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SAPMATP6_W52Page 1/6
Taken from Learning Maths Book 6B
Revision: Challenging Word Problems
Write your answers on the lines provided.
1. Findthecircumferenceofaquadrantofdiameter56cm.(Takeπ= 22___ 7 )
2. If 1__ 4 ofthestudentswholikegreenand 1__ 9 ofthestudentswholikeblueandhalfofthenumber
ofstudentswholikeyellowareboys,howmanygirlsarethereintheclass?
3. Thebaseofatriangleis5cmanditsheightis10cm.Thetrianglehasthesameareaasasquare.Findtheperimeterofthesquare.
4. Thefigurebelowismadeupoftwoidenticalsquaresofside7cm.Whatistheareaoftheshadedregion?
5. Halfofacircleisplacedinsidearectanglemeasuring14cmby22cm.Findtheareaoftheshadedpart.(Takeπ=3.14)
P6 MATHS
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1. 44 cm
�d___ 4 = 22___ 7 ×56× 1__ 4 =44cm
2. 5 ___36Boyswholikegreen= 1__ 4 = 9___
36
Boyswholikeblue= 1__ 9 = 4___36
Boyswholikeyellow= 18___36
Girls=1– 9___ 36 – 4___ 36 – 18___ 36 = 5___36
3. 20 cmAreaoftriangle= 1__ 2 ×5×10=25cm2
Lengthofsquare= √___
25 =5cmPerimeterofsquare=4×5=20cm
4. 73.5 cm2
Areaoftheunshadedsquare=3.5×3.5=12.25cm2
Areaoftheshadedregion=6×12.25=73.5cm2
5. 231.07 cm2
Areaoftherectangle=14×22=308cm2
Areaofthesemicircle= 1__ 2 ×3.14× 14___ 2 × 14___ 2 =76.93cm2
Areaoftheshadedpart=308–76.93=231.07cm2
Solutions to Revision: Challenging Word Problems
SAPMATP6_W52Page 2/6
Taken from Learning Maths Book 6B
P6 MATHS
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SAPMATP6_W52Page 3/6
Taken from Learning Maths Book 6B
Revision: Challenging Word Problems
Do these word problems. Show your working clearly in the space provided.
1. Thisfigurebelowismadeupoftwoidenticaltriangles.Thebaseofthetriangleis36cmanditsheightis30cm.Findtheareaofeachsmalltriangleiftheshadedareais720cm2.
2. Six similar cubes are stacked up as shown below. The total area of the shaded part is600cm2.Findthevolumeofeachcube.
1.Areaofeachtriangle= 1 _ 2 ×36×30=540cm2
Areaof6smalltriangles=540+540–720=360cm2
360÷6=60cm2
Theareaofeachsmalltriangleis60 cm2.
2.Areaoftheshadedpartofeachcube=600÷6=100cm2
Sideofeachcube= √___
100 =10cm10×10×10=1000cm3
Thevolumeofeachcubeis1000 cm3.
Solutions:
P6 MATHS
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SAPMATP6_W52Page 4/6
Taken from Learning Maths Book 6B
Revision: Challenging Word Problems
Do these word problems. Show your working clearly in the space provided.
1. Georgesetoffat12noon,drivingatauniformspeedof80km/handreachedhisdestinationat10pm.CalvinsetoffthesametimeasGeorgeandreachedthesamedestination2hoursearlierthanGeorge.HowfarwasGeorgefromhisdestinationwhenhewas120kmapartfromCalvin?
2. MikeandKensharedsomestamps. 1__ 5 ofKen’sstampswere 1__ 3 ofMike’sstamps.IfMikegave
Ken24stamps,KenwouldhavethriceasmanystampsasMike.Findthenumberofstamps
eachofthemhadinthebeginning.
3. Foursimilarpensandthreesimilarpencilscost$18.Foursimilarrulersandthreesimilarpencilscost$12.Thecostofthreesuchpensandthreesuchrulersis$13.50.Findthecostofeachpencil.
P6 MATHS
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1. From12noonto10pm,itwas10hours.80×10=800kmThedistancewas800km.10–2=8hCalvintook8htocompletethejourney.800÷8=100km/hCalvinspeedwas100km/h.
Time DistancetravelledbyGeorge DistancetravelledbyCalvin Difference1pm 80km 100km2pm 160km 200km3pm 240km 300km4pm 320km 400km5pm 400km 500km6pm 480km 600km 120km7pm 560km 700km 140km8pm 640km 800km 160km9pm 720km
10pm 800km
800–480=320kmGeorgewas320 kmawayfromhisdestinationwhenhewas120kmapartfromCalvin.
2. Mike
Ken
?
24
?
3×24=72Mikehad72stampsinthebeginning.5×24=120Kenhad120stampsinthebeginning.
3. 12rulers+9pencils=3×$12=$3612pens+12rulers=4×$13.50=$54
12pens+9pencils=3×$18=$54(12rulers+9pencils)+(12pens+12rulers)–(12pens+9pencils)=$36+$54–$54
24rulers=$361ruler=$36÷24=$1.50
(4×$1.50)+3pencils=$123pencils=$12–$6=$61pencil=$6÷3=$2
Thecostofeachpencilwas$2.
SAPMATP6_W52Page 5/6
Taken from Learning Maths Book 6B
Solutions to Revision: Challenging Word Problems
P6 MATHS
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SAPMATP6_W52Page 6/6
Taken from Mental Mathematics Book 6
Mental Maths: GENERAL REVISION 20
Do these sums mentally.
1. 403×99=
2. 140×715=
3. √_____
3025 =
4. 59×99=
5. 30157×11=
6. 48×1 1__3 =
7. 492=
8. 3√_______
64000 =
9. 582–572=
10. 36+38+40+42+44+46+48=
Solutions:1.398972.1001003.554.58415.3317276.647.24018.409.11510.294
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