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## Wednesday, April 15, 2009

### Rosyth Sch 2006 PSLE Math Prelim Q44

A rectangular tank 2.5 m long and 1.2 m wide is filled with water from two taps. Tap A fills it with water at the rate of 12 litres per minute and Tap B fills it up with water at the rate of 15 litres per minute. Both taps are turned on at the same time. What is the height of the water in the tank after 8 minutes?

Solution

Tap A ----- 12 litres per min
Tap B ----- 15 litres per min
Total ----- 27 litres per min

1 min ----- 27 litres
8 min ----- 8 x 27 litres = 216 litres or 216 000 cubic cm

Volume = Length x breadth x height
216 000 cubic cm = 250 cm x 120 cm x height
height = 216 000 cubic cm divided by (250 x 120) square cm

xiaorou said...

Questions from cbox

The ratio of the number of X Y is 7:3.195 oranges from Box X transfered to Box Y,then number of orange in Box X became 2/3 of the number of orange in Box Y.If 3/5 of the oranges in Box Y were sold, how many oranges were left in Box Y?

Grace had 40 red and 48 green beads. SHe used all the beads to make into flower bookmarks. The number of red and green beads used for each bookmark was the same. What is the maximum number of bookmarks that she could make?

Alan and Roy decided to meet and then walk to a park together. Alan cycled from school and reached the meeting place 10 minues before Roy. Roy skated from school and took 25 minutes. Both of them then walked to the park at 1/6 of the average speed by 25% for the last 15 minutes. (1) If they walked 240 m in the last 15 minutes, what was the average speed at which Roy skated to the meeting place? (2) What was the average speed at which Alan cycled to the meeting place?

Tong Ming and Wei De bought 2 fruits, apples and oranges, from a fruit stall. Altogether, they bought 30 more apples than orange. Tong Ming bought 2/5 of the total number of fruits. Wei De bought 2/3 of the apples ajnd 1/2 of the orange. How many fruits did they buy altogether? Ans : 150

Jerry bought some wallets at \$4 each and later sold all of them at \$9 each. His customers bought either 1 or 2 wallet from him. Customers who bought 2 wallets from him. Customers who bought 2 wallets from his would be given one wallet for free. At the end of the sale, Jerry paid away 198 free wallets and mde \$2628 above what he had paid for them at first. How many of his customers only bought one wallet from him? Answer : 288

When Mrs Lee was 40 years old, her son was twice her daughter’s age. Mrs Lee will be twice her son’s age when her daughter is 28 years old. How old will Mrs Lee be when her daughter is 20 years old?

Last year, the ratio of the number of boys to the number of girls in a chess club was 5 : 3. After 5 boys and 9 girls joined the club this year, the ratio became 5 : 4. How many girls were there in the chess club at first?

consolidated said...

Consolidated problems from Cbox
Jack: Jack picks Jane up from her house every morning at 7am. Yesterday Jane left for work at 6am. She cycled to work taking the same road that Jack normally takes.When jane met Jack on the way she boarded the van and arrived at her office 50 minutes earlier than usual. Given that Jack drove at a constant speed and decided to arrive at Jane house at 7am sharp, how long had Jane cycled before she was picked up by Jack ?

John: John cycled at an average speed of 10km/h from his house to the park. On reaching the park, he cycled back home along the same route at an average speed of 8km/h. He took 1h 12 mins for the entire journey. How long did he take to cycle from the park back to his home?

Cycling: D = S1 x T1 = S2 x T2. T2 / T1 = S1 / S2 = 10 / 8 = 5 / 4. 1 h 12 mins = 72 mins. T2 = 5 / 9 x 72 = 40 mins.

Consolidated said...

Saving from cbox

SOS: Amy took 3 hours to travel from Town P to Town Q. Pansy took 2 hours to travel from Town Q to Town P. Both of them started travelling at 9am. At what time did they pass each other?

Helping SOS: Amy : 3 h -> 1 journey. 1 h -> 1/3 journey.Pansy: 2 h-> 1 journey. 1 h - 1/2 journey.Total distance travelled in 1 h-> 1/3+ 1/2 = 5/6 journey. Time required to meet-> 6/5 = 1-1/5 h.From 9 am, add 1-1/5 h (72 min) -> 1012 am.

A met P: D = SA x TA = SP x TP. SA : SP = TP : TA = 2 : 3. When A met P, distance is a constant, so T = 2/5 x 3 (or 3/5 x 2) = 6/5 h = 1 h 12 min. Using timeline 9 am.....1 h 12 min .... 10.12 am.
A met P: Sorry it should be T is same.

Ji Ho the Man: Hello. I'm new here. Just asking how do you solve two questions: A beaker of water mixed with wood chips, salt, and iron filings are required to be filtered. How should you separate each of them?

Mixture: First use a magnet to attract the iron filings to separate them from the mixture. Then pour the remaining mixture into the filter, wood chips will be filtered out. Heat up the remaining mixture, you will get salt.

SOS: A bus and a car travelled from Town X to Town Y. The bus left Town X at 10.48pm and it took 5 hours to reach Town Y. The car started 30 minutes later than the bus and it took 4 hours to reach Town Y. At what time did the car catch up with the bus?

Next day: D = SB x TB = SC x TC. SC = TB / TC x SB = 5/4 SB = 1.25 SB. In 30 min, DB1 = SB x TB1 = 0.5 SB (=Dd). SC - SB = 1.25 SB - SB = 0.25 SB. 0.25 SB --> 1 h. 0.5 SB --> 2 h. Using timeline 10.48 pm.....+30 min + 2 h ..... 1.18 am.

consolidated said...

Consolidated from Cbox, just to prevent them from erasing from memory.

P6 Student: BTW Q45: A car was on its way from Town Y to Town Z.After covering 2/7 of his journey,it overtook a lorry which was travelling at an average speed of 65 km/h.4 hours later, the car reached Town Z, but the lorry was still 60 km away from Town Z.(a) Find the average speed of the car (b)Find the distance between Town Y and Town Z Answer is 80 km/h & 448 km

KKhing: Sam and Colin drove from City A to City B. Sam reached City B in 16 hours. Colin left 6 hours later than Sam and reached City B at the same time as Sam. The difference between their average speed was 30 km/h. (a) Find the distance Sam had travelled when Colin left City A. (b) Find the distance between the 2 cities.

