Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
Make the best use of the initiative. All the best!
To Know More about Ace the Prelims (ATP) 2021 – CLICK HERE
0 of 5 questions completed
Questions:
To view Solutions, follow these instructions:
You have already completed the test before. Hence you can not start it again.
Test is loading...
You must sign in or sign up to start the test.
You have to finish following test, to start this test:
0 of 5 questions answered correctly
Your time:
Time has elapsed
You have scored 0 points out of 0 points, (0)
| Average score |
|
| Your score |
|
| Pos. | Name | Entered on | Points | Result |
|---|---|---|---|---|
| Table is loading | ||||
| No data available | ||||
Two shopkeepers announce the same price of Rs. 700 for a shirt. The first offers successive discounts of 30% and 6% while the second offers successive discounts of 20% and 16%. The shopkeeper that offers better discount, charges how many rupees less than the other shopkeeper.
Solution (d)
We know that the formula for successive discounts = a+b-ab/100
The net discount given by first shopkeeper is 30 + 6 – 30×6/100 = 34.2%.
The net discount by second shopkeeper is 20+16 – 20×16/100 = 32.8%.
Difference between them = 34.2 – 32.8 = 1.4
So, required answer is 1.4% of 700 = 9.80 Rs
Solution (d)
We know that the formula for successive discounts = a+b-ab/100
The net discount given by first shopkeeper is 30 + 6 – 30×6/100 = 34.2%.
The net discount by second shopkeeper is 20+16 – 20×16/100 = 32.8%.
Difference between them = 34.2 – 32.8 = 1.4
So, required answer is 1.4% of 700 = 9.80 Rs
If n is any positive integer, then (n3 – n) is
Solution (c)
(n3 – n) can be written as, (n-1)*n*(n+1) (where n is a positive integer) i.e. product of three consecutive integers.
Hence, for any number n=2 or >2, the product will have a factor of 6 in it.
When there two prime numbers in a product, then the third number will always be divisible by 6
Or the product will always have a factor of 6 in it.
Solution (c)
(n3 – n) can be written as, (n-1)*n*(n+1) (where n is a positive integer) i.e. product of three consecutive integers.
Hence, for any number n=2 or >2, the product will have a factor of 6 in it.
When there two prime numbers in a product, then the third number will always be divisible by 6
Or the product will always have a factor of 6 in it.
In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 are not defective. What is the percentage of defective products in the stockpile?
Solution (a)
Let’s say total products manufactured by M1, M2 and M3 are 100.
So, M1, M2, and M3 produced 40, 30 and 30 products respectively.
Defective pieces for M1 = 120/100
Defective pieces for M2 = 30/100
Defective pieces for M3 = 150/100
So, total defective pieces are = (120 + 30 + 150)/100 = 3% of total products.
Solution (a)
Let’s say total products manufactured by M1, M2 and M3 are 100.
So, M1, M2, and M3 produced 40, 30 and 30 products respectively.
Defective pieces for M1 = 120/100
Defective pieces for M2 = 30/100
Defective pieces for M3 = 150/100
So, total defective pieces are = (120 + 30 + 150)/100 = 3% of total products.
There are six boxes numbered 1, 2, 3, 4, 5, 6. Each box is to be filled up either with a white ball or a black ball in such a manner that at least one box contains a black ball and all the boxes containing black balls are consecutively numbered. The total number of ways in which this can be done is
Solution (b)
Total ways when all 6 boxes have only black balls = 1
Total ways when 5 boxes have black balls = 2
Total ways when 4 boxes have black balls = 3
Total ways when 3 boxes have black balls = 4
Total ways when 2 boxes have black balls = 5
Total ways when only 1 box has a black ball = 6
So, total ways of putting a black ball such that all of them come consecutively = (1+2+3+4+5+6) = 21
Solution (b)
Total ways when all 6 boxes have only black balls = 1
Total ways when 5 boxes have black balls = 2
Total ways when 4 boxes have black balls = 3
Total ways when 3 boxes have black balls = 4
Total ways when 2 boxes have black balls = 5
Total ways when only 1 box has a black ball = 6
So, total ways of putting a black ball such that all of them come consecutively = (1+2+3+4+5+6) = 21
Suraj got an order for the manufacture of 480 Denim Shirts. He brought 12 sewing machines and appointed some expert tailors to do the job. However, many didn’t report to duty. As a result, each of those who did, had to stitch 32 more shirts than originally planned by Suraj, with equal distribution of work. How many tailors had been appointed earlier and how many had not reported for work?
Solution (c)
Suppose, he appointed x persons and y of them didn’t come, then work done by each of them increases by 32.
So, 480/(x-y) – 480/x = 32
Now, we can check options by putting in the above equation
x=10 and y=4 satisfy the above condition.
Hence option c is the correct answer.
Solution (c)
Suppose, he appointed x persons and y of them didn’t come, then work done by each of them increases by 32.
So, 480/(x-y) – 480/x = 32
Now, we can check options by putting in the above equation
x=10 and y=4 satisfy the above condition.
Hence option c is the correct answer.
