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FIFTH GRADE /ADDITION / TEST 1.3

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.

 THE ADDITION OPERATION IS,

  TO COUNT FORWARD FROM  A SUMMAND

 ♦ 
AS MANY AS ANOTHER SUMMAND
OR BETTER SAID

   ♦  IT IS THE METHOD(S)   
OF FINDING THE RESULT OF THIS  UPCOUNTING.
1. Which ones of the following addition operations do have error ?

43587923 + 27814682 = 71402605 
  1542834756 + 2358714637 = 3901549393
  753951468 + 365984247 = 1119935715


a) only I
b) all
c) none
d) I and II
No errors of multiples of 9 .
2. Which choice gives A if [(1+9)+(101+109) + (Ax6+25)] = 220 + (102+210) + 25 ?

a) 549
b) 442
c) 52
d) 74
Drop equal quantities that are on both sides of the equality like (1+9)+(101+109) and 220 and 25 at both sides.
3. What will be the sum of two triangular numbers whose difference is 30 ?

!


a) 900
b) 1050
c) 750
d) 445
nth triangular number is the sum of natural numbers from 1 to n.
If their difference is 30, then the last one is obtained to the first by adding 30. The greater number's last element should be 30.
15 is the 5th and 45 is 9th triangular numbers. Their difference is also 30 but their sum is not among the choices.
4. Jacksons had a garden party and 61 guests had attended. How many handshakes should have occurred among the guests if each guest shaked hands with every guest ?

!


a) 644
b) 1074
c) 2054
d) 1830
First hand shakes hands with 60 other guests, 2nd makes an additional 59 handshakes etc. .
5. Which one of the following statements is wrong ?

a) The sum is conserved if the order of the summands changes .
b) In case of additions without any carry, the choce "c" is valid.
c) The digit sum of the summands and sum are equal.
d) Even if carry is present, mod 9 sums of all summands and the sum are equal.
Mod 9 sums are summation discarding all 9s, i.e. casting out 9s .
You are supposed to select between two choices .
y
6. The sum of a 2-digit number with consecutive digits and another 2-digit number having same digits in both of the places is evaluated and it is observed that the digit sums of the numbers and the sum are not equal.
If the first summand is the least 2 digit number with consecutive digits, what can the other summand's minimal value be ?

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



a) 77
b) 88
c) 99
d) 75


Heve please a look at ry1.
ClickN KIDS Beginning Reading & Spelling Programs Teacher Created Resources ClickN KIDS Beginning Reading & Spelling Programs
7. The numbers 1577 and 2367 are calculated with a calculator displaying 5 instead of 4. What will be the difference of mod 9 sum (casting out 9s) of the both summands and the number displayed by the machine ?

a) 2
b) 4
c) 3
d) 5
38 in Mod 9 is 2 (1+5+7+7+2+3+6+7=38)).
The machine will show 3955.
8. We are asked to form two numbers by using every digit except 0 only once. Which of the choices gives the maximum sum available this way ?

a) 98765433
b) 98765432
c) 98765431
d) 108765432
98765432 and 1 or 98765431 and 2 .
9. Given that A,B and C are digits and C1A + 3BA + D1A = 741 holds. Which of the following values may be taken by C and D ?

a) 7 and 3
b) 1 and 3
c) 1 and 7
d) 4 and 5


Write please in the form:
C1A
3BA
D1A
741
10. Given that A,B and C are digits and 1A4 + 2A3 + 3BC = 804 holds. Which of the following values may be taken by A, B and C ?

a) 5 , 6 and 7
b) 5 , 8 and 7
c) 6 , 3 and 7
d) 5 , 9 and 7
Write please in the form:
1A4
2A3
3BC
804
11. (3:6x2 + 762:6) + (156493 + 414547) = ?

a) 571168
b) 453481
c) 548529
d) 341270
If no parenthesis are used to define the correct order of operations first are multiplications and divisions evaluated. Then are addition and subtraction.
In case of operations of same weight, the one at left has priority.
y
12. The sum of a 2-digit number with consecutive digits and another 2-digit number having same digits in both of the places is evaluated and it is observed that the digit sums of the numbers and the sum are not equal
How many cases are posssible ?


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

RY1 RY2



a) 24
b) 35
c) 17
d) 44
Please have a look at ry1 .
13. Which of the following statements are/is true ?