Complete: To P6 student ---Please show the complete question --- The ratio of the number of buns to the number of cakes in a bakery was 4:1. After another 63 buns and 46 cakes were baked, the ratio became 3:1. How many buns and cakes were there in the bakery at first?

Question: Elieen, Gideon and Fiona have some money. Eileens money is 25% more than Gideons and Gideon's money is 60% more than Fiona's. Then, Eileen and Fiona gave some money in the ratio of 3:1, and Eileen left with \$910 and Fiona is left with \$560. How much money did Gideon have at first?

Money: Before: E-> 125/100x160=200%, G->160%, F->100%.After: E-> \$910, G->?, F->\$560. Let the amount of money that Fiona gave away be unit.\$910+3 units=2(\$560+1 unit). \$910+3 units=\$1120+2 units. 1 unit=\$210. Amount of money that Fiona had at first-> 210+560= \$770. Amount of money that Gideon had at first-> 160/100x770 = \$ 1232.

consolidated said...

From Cbox

KKhing: Car A and Car B left Town Y at the same time heading in the opposite directions.Car A headed for Town Z while Car B left for Town X. The speed of Car B was 20 km/h faster than Car A. After 1/2 h, Car A had completed 2/3 of its journey while Car B had completed 1/2 of its journey. The two cars were also 110 km apart.(a) Calculate the speed of Car A (b) How far was Car B from Town X when car A reached its destination?

KKhing: Mr Lee and Mr Tan both drove from Town X to Town Y.Mr Lee started his journey at 10.00 am travelling at an average speed of 75 km/h. Some time later, Mr Tan started his journey. At 12.00 noon, Mr Tan overtook Mr Lee.When Mr Tan reached Town Y at 2.00 pm, Mr Lee was still 50 km from Town Y. Find (a) Mr Tan's average speed (b) The time at which Mr Tan started his journey

KKhing: Can someone help me? Tks. Town A and Town B were 600 km apart. At 10.45 am,a lorry traveling at a uniform speed left Town A for Town B. At the same time, a taxi set off from Town B to Town A at a uniform speed which was 12 km/h faster than that of the lorry's. The two vehicles met at 3.45 pm.Find the speed of the taxi.

Car: 4 h --> 60 km. 1 h --> 60 / 4 = 15 km. a) SC = 65 + 15 = 80 km/h. 5/7D = 80 x 4 = 320 km. b) D = 7/5 x 320 = 448 km.

consolidated said...

consolidated from Cbox

A van left Metropolis Town and travelled towards Gotham Town at an average speed of 80 km/h. An hour later, a car left Metropolis Town and travelled towards Gotham Town at an average speed of 100 km/h. The car passed Keystone Town, which was halfway between Metropolis Town and Gotham Town, 30 minutes earlier than the van. The car reached Gotham Town at 10 pm. At what time did the van leave Metropolis Town for Gotham Town?

At 9.00 am, Mr Lee's car passed a certain point X, travelling at an average speed of 60 km/h. At 10.30 am, Mr Chua's car started off from point X at an average speed of 80 km/h in pursuit of Mr Lim's car. (a) At what time did Mr Chua's car overtake Mr Lim's car (b) How far was Mr Chua's car from point X when it overtook Mr Lim's car

consolidated said...

from cbox

in my science paper, there was a question that asked why the temperature of ice would not rise until it is completely melted. i answered that it was because ice will not gain heat till it melt. This answer was not accepted. The model answer is that the heat from the melting ice was used to melt the remaining ice. Are there any other answers that are expected in PSLE?

Anonymous said...

In need: Q fm Nanyang 2006 Prelim Q46:coach--\$30, train--\$15,bus--\$10. The ratio of the no. of coach tickets sold to the no. of train tickets sold was 5:6. The ratio of the no. of train tickets sold to the no. of bus tickets sold was 4:11. The amount of money collected from the sale of the bus tickets was \$270 more than that collected from sales of coach tickets. How much more money was collected from the sale of coach tickets than the train tickets???

To "In need": C : T 5 : 6 (Multiply by 2) 10: 12 T : B 4 : 11 (Multiply by 3) 12 : 33 C : T : B 10 : 12 : 33 Amount collected Coach tickets-> 10 units x \$ 30 = \$ 300 units Train tickets-> 12 units x \$ 15 = \$ 180 units Bus tickets -> 33 units x \$ 10 = \$ 330 units \$330 units - \$ 300 units = \$ 270 30 units = \$ 270 1 unit = \$ 9 Difference between money collected from sale of coach tickets and train tickets-> \$300 units - \$ 180 units = \$ 120 units = 120 x 9 = \$ 1080

Help: bus and a car travelled from Town X to Town Y. The bus left Town X at 1048pm and it took 5 hours to reach Town Y. The car started 30 min later than the bus and it took 4 hours to reach Town Y. At what time did the car catch up with the bus?

Car: To Help: This qn sounds familiar and pls advise the source of your question. Ratio of time taken for bus and car is 5 : 4. Speed for bus and car will be 4 : 5. 1/2 x 4 = 2 (distance travelled by bus). Difference in speed-> 5-4 = 1 unit. 2/1 = 2 h.Use timeline to add 2 h to 11.18 pm will be 1.18 am.

Anonymous said...

In need: Q fm Nanyang 2006 Prelim Q46:coach--\$30, train--\$15,bus--\$10. The ratio of the no. of coach tickets sold to the no. of train tickets sold was 5:6. The ratio of the no. of train tickets sold to the no. of bus tickets sold was 4:11. The amount of money collected from the sale of the bus tickets was \$270 more than that collected from sales of coach tickets. How much more money was collected from the sale of coach tickets than the train tickets???

