RE 4 - Arithmetic Revision Exercise | Arithmetic - Revision
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Answer the next 2 questions based on the information given below.
Everyday, Aakash's driver takes the car from Aakash's home at certain time, reaches his office exactly at 7 p.m., picks up Aakash and drives him home, travelling at a constant speed throughout. One day, Aakash left his office at 5 p.m. and started walking on the usual route towards his home. His driver picked him on the way and drove him home. Aakash noted that he reached home 40 minutes earlier than his usual time.
How long did Aakash walk?
- (a)
20 minutes
- (b)
40 minutes
- (c)
100 minutes
- (d)
80 minutes
Answer: Option C
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Explanation :
The driver saved 40 minutes while travelling both ways.
⇒ He saved 20 minutes one way.
∴ Driver must have picked Aakash 20 minutes before normal time i.e., at 6:40 pm.
⇒ Aakash walked from 5 pm till 6:40 pm.
∴ Aakash walked for 100 minutes.
Hence, option (c).
Workspace:
What was the speed with which Aakash walked?
- (a)
2 km/hr
- (b)
3 km/hr
- (c)
5 km/hr
- (d)
Data insufficient
Answer: Option D
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Explanation :
Since distance between home and office is not given, we cannot calculate the speed of Aakash.
Hence, option (d).
Workspace:
Abhinav accidently spilled some water into a can of juice. In order to compensate for the spilling, he decided to replace 100 ml of the mixture by 100 ml of pure juice such that the ratio of water to juice in the mixture becomes 9 : 25. If the can initially had 600 ml of pure juice, then how much water was spilled into the can?
- (a)
25 ml
- (b)
150 ml
- (c)
50 ml
- (d)
None of the above
Answer: Option D
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Explanation :
Let the amount of water that spilled into the can be x ml.
Then the total volume of the mixture = (600 + x)ml.
Total volume of juice in the can after 100 ml of mixture was taken out from the can
= 600 - × 100
Now, total volume of juice in the can after 100 ml of juice was poured into the can
= + 100
Also, total volume of water is the can in the final mixture
= x - × 100
∴ Ratio = =
Solving this we get, x = 200.
Note: We can also check for options here. None of the three options satisfies the above equation.
Hence, option (d).
Workspace:
A man can do a piece of work in 20 days. He starts the work on Monday, and on each successive, starting from 2nd day, a man, of half the efficiency of the man who started working on the previous day, joins the existing member(s). On which day in the week will the work be completed?
- (a)
Friday
- (b)
Thursday
- (c)
Saturday
- (d)
Wednesday
Answer: Option B
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Explanation :
Let the efficiency of man be x unit’s per day.
Day 1: Only 1 person is working
Work done = x
Day 2: 2 persons are working
Work done = x + =
Day 3: 3 persons are working
Work done = x + + =
...
Day n: n persons are working
Work done = = =
∴ Total work done in n days = + + + .... +
=
One man can complete the work in 20 days, i.e., total work done = x × 20 = 20x
⇒ x ≥ 20x
⇒ 2n - ≥ 20
⇒ 2n - ≥ 20
⇒ 2n + ≥ 22
⇒ n + ≥ 11
⇒ n ≥ 11
11th day from Monday will be Thursday.
Hence, option (b).
Workspace:
A and B started running simultaneously around a circular track from the same point in opposite direction. A and B take 28 mins and 35 mins to complete one round. After what time (approximately) will they be diametrically opposite to each other for the first time?
- (a)
80 min
- (b)
70 mins
- (c)
75 min
- (d)
None of these
Answer: Option B
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Explanation :
Let the length of the track be LCM (28, 35) = 140 meters.
⇒ Speed of A = 140/28 = 5 m/sec
⇒ Speed of B = 140/34 = 4 m/sec
Relative speed of A and B = 5 – 4 = 1 meter/min
Time taken to cover half the track = min = 70 minutes.
Hence, option (b).
Workspace:
There are fourteen positive integers less than 1000 out of which 2 have equal values, 3 others also have equal values, 4 others have equal values and the remaining 5 are distinct. If one of these integers is equal to the average of all these ten integers and is less than 50, then what can be the largest possible value of one of the ten integers?
Answer: 606
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Explanation :
Let the numbers be - a, a, b, b, b, c, c, c, c, d, e, f, g and h.
Since we need to maximise one of these numbers, we need to minimise others and maximise the average.
∴ Let us assume c = 1, b = 2, a = 3, d = 4, e = 5, f = 6, g = average of all the numbers and h is the highest number.
Since average is less than 50, highest value of g is 49.
