X + y) years; husband's age = (10y + x) years.
Therefore (10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)
⇔ (9y-9x) = (1/11)(11y + 11x) = (x + y) ⇔ 10x = 8y ⇔ x = (4/5)y
Clearly, y should be a single-digit multiple of 5, which is 5.
So, x = 4, y = 5.
Hence, woman's age = 10x + y = 45 years.
Answer: Option D. -> None of these
Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.
Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.
Question 243. A girl counted in the following way on the fingers of her left hand : She started by calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5 and then reversed direction calling the ring finger 6, middle finger 7 and so on. She counted upto 1994. She ended counting on which finger ?
Answer: Option B. -> Index finger
Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....
i.e. numbers of the form (8n + 1).
Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.
Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....
i.e. numbers of the form (8n + 1).
Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.
Answer: Option D. -> 90
Let number of notes of each denomination be x.
Then, x + 5x + 10x = 480 ⇔ 16x = 480 ⇔ x = 30.
Hence, total number of notes = 3x = 90.
Let number of notes of each denomination be x.
Then, x + 5x + 10x = 480 ⇔ 16x = 480 ⇔ x = 30.
Hence, total number of notes = 3x = 90.
Answer: Option C. -> 6
Since B and D are twins, so B = D.
Now, A = B + 3 and A = C - 3.
Thus, B + 3 = C - 3 ⇔ D + 3 = C-3 ⇔ C - D = 6.
Since B and D are twins, so B = D.
Now, A = B + 3 and A = C - 3.
Thus, B + 3 = C - 3 ⇔ D + 3 = C-3 ⇔ C - D = 6.
Answer: Option C. -> 24 m
Each row contains 12 plants.
There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left.
Therefore Length = (22 + 2) m = 24 m.
Each row contains 12 plants.
There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left.
Therefore Length = (22 + 2) m = 24 m.
Answer: Option B. -> 29
Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.
Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.
Answer: Option D. -> 120 rolls
Number of cuts made to cut a roll into 10 pieces = 9.Therefore Required number of rolls = (45 x 24)/9 = 120.
Number of cuts made to cut a roll into 10 pieces = 9.Therefore Required number of rolls = (45 x 24)/9 = 120.
Answer: Option B. -> 16 years
Manick's present age = 12 years, Rahul's present age = 4 years.
Let Manick be twice as old as Rahul after x years from now.
Then, 12 + x = 2 (4 + x) ⇔ 12 + x = 8 + 2x ⇔ x = 4.
Hence, Manick's required age = 12 + x = 16 years.
Manick's present age = 12 years, Rahul's present age = 4 years.
Let Manick be twice as old as Rahul after x years from now.
Then, 12 + x = 2 (4 + x) ⇔ 12 + x = 8 + 2x ⇔ x = 4.
Hence, Manick's required age = 12 + x = 16 years.
Answer: Option B. -> 8 times
L.C.M. of 6, 5, 7, 10 and 12 is 420.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.
L.C.M. of 6, 5, 7, 10 and 12 is 420.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.
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