To "In need": C : T 5 : 6 (Multiply by 2) 10: 12 T : B 4 : 11 (Multiply by 3) 12 : 33 C : T : B 10 : 12 : 33 Amount collected Coach tickets-> 10 units x \$ 30 = \$ 300 units Train tickets-> 12 units x \$ 15 = \$ 180 units Bus tickets -> 33 units x \$ 10 = \$ 330 units \$330 units - \$ 300 units = \$ 270 30 units = \$ 270 1 unit = \$ 9 Difference between money collected from sale of coach tickets and train tickets-> \$300 units - \$ 180 units = \$ 120 units = 120 x 9 = \$ 1080

Help: bus and a car travelled from Town X to Town Y. The bus left Town X at 1048pm and it took 5 hours to reach Town Y. The car started 30 min later than the bus and it took 4 hours to reach Town Y. At what time did the car catch up with the bus?

Car: To Help: This qn sounds familiar and pls advise the source of your question. Ratio of time taken for bus and car is 5 : 4. Speed for bus and car will be 4 : 5. 1/2 x 4 = 2 (distance travelled by bus). Difference in speed-> 5-4 = 1 unit. 2/1 = 2 h.Use timeline to add 2 h to 11.18 pm will be 1.18 am.

Anonymous said...

In need: Q fm Nanyang 2006 Prelim Q46:coach--\$30, train--\$15,bus--\$10. The ratio of the no. of coach tickets sold to the no. of train tickets sold was 5:6. The ratio of the no. of train tickets sold to the no. of bus tickets sold was 4:11. The amount of money collected from the sale of the bus tickets was \$270 more than that collected from sales of coach tickets. How much more money was collected from the sale of coach tickets than the train tickets???

To "In need": C : T 5 : 6 (Multiply by 2) 10: 12 T : B 4 : 11 (Multiply by 3) 12 : 33 C : T : B 10 : 12 : 33 Amount collected Coach tickets-> 10 units x \$ 30 = \$ 300 units Train tickets-> 12 units x \$ 15 = \$ 180 units Bus tickets -> 33 units x \$ 10 = \$ 330 units \$330 units - \$ 300 units = \$ 270 30 units = \$ 270 1 unit = \$ 9 Difference between money collected from sale of coach tickets and train tickets-> \$300 units - \$ 180 units = \$ 120 units = 120 x 9 = \$ 1080

Help: bus and a car travelled from Town X to Town Y. The bus left Town X at 1048pm and it took 5 hours to reach Town Y. The car started 30 min later than the bus and it took 4 hours to reach Town Y. At what time did the car catch up with the bus?

Car: To Help: This qn sounds familiar and pls advise the source of your question. Ratio of time taken for bus and car is 5 : 4. Speed for bus and car will be 4 : 5. 1/2 x 4 = 2 (distance travelled by bus). Difference in speed-> 5-4 = 1 unit. 2/1 = 2 h.Use timeline to add 2 h to 11.18 pm will be 1.18 am.

Anonymous said...

Sorry, duplicated posts

Anonymous said...

No Algebra if possible: This Qn is from another blog : A motorist takes 1/2 h less than the usual time to travel from Town K to Town J when he increases his average speed by 10%. For a distance of 100km, he will take 1/3 h less than the usual time if his speed is 20% more than the average speed from Town K to Town J. Find the distance between Town K and Town J.

Constant distance: 1 : 1.1 = T - 0.5 : T. T = 1.1 T - 0.55. T = 5.5 h. 1 : 1.3 = T1 -1/3 : T1. T1 = 1.2 T1 - 0.4. T1 = 2 h. 2 h -- 100 km. 5.5 h -- 5.5/2 x 100 = 275 km.

Anonymous said...

Another method: Scenario A -> Speed (110%) & time is 1/2 h less. Scenario B -> Speed (120% for first 100 km & 100% for the rest of remaining distance) & time is 1/3 h less.Scenario C -> Speed (100%). Ratio of speed A : B : C 110 : 120 : 100 11 : 12 : 10 Common multiple of 11,12,10 is 660 Ratio of time A : B : C 60 : 55 : 66 C-A -> 66-60 = 6 units 6 units-> 1/2 h 60 units-> 1/2 x 10 = 5 h (A) Speed A : 11 units Time A : 5 h Distance = 11 x 5 = 55 units For scenario B-> 5+1/2-1/3 = 5 and 1/6 = 31/6 h (55 units - 100 km)/10 + 100 km/12 = 31/6 (330 units - 600 km + 500 km)/60 = 31/6 1980 units - 600 km = 1860 units 120 units = 600 km 1 unit = 5 km 55 units = 5 x 55 = 275 km Distance between the two towns is 275 km.

Constant distance: D = S x T = 1.1 S x (T - 0.5). S/1.1 S = (T -0.5)/T. 1 : 1.1 = T - 0.5 : T (T is time for complete distance). Similarly, 1 : 1.2 = T1 - 1/3 : T1 (T1 is time for 100 km).

Anonymous said...

Full Solution by "CD": Ratio of speed 100 % : 110 % ->1 : 1.1. Ratio of time -> T-0.5 : T where T is the time to reach the destination while travelling at normal speed. Cross multiply them becomes : T = 1.1(T-0.5). T = 1.1T-0.55. 0.1T = 0.55. T = 5.5 hours. Ratio of speed for 100 % : 120 %-> 1 : 1.2. Ratio of time ->T1-1/3 : T1 where T1 is the time to travel first 100 km of the journey. Cross multiply them becomes T1 = 1.2(T1-1/3). 3T1 = 1.2(3T1-1). 3T1 = 3.6T1 - 1.2. 0.6T1 = 1.2. T1 = 2 hours. 2 hours-> 100 km. 5.5 hours -> 100/2 x 5.5 = 275 km. Distance between the two towns is 275 km.

Anonymous said...

enlightenment needed: ahmad has 50% more money than Bernard. Bernard has 20% more money than Clement. after Ahmad gives Bernard and Clement a total of \$350, and Bernard gives Clement \$400, the ratio of Ahmad's money to Bernars money to Clements money became 1:1:3. How much money did ahmad give to bernard?

