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Number System: Level 3 Test -2

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Question 1
Each of the following questions is followed by two statements. MARK, If R is an integer between 1 & 9, P – R = 2370, what is the value of R? I. P is divisible by 4. II. P is divisible by 9.
A
if the question can be answered with the help of statement I alone,
B
if the question can be answered with the help of statement II alone,
C
if both, statement I and statement II are needed to answer the question,
D
if the statement cannot be answered even with the help of both the statements.
Question 1 Explanation: 
To solve this question, we should have a good  command on numbers. Look carefully , if P-R =2370 then P lies between 2371 and 2379. So if we implement all the logical conditions, it can be concluded that P may have two options 2372 and 2376, because the first statement tells us about the divisibility of the result with 4 and it will end with two options for R: one is 2 and the second is 6 ,and the result formed with this is divisible by 4, this is all about the 1st option statement . So this statement could not satisfy a specific answer. But when we shift to the other statement, the second one we find that the only integer left for R is 6 and by this value automatically we have the value for P which is 2376.  So from the 2nd statement alone, we can have a unique value for R.
Question 2
A man distributed 43 chocolates to his children. How many of his children are more than five years old?I. A child older than five years gets 5 chocolates. II. A child 5 years or younger in age gets 6 chocolates.
A
if the question can be answered with the help of statement I alone,
B
if the question can be answered with the help of statement II alone,
C
if both, statement I and statement II are needed to answer the question,
D
if the statement cannot be answered even with the help of both the statements.
Question 2 Explanation: 
By looking at the first statement, we can analyse that the chocolates are given to those who are older than 5. Let the number of children bee 5x, where x is the number of children older than 5 years. Now from this we cannot conclude anything more, so we have to shift to the next statement and thus, option 1 and option 2 are ruled out. Now by the second statement, 6 chocolates are given to the children who are 5 years or younger, so let this number be 6y where y are those number of children who are younger 5 years or younger. So the equation thus formed by using both conditions is : 5x +6y=43. Since x and y have to be integers, there is only one pair of values that satisfies above equation, viz. a = 5 and b = 3.
Question 3
X, y, and z are three positive odd integers, Is x + z divisible by 4?I. y − x = 2 II. z − y = 2
A
if the question can be answered with the help of statement I alone,
B
if the question can be answered with the help of statement II alone,
C
if both, statement I and statement II are needed to answer the question
D
if the statement cannot be answered even with the help of both the statements.
Question 3 Explanation: 
In this type of question, we don’t have any values of x,y,z. And therefore we have to solve this using logic. We have two equations and we can see that question is asking for z+x, which actually is the hint to solve this question. We can say x-y=-2 z-y= 2 From this, z+x=2y . In order to make R.H.S. equal to the L.H.S. we have to put the relevant values so we can conclude that when r.h.s and l.h.s will equal the z+x will be divisible by 4 If we put x=0 and z=4, the result is 0+4=4 and the value of y will be 2. So the condition satisfies that 4 is divisible by 4.
Question 4
If m and n are integers divisible by 5 such that m>n, which of the following is not necessarily true?
A
m – n is divisible by 5
B
m^2 – n^2 is divisible by 25
C
m + n is divisible by 10
D
None of these
Question 4 Explanation: 
Students please note that the best way to solve this is the method of simulation, e.g. let m = 10 and n = 5. Hence, m – n = 5, which is divisible by 5. m2 – n2 = 100 – 25 = 75, divisible by 25. m + n = 10 + 5 = 15 is not divisible by 10. Hence, the answer is (c). Note that for the sum of two multiples of 5 to be divisible by 10, either both of them should be odd (i.e. ending in 5) or both of them should be even (i.e. ending in 0).
Question 5
One bacterium can divide into eight bacteria of the next generation. But due to low mortality ratio only 50% survives and remaining 50% dies after producing next generation. If the seventh generation Number is 4,096 million, what is the number of bacteria in first generation?
A
1 million
B
2 million
C
4 million
D
8 million
Question 5 Explanation: 
Let us try and find a pattern. Let there be x bacteria in the first generation. Hence, n1 = x. But only x/2 among them will be able to produce the next generation. Andthey would give rise t8 = 4x bacteria. Hence, n2 = 4x. Of these only 2x will give rise to next generation. So number of bacteria in the third generation = 8(2x) = 16x. So n3 = 16x. Similarly, we would find that n4 = 64x. If we observe: n1 = x, n2 = 4x, n3 = 16x, n4 = 64x, this will form a GP with a = x and r = 4. The seventh term of this GP = 4096. So we can write 4096 = x(4)6 = x(2)12 = 4096x. Hence, x = 1, i.e. 1 million.
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