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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 â€“ R = 2370, what is the value of R?
I. P is divisible by 4.
II. P is divisible by 9.

if the question can be answered with the help of statement I alone, | |

if the question can be answered with the help of statement II alone, | |

if both, statement I and statement II are needed to answer the question, | |

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 Â 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.Â 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.

if the question can be answered with the help of statement I alone, | |

if the question can be answered with the help of statement II alone, | |

if both, statement I and statement II are needed to answer the question, | |

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 âˆ’ x = 2
II. z âˆ’ y = 2

if the question can be answered with the help of statement I alone, | |

if the question can be answered with the help of statement II alone, | |

if both, statement I and statement II are needed to answer the question | |

if the statement cannot be answered even with the help of both the statements. |

Question 3 Explanation:

In this type of question, we donâ€™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?

m â€“ n is divisible by 5 | |

m^2 â€“ n^2 is divisible by 25 | |

m + n is divisible by 10 | |

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 â€“ n = 5, which is divisible by 5. m2 â€“ n2 = 100 â€“ 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?

1 million | |

2 million | |

4 million | |

8 million |

Question 5 Explanation:

Let us try and find a pattern. Let there be x bacteria in the first generation. Hence, n

^{1}= x. But only x/2 among them will be able to produce the next generation. Andthey would give rise t8 = 4x bacteria. Hence, n^{2}= 4x. Of these only 2x will give rise to next generation. So number of bacteria in the third generation = 8(2x) = 16x. So n^{3}= 16x. Similarly, we would find that n^{4}= 64x. If we observe: n1 = x, n2 = 4x, n^{3}= 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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