Chocolate Cake: Ratio of A:B-> 150:100 ie 3:2.Ratio of B : C->120:100 ie 6:5.Before scenario#A:B:C->9:6:5.After scenario#A:B:C-> 1:1:3 ie 4:4:12. 5 units(9-4)->\$350.12 units->\$840.7 units(12-5)->\$490.490-400=90.350-90=260.

Chocolate Cake: Ratio of A:B-> 150:100 ie 3:2.Ratio of B : C->120:100 ie 6:5.Before scenario#A:B:C->9:6:5.After scenario#A:B:C-> 1:1:3 ie 4:4:12. 5 units(9-4)->\$350.12 units->\$840.7 units(12-5)->\$490.490-400=90.350-90=260.Ahmad gave \$260 to Bernard

B C: A : B = 3 : 2 = 9 : 6. B : C = 6 : 5. A : B : C = 9 : 6 : 5. AN : BN : CN = 1 : 1 : 3 = 4 : 4 : 12. 5 u -- \$350. 1 u -- 350 / 5 = \$70. 7 u -- 7 x 70 = \$490 [400 from Bernad]. 490 -- 400 = 90 [from Ahmad]. 350 -- 90 = \$260.

Anonymous said...

urgent: MR kabil is going to deliver some vegetable from his home to a market in town. If je travelled at an average speed of 20 km/h he wil reach the market at 1pm. If he travelled at and average speed of 30km/h, he will reach the market at 11am. What speed should he travel at to reach the market at 12nn

help me: The no. of 20cent coins in a box is 1/3 the no. of 50cent coins. 8 fifty cent coins were exchanged for 20cent coins and put in the box. the no. of 50 cent coins became 4/7 the no. of 20 cent coins. how much money was there in the box

Square: Diagonal x Diagonal = 2 x Area of Square.

To "urgent": 20 km/h -> 1 pm. 30 km/h -> 11 am. 30 x T = 20 x ( T+2 ). T = 4 hours. 11 am -> 4 hours at 30 km/h .12 pm-> 30 x 4/5 = 24 km/h.

Old messages: 8 fifty cent coins were exchanged for 20cent coins and put in the box. the no. of 50 cent coins became 4/7 the no. of 20 cent coins. ==> New 50-cent - 4 parts, New 20-cent - 7 parts. So x 7 for 50-cent and x 4 for 20-cent. Then equate them.

Anonymous said...

Red Swastika 2008 CA1 Q41: Auntie May cooked 520 fishballs and sotong balls altogether.She sold 75% of the fishballs and 50% of the sotong balls.She was left with 180 fishballs and sotong balls altogether.How many fishballs did she cook at first?

curious boy: 60% of the admission tickets to a show are reserved for children. The ratio of admission tickets for men to women is 5:7. During the school, the number of admission tickets are increased by 20%. the number of admission tickets reserved for children has increased to 1080. How many tickets are reserved for adults during the holidays? and how many tickets are reserved for men only before the school holidays?

PSLE 2007 Maths Qn: At first, Shop A had 156 kg of rice and Shop B had 72 kg of rice.After each shop sold the same quantity of rice, the amount of rice that Shop A had was 4 times that of Shop B.How many kilograms of rice did Shop A sell ?

To curious boy: Ratio of Men : Women -> 5 : 7.Ratio of Children : Adults-> 60 : 100 ie 15 : 25. (a) 20/100 x 15 = 3. 3 units->1080. 25 units-> 9000.120/100 x 9000 = 10800. (b) 3 units-> 1080.25 units-> 9000. 5/12 x 9000 = 3750.

Anonymous said...

Blur Blur: So happy, finally get to draw semi circle with Word 2007.First.insert shape. Choose circle. Then choose square. Overlap square over circle to see a semi circle. Right click on square, select no line. Then insert line and draw to complete semi circle.

Anonymous said...

Solar System: The sun is 400 times bigger than the moon. But the moon is 1/400 times nearer to Tricia than the sun. So the sun appeared to be the same size as the moon.

Anonymous said...

PSLE kids: Any simpler way of doing this question --- Rooney has five more 50 cent coins than 20 cent coins. After he used eight 50 cent coins, the value of 50 cent conis is \$1.50 more than the value of 20 cent coins. How many coins did he have at first?

Another method: Draw a model for one block of 20-cent coin (1 unit) and one bigger block to represent 1 unit+5 for 50-cent coin.After using 8 each of 50 cent coins, draw another model to show 20-cent coins remaining as 1 block (1unit) and a smaller block for 50 cent coins (unit+5-8 that is unit-3).Value of 50 cent coins is \$1.50 more than value of 20 cent coins.Value of 20 cent coins will be 20 cents x u -> 20 cents units.Value of 50 cent coins will be 50 cents x (units-3) which will be 50 units - 150.Equate them.20u+150=50(u-3).20u+150=50u-150.30u=300.u=10.The rest is simple, try it out.Have fun.

Anonymous said...

help!: In class of 42 pupils , 2/5 of the boys and 75% of the girls weear spectacles. Given that there are 14 pupils who did not wear spectacles, find the no. of boysa who wear spectacles.

enlightenment needed: At 12nn, a motorist left town A travelling towards town B at a constant speed. 2 hrs later, a van started from town A along the same road. The van drivver overtook the motorist at 5pm. The speed of the van was 40km/h faster than the motorist. What is the speed of the motorist?

questioner: Mr tan and Mr lee each reared a certain no. of chicken in their farm. If mr tan sold 60 chicken per day and mr lee sold 30 per day, Mr tan would have 300 left when mr lee has none.If mr tan sold 30 chicken per day and mr lee sold 60 per day, Mr tan would have 930 left when mr lee has none. Find the no. of chickens mr tan has in his farm.

Anonymous said...

Help needed: Alan had 25% more marbles than Chris. Chris had 60% more marbles than Ben. During a game, Alan and Ben lost some marbles to Chris in the ratio 3:1. In the end, Alan and Ben had 780 and 480 marbles left respectively. How many marbles did Alan have at first?

Use same concept: Use same concept to solve.Change of names only.Ratio of A:C ->125:100 ie 5:4.Ratio of C:B-> 160:100 ie 8:5.Ratio of A:C:B->10:8:5.Ratio of marbles lost->A:B was 3:1 ie 6:2.(10+5)-(6+2)=7. 7 units-> 1260. 1 unit-> 180. 3 units-> 540. 540+780 = 1320.At first, Alan had 1320 marbles.

