- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.
Number System: Level 2 Test - 6
Please wait while the activity loads. If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.
Congratulations - you have completed Number System: Level 2 Test - 6.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1 |
The total tractor production in a state is 294000, out of which 150000 are made by Mahindra & Mahindra. While Out of every 1000 Mahindra tractors, 98 are red in color, but only 53% of the total tractor production is red. Find the total percentage of non-Mahindra tractors that are red out of total non-Mahindra tractors.
50.25% | |
81.25% | |
98.30% | |
61.40% |
Question 1 Explanation:
Total production of red tractor = 294000 x 53% = 155820
Number of red tractors of Mahindra make = 150000/1000x98 = 14700
Number of non-Mahindra tractors = 294000 -150000 = 144000
Therefore, Number of non-Mahindra red tractors = 155820-14700 = 141120
Hence, required percentage = (141120/144000) x 100 = 98%
Number of red tractors of Mahindra make = 150000/1000x98 = 14700
Number of non-Mahindra tractors = 294000 -150000 = 144000
Therefore, Number of non-Mahindra red tractors = 155820-14700 = 141120
Hence, required percentage = (141120/144000) x 100 = 98%
Question 2 |
The highest score in an innings is 2/9th of the total score in the inning and the next highest score is 2/9 of the remainder score. The difference of two scores is 8 runs. Find the total score in the innings?
152 | |
162 | |
142 | |
132 |
Question 2 Explanation:
Let the total score in the innings be x.
Then, highest score = 2/9 x
then Next highest score= 2/9 (x -2/9 x)
=2/9x-2/9 (x-2/9x) =8
= x- x+2/9x = 8x9/2
x=8x9x9/2x2 = 162
Then, highest score = 2/9 x
then Next highest score= 2/9 (x -2/9 x)
=2/9x-2/9 (x-2/9x) =8
= x- x+2/9x = 8x9/2
x=8x9x9/2x2 = 162
Question 3 |
In a class, 50 students play Hockey, 20 students Football , while 10 students play both hockey Football. How many student play at least one of these two games?
10 | |
80 | |
50 | |
60 |
Question 3 Explanation:
Number of students who play Hockey = 50
Number of students who play Football = 20
Number of students who play both games = 10
Number of students who play only hockey = 50 -10= 40
Number of students who play only Football = 20 -10 = 10
:. Number of students who play at least one game =40+ 10+10=60
Number of students who play Football = 20
Number of students who play both games = 10
Number of students who play only hockey = 50 -10= 40
Number of students who play only Football = 20 -10 = 10
:. Number of students who play at least one game =40+ 10+10=60
Question 4 |
The natural numbers are divided into groups of (1), (2, 3), (4, 5, 6), (7, 8, 9, 10) and so on. Find the sum of numbers in the 50th group of this series.
1225 | |
24505 | |
62525 | |
52650 |
Question 4 Explanation:
Let S = 1 + 2 + 4 + 7 + ... + Tn or =1+2+4+ ... +Tn_1+Tn
Subtracting, we get 0=1 + [1 + 2+ 3 + ... (n -1)] - Tn
TN=1+2+3+………….. +(n-l)+1 = n (n -1)/2 + 1
First number of 50th term
= 50 x 49/2+ 1 = 1226
Sum of numbers of 50th term = 1226 + 1227 +…………….. Upto 50th term
= 50/2 [2 x 1226 + (50 - 1) x 1]
= 25 x 2501= 62525
Subtracting, we get 0=1 + [1 + 2+ 3 + ... (n -1)] - Tn
TN=1+2+3+………….. +(n-l)+1 = n (n -1)/2 + 1
First number of 50th term
= 50 x 49/2+ 1 = 1226
Sum of numbers of 50th term = 1226 + 1227 +…………….. Upto 50th term
= 50/2 [2 x 1226 + (50 - 1) x 1]
= 25 x 2501= 62525
Question 5 |
A man decided to pay off a debt of Rs 3600 by 40 annual instalments that are in AP. When 30 of the installments are paid he dies leaving one-third of the debt unpaid. The value of the 8th installment is
Rs 35 | |
Rs 50 | |
Rs 65 | |
None |
Question 5 Explanation:
Let the first instalment be 'a' and the common difference between any two consecutive instalments be 'd’. Using the formula for the sum of an AP,
Sn = n/2 [2a + (n -1) d]
we have, 3600= 40/2[2a+ (40-1) d] =20(2a+ 39d)
=> 180 = 2a+ 39d ... (i )
Again, 2400 = 30 [2a + (30 -l) d] = 15(2a + 29d)
=> 160 = 2a + 29d ... (ii)
Solving Eqs. (i) And (ii), 20 = 10d => d = 2
Therefore, 180 = 2a + 39 x 2 => 2a=102=>a=51
Value of 8th instalment
=51+ (8-1) x2=51+14 =Rs 65
=> 180 = 2a+ 39d ... (i )
Again, 2400 = 30 [2a + (30 -l) d] = 15(2a + 29d)
=> 160 = 2a + 29d ... (ii)
Solving Eqs. (i) And (ii), 20 = 10d => d = 2
Therefore, 180 = 2a + 39 x 2 => 2a=102=>a=51
Value of 8th instalment
=51+ (8-1) x2=51+14 =Rs 65
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
List |
You have completed
questions
question
Your score is
Correct
Wrong
Partial-Credit
You have not finished your quiz. If you leave this page, your progress will be lost.
Correct Answer
You Selected
Not Attempted
Final Score on Quiz
Attempted Questions Correct
Attempted Questions Wrong
Questions Not Attempted
Total Questions on Quiz
Question Details
Results
Date
Score
Hint
Time allowed
minutes
seconds
Time used
Answer Choice(s) Selected
Question Text
All done
Need more practice!
Keep trying!
Not bad!
Good work!
Perfect!