IASbaba’s Daily CSAT Practice Test
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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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Question 1 of 5
1. Question
If x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is
Correct
Solution (c)
If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)
If x = 1 then 2*1+12 = 14 is divisible by 1
If x = 2 then 2*2+12 = 18 is divisible by 2
If x = 3 then 2*3+12 = 18 is divisible by 3
If x = 4 then 2*4+12 = 20 is divisible by 4
If x = 6 then 2*6+12 = 24 is divisible by 6
If x = 12 then 2*12+12 = 36 is divisible by 12
Hence, option c is correct.
Incorrect
Solution (c)
If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)
If x = 1 then 2*1+12 = 14 is divisible by 1
If x = 2 then 2*2+12 = 18 is divisible by 2
If x = 3 then 2*3+12 = 18 is divisible by 3
If x = 4 then 2*4+12 = 20 is divisible by 4
If x = 6 then 2*6+12 = 24 is divisible by 6
If x = 12 then 2*12+12 = 36 is divisible by 12
Hence, option c is correct.

Question 2 of 5
2. Question
An outgoing batch of students wants to construct an auditorium worth Rs 42,00,000 for their college. If the teachers offer to pay 50% more than the student’s contribution and an external benefactor give three times the teachers contribution, how much should the teachers donate?
Correct
Solution (a)
Let’s assume the students contributed X Rs.
So, the teachers contributed 3X/2 Rs
And external benefactor contributed 9X/2 Rs
Summation of all contribution = 42,00,000 = 7X
Or, X = 6,00,000 Rs
Hence, teachers contributed 9,00,000 Rs
Incorrect
Solution (a)
Let’s assume the students contributed X Rs.
So, the teachers contributed 3X/2 Rs
And external benefactor contributed 9X/2 Rs
Summation of all contribution = 42,00,000 = 7X
Or, X = 6,00,000 Rs
Hence, teachers contributed 9,00,000 Rs

Question 3 of 5
3. Question
The cost of a diamond varies directly as the square of its weight. While showing to the customer, this diamond broke into four pieces with weights in the ratio 1:2:3:4. When the pieces were sold, the merchant got Rs 70,000 less. Find the original price of the diamond.
Correct
Solution (c)
Let the original weight of the diamond be 10k gm. So, after breaking into 4 pieces, the parts of the diamond weigh k gram, 2k gram, 3k gram, and 4k gram respectively.
The price of the diamond varies directly in proportion to the weight. Let us assume, P =C*W2 the where C is a constant and W is the weight of the diamond.
Therefore, the original price is C*10k*10k = 100k2 *C
The new weight is Ck2 + C(2k)2 + C(3k)2 + C(4k)2 = 30k2C
The decrease in the price is 70,000 Rs. So, 100k2C − 30k2C = 70,000
Or, k2C = 1,000
Therefore the original price = 100 k2C = 1,00,000 Rs
Incorrect
Solution (c)
Let the original weight of the diamond be 10k gm. So, after breaking into 4 pieces, the parts of the diamond weigh k gram, 2k gram, 3k gram, and 4k gram respectively.
The price of the diamond varies directly in proportion to the weight. Let us assume, P =C*W2 the where C is a constant and W is the weight of the diamond.
Therefore, the original price is C*10k*10k = 100k2 *C
The new weight is Ck2 + C(2k)2 + C(3k)2 + C(4k)2 = 30k2C
The decrease in the price is 70,000 Rs. So, 100k2C − 30k2C = 70,000
Or, k2C = 1,000
Therefore the original price = 100 k2C = 1,00,000 Rs

Question 4 of 5
4. Question
A bought 5 cars, 7 tractors and 4 buses. B bought 6 cars, 8 buses and 14 tractors for an amount which was half more than what A had paid. What per cent of the total amount paid by A was paid for the cars?
Correct
Solution (b)
Let the cost of car, tractor and bus be x, y, and z respectively.
5x+7y+4z = Aà(1)
6x+8z+14y = 3A/2
4x + 16/3 z + 28/3 y = Aà(2)
Comparing two equations
5x+7y+4z = 4x+16/3 z + 28/3 y
x = 7/3 y + 4/3 z
3x = 7y+4z
Now required percentage = [5x/(5x+7y+4z)]*100 = 5x/(5x+3x) = 62.5%
Hence, option b is correct.
Incorrect
Solution (b)
Let the cost of car, tractor and bus be x, y, and z respectively.
5x+7y+4z = Aà(1)
6x+8z+14y = 3A/2
4x + 16/3 z + 28/3 y = Aà(2)
Comparing two equations
5x+7y+4z = 4x+16/3 z + 28/3 y
x = 7/3 y + 4/3 z
3x = 7y+4z
Now required percentage = [5x/(5x+7y+4z)]*100 = 5x/(5x+3x) = 62.5%
Hence, option b is correct.

Question 5 of 5
5. Question
Each odd digit in the number 5263187 is substituted by the next higher digit and each even digit is substituted by the previous lower digit and the digit so obtained are rearranged in ascending order. which of the following will be the third digit from the left end after the rearrangement?
Correct
Solution (b)
On replacing the digits as per the question we will get 6154278
Now on rearranging this, we get 1245678
So, the third digit from the left is 4, which is option b.
Incorrect
Solution (b)
On replacing the digits as per the question we will get 6154278
Now on rearranging this, we get 1245678
So, the third digit from the left is 4, which is option b.