
By: achieve_goal (offline) Tuesday, June 04 2013 @ 01:24 AM CDT (Read 855 times)



achieve_goal 
Q1.mrs Liu needed to type a 20page report to submit to her boss.She typed at a rate of
50 words per minutes for the first 8 pages.She slowed down to a rate of words per
minute for the remaining pages.On average, the first 8 pages had 500 words each
and the rest of the pages had 200 words each. How long did Mrs Liu take to type the
entire report? Give the answer in hours and minutes.
Q2.To prepare for the basketball challenge, James practised throwing the ball into the basket.
He threw 80 times in total.For the first 60 throws, the ball went through the basket
2times out of every 5 throws. For the remaining throws, he managed to score
85% of the throws. How many times did his ball miss the basket?
Q3. Car A and B started travelling at 9.45 from the opposite ends of a
straight road.Car Atravelled at an average speed of 40km/h. Caar B travelled 10km/h faster and took 1.5 hours less than Car. A to travel the entire road.Find the tim e when both cars met.

Regular Member
Registered: 09/14/11 Posts: 102





By: echeewh (offline) Wednesday, June 05 2013 @ 04:02 AM CDT



echeewh 
Hey there,
Appreciate that it will be good if you can help to provide the <Answerkey> for verification purpose. Thankyou.
Following pls find my worked solutions:
Q1.
Clarification:
Is the rate missing for the remaining 12 pages of the report? Kindly check.
Your statement:
She slowed down to a rate of words per minute for the remaining pages.
=========
Q2.
<1st 60 throws>
60 ÷ 5 = 12 groups/sets (of 2 into basket)
12 × 2 = 24 times (into basket)
Missed basket = 60  24 = 36 times
<remaining 20 throws>
Missed basket = 15%
(15/100) × 20 = 3
Total misses = 36 + 3 = 39
=========
Q3.
*Assume both cars started at 9.45 am {missing am or pm in question}
s(A) : s(B )
40 : 50
4 : 5
Given that distance is common,
t(A) : t(B )
5 : 4
Given that Car B took 1.5 hours less than Car A to travel entire road,
t(A)  t(B ) = 1.5 hrs
5u  4u > 1.5 hrs
1u > 1.5 hrs
Hence,
t(A) = 5u
5u > 5 × 1.5 = 7.5 hrs
total distance = s(A) × t(A)
= 40 × 7.5 = 300 km
Time taken to meetup
= 300 ÷ (40 + 50)
= 300 ÷ 90
= 3 1/3 hr
= 3 hr 20 min
Time when both cars met
= 09 45 + 3 hr 20 min
= 13 05
= 1.05 pm
=========
Trust this helps.
Do let me know again if this is different from your <Answerkey> or if there's further clarification.
Cheers,
Edward

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



