Late in the day but I need help !!
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Late in the day but I need help !!
DD has suddenly panicked about her problem with square brackets and how she can usually suss out 4 of say 8 but the other 4 can be an educated guess....why now!!!
Found some examples and lo and behold WE can't do these three, help please before she goes to bed
(57 [27] 9) (60 [30] 10)
(54 [?] 8 )
(10 [15] 45) (12 [18] 54)
(8 [?] 36)
(3 [18] 9) (5 [50] 15)
(7 [?] 12)
Thanks loads
Ambridge x
Found some examples and lo and behold WE can't do these three, help please before she goes to bed
(57 [27] 9) (60 [30] 10)
(54 [?] 8 )
(10 [15] 45) (12 [18] 54)
(8 [?] 36)
(3 [18] 9) (5 [50] 15)
(7 [?] 12)
Thanks loads
Ambridge x
Dear Ambridge
Some possible Answers and methods:
What are the answers given by the author? If M/C what are the choices given?
57 [27] 9) (60 [30] 10)
(54 [?] 8 )
57 + 9 = 66 divide by 2 = 33 – 6 = 27
60 + 10 = 70 divide by 2 = 35 – 5 = 30
54 + 8 = 62 divide by 2 = 31 – 4 = 27
or
57 - 9 = 48 - 21 = 27
60 - 10 = 50 - 20 = 30
54 - 8 = 46 - 19= 27
(10 [15] 45) (12 [18] 54)
(8 [?] 36)
45 – 10 = 35 subtract 20 [number on the left x 2]
54 – 12 = 42 subtract 24 [number on the left x 2]
36 – 8 = 28 subtract 16 = 12
(3 [18] 9) (5 [50] 15)
(7 [?] 12)
3 x 9 = 27 subtract 9 [ the number on the right]
5 x 15 = 65 subtract 15 [the number on the right]
7 x 12 = 84 subtract 12 = 72
Edit to add:
Please note the above is incorrect [5 x 15 is of course not 65!]
9 minus 3 = 6 x 3 [number on the left]
15 minus 5 = 10 x 5 [ number on the left]
12 minus 7 = 5 x 7 = 35
Some possible Answers and methods:
What are the answers given by the author? If M/C what are the choices given?
57 [27] 9) (60 [30] 10)
(54 [?] 8 )
57 + 9 = 66 divide by 2 = 33 – 6 = 27
60 + 10 = 70 divide by 2 = 35 – 5 = 30
54 + 8 = 62 divide by 2 = 31 – 4 = 27
or
57 - 9 = 48 - 21 = 27
60 - 10 = 50 - 20 = 30
54 - 8 = 46 - 19= 27
(10 [15] 45) (12 [18] 54)
(8 [?] 36)
45 – 10 = 35 subtract 20 [number on the left x 2]
54 – 12 = 42 subtract 24 [number on the left x 2]
36 – 8 = 28 subtract 16 = 12
(3 [18] 9) (5 [50] 15)
(7 [?] 12)
3 x 9 = 27 subtract 9 [ the number on the right]
5 x 15 = 65 subtract 15 [the number on the right]
7 x 12 = 84 subtract 12 = 72
Edit to add:
Please note the above is incorrect [5 x 15 is of course not 65!]
9 minus 3 = 6 x 3 [number on the left]
15 minus 5 = 10 x 5 [ number on the left]
12 minus 7 = 5 x 7 = 35
Thank you Patricia, the third one I should have done easily, wasn't thinking by then but the first two, I think are hard.
Anyway, will print out your answers and she can read them over her cornflakes - am just disappointed or surprised that she is having a mini wobble at this late hour - but at just 10 I suppose she just can!
Thanks again
Ambridge x
Anyway, will print out your answers and she can read them over her cornflakes - am just disappointed or surprised that she is having a mini wobble at this late hour - but at just 10 I suppose she just can!
Thanks again
Ambridge x
(57 [27] 9) (60 [30] 10)
(54 [?] 8 )
Why couldn't it just be multiply the number on the right by 3 to get the middle number (in brackets), and the number on the left is 30 more than the one in brackets?? It gives you the answer 24...rather than 27, but works for the other numbers. Do we KNOW that the answer should be 27?
I confess we only really concentrated on VR last year, so this is not my strength!! I, of course, bow down to your superiority Patricia, I am just interested...
(54 [?] 8 )
Why couldn't it just be multiply the number on the right by 3 to get the middle number (in brackets), and the number on the left is 30 more than the one in brackets?? It gives you the answer 24...rather than 27, but works for the other numbers. Do we KNOW that the answer should be 27?
I confess we only really concentrated on VR last year, so this is not my strength!! I, of course, bow down to your superiority Patricia, I am just interested...
Dear All
Yes, perhaps I should have made it clear that there can a variety of answers and methods, perhaps Ambridge could enlighten us with the correct answer [and if its M/C what choices were given] I have edited my original post accordingly.
Something I have learnt from these type of questions is that if the paper is in standard form, then the author really has to be very careful NOT to include number combinations that can arrive at more than 1 answer. In addition if its multiple choice, the author must ONLY give one of the answers.
GL assessment [formely NFER] state: the 3 numbers in each group are related, so yes you do need to use the numbers on both sides of the middle number.
Patricia
Yes, perhaps I should have made it clear that there can a variety of answers and methods, perhaps Ambridge could enlighten us with the correct answer [and if its M/C what choices were given] I have edited my original post accordingly.
Something I have learnt from these type of questions is that if the paper is in standard form, then the author really has to be very careful NOT to include number combinations that can arrive at more than 1 answer. In addition if its multiple choice, the author must ONLY give one of the answers.
GL assessment [formely NFER] state: the 3 numbers in each group are related, so yes you do need to use the numbers on both sides of the middle number.
Patricia