Anonymous said...

Math: A : C = 5 : 4 = 10 : 8 (x 2, so that C is now the same). C : B = 8 : 5. 10 units - 3 parts --> 780. 5 units - 1 part --> 480 => 15 units - 3 parts --> 1440 (x 3 to get 3 parts, same as earlier statement). 5 units --> 1440 - 780 = 660. A:10 units --> 10/5 x 660 = 1320 marbles

Anonymous said...

P6 BOY: Ally, Billy, Chitra and Dawn have some stamps. The number of stamps Ally has is 1/3 of the total number of stamps Billy, Chitra and Dawn have. The number of stamps Billy has is 1/4 of the total number of stamps Ally, Chitra and Dawn have. The number of stamps Chitra has is 1/5 of the total number of stamps Ally, Billy and Dawn have. If Dawn has 460 stamps, how many stamps do the four of them have altogether.

Anonymous said...

mona has 60% more stamps than HL. K has 15% less stamps than Mona. diff. in K and HL is 216 stamps. how many stamps does Mona have?

Anonymous said...

There are two circles, A and B. The circumference of circle A is 30pie cm more than the circumference of circle B. Find the diameter of circle A?

Circle: Diameter of circle A = 30 cm + Diameter of circle B.

Pls help: 2009 X 2009 X 2009 X 2009.............2009 ( 2009 terms ) What is the one's digit of the value?

Odd and Even: 9 x 9 = 81 (End with 1). 81 x 9 = 729 (End with 9). When repeated, the ending digits will be 1, 9, 1, 9, etc. So even 2009 --> 1, Odd 2009 --> 9.

Think so -- Pls help: I think the answer is "9". Those to the power of odd numbers (1,3,5,....) end in "9" and powers in even numbers (2,4,6,.....) end in "1".Since 2009 to the power of 2009 is an odd number (power), answer should be "9".

Anonymous said...

Model Drawing: 3 Oranges --> \$2.10. 1 Orange --> 2.10/3 = \$0.70. 1 Mango --> 0.70 + 2.10 = \$2.80. 8 Mangoes --> 8 x 2.80 = \$22.40. Cost of Oranges --> 30.80 - 22.40 = \$\$8.40. No of oranges = 8.40/0.70 = 12. 12/20 = 3/5.

Anonymous said...

Topic on Speed: If they started at different time, distance is constant

Technically speaking, it will be easily for us to solve questions that started at different time without using ratio. These questions are easy as compare to the qusetions that started at the same time.

A better way to help you understand better is to draw a table and list down their speed. You will realise that the speed is not proportional unlike those who set off at the same point and at the same time.

Anonymous said...

In 1 hour, Amy covered 1/3 of the distance and Pansy covered 1/2 of the distance. Together, both PRETTY girls covered 1/3 + 1/2 = 5/6 of the distance. They travelled 5/6 of the distance in 1 hour.So, they could cover 1/6 of the distance in 1/5 hour.The time to cover the total distance between Towns P & Q together would be 6 x 1/5 hour = 1 and 1/5 hour ie 1 h 12 min. Use timeline. 9 am -----------1h 12 min...............10:12 am. They passed each other at 10.12 am.

Anonymous said...

2008 PSLE Speed question: At 10.15 am., Alice left Town X for Town Y driving at a speed of 80 km/h throughout her journey. At 11.15 a.m., Bobby also left Town X for Town Y driving at a certain speed. He kept to the same speed throughout his journey. At 1.15 p.m., both of them passed a petrol station that was 100 km away from Town Y. (a) What was Bobby's driving speed ? (b) How many minutes earlier did Bobby reach Town Y than Alice ?

2007 PSLE Speed qn: David and Michael drove from Town A to Town B at different speeds. Both did not change their speeds throughout their journeys. David started his journey 30 minutes earlier than Michael. However, Micheal reached Town B 50 minutes earlier than David. When Michael reached Town B, David had travelled 4/5 of the journey and was 75 km away from Town B. (a) What was the distance between Town A and Town B? (b) How many kilometres did David travel in 1 hour? (c) What was the time taken by Michael to travel from Town A to Town B?

2006 PSLE Speed qn: At 0900, a lorry started from Town X and travelled towards Town Y at a speed of 55 km/h for the whole journey. At 1100, a car started from Town Y and travelled towards Town X. The speed of the car remained the same until it passed the lorry at 1300. At this point, the lorry had travelled 5/9 of the journey. After passing the lorry, the car decreased its speed by 8 km/h and travelled at the new speed for the remaining journey. At what time did the car reach Town X ?

2005 PSLE Speed qn: Raju and Gopal took part in their school's walkathon which started at 7.30 am. Raju's average speed was 20 m/min faster than Gopal. When Raju completed the walkathon in 30 minutes, Gopal had only walked 5/6 of the distance. (a) What time was it when Gopal completed the walkathon? (b) Find Raju's average speed for the walkathon in m/min?

Topic on Speed said...

2008 PSLE Speed question: At 10.15 am., Alice left Town X for Town Y driving at a speed of 80 km/h throughout her journey. At 11.15 a.m., Bobby also left Town X for Town Y driving at a certain speed. He kept to the same speed throughout his journey. At 1.15 p.m., both of them passed a petrol station that was 100 km away from Town Y. (a) What was Bobby's driving speed ? (b) How many minutes earlier did Bobby reach Town Y than Alice ?

Solution:
(a)
Time taken by Alice = 13.15 – 10.15
= 3 hrs

Distance covered by Alice
= 3 hrs x 80 km/h
= 240 km

Since Alice has a headstart of 1 hr, time taken for Bobby to travel the same distance is 2 hrs ( 3 – 1). To find the speed of Bobby, Distance travelled by both is constant.

