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Question 1 of 5
1. Question
Two men set out at the same time to walk towards each other from points X and Y; 108 km apart. The first man walks at the speed of 6 kmph while the second one walks 4km in the first hour, 4.5km in second hour, 5 km in the third hour, and so on. Two men will meet in how many hours?
Correct
Solution: c) 9
Explanation:
Total distance covered by both in first hour = 6 + 4 = 10km
Total distance covered by both in second hour = 6 + 4.5 = 10.5 km
Total distance covered by both in third hour = 6 + 5 = 11km
By observing above calculations we can conclude that the distance covered by the two is in Series with difference of 0.5km.
To get the sum of the series as 108km, we have to apply the concept of sum of series.
i.e. 108 = (n/2) × (2a+(n-1)d) (d = 0.5, a = 10, n = hours)
108 = (n/2) × (2×10 + (n – 1) × 0.5)
108 = (n/2) × (20 + (n – 1) × 0.5)
108 × 2 = n (20 + (n-1)/2)
216 = (20n + n (n-1)/2)
216 – 20n = (n2 – n)/2
432 – 40n = n2 – n
n2 + 39n – 432 = 0
n2 + 48n – 9n – 432 = 0
n (n + 48) – 9 (n + 48) = 0
(n – 9) (n + 48) = 0
n – 9 = 0 or n + 48 = 0
n = 9
Incorrect
Solution: c) 9
Explanation:
Total distance covered by both in first hour = 6 + 4 = 10km
Total distance covered by both in second hour = 6 + 4.5 = 10.5 km
Total distance covered by both in third hour = 6 + 5 = 11km
By observing above calculations we can conclude that the distance covered by the two is in Series with difference of 0.5km.
To get the sum of the series as 108km, we have to apply the concept of sum of series.
i.e. 108 = (n/2) × (2a+(n-1)d) (d = 0.5, a = 10, n = hours)
108 = (n/2) × (2×10 + (n – 1) × 0.5)
108 = (n/2) × (20 + (n – 1) × 0.5)
108 × 2 = n (20 + (n-1)/2)
216 = (20n + n (n-1)/2)
216 – 20n = (n2 – n)/2
432 – 40n = n2 – n
n2 + 39n – 432 = 0
n2 + 48n – 9n – 432 = 0
n (n + 48) – 9 (n + 48) = 0
(n – 9) (n + 48) = 0
n – 9 = 0 or n + 48 = 0
n = 9
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Question 2 of 5
2. Question
Ajay decides to sell milk at cost price. In what ratio should he mix water and milk so that he can earn 20% profit?
Correct
Solution: C) 1:5
Let cost price of pure milk = Rs 1 per litre.
Now he adds some water in this milk.
Then selling price of mixture of milk and water is also = Rs 1 per litre
Suppose he had only 1 litre milk costing Rs 1.
To earn profit of 20 % he has to earn Rs 1.2
As mixture also costs Rs 1 per litre means to earn Rs 1.2 he has sold total 1.2 litre of mixture .
That means he added 0.2 litre of milk in 1 litre.
Thus, ratio is 0.2: 1 = 1:5 .Hence, option (c) is correct.
Incorrect
Solution: C) 1:5
Let cost price of pure milk = Rs 1 per litre.
Now he adds some water in this milk.
Then selling price of mixture of milk and water is also = Rs 1 per litre
Suppose he had only 1 litre milk costing Rs 1.
To earn profit of 20 % he has to earn Rs 1.2
As mixture also costs Rs 1 per litre means to earn Rs 1.2 he has sold total 1.2 litre of mixture .
That means he added 0.2 litre of milk in 1 litre.
Thus, ratio is 0.2: 1 = 1:5 .Hence, option (c) is correct.
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Question 3 of 5
3. Question
Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again?
Correct
Solution: B) 4 hours 48 minutes
L.C.M of 18, 24 and 32 = 288
Hence, they would chime after every 288 min. or 4 hrs 48 min.Therefore, option (b) is correct.
Incorrect
Solution: B) 4 hours 48 minutes
L.C.M of 18, 24 and 32 = 288
Hence, they would chime after every 288 min. or 4 hrs 48 min.Therefore, option (b) is correct.
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Question 4 of 5
4. Question
A report consists of 10 sheets each of 110 lines and each such line consists of 65 characters. This report is retyped into sheets each of 65 lines such that each line consists of 22 characters. The approximate percentage change in number of sheets are
Correct
Solution: C) 400
Total volume of first report = (110 × 10 × 65) = 71500
Let n be the number of pages.
Therefore, the new volume will be (65 × 22 × n) = 1430n
As the volume remains same, we can say that 71500 = 1430n
è n = 50 sheets.
Clearly, the sheets increased by (50 –10 = 40)Therefore, percentage increase in number of sheets = (40/10) * 100 = 400 %
Hence, option (c) is correct.
Incorrect
Solution: C) 400
Total volume of first report = (110 × 10 × 65) = 71500
Let n be the number of pages.
Therefore, the new volume will be (65 × 22 × n) = 1430n
As the volume remains same, we can say that 71500 = 1430n
è n = 50 sheets.
Clearly, the sheets increased by (50 –10 = 40)Therefore, percentage increase in number of sheets = (40/10) * 100 = 400 %
Hence, option (c) is correct.
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Question 5 of 5
5. Question
Carpenter A can make a chair in 8 hours, carpenter B in 7 hours and carpenter C in 6 hours. If each carpenter works for 8 hours per day, how many chairs will be made in 21 days?
Correct
Solution: A) 73
It is given that:
Carpenter A can make a chair in 8 hours.
Carpenter B can make a chair in 7 hours.
Carpenter C can make a chair in 6 hours
Now, if each carpenter works for 8 hours per day for 21 days, then total number of hours worked is 21 * 8 or 168 hours by each carpenter.
Now , in 168 hours Carpenter A can make (168/8) chairs = 21 chairs.
In 168 hours Carpenter B can make (168/7) chairs = 24 chairs.
In 168 hours Carpenter C can make (168/6) chairs = 28 chairs.
Therefore, total number of chairs done together = 21+24+28 = 73 chairs
Hence, option (a) is correct.
Incorrect
Solution: A) 73
It is given that:
Carpenter A can make a chair in 8 hours.
Carpenter B can make a chair in 7 hours.
Carpenter C can make a chair in 6 hours
Now, if each carpenter works for 8 hours per day for 21 days, then total number of hours worked is 21 * 8 or 168 hours by each carpenter.
Now , in 168 hours Carpenter A can make (168/8) chairs = 21 chairs.
In 168 hours Carpenter B can make (168/7) chairs = 24 chairs.
In 168 hours Carpenter C can make (168/6) chairs = 28 chairs.
Therefore, total number of chairs done together = 21+24+28 = 73 chairs
Hence, option (a) is correct.
