Solve the Question:
1. The distance between the school and Reena’s house is 1 km 480 m. Everyday she walks
both ways. What distance does she cover in 6 days of a week?
2. To stitch a pant 1 m 15 cm cloth is needed. Out of 36 m cloth, how many pants can be
stitched and how much cloth will remain?
3. The capacity of a water tank is 300 litres. Express its capacity in millilitres.
4. Round the given numbers to the nearest tens.
(a) 48
(b) 59
(c) 64
5. Estimate the following products:
(а) 86 x 316
(b) 898 x 786
6. There are two factories located at place P and the other at place Q. From these
factories, a certain commodity is to be delivered to each of the depots situated at A, B
and C. Weekly production of commodity by P and Q are 120 kg and 150 kg respectively.
Weekly requirement of commodity by A, B and C are 80 kg, 90 kg and 100 kg
respectively. P delivers 60 kg to A, 40 kg to B and 20 kg to C. How much amount of the
commodity should Q deliver to A, B and C to meet their requirement? If the rate of the
commodity is ? 20 per kg, find the total amount to be paid to P and Q.
Answers
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- DISTANCE FROM SCHOOL-1KM480M
- THEREFORE DISTANCE COVERED IN 1 DAY=1480M×2=2960M=2KM960M
- THEREFORE,DISTANCE COVERED IN 6 DAYS=2960×6=17760m=17km760m.
2 .LENGTH OF CLOTH NEEDED TO STICH 1 PANT=1M 15CM
THEREFORE NO.OF PANTS MADE WITH 36M CLOTHS=3600CM/150CM=24pants
3. CAPACITY OF TANK-300L
1L=1000ML
THEREFORE,
300L=300×1000
=3,00,000ML
4.ROUND TO NEAREST TENS
- 50
- 60
- 60
5.ESTIMATE
- 90×320=28800
- 900×790=711000
6.
- .COMODITY MADE IN 1 WEEK BY P=120KG
- """""""""BY Q=150KG
- REQUIREMENTS OF A=80KG
- """"OF B=90KG
- """"OF C=100KG
- COMMODITY GIVEN TO A BY P=60KG
- """"""""B BY P=40KG
- """"""""C BY P=20KG
- THEREFORE,
- Q SHOULD GIVE 20 KG TO A,50KG TO B & 80 KG TO C RESPECTIVELY.
- COST PER KG=20
- THEREFORE,COST GIVEN TO P=120×20=2400
- COST GIVEN TO Q=150×20=3000.
- TOTAL AMOUNT PAID TO P&Q=5400
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