Speed of Bobby = 240 / 2
= 120 km/h

(b) Time taken for Bobby to reach Town Y after passing Alice
= 100 /120
= 5/6 hrs

Time taken for Alice to reach Town Y after passing Bobby
= 100 / 80
= 1 1/4 hrs

1 1/4 - 5/6 = 5/12 hrs or 25 mins earlier for Bobby

2007 PSLE Speed qn: David and Michael drove from Town A to Town B at different speeds. Both did not change their speeds throughout their journeys. David started his journey 30 minutes earlier than Michael. However, Michael reached Town B 50 minutes earlier than David. When Michael reached Town B, David had travelled 4/5 of the journey and was 75 km away from Town B. (a) What was the distance between Town A and Town B? (b) How many kilometres did David travel in 1 hour? (c) What was the time taken by Michael to travel from Town A to Town B?

Solution:

1 unit ----> 75 km
5 units ----> 5 x 75
= 375 km

(a) Distance between Town A and B is 375 km.

Td = 5/6 hrs (This Td is the extra time for David to cover the remaining distance of 1 unit.)

5/6 hrs -----> 75 km
1 hr -----> 75 x 6/5
= 90 km/h
(b) Thus, speed of David is 90 km/h

Distance from Town A to Town B = 375 km [from part (a)]

Time taken for David to cover 375 km
= 375 / 90
= 4 1/6 hrs

Time taken for Michael to cover 375km
= 4 1/6 – 1/2 - 5/6
= 2 5/6 hr or 2 hrs 50 mins

2006 PSLE Speed qn: At 0900, a lorry started from Town X and travelled towards Town Y at a speed of 55 km/h for the whole journey. At 1100, a car started from Town Y and travelled towards Town X. The speed of the car remained the same until it passed the lorry at 1300. At this point, the lorry had travelled 5/9 of the journey. After passing the lorry, the car decreased its speed by 8 km/h and travelled at the new speed for the remaining journey. At what time did the car reach Town X ?

Solution:

First 4 hrs (from 0900 – 1300)
= 55 x 4
= 220 km for lorry

Distance from Town X to Town Y
= 9/5 x 220
= 396 km

Distance covered by car
= 396 – 220
= 176 km

Distance covered by car for first 2 hrs (from 1100 to 1300)
= 176 / 2
= 88 km/h

New speed = 88 km/h – 8
= 80 km/h

Time taken to for car to cover 220 km
= 220 / 80
= 11 / 4
= 2 3/4 hrs

Time taken for Car to reach Town X
= 1300 + 2 hrs 45 mins
= 1545 hrs

2005 PSLE Speed qn: Raju and Gopal took part in their school's walkathon which started at 7.30 am. Raju's average speed was 20 m/min faster than Gopal. When Raju completed the walkathon in 30 minutes, Gopal had only walked 5/6 of the distance. (a) What time was it when Gopal completed the walkathon? (b) Find Raju's average speed for the walkathon in m/min?

Solution:

1 min -----> 20 m more
30 mins -----> 600 m more

1 unit -----> 600 m
6 units -----> 600 x 6
= 3600 m

Speed of Gopal = 3000 / 30
= 100 km/h

Time taken for Gopal = 3600 / 100
= 36 mins

(a) Thus, 0730 + 0036 = 0806 hrs

(b) Speed of Raju = 100 + 20
= 120 km/h

Alternative, S = D / T ;

Speed of Raju = 3600 / 30
= 120 km/h

Topic on Speed said...

Mr Kabil is going to deliver some vegetable from his home to a market in town. If he travelled at an average speed of 20 km/h he will reach the market at 1pm. If he travelled at an average speed of 30km/h, he will reach the market at 11am. What speed should he travel at to reach the market at 12noon?

Note: The above questions can be solved using several methods such as Guess and Check method, logic reasoning, Ratio method and algebra. The use of guess and check will not be shown here as the method is tedious and time consuming.

Solution:

Method 1: Logic Reasoning

Td = 2 hr (1300 – 1100)

Note: Td is the extra time for Mr Kabil to cover the remaining distance at a speed of 20 km/h.

Dd = 20 km/h x 2 h
= 40 km more

Note: If there is a difference in distance, it will create difference in speed or vice versa.

Sd = 30 – 20
= 10 km/h more

10 km more -----> 1 h
40 km more -----> 40/10
= 4 hrs

Note: It took Mr Kabil 4 hrs to cover 40 km more. In other words, it took him 4 hrs to cover the whole journey.

Distance from home to market at a speed of 30 km/h
= 30 km/h x 4 hrs
= 120 km

Note: Distance is constant although he travelled at different speeds.

New time = 4 + 1 = 5 hr

Speed = 120 / 5
= 24 km/h

Method 2: Ratio method

For constant distance, D ;

T2 : T1 = S1 : S2
= 2 : 3

Note: It is important that the ratio of speed and time are reverse for constant distance.
However, most students are more comfortable by expressing the ratio of speed and time separately as follow;

S1 : S2
20 : 30
= 2 : 3

T1 : T2
3 : 2

Basically, both are correct although they are expressed differently.

Td = 3 -2
= 1 unit
1 unit -----> 2 hrs (1300 – 1100)

3 units -----> 6 hrs(3 units x 2 hrs)

Distance from home to market at a speed of 20 km/h (1300 hr)
= 20 km/h x 6 hrs
= 120 km

Alternatively,

2 units -----> 4 hrs (2 units x 2 hrs)

Distance from home to market at a speed of 30 km/h (1100 hr)
= 30 km/h x 4 hrs
= 120 km

T1 = 4 + 1 (1100 + 0100 = 1200) or
T2 = 6 – 1 (1300 – 0100 = 1200)

Note : Both T1 and T2 yield the same time, 5 hr.

Speed = 120 / 5
= 24 km/h

Method 3: Algebra method

Let t be the original time to cover the distance at a speed of 30km/h.

D1 = S1 x T1
= 30 x t
= 30t

and D2 = S2 x T2
= 20 x (t + 2)

For constant distance,

D1 = D2
30t = 20t + 40
10t = 40
t = 4

D1 = 30 x 4

or D2 = 20 x (4 + 2) ;

both will yield the same distance, 120 km.

Speed = 120 / 5
= 24 km/h

Alternatively,

Let t be the original time to cover the distance at a speed of 20km/h.

