Online exam in Number System For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Number System -Test 2
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Subject :- Quantitative Aptitude
Chapter :- Number System-Test 1
Questions :- 25
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So, from this, five odd numbers are 111, 113, 115, 117, 119.
By this, we cannot determine the next set of odd numbers.
Question 6 of 25
6. Question
1 points
If the numerator of a fraction is increased by 200% and the denominator is increased by 350%. The resultant fraction is 5/12. What was the original fraction ?
If the numerator of a fraction is increased by 400% and the denominator is increased by 500%. The resultant fraction is 20/27. What was the original fraction ?
If the numerator of a faction is increased by 200% and the denominator of the fraction is increased by 150%, the resultant fraction is 7/10. What is the original fraction ?
LCM of two numbers is 120 and their HCF is 10. Which of the following can be the sum of those two numbers ?
Correct
Here, LCM of two numbers = 120
HCF of these two numbers = 10
Let the two numbers be 10x and 10y
Then, LCm of 10x and 10y = 120
10xy = 120
xy = 12
Here, xy can be (3,4)
So, numbers are 10*3 and 10*4
Sum of these two numbers = 30+40 = 70
Incorrect
Here, LCM of two numbers = 120
HCF of these two numbers = 10
Let the two numbers be 10x and 10y
Then, LCm of 10x and 10y = 120
10xy = 120
xy = 12
Here, xy can be (3,4)
So, numbers are 10*3 and 10*4
Sum of these two numbers = 30+40 = 70
Question 13 of 25
13. Question
1 points
Find the maximum number of girls, among whom 2923 bags and 3239 eyeliners can be distributed in such a way that each girl gets the same number of bags and eyeliners.
Correct
Here, we have to find maximum number. Hence, here we have to find out the HCF of 2923 and 3239.
So, find the HCF of these two numbers by factorization method.
i.e., 2923 = 1, 37, 79
3239 = 1, 41, 79
Therefore, HCF = 79
Hence, maximum number of girls = 79
Incorrect
Here, we have to find maximum number. Hence, here we have to find out the HCF of 2923 and 3239.
So, find the HCF of these two numbers by factorization method.
i.e., 2923 = 1, 37, 79
3239 = 1, 41, 79
Therefore, HCF = 79
Hence, maximum number of girls = 79
Question 14 of 25
14. Question
1 points
The sum of two numbers is 45 and their product is 500. Their HCF is
Correct
Here, the sum of two numbers = 45
or x+y = 45
And x*y = 500
So, going through the options 5 is the only common factor which satisfies both the equations.
Incorrect
Here, the sum of two numbers = 45
or x+y = 45
And x*y = 500
So, going through the options 5 is the only common factor which satisfies both the equations.
Question 15 of 25
15. Question
1 points
LCM of two numbers is 28 times their HCF. The sum of HCF and LCM is 1740. If one of these numbers is 240, the sum of digits of the other number is
HCF of two numbers each of 4 digits is 103 and their LCM is 19261. Sum of the numbers is
Correct
Here, HCF of two 4 digits numbers = 103
Therefore, Let the numbers be 103a and 103b.
And LCM of 103a and 103b = 19261
or 103(a*b) = 19261
$\Rightarrow a*b=187$
or a= 11 or 17 and b = 11 or 17
Therefore, Numbers = 103*11 = 1133
= 103*17 = 1751
Therefore, Sum of these two numbers = 2884
Incorrect
Here, HCF of two 4 digits numbers = 103
Therefore, Let the numbers be 103a and 103b.
And LCM of 103a and 103b = 19261
or 103(a*b) = 19261
$\Rightarrow a*b=187$
or a= 11 or 17 and b = 11 or 17
Therefore, Numbers = 103*11 = 1133
= 103*17 = 1751
Therefore, Sum of these two numbers = 2884
Question 17 of 25
17. Question
1 points
When a number is divided by 2,3,4,5 or 6 remainder in each case is 1. But the number is exactly divisible by 7. If the number lies between 250 and 350, the sum of digits of the number will be
Correct
Here, first of all we have to find the LCM of 2,3,4,5 and 6 = 60
Now, 6*5+1 = 7k
or 7k = 301
k = 43
So, sum of the digits of 301 = 3+0+1 = 4
Incorrect
Here, first of all we have to find the LCM of 2,3,4,5 and 6 = 60
Now, 6*5+1 = 7k
or 7k = 301
k = 43
So, sum of the digits of 301 = 3+0+1 = 4
Question 18 of 25
18. Question
1 points
Amit, Sucheta and Neeti start running around a circular track and complete one round in 18s, 24s and 32s, respectively.
In how many seconds, will the three meet again at the starting point as they all have started running at the same time ?
Correct
Here, to find out the time after which they meet again at the starting point
= LCM of 18, 24 and 32
= 288 s
Therefore, After 288 s they will meet with one another.
Incorrect
Here, to find out the time after which they meet again at the starting point
= LCM of 18, 24 and 32
= 288 s
Therefore, After 288 s they will meet with one another.
Punit, Vinit and Ajit begin to jog around a stadium. They complete their revolutions in 45s, 54s and 36s respectively. After how many seconds, will they be together at the starting point ?
Correct
Here, to find out the time after which they will meet one another at the starting point, we have to find LCM of 45s, 36s and 54s.
So, LCM of 45, 36 and 54 = 540s
Incorrect
Here, to find out the time after which they will meet one another at the starting point, we have to find LCM of 45s, 36s and 54s.
So, LCM of 45, 36 and 54 = 540s
Question 21 of 25
21. Question
1 points
The sum of 8 consecutive odd numbers is 656. Also, average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number ?
The sum of three consecutive odd numbers and three consecutive even numbers together is 231. Also, the smallest odd number is 11 less than the smallest even number. What is the sum of the largest odd number and the largest even number ?
Correct
Here, let the three consecutive odd numbers be x-2 and x+2 and three even numbers be y-2, y and y+2.
If the numerator of a fraction is increased by 1/4 and the denominator is decreased by 1/3, the new fraction obtained by 33/64. What was the original fraction ?