Time and Work class-8th(Republished)

Class 8th Time and Work-1

8th Time and work Test paper solved 2018-1

Download File

8th Time and work Test paper solved 2018-2

Download File

For more Download Class VIII Mathematics

VIII Algebra VIII Commercial Maths VIII Geometry & Mensuration

1. Ramdin can reap a field in 30 days. What part of the field would he have reaped in 25 days?

2. If 12 men working 8 hours a day can construct a shed in 5 days, then in how many days will 8 men working 10 hours a day complete the same job?

3. Three ants separately can gobble a dead grasshopper in 3, 4 and 6 days respectively. How many days will they take together to finish off the poor chap? If the grasshopper weighs 63 grams, find the share of each.

4. If a chicken lays an egg and a half in a day. how long does it take to get a dozen eggs from a chicken?

5. A dozen pens cost 15 rupees. How much would a hundred pens cost?

6. 400 grams cake costs 80 rupees. How much would a 1.5 kg cake cost?

7. It rained 80 mm in first 20 days of April. What would be the total rainfall in April?

8. 50 grams of fish food is sufficient for 8 fish for a day. How much would 12 fish require in a day?

9. 50 grams of fish food is sufficient for 8 fish for a day. On the same amount of food, for how many days can 2 fish survive?

10. A man builds a chair in 4 days, working 6 hours a day. How many hours should he work in a day if he wants to do it in 3 days instead of 4 days?

11. Four men working together take 3 hours to paint a wall. How much time would 6 men take to do the same job?

12. A man earns Rs 18000 in 3 months.

(a) How much time would he take to earn Rs 30000?

(b) How much will he earn in 7 months?

13. Ram can build a table in 6 days while Shyam requires 8 days for the same work. Working together,

(a) how many days would they require to construct 7 tables?

(b) How many tables can they construct in 48 days?

14. A man and a woman working together can sow a field in 20/9 days. The man alone can do it in 4 days. How much time would the woman working alone take to sow the field?

Answers:

1. 5/6th 2. 6 days 3. 4/3 days, 28 grams, 21 grams, 14 grams

4. 8 days 5. Rs. 125 6. Rs. 300 7. can not say

8. 75 grams 9. 4 days 10. 8 hours per day

11. 2 hours 12. (a) 5 months (b) Rs 42000

13. (a) 24 days (b) 14 tables 14. 5 days

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall? Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100 U = - 33.3 cm Therefore, the distance of the object from the wall x = 50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca

CBSE ADDA

Search This Blog