D1 = S1 x T1
= 20 x t
= 20t

D2 = S2 x T2
= 30 x (t – 2)

since distance is constant;

D1 = D2
20t = 30t – 60
10t = 60
t = 6

D1 = 20 x 6

or D2 = 30 x (6 – 2) ;

both will yield the same distance, 120 km.

Speed = 120 / 5
= 24 km/h

Anonymous said...

To Topic on Speed: Please help ----- Andy increases his speed by 25km/h. When he does so, the time taken to travel on a 25km trip decreases by 10 mins. What is his initial speed?

Topic on Speed said...

To Anonymous----> Please provide your nick. Many people are using Anonymous and can be very confusing as to whether the Anonymous is the same person.

This question had been solved by other students. Please roam around and check...

Hint: Such question can be solve using guess and check or algebra.

Ratio method may not be applied as there are missing hints such as speed or time....

Solve them using units.

Anonymous said...

if ann spent \$25 each week and Ben spent \$75 each week, Ann would still have \$1350 when ben finished all his money. If ann apen\$75 each week and Ben spent \$25 each week, ann would still have \$150 left when ben had finished all his money. How much did ann have at first???

To P6: Algebraic method but not advisable to use at P6 level. Scenario 1 : \$ spent in A days. Ann: 1 unit - 25A=1350. Ben : 1 part - 75A = 0. Scenario 2 : Spent in B days. Ann : 1 unit - 75 B = 150. Ben : 1 part - 25 B = 0. 1 part = 75A = 25 B. 3A=B. 1 unit = 25A+1350=150+75B. 25A+1350=150+75(3A). 25A+1350=150+225A. 200A=1200. A= 6. Ann-> 25A+1350. 150+1350=1500. At first, Ann had \$ 1500.

Anonymous said...

Mr Observer travels in his Mercedes-benz at a constant speed towards his office. If he increases the car's speed by 15 km/h, the time required is in a ratio of 6 : 5. If he reduces the car's speed by 15 km/h, it will need another 105 minutes to arrive at his office. Find the distance travelled by Mr Observer's car.

Mr Observer's car departed from Town A and travelled towards Town B. At the same time, his colleague's car started its journey in the direction of Town A from Town B. The two cars passed each other 64 km away from Town B and carried on with their journeys. Both men made an U-turn upon arriving at their own destinations so as to return to their own towns. They passed each other this time 52 km away from Town A. Find the distance between the two places where they passed each other.

Mr Observer was walking at a constant speed of 5 km/h towards home that was 40 km away because his car had broken down. He got a ride from a motorist with a kind soul who drove at 25 km/h in the same direction. Both men arrived at Mr Observer's home 2.5 hours later. How far had Mr Observer walked before he took a ride from the kind motorist?

Mr Observer and his colleague embarked on a moving escalator at the 1st floor. Impatient with the speed of the escalator, Mr Observer walked at a speed of 3 steps per second while his colleague walked at a speed of 2 steps per second. How many steps had the escalator when it was stationary? It was noted that Mr Observer took 20 seconds to arrive at the second floor and his colleague took 25 seconds to arrive at the same floor.

Mr Observer drives a Mercedes-benz at a constant speed of 108 km/h to his office in downtown every weekday morning. He drives back home at a constant speed of 72 km/h for his return trip. What is his average speed for the round-trip?

Topic on Speed said...

Mr Observer travels in his Mercedes-benz at a constant speed towards his office. If he increases the car's speed by 15 km/h, the time required is in a ratio of 6 : 5. If he reduces the car's speed by 15 km/h, it will need another 105 minutes to arrive at his office. Find the distance travelled by Mr Observer's car.

Solution: Ratio method

Case 1
For constant distance, D1 ;
S2 : S1 = T1 : T2
= 6 : 5

S2 denotes increased speed and
S1 denotes initiate (original) speed.
Alternatively, this can also be express as follow;

T1 : T2
6 : 5
S1 : S2
5 : 6

Sd = 15 km/h

1 unit -----> 15 km/h
5 units -----> 75 km/h (15 x 5)

Original speed = 75 km/h

Case 2

Reduced speed = 75 – 15
= 60 km/h

Td = 105 mins or 7/4 hr

Dd = 60 km/h x 7/4 hr
= 105 km

Sd = 15 km/h (75 – 60)

15 km more -----> 1 h

105 km more -----> 105 / 15
= 7 hr

Distance travelled by Observer
= 75 km/h x 7 hr
= 525 km

Mr Observer was walking at a constant speed of 5 km/h towards home that was 40 km away because his car had broken down. He got a ride from a motorist with a kind soul who drove at 25 km/h in the same direction. Both men arrived at Mr Observer's home 2.5 hours later. How far had Mr Observer walked before he took a ride from the kind motorist?

Solution: Guess and Check

Walk Time Ride time Total
km (h) km (h) (h)
5 1 h 35 1.4 2.4 Wrong
5.625 1.125 34.375 1.375 2.5

Topic on Speed said...

Mr Observer and his colleague embarked on a moving escalator at the 1st floor. Impatient with the speed of the escalator, Mr Observer walked at a speed of 3 steps per second while his colleague walked at a speed of 2 steps per second. How many steps had the escalator when it was stationary? It was noted that Mr Observer took 20 seconds to arrive at the second floor and his colleague took 25 seconds to arrive at the same floor.

Solution: Logic reasoning

S1 denotes the constant speed of the escalator.

The escalator is moving for a certain time while Observer and Colleague both started to walk at a later time;

Distance travelled by escalator
= S1 x 20
= 20 S1

Distance travelled by Observer alone
= 3 x 20
= 60

Combine distance (escalator + Observer), D1
= 20 S1 + 60

Distance travelled by escalator
= S1 x 25
= 25 S1

Distance travelled by Colleague alone
= 2 x 25
= 50

Combine distance (escalator + Colleague), D2
= 25 S1 + 50

For constant distance (number of steps remain constant),

D1 = D2
20 S1 + 60 = 25 S1 + 50
5 S1 = 10
S1 = 2

Number of steps
= 20 x 2 + 60
= 100 steps

or

Number of steps
= 25 x 2 + 50
= 100 steps

Topic on Speed said...

