Challenge maths - DS finding very difficult - please help
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Challenge maths - DS finding very difficult - please help
before you ask why we are doing them - this is what set for hwk - any thoughts please
Tickets for a school play cost £3 for adults and £1 for children. The total amount collected from ticket sales was £1320. The play was staged in a hall seating 600 but the hall was not completely full. What was the smallest possible number of adutls at the play?
358, 359, 360, 361, 362
the grand old duke of york marched his men etc. By 2pm they were one third of the way up. By 4pm they were three quarters of the way up. When did they set out?
Tickets for a school play cost £3 for adults and £1 for children. The total amount collected from ticket sales was £1320. The play was staged in a hall seating 600 but the hall was not completely full. What was the smallest possible number of adutls at the play?
358, 359, 360, 361, 362
the grand old duke of york marched his men etc. By 2pm they were one third of the way up. By 4pm they were three quarters of the way up. When did they set out?
Re: Challenge maths - DS finding very difficult - please hel
358 = smallest number of adults because the children's tickets are £1 all the rest could be children
Divide the distance into twelfths because this can be divided easily into both thirds and quarters.
The distance between 1/3 and 3/4 of the way up is 5 twelfths. Thus 5 twelfths of the distance takes 2 hours; so one twelfth takes 24 mins (120 ÷ 5) ; so one third (4 twelfths) of the distance must take 96 mins. 96 mins before 2pm is 12:24
I think the soldiers set out at 12:24 but not certain.
This supposes of course that they are not getting more tired as they march up.
One of the word puzzles my class were doing last week read "Sarah and Jo saved up a total of £26 how much did they each save?
One of my pupils answered: Sarah saved £24 and Jo saved £2. Perfectly fine with the way the Q is worded
Divide the distance into twelfths because this can be divided easily into both thirds and quarters.
The distance between 1/3 and 3/4 of the way up is 5 twelfths. Thus 5 twelfths of the distance takes 2 hours; so one twelfth takes 24 mins (120 ÷ 5) ; so one third (4 twelfths) of the distance must take 96 mins. 96 mins before 2pm is 12:24
I think the soldiers set out at 12:24 but not certain.
This supposes of course that they are not getting more tired as they march up.
One of the word puzzles my class were doing last week read "Sarah and Jo saved up a total of £26 how much did they each save?
One of my pupils answered: Sarah saved £24 and Jo saved £2. Perfectly fine with the way the Q is worded
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Re: Challenge maths - DS finding very difficult - please hel
If the seats were fully taken there would have been 360 adults in the hall.
We had 600 seats and total amount collected from the ticket sale was 1320.
360 * 3 + (600 - 360) * 1 = 1320
Since the seats are not full, I would say the smallest possible number of adults is 361 - leaving two seats vacant.
1320 - 361 x 3 = 1320 - 1083 = 237;
361 + 237 = 598 < 600.
We had 600 seats and total amount collected from the ticket sale was 1320.
360 * 3 + (600 - 360) * 1 = 1320
Since the seats are not full, I would say the smallest possible number of adults is 361 - leaving two seats vacant.
1320 - 361 x 3 = 1320 - 1083 = 237;
361 + 237 = 598 < 600.
Re: Challenge maths - DS finding very difficult - please hel
yes I forgot about the capacity of the hall
Re: Challenge maths - DS finding very difficult - please hel
The first one is one of those times when algebra really is the easiest way. Let a and c be the numbers of adults and children, and translate what we're given:
Subtracting these (which flips the <):
2a > 720
which is equivalent to a > 360, so a must be at least 361.
In 2 hours they marched 3/4 - 1/3 = 5/12 of the way up, so they would do 5/24 of a hill in one hour and it would take them 24/5 hours to march the whole way up. At 2pm they were 1/3 of the way up; to get there would have taken them 1/3*24/5 = 8/5 hours = 1 hour 36 minutes, so they started at 12:24.
3a + c = 1320Minesh wrote:Tickets for a school play cost £3 for adults and £1 for children. The total amount collected from ticket sales was £1320.
a + c < 600Minesh wrote:The play was staged in a hall seating 600 but the hall was not completely full.
Subtracting these (which flips the <):
2a > 720
which is equivalent to a > 360, so a must be at least 361.
aargh has the answer to this, but here's a slightly different method:Minesh wrote:the grand old duke of york marched his men etc. By 2pm they were one third of the way up. By 4pm they were three quarters of the way up. When did they set out?
In 2 hours they marched 3/4 - 1/3 = 5/12 of the way up, so they would do 5/24 of a hill in one hour and it would take them 24/5 hours to march the whole way up. At 2pm they were 1/3 of the way up; to get there would have taken them 1/3*24/5 = 8/5 hours = 1 hour 36 minutes, so they started at 12:24.
Re: Challenge maths - DS finding very difficult - please hel
But if your child is finding it difficult TELL THE SCHOOL!
Re: Challenge maths - DS finding very difficult - please hel
Alternatively, you could work back from the answers offered:Minesh wrote:before you ask why we are doing them - this is what set for hwk - any thoughts please
Tickets for a school play cost £3 for adults and £1 for children. The total amount collected from ticket sales was £1320. The play was staged in a hall seating 600 but the hall was not completely full. What was the smallest possible number of adults at the play?
358, 359, 360, 361, 362
358 adults means 1320-3*358 = 246 children, total 604 (overfull)
359 adults means 1320-3*359 = 243 children, total 602 (overfull)
360 adults means 1320-3*360 = 240 children, total 600 (completely full)
361 adults means 1320-3*361 = 237 children, total 598 (not completely full)
362 adults means 1320-3*362 = 234 children, total 596 (not completely full)
so 361 is the smallest possible.