I: Addition is the simplest among the 4 basic operations.
II: Regarding operational order, addition and subtraction have same priority.
III: The based used as we add is not of any importance.
IV: The sign of addition is "/" .


a) all
b) I , II and IV
c) I and II
d) I and III
Without base or with small bases like 2, the addition of very big numbers is a burden .
14. What is the probabilty of having a prime sum if two randomly selected but different prime numbers less then 10 are added ?

!


a) 1/2
b) 2/3
c) 2/7
d) 1/3
The possible cases : 2+3 , 2+5, 2+7, 3+5, 3+7 and 5+7 .
15. What is the probabilty of having a prime sum if two randomly selected prime numbers less then 10 are added ?

a) 2/7
b) 1/3
c) 1/2
d) 1/5
This time we have : 2+2, 2+3, 2+5, 2+7, 3+3, 3+5, 3+7, 5+5, 5+7 and 7+7.
16. Which of the following addition operations are wrong ?

  I : 12211332 + 1117291 = 13328624
  II : 60941197 + 27092344 = 88033561
  III : 305034666 + 1752248025 = 1752248025

a) only I
b) all
c) I and II
d) only III
Try the rule for even-odd, then evaluate the sums of units' places ... .
If you count mod 3 or mod 9 and compare the results of both sides of an equation ... .
17. How many meters is the front view average height of the iceberg seen in the figure below ?

!


a) 8
b) 13
c) 11
d) 12
The empty region in the middle is not a part of the iceberg .
y
18. There is a method to evaluate the squares of number greater than 51 and less than 100 :
   1- Subtract the number (say 87) from 100 (13).
   2- Subtract from the number its difference from 100 and multiply it with 100 [(87-13)x100=74x100=7400] .
   3- Find the square of the number's difference from 100 (13x13=169)
   4- Add the rsults obtained in 2nd and 3rd steps (7400+169=7569) .

Which of the choices below gives this method applied to the number 77 ?


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


a) 23 , 54 , 529 , 5400 + 529 = 5929=77x77
b) 23 , 5400 , 529 , 5400 + 529 = 5929=77x77
c) 13 , 5400 , 529 , 5400 + 529 = 5929=77x77
d) 13 , 54 , 529 , 5400 + 529 = 5929=77x77
Learn this method, it may be useful especially for square of numbers from 90 to 99 .
to learn why it works, have a look at ry1.
19. Given are the natural numbers 7, 13 , 30, 43, 50, 57 and 60 . Using any of these numbers only once and summing them, in how many different ways can we obtain 100 ?

a) 3
b) 2
c) 4
d) 6
e.g. 7+13+30+50 = 100 ... .
20. Which is the best way of adding 35455, 55549, 53545 ve 72475 ?

a) (35455+53545) + (55549+72451)
b) (35455+53545) + (55549+72451) + 424
c) (35455+53545) + (55549+72451) + 24
d) (35455+53545) + (55549+72451) + 42
Try to obtain "000" at the end in each of paranthesis.
21. Which is the maximal sum obtained via addition of any two numbers from the raws and the columns ?

!


a) 19556777
b) 19556673
c) 19556763
d) 19556663
It should be easy for you to find the right summands .
y
22. Interpret the figure and find the correct choice .

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

!
RY1 RY2



a) 22
b) 33
c) 27
d) 39


trace of a matrice is an important concept, it should be learned as early as possible .
23. 2010 FIFA World Cup took place in South Africa from 11 June to 11 July 2010. Thirty-two teams were selected for participation via a worldwide qualification tournament that began in August 2007. Each team had 18 football players .
If we want to calculate the total number of players without any multiplication, what will be the minimal number of required additions ?

!


a) 5
b) 4
c) 7
d) 9
32x18 = 18x32 means we need the sum of 18 terms 32+ ... +32 (17 additions required).
Grouping into 3 each having 6 teams requires 5 additions for each group and ... .
24. (143672 + 15498)+(990000+10000)+(99+1) = ?

a) 1159270
b) 159270
c) 1150270
d) 115927
The operations in the last two parenthesis are simple .
25. 427000 + 189000 + 56000 + 328000 = ?

a) 550000
b) 950000
c) 900000
d) 1000000
(400 + 100 + 300) + (20 + 80 + 50 + 20) + (7 + 9 + 6 + 8) = 800 + 170 + ...
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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