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# Exercise 1.5

7th Maths : Term 1 Unit 1 : Number System : Exercise 1.5 ; Text Book Back Exercises Questions with Answers, Solution

Exercise 1.5

1. One night in Kashmir, the temperature is −5º C. Next day the temperature is 9º C. What is the increase in temperature?

9°C – (–5°C) = 9°C + 5°C = 14°C

2. An atom can contain protons which have a positive charge ( +) and electrons which have a negative charge ( −) . When an electron and a proton pair up, they become neutral (0) and cancel the charge out. Now, Determine the net charge:

(i) 5 electrons and 3 protons → −5 + 3 = −2 that is 2 electrons  (ii) 6 Protons and 6 electrons →  +6 – 6 = 0

(iii) 9 protons and 12 electrons → +9 –12 = –3 that is electrons   (iv) 4 protons and 8 electrons → +4 – 8 = –1 that is electrons    (v) 7 protons and 6 electrons → +7 – 6 = +1 that is proton (+)

3. Scientists use the Kelvin Scale (K) as an alternative temperature scale to degrees

Celsius (°C) by the relation T º C = ( T + 273) K.

Convert the following to kelvin:

(i) −275° C (ii) 45° C (iii) −400° C (iv) −273° C

(i) –275° C

(–275 + 273) K = –2° K

(ii) 45°C

(45 + 273) K = 318° K

(iii) –400° C

(–400 +273) K = 127° K

(iv) –273°C

(–273 + 273)K = 0° K

4. Find the amount that is left in the student’s bank account, if he has made the following transaction in a month. His initial balance is ₹ 690.

(i) Deposit ( +) of ₹ 485

(ii) Withdrawal ( −) of ₹ 500

(iii) Withdrawal ( −) of ₹ 350

(iv) Deposit ( +) of ₹ 89

(v) If another ₹ 300 was withdrawn, what would the balance be?

(i) Deposit (+) of ₹ 485 ?

₹ 690 + ₹ 485 = ₹ 1175

(ii) Withdrawal (–) of ₹ 500 ?

₹ 1175 – ₹ 500 = ₹ 675

(iii) Withdrawal (–) of ₹ 350

₹ 675 – ₹ 350 = ₹ 325

(iv) Deposit of (+) ₹ 89

₹ 325 + ₹ 89 = ₹ 414

(v) If another ₹ 300 was withdrawn, what would the balance be?

₹ 414 – ₹ 300 = ₹ 114

The balance would be = ₹ 114

5. A poet Tamizh Nambi lost 35 pages of his ‘lyrics’ when his file had got wet in the rain. Use integers, to determine the following:

(i) If Tamizh Nambi wrote 5 page per day, how many day’s work did he lose?

35 ÷ 5 = 7 days

(ii) If four pages contained 1800 characters, (letters) how many characters were lost?

4 pages contained characters = 1800

35 pages contained characters = [ 1800 × 35 ] / 4 = 15750

15750 characters were lost

(iii) If Tamizh Nambi is paid ₹250 for each page produced, how much money did he lose?

Paid money for 1 page = ₹ 250

Paid money for 35 pages = ₹ 250 × 35 = ₹ 8750

He lost = ₹ 8750

(iv) If Kavimaan helps Tamizh Nambi and they are able to produce 7 pages per day, how many days will it take to recreate the work lost?

Number of days it will take to recreate the work lost = 35 ÷ 7 = 5 days

(v) Tamizh Nambi pays kavimaan ₹ 100 per page for his help. How much money does kavimaan receive?

Kavimaan receives money = ₹ 35 × 100 = ₹ 3500

6. Add 2 to me. Then multiply by 5 and subtract 10 and divide now by 4 and I will give you 15! Who am I?

Add 2 to me, multiply by 5 and subtract 10 and divide by 4 = 15

[ (x+2)5 – 10 ] / 4 = 15

5x + 10 – 10 = 60

5x = 60

x = 60 / 5 = 12

The number is 12.

7. Kamatchi, a fruit vendor sells 30 apples and 50 pomegranates. If she makes a profit of ₹ 8 per apple and loss ₹ 5 per pomegranate, what will be her overall profit(or)loss?

Number of apples = 30

Profit on apples = ₹ 30 × 8 = ₹240

Number of pomegranates = 50

Loss on pomegranates = ₹ 50 × 5 = ₹ 250

Loss = 250

Profit = ₹ 240

Over all loss = ₹ 10

8. During a drought, the water level in a dam fell 3 inches per week for 6 consecutive weeks. What was the change in the water level in the dam at the end of this period?

Water level decrease per week = 3 inches

Water level decrease in 6 weeks = 3 × 6 = 18 inches

Water level decrease 18 inches in 6 weeks

9. Buddha was born in 563 BC(BCE) and died in 483 BC(BCE). Was he alive in 500 BC(BCE)? and find his life time. (Source: Compton’s Encylopedia) Buddha was bom in 563 BC (BCE) died in 483 BC (BCE)

500 BC (BCE) is in between 563 BC and 483 BC

Yes. He alive in 500 BC (BCE)

Life: 80 year.

10. A young boy has got fever. The following chart shows how his temperature varies each day in a week. M→up 3° C, T→up 2° C, W→ Down 2° C, Th→ up 2° C, F→ Down 1° C, S→ up 2° C What was the average daily variation in his temperature?

(Average = Total change in temperature / Number of days)

Total change in temperature = +3 +2 –2 + 2 –1 + 2

= 9 – 3 = 6

Number of days = 6

Average = Total change in temperature / Number of days

= 6 / 6 = 1

Average = 1

Exercise 1.5

1. 14˚c

2. (i) −2 (ii) 0 (iii) −3 (iv) −4 (v) 1

3. (i) −2 k (ii) 318°k (iii) −127° k (iv) 0°k

4. (i) ₹1175 (ii) ₹675 (iii) ₹325 (iv) ₹414 (v) ₹114

5. (i) 7 days (ii) 15,750 characters (iii) ₹8750 (iv) 5 days (v) ₹3,500 6. 12

7. Loss of ₹10

8. decrease of 18 inches

9. yes; 80 years

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7th Maths : Term 1 Unit 1 : Number System : Exercise 1.5 | Questions with Answers, Solution | Number System | Term 1 Chapter 1 | 7th Maths