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FIFTH GRADE /DIVISION / TEST 1.2

MATHEMATICS IS EASY. PROVIDED THAT ONE WORKS ENOUGH.
ENOUGH IS SKILL DEPENDENT, HAVE TO FIND "ENOUGH" FOR YOUR SIZE.

SOME OF OUR QUESTIONS ARE REALLY HARD, FOR THOSE THAT ARE SKILLED IN MATH.
SKIP THEM IN CASE NOT FOR YOU.

IF YOU DESIRE SO, WE CAN DISCUSS HARD/INTERESTING QUESTIONS IN OUR FORUM.

 Method(s) of DIVISION of INTEGERS   is  (are)
 ♦ 
METHOD(s) of DETERMINING the RESULT  (QUOTIENT) of
 ♦ HOW MANY TIMES A NATURAL NUMBER  CALLED a  DIVISOR

 ♦ CAN BE SUBTRACTED FROM / IS CONTAIN-  ED IN ANOTHER NATURAL NUMBER CALLED
 a DIVIDEND


 and WHICH NATURAL NUMBER (REMAINDER,  LESS THAN DIVISOR) STAYS BEHIND.
  Ref.-1
1. How many of the following numbers leave a remainder of 1 when divided by 4 ?

248    321   436  
555   660   770


a) 1
b) 2
c) 3
d) 4
A number is divisible by 4 if the number formed by the last two digits in the number is divisible by 4 .
2. What should A be if [(1+9)x(1+9) + (Ax6-25)] = 120 - (1+12) + 10 ?

a) 7
b) 17
c) 9
d) 21
It should be that 75 + 6xA = 117 .
3. A is a digit and the number 142A divided by 4 gives a remainder. Which choice can not replace A under these conditions ?

a) 1
b) 5
c) 8
d) 9
A = 8 can not have a remainder .
4. B is a digit and A is a 2-digit number within the natural number 3BA2 which is supposed to be divisible by 4. How many choices does there exist for A ?

a) 12
b) 50
c) 99
d) 20
Write A as a 2-digit number CD, then the number happens to be 3BCD2. C can be any digit for which there are 10 choices. D can be 1, 3, 5, 7 and 9.
5. Which of the following numbers will give a remainder of 7 if divided by 4 ?

a) 10747
b) 345409
c) 78787
d) max. possible remainder is 3.
Remainder should at most be "divisor - 1" .
y
6. Martin is a good student. He learned the rule that divisibility by 4 of the number formed by the last two digits of any number implies the number's divisibility by 4 and he even proved it .
Which of the followings may be the proof of Martin ?

In case of need for tips get the cursor on "y" or for help click at "ry1/2" .
RY1 RY2



a) He began with base-10 expansion of any number and saw that all of the elements except for the last 2 are divisible by 4 whatever coefficients they may have .
b) He used trial and error method .
c) He looked for the proof in math. books.
d) He accepted the divisibility of all former digits in front of the last 2 .


Have please a look at ry1 .
ClickN KIDS Beginning Reading & Spelling Programs Teacher Created Resources ClickN KIDS Beginning Reading & Spelling Programs
7. How many of the following statements is/are true ?

  I : For divisibility by 5, the last digit should be either 0 or 5 .
  II : If the last digit is 0 or 5, the natural number is divisible by 5 .
  III : For divisibility by 8, the number formed by the last 2 digits should be a multiple of 8 .
  IV : If the number formed by the last 2 digits is a multiple of 8, then the whole number is divisible by 8 .
  V : All claims given above may be proven by the methods seen in Q6 or Q12 .
  VI : The methods seen in Q6 or Q12 are not enough to show the validitiy of the first 4 claims .

a) 5
b) 2
c) 4
d) 3
I and II are true . V or VI must be true.
8. Arthur has a made-in-china calculator whose key-3 has gone defect.
What can he do to evaluate 35631/333 ?

!


a) The dividend should be replaced by another number.
b) The divisor should be replaced by another number.
c) The dividend should be multiplied by 2 and the divisor should be 666 .
d) The divisor should be multiplied by 2 and the dividend should be 666 .
If both dividend and divisor is multiplied by the same number, the quotient does not change .
9. How many by 2, and by 3, and also by 5 divisible numbers are there between 51 and 821 ?

a) 24
b) 25
c) 26
d) 27


The numbers should be multiples of 30.
821 - 51 = 770 and 770/30 = ...
The first number is 60, and the last is 810. Difference is 750 which is 25 x 30 . But be careful about intervals and points ...
10. What should the digits A , B , C and D respectively be ?

!


a) 8 , 8 , 5 and 2
b) 5 , 8 , 8 and 2
c) 5 , 8 , 5 and 2
d) 5 , 8 , 5 and 5
124 = 23 x C + 9 implies C = 5 . This solves.
11. (2500 : 5 + 28) : ( 88 : 44 + 1) = ?

a) 176
b) 233
c) 158
d) 311
First are divisions due: 2500/5 = 500 , 88/44 = ... .
y
12. Martin of question 6 has told his teacher about his proof of divisibility by 4 and :
- Excellent, are they only these numbers that are divisible (by 4) ?
- No idea, Sir.
- In order to know you have to show that if a number is divisible by 4, the number formed by the last two digits should be "00" or a multiple of 4.
How is this claim to be proven ?


