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Number System: Level 2 Test -4

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Question 1
Two painters undertake to do a job. The one painter begins working one hour after the first. Three hours after the first painter has begun working, there is still 9/20 of the work to be done. When the assignment is completed, it turns out that each painter has done half the work. How many hours would it take each one to do the whole job individually?
12 hr and 8 hr
8 hr and 5.6 hr
10 hr and 8 hr
5 hr and 4 hr
Question 1 Explanation: 
Let the first painter takes X hours and the second Y hours to do the whole job.
When the first was busy typing for 3 hr, the second was busy only for 2 hr.
Both of them did 11/20 of the whole work.
3/x +2/y =11/20
When the assignment was completed, it turned out that each painter had done half the work.
Hence, the first spent X/2 hr, and the second, Y/2 hr.
And since the first had begun one hour before the second,
we have x/2 –y/2=1 Solving the above equations,
we have: X = 10 hr, Y = 8 hr.
Question 2
A student instead of finding the 5/8th value of a number, found 5/18th of the number. If the difference between the original value and the new value is 550, find the number.
Question 2 Explanation: 
This equation is very straightforward. If the number is 'x', then 5x/8- 5x/18=550 .
On solving this equation, we get x = 1584.
Hint: Students please note that if the difference in5/8 & 5/18 of a number is 550,
then the difference in1/8 and 1/18 of the number should be 110.
If we express this as an equation, we get x/8-x/18=110
or 10x = 110 × 18 × 8 or x = 11 × 18 × 8
You can further proceed from here in two ways:
(i) the last digit of the required answer should be (1 × 8 × 8) = 4,
(ii) number should be divisible by 11.
In both cases, the answer that is obtained from the given choices is 1584.
Question 3
P3 is odd. Which of the following statement(s) is(are) true?I) P is odd      ll )  Pis odd    lll )  P2 is even
I only
II only
I and II
I and III
Question 3 Explanation: 
If P3 is odd, then P should also be odd.
Hence, P2 should also be odd.
And P2   will again be odd and not even.
So only I and II are true.
Question 4
Number of students who have opted for subjects A, B and C are 60, 84 and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions?
Question 4 Explanation: 
HCF of 60, 84 and 108 is 12.
Hence, 12 students should be seated in each room.
So for subject A we would require 60/15= 5 rooms,
for subject B we would require 84/12= 7
rooms and for subject C we would require 108/12 = 9 rooms.
Hence, minimum number of rooms required to satisfy our condition = (5 + 7 + 9) = 21 rooms
Question 5
If I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 5:6:8. If the total amount with me is Rs. 210, find the number of one-rupee coins.
Question 5 Explanation: 
Since the number of coins are in the ratio 2.5 : 3 : 4,
the values of the coins will be in the ratio
(1 × 2.5) : (0.5 × 3) : (0.25 × 4) = 2.5 : 1.5 : 1 or 5 : 3 : 2 Since they totally amount to Rs. 210,
if the value of each type of coins are assumed to be 5x, 3x and 2x,
the average value per coin will be 210/10x
So the total value of one-rupee coins will beRs. 105
So the total number of one-rupee coins will be 105.
or from the given ratio 5x1+6x0.5+8x0.25=21
from the ratio 5x21=105
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