
By: Kuro (offline) Sunday, May 18 2014 @ 09:54 AM CDT (Read 2838 times)



Kuro 
Can help solve the following questions? (No answer key given)
Q1. One machine took 70 minutes while another took 100 minutes to print the same number of copies of a newspaper. The faster machine printed 6 more copies of the newspaper per minute than the slower one.
(a) The slower machine completed the job at 1pm. At what time was the printing started?
(b) What was the total number of copies printed by the two machines?
Q2. Leon had a bottle of apple juice. He used an equal amount of juice each day. At the end of the 12th day, 1/3 of the bottle was left. At the end of the 14th day the amount of juice left was 200ml. What was the amount of juice he had at first?

Junior
Registered: 12/31/06 Posts: 27





By: shrinandhan (offline) Sunday, May 18 2014 @ 08:48 PM CDT



shrinandhan 
problem  1
a) 11.20 a.m
We first covert 100 mins into hours and mins. so that will be 1hr and 40 mins.then we find the time that is 1 hr and 40 mins before 1 p.m. i.e., 11.20 a.m
b) 2800 copies
find the LCM of 70 and 100 which is 700. we then divide 700 be 70 and 100.(time taken for printing by the two printers)
But the answers are 10 and 7 respectively. their difference is only 3. in the problem it says the difference is 6, which
is twice the answer we got. so we multiply 700 by 2. the answer the 1400 for 1 machine. since there are 2 machines, we multiply 1400 by 2 to get the answer.
Problem  2
900 ml, he had at first.
use algebra in equations, where x = the total amount of juice he had at first. let y = no. of ml he used per day.
x  14 y = 200
x  12y = 1/3x

Newbie
Registered: 08/17/10 Posts: 1





By: echeewh (offline) Tuesday, May 20 2014 @ 11:10 PM CDT



echeewh 
Hey <Kuro>
Following are my worked solutions, which make use of a slightly different approach to solve <Fractions>:
Q1
We can make use of Rate and Fractions to solve this question as follows:
(a)
Let the slower machine be B.
Given that B took 100 mins to print same number of copies and it took B to complete job at 1pm, we have ...
13 00  1 hr 40 mins = 11 20
Hence, B started printing at 11.20am
(b)
Machine A printed (1/70)total in 1 min
Machine B printed (1/100)total in 1 min.
Given that A printed 6 more copies of the newspaper per minute than B, we have ...
(1/70)total  (1/100)total = 6
(10/700)total  (7/700)total = 6
(3/700)total = 6
Total = 6 × (700/3) = 1400
Hence, total copies (A + B ) = 2 × 1400 = 2800 copies
========
Q2
(2/3)bottle was used after 12 days.
Given that equal amount of juice was used each day, we have ...
(1/3)bottle was used in 6 days.
After another 2 days, 200ml was left. This implies that this 200ml would be consumed in another 4 days { as (1/3)bottle was left after 12 days }
4 days > 200
18 days (whole bottle) > (200 ÷ 4) × 18 = 900ml
Hence, amount of juice at first was 900ml
==========
Trust these help.
Do let me know if these are different from your <Answerkey> or if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623