In case of need for tips get the cursor on "y" or for help click at "ry1/2".

RY1 RY2



a) He proved that the difference of natural numbers are natural numbers.
b) He did not see any reason for a proof.
c) By proving "If a natural number is a multiple of 4, the number formed by the last two digits (keeping order) is also a multiple of 4."
d) By proving the Drichlet Theorem.
ry1 gives the solution .
13. What will the remainder be if among the nearest primes to 32, the greater one is divided by the smaller ?

!


a) 5
b) 4
c) 3
d) 2
Be careful about the nearest primes .
14. We know natural numbers and we know the 4 basic operations. When we multiply a natural number by an another one, the product is a natural number but when we divide one by another, the quotient is not a natural number; there remains something we call remainder.
Which term is the most appropriate one to describe this situation, what should be put instead of the dots in "Natural numbers are ... under multiplication but not under division." ?

a) open
b) shut
c) closed
d) complete
Just to check your common sense .
15. Suppose you have a grassy field of 144 000 m2, and 48 cows eat grass at a constant rate. Assume for a while that the grass does not grow.
How big is the area cleared in a day by a cow if all the grass is cleared off the field in 90 days ?

!


a) 33 1/3
b) 35 2/3
c) 33.5
d) 33
144 000 / (48x90) = ?
16. Suppose you have a grassy field of 144 000 m2, and 120 cows eat grass at a constant rate. Assume for a while that the grass does not grow.
How big is the area cleared in a day by a cow if all the grass is cleared off the field in 30 days ?

a) 50
b) 40
c) 45
d) 30
144 000 / (120x30) = ?
17. Suppose that the cows of question 15 and 16 are the same cows and 120 cows for a month can be fed due to the fact that the grass keeps growing. Which area per cow should have been added per day to the field to compensate this growth ?

a) 6 2/3 m2
b) 6 m2
c) 6 1/3 m2
d) 7 m2
The cows of question 16 need 40 m2 area per day whereas the other require 33 1/2 .
y
18. In our previous questions we ignored the initial grass on the field which should exist. Suppose you have a grassy field, and cows eat grass at a constant rate. Keep in mind, the grass also keeps growing continuously.
All the grass off the field can be cleared by 48 cows in 90, and by 120 cows in 30 days.
What amount of initial grass should have been on the field ?


In case of need for tips get the cursor on "y" or for help click at "ry1/2" .
RY1 RY2


a) 240
b) 270
c) 310
d) 340
Please have a look at ry1.
19. Suppose that the conditions of Question-18 is valid. What is the number of cows that will consume exactly the daily growth ?

a) 12
b) 6
c) 9
d) 18
Take the case of 120 cows for 30 days. Initiaally grass from 270 days is on the field. 30 days added means grass of 300 days.
The grass of 300 days is consumed by 120 cows in 30 days means it will be consumed by ... cows in 300 days.
20. Which of the following equations proves that 12 cows is also the right number for the case of 48 cows and 90 days ?

a) 360/90 = 4 and 90/4 = 22.5
b) 360/90 = 4 and 48/4 = 12
c) 450/90 = 5 and 48/4 = 12
d) 450/90 = 5 and 60/5 = 12
Now there is 360 days' grass on the field which is consumed by 48 cows in 90 days . It will be consumed by ... cows 360 days.
21. Which of the following figures may be the graph of the grass on the field under the conditions of Question-18 with 24 cows on the field ?

!


a) A
b) B
c) both
d) neither
In 270 days the whole field will be cleared off .
y
22. Below in the figure shown the population and resources graphs of The Limits to Growth. Why does population begin to decay with delay ?

!


In case of need for tips get the cursor on "y" or for help click at "ry1/2" .

RY1 RY2



a) people are careless
b) people are near sighted
c) people think themselves
d) people are reactive


Details are given in ry1 .
23. Water flowing under the bridge may be around 750 000 m3/hour. Assuming that the number is exact, what will be the flow (per second) of the water be ?

!


a) 300 1/6 m3
b) 250 3/7 m3
c) 208 1/3 m3
d) 190 4/9 m3
750 000/3600 = 7 500/36 = 2500/12 = 2400/12 + 100/12 = 200 + ...
24. (axb + bxc)/a = ? , for a=3, b=4 and c=5

a) 9.333
b) 10.666...
c) 5.888
d) 12.222
(12+20)/3 = 32/3 =10+2/3 ... .
25. 101110111011/5055 ?

a) 20002000.2
b) 200020002
c) 10001000.1
d) 100010001
101110111011/5055 = (2/10) x (101110111011/1011)= 2/10 x 100000000+10000+1) = (2/10)x100010001 = 200020002/10 = ?
This is the feedback!

THANK YOU FOR YOUR VISIT; HOPE WE COULD BE OF USE.
FOR MUTUAL BENEFIT YOU MAY USE OUR FORUM.
© 2011 , İsmail GERMAN. ALL RIGHTS RESERVED
PLEASE PREFER GIVING LINK IF THAT SUFFICES

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