∴ Sum of all 14 numbers = 14 × 49 = 686
⇒ 686 = 3 + 3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 4 + 5 + 6 + 49 + h
⇒ h = 606
Hence, 606.
Workspace:
Once upon a time, in a jungle, a rabbit and a tortoise planned a 5-kilometre. The rabbit dashed off and soon took a huge lead. He realized that even after taking a rest for ‘x’ mins, he would still beat the tortoise by 10 mins, he sat under a tree and went to sleep. Meanwhile the tortoise kept walking slowly and went past the sleeping rabbit. When the rabbit woke up, he realized that he had slept for (20 + x) mins, and immediately started running towards the target at a speed 5/3 times his original speed. The race eventually ended in a dead heat. If the ratio of the original speed of the rabbit to that of the tortoise was 5 : 1 and the rabbit overstretched his nap by 1 2/3 x mins, then how long did the rabbit take to complete the race?
- (a)
24 mins 30 seconds
- (b)
28 mins 30 seconds
- (c)
28 mins
- (d)
Cannot be determined
Answer: Option B
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Explanation :
The rabbit overstretched his sleeping time by 1x mins.
∴ x + 1x = 20 + x
⇒ x = 12 mins
Let the speed (in km/min) of the rabbit and tortoise be ‘5s’ and ‘s’ respectively.
Had the rabbit not overstretched his nap, he would have beaten the tortoise by 13 min.
∴ + 10 + 12 =
⇒ s =
Hence, the time taken by the tortoise to complete the race = = = 28 mins 30 seconds.
This is also the time taken by Rabbit to complete the race since the race ended in a dead heat.
Hence, option (b).
Workspace:
Aman, Baman and Chaman invested some amount in a partnership. The amounts invested by the three are in the ratio of a : b : c and the profits earned by them are in the ratio of c : b : a. Find the ratio of the time period for which the amount is invested by them respectively.
- (a)
a2 : ac : c2
- (b)
bc : ca : ab
- (c)
c2 : ac : a2
- (d)
a2 : b2 : c2
Answer: Option C
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Explanation :
Profit ∝ Investment × Time
⇒ PA : PB : PC = IA × TA : IB × TB : IC × TC
⇒ c : b : a = a × TA : b × TB : c × TC
⇒ TA : TB : TC = : 1 :
⇒ TA : TB : TC = c2 : ac : a2
Hence, option (c).
Workspace:
A, B and C entered a room and saw a bowl full of apples. Firstly, A ate 5/6th of all the apples and four more apples. Then, B ate 4/5th of the remaining apples and two more apples. Finally, C ate half of the remaining apples and one more apple. To their surprise, one apple was still left in the bowl. Find the difference between the number of apples eaten by A and that eaten by B and C together.
Answer: 125
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Explanation :
We can solve this question with reverse approach.
Let the number of apples before C ate is x.
C ate half of the apples hence number of apples left is x/2.
Now, C ate one more apple, hence number of apples left is x/2 – 1 =1
∴ Number of apples left before C ate = (1 + 1) × 2/1 = 4
C ate half of this + 1 apples i.e., 3 apples.
Similarly, number of apples left before B ate = (4 + 2) × 5/1 = 30
B ate 4/5th of this + 2 apples i.e., 26 apples.
Similarly, number of apples left before A ate = (26 + 4) × 6/1 = 180
A ate 5/6th of this + 4 apples i.e., 154 apples.
Number of apples eaten by A = 154
Number of apples eaten by B and C = 26 + 3 = 29
∴ Required difference = 154 – 29 = 125
Hence, 125.
Workspace:
Vidhan has a bottle full of milk. He pours half the content of the bottle into an empty can, then fills the bottle completely with water and mixes thoroughly. Then he repeats this process for 48 times more. Afterward, he pours the whole content of the bottle into the can. If the capacity of the bottle is 10 liters, then what is the ratio of the volume of the mango juice to the volume of water in the can?
- (a)
2 : 49
- (b)
1 : 25
- (c)
1 : 24
- (d)
None of these
Answer: Option A
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Explanation :
Since finally the whole content of the bottle is transferred into the can, total milk transferred in the can = amount of milk present initially in the bottle = 10 litres.
Amount of water in the can = amount of water added in the bottle.
Now, Vidhan adds 5 litres of water a total of 49 times. Hence, amount of water added in the bottle = 5 × 49 = 145 litres.
∴ Amount of water in the can = 245 litres.
⇒ Ratio of amount of milk and water in the can = 10 : 245 = 2 : 49.
Hence, option (a).
Workspace:
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