Mr Observer was walking at a constant speed of 5 km/h towards home that was 40 km away because his car had broken down. He got a ride from a motorist with a kind soul who drove at 25 km/h in the same direction. Both men arrived at Mr Observer's home 2.5 hours later. How far had Mr Observer walked before he took a ride from the kind motorist?

Solution:

Method 1: Guess and Check

Walk Time Ride time Total
km (h) km (h) (h)
5 1 h 35 1.4 2.4 Wrong
5.625 1.125 34.375 1.375 2.5

Thus, 5.625 km is for walking.

Method 2 : Logic Reasoning

Assuming that Observer walk at a speed of 5 km/m for 2.5 hr from the start to the end;

Walking distance = 5 x 2.5
= 12.5 km

He can only walk for a distance of 12.5 km at the expected time of 2.5hr

Time taken to cover the distance of 40 km at a speed of 5 km/h
= 40 / 5
= 8 hr for walking

Dd = 40 – 12.5
= 27.5 km riding

Sd = 25 – 5
= 20 km/h more

20 km more -----> 1 h
27.5 km more -----> 27.5 / 20
= 1.375 hr for riding

2.5 – 1.375 = 1.125 hr for walking

Distance for walking
= 5 km/h x 1.125 hr
= 5.625 km

Alternatively,

Assuming that Observer ride at a speed of 25 km/h for 2.5 hr from the start to the end;

Riding distance
= 25 km/h x 2.5 hr
= 62.5 km

Dd = 62.5 – 40
= 22.5 km

Note: If he ride from the start, he can cover a distance of 62.5. This will allow him to save 22.5 km from walking.

Sd = 25 – 5
= 20 km /h more

20 km less (save) ----->1 hr
22.5 km less (save) -----> 22.5/20
= 1.125 hr save for walking

Distance for walking
= 5 km/h x 1.125
= 5.625 km

Anonymous said...

KS: After both cars had passed each other, reduced speed for Mr Observer was 5/6x6= 5 units. Reduced speed for his friend was 3/4x5=15/4 units.After passing each other, Mr Observer travelled 5/5 = 1 hour to reach Town B. Why 5/5? Topic on Speed: To KS -----> why 5/5? Simple....Constant distance, ratio of speed, S1 : S2 = 6 : 5 and ratio of time is the reverse of speed, T1 : T2 = 5 : 6.....5 can be in units or hr. I still can't understand how you could have found the time when no actual speed was given.

Anonymous said...

Potential PSLE 2009 Question on Volume A rectangular tank, measuring 60 cm by 50 cm and 40 cm is completely filled with water to the brim. At the bottom of the tank, there are 9 cubes each of side 10 cm. Four similar cubes are on top of these 9 cubes and one similiar cube is on top of the four cubes. A tap at the bottom of the tank can drain water out at 30 ml/s. Find the time to drain the water out of the tank a) completely. [1] b) till the water just reaches the top of the bottom layer of cubes. [1] c) What is the time taken for the water to reach the level when the top cube is half submerged in water when the tap was on at 9.15 a.m.? [3] (Note: To ignore fraction of a second. Answers for a) and b) are to be given in hour and minutes and answer to c) to be given in 24 hour clock). Ans: a) 58 min 53 s b) 47 min 13 s c) 0923.

KS: a) Volume of tank = 60 x 50 x 40 = 120 000 ml. 1 cube --> 10 x 10 x 10 = 1000 ml. 14 cubes --> 14 x 1000 = 14 000 ml. Volume of water = 120 000 - 1 400 = 106 000 ml. Time = 106 000 / 3 second = 9 h 48 min 53 1/3 s. b) 60 x 50 x 30 = 90 000 ml. 5 cubes --> 9 x 1000 = 9 000 ml. Volume of water drained out = 90 000 - 5000 = 85 000 ml. Time = 85 000 / 3 = 7 h 52 min 13 1/3 s. c) 60 x 50 x 15 = 15 000 ml. 1/2 cube --. 1/2 x 1000 = 500 ml. Volume of water drained out = 15 000 - 500 = 14 500 ml. Time = 14 500 / 3 = 1 h 20 min 33 1/3 s. Using timeline 9.15 am .......... 1 h 20 min = 10.35 a.m.

Topic on Speed said...

There were 2 identical flights of steps. For the first flight of steps, Peter walked up some steps and ran 7 steps in 67 seconds. For the second flight of steps, he walked up some steps and ran up 15 steps in 43 seconds. How long would Peter take to walk up both flights of steps?

Solution : Logic

Difference in steps
= 15 – 7
= 8

Difference in time
= 67 – 43
= 24 s

Note: It took Peter 24 s more to walk up 8 steps than running 8 steps.

Thus,

8 steps -----> 24s

1 step -----> 3 s

It took Peter 3 s more to walk up one step.

1 step -----> 3s

7 steps -----> 21 s

It took Peter 21 s more to walk up 7 steps.

Time taken
= time of running + time of walking
= 67 s + 21 s more
= 88 s for walking up 1 flight of steps

Therefore,

total time
= 2 x 88
= 176 s

Alternatively,

1 step -----> 3s

15 steps -----> 45 s

It took Peter 45 s more to walk up 15 steps.

Time taken
= time of running + time of walking
= 43 s + 45 s more
= 88 s for walking up 1 flight of steps

Therefore,

total time
= 2 x 88
= 176 s

Both yield the same total time.

Topic on Speed said...

Three pieces of rubber hoses of lengths 56 cm, 84 cm and 112 cm are to be cut into pieces of equal length. What is the longest possible length for the three pieces to be cut without any wastage or leftover? [5]

Solution : Listing

List all the factors of 56.

1, 2, 4..., 28

Thus, 28 x 2 = 56

If each piece is cut into 28 cm,

we will get 2 equal pieces of 28 cm.

Likewise,

84 / 28 = 3 pieces of equal length
112 / 28 = 4 pieces of equal length

Therefore,

the highest length is 28 cm without any wastage or left over.