Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
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National School of Drama has appointed 3 instructors for music, dance, and painting. Music instructor takes session from 12 noon to 4:00 pm on Monday, Thursday and Sunday. The sessions of dance instructor are scheduled on Tuesday, Thursday, Wednesday and Sunday between 10:00 am to 2:00 pm. The 9:00 am to 12:00 noon slot on Tuesday, Friday and Thursday and also 2:00 pm to 4:00 pm slot on Wednesday, Saturday and Sunday is filled up by Painting Instructor. On which day(s) of a week the dance and painting sessions are simultaneously held?
Solution (c)
According to the given information we can classify the working of instructors as explained below
Music Instructor works from 12-4 pm on Monday, Thursday and Sunday.
Dance Instructor works from 10 am-2 pm on Tuesday, Wednesday, Thursday and Sunday
Painting Instructor works from – 9 am – 12 noon on Tuesday, Thursday and Friday and also, 2 – 4 pm on Wednesday, Saturday and Sunday
After this classification, it is clear that Tuesday and Thursday are the days of the week the dance and painting sessions are simultaneously held.
Solution (c)
According to the given information we can classify the working of instructors as explained below
Music Instructor works from 12-4 pm on Monday, Thursday and Sunday.
Dance Instructor works from 10 am-2 pm on Tuesday, Wednesday, Thursday and Sunday
Painting Instructor works from – 9 am – 12 noon on Tuesday, Thursday and Friday and also, 2 – 4 pm on Wednesday, Saturday and Sunday
After this classification, it is clear that Tuesday and Thursday are the days of the week the dance and painting sessions are simultaneously held.
Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hours, 10 hours and 15 hours respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at 3/4th of their normal rates for the first hour after which they supplied water at the normal rate. Pipe C supplied water at 2/3rd of its normal rate for first 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled.
Solution (d)
Let the capacity of the tank be 60 litres.
Capacity of the first pipe = 12 l/hr
Capacity of the second pipe = 6 l/hr
Capacity of the third pipe = 4 l/hr
In 2 hour first pipe fills (9 + 12) l = 21 l
In 2 hour second pipe fills (4.5 + 6) = 10.5 l
In 2 hour third pipe fills (16/3) l
In 2 hour tank filled = (21 + 10.5 + 5.33) l = 36.83 l
Tank left to be filled = (60 – 36.83) l = 23.17 l
Time required = (23.17/22) hr = 1.05 hour
Total time = 3.05 hour
Hence, option d is the correct answer.
Solution (d)
Let the capacity of the tank be 60 litres.
Capacity of the first pipe = 12 l/hr
Capacity of the second pipe = 6 l/hr
Capacity of the third pipe = 4 l/hr
In 2 hour first pipe fills (9 + 12) l = 21 l
In 2 hour second pipe fills (4.5 + 6) = 10.5 l
In 2 hour third pipe fills (16/3) l
In 2 hour tank filled = (21 + 10.5 + 5.33) l = 36.83 l
Tank left to be filled = (60 – 36.83) l = 23.17 l
Time required = (23.17/22) hr = 1.05 hour
Total time = 3.05 hour
Hence, option d is the correct answer.
In one of the Taj hotels, the rooms of a row are numbered consecutively from 1 to 49. Assuming that there is a value of ‘x’ such that the sum of the numbers of the rooms preceding the room numbered ‘x’ is equal to the sum of the numbers of the rooms following it. Then what will be the value of ‘x’?
Solution (c)
It is given that sum of the first (x-1) numbers is equal to sum of the numbers from (x+ 1) to 49
Or we can say that the sum of (x-1) numbers = sum of first 49 numbers – sum of first x numbers
We know that the sum of first n consecutive numbers is given by n(n+1)/2
Likewise, x(x-1)/2 = 49*50/2 – x*(x+1)/2
On solving, we get = 35
Hence, option c is correct.
Solution (c)
It is given that sum of the first (x-1) numbers is equal to sum of the numbers from (x+ 1) to 49
Or we can say that the sum of (x-1) numbers = sum of first 49 numbers – sum of first x numbers
We know that the sum of first n consecutive numbers is given by n(n+1)/2
Likewise, x(x-1)/2 = 49*50/2 – x*(x+1)/2
On solving, we get = 35
Hence, option c is correct.
In order to start a new enterprise, Mr. Vijay has borrowed a total of Rs 60,000 from two money lenders with the interest being compounded annually, to be repaid at the end of two years. He repaid Rs 38,800 more to the first money lender compared to the second money lender at the end of two years. The first money lender charged an interest rate, which was 10% more than what was charged by the second money lender. Had Mr. Vijay instead borrowed Rs 30,000 from each at their respective initial rates for two years, he would have paid Rs 7,500 more to the first money lender compared to the second. Then money borrowed by him from first money lender is
Solution (c)
Let the interest on the second part be r %
Then, the rate on the first part = (r + 10) %
We know that for compound interest, A = P(1+r/100)2
It is given that, 30,000[1 + (r+10)/100]2 – 30,000[1 + r/100]2 = 7,500
On solving, we get = 20%
Let the first part be ‘a’ Rs
Then, the second part = Rs (60000 – a)
a[1 + (20+10)/100]2 – (60,000 – a)[1 + 20/100]2 = 38,800
On solving, we get a = Rs 40,000
Hence, option c is correct.
Solution (c)
Let the interest on the second part be r %
Then, the rate on the first part = (r + 10) %
We know that for compound interest, A = P(1+r/100)2
It is given that, 30,000[1 + (r+10)/100]2 – 30,000[1 + r/100]2 = 7,500
On solving, we get = 20%
Let the first part be ‘a’ Rs
Then, the second part = Rs (60000 – a)
a[1 + (20+10)/100]2 – (60,000 – a)[1 + 20/100]2 = 38,800
On solving, we get a = Rs 40,000
Hence, option c is correct.
The ratio of metal 1 and metal 2 in alloy A is 3:4. In alloy B the same metals are mixed in the ratio 5:8. If 26 kg of alloy B and 14 kg of alloy A are mixed, then find out the ratio of metal 1 and metal 2 in the new alloy.
Solution (c)
The ratio of metal 1 and metal 2 in alloy A is 3:4. Therefore, we can say that 14 kg of alloy A contains (3/7)*14 = 6 kg of metal 1 and (4/7)*14 = 8 kg of metal 2.
The ratio of metal 1 and metal 2 in alloy B is 5:8. Therefore, we can say that 26 kg of alloy B contains (5/13)*26 = 10 kg of metal 1 and (8/13)*26= 16 kg of metal 2.
Hence, total weight of metal 1 in the new alloy = 6 + 10 = 16 kg
Total weight of metal 2 in the new alloy = 8 + 16 = 24 kg
Therefore, the ratio of metal 1 and metal 2 in the new alloy = 16:24 = 2:3.
Hence, option c is correct.
Solution (c)
The ratio of metal 1 and metal 2 in alloy A is 3:4. Therefore, we can say that 14 kg of alloy A contains (3/7)*14 = 6 kg of metal 1 and (4/7)*14 = 8 kg of metal 2.
The ratio of metal 1 and metal 2 in alloy B is 5:8. Therefore, we can say that 26 kg of alloy B contains (5/13)*26 = 10 kg of metal 1 and (8/13)*26= 16 kg of metal 2.
Hence, total weight of metal 1 in the new alloy = 6 + 10 = 16 kg
Total weight of metal 2 in the new alloy = 8 + 16 = 24 kg
Therefore, the ratio of metal 1 and metal 2 in the new alloy = 16:24 = 2:3.
Hence, option c is correct.
