# Extra Curriculum Support

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## Sample Papers

**1. If there is a discount of 40% on an article costing Rs 7000, then the priceafter discount is.**

A. Rs 4500

B. Rs 4200

C. Rs 4400

D. Rs 4600

**Solution :**

**2. How many feet long strips of ribbon can be cut from a ribbon that is feet long?**

**Solution :**

A football team gains 8 yards on one play, loses 5 yards on the next play, and then gains 12 yards on the third play. The net gain or loss in yards is calculated as follows:

Net Gain/Loss = Gain on 1st play - Loss on 2nd play + Gain on 3rd play Net Gain/Loss = 8 - 5 + 12 = 15 yards

The team's net gain is 15 yards.

**A submarine starts at 50 meters below sea level, ascends 20 meters, and then descends 35 meters. What is its final depth relative to sea level?**

**Solution :**

A submarine starts at 50 meters below sea level, ascends 20 meters, and then descends 35 meters. The final depth relative to sea level is calculated as follows:

Starting Depth = -50 meters (below sea level) After Ascending 20 meters = -50 + 20 = -30 meters After Descending 35 meters = -30 - 35 = -65 meters

The submarine's final depth relative to sea level is 65 meters below sea level.

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## Value Based

**1.Community Project Fundraising.**

**Your class is raising funds for a community project. On the first day, you raised ₹250, but on the second day, you had to spend ₹150 on supplies. Write an integer to represent the total amount of money raised so far. Explain why it is important to keep track of both positive and negative amounts in fundraising.**

**Solution:**

₹100

Explanation: The total amount raised so far is ₹250 (positive) minus ₹150 (negative), which equals ₹100. Keeping track of both positive and negative amounts is important in fundraising to ensure accurate accounting of money raised and spent.

**2.Weather Monitoring.**

**The temperature in your city was recorded as 10°C in the morning. By evening, it dropped by 7°C. Represent the evening temperature as an integer and discuss why monitoring temperature changes is important for weather forecasting and daily planning.**

**Solution:** 3°C

Explanation: The evening temperature is 10°C - 7°C = 3°C. Monitoring temperature changes is important for weather forecasting and daily planning to prepare for appropriate clothing and activities.

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## Hots

**1.In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, -5, -10, 15 and 10, what was his total at the end?**

**Solution:**

Given scores are 25, -5, -10, 15, 10

Marks given for correct answers = 25 + 15 + 10 = 50

Marks given for incorrect answers = (-5) + (-10) = -15

∴ Total marks given at the end = 50 + (-15) = 50 – 15 = 35

**2.At Srinagar temperature was -5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?**

**Solution:**

Initial temperature of Srinagar on Monday = -5°C

Temperature on Tuesday = -5°C – 2°C = -7°C

Temperature was increased by 4°C on Wednesday.

∴ Temperature on Wednesday = -7°C + 4°C = -3°C

Hence, the required temperature on Tuesday = -7°C

and the temperature on Wednesday = -3°C

**3.Mohan deposits ₹ 2,000 in a bank account and withdraws ₹ 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.**

**Solution:**

The deposited amount will be represented by a positive integer i.e., ₹ 2000.

Amount withdrawn = ₹ 1,642

∴ Balance in the account = ₹ 2,000 – ₹ 1,642 = ₹ 358

Hence, the balance in Mohan’s account after the withdrawal = ₹ 358

**4.Fill in the blanks to make the following statements true:**

**(i) (-5) + (-8) = (-8) + (…)**

**(ii) -53 + … = -53**

**(iii) 17 + … = 0**

**(iv) [13 + (-12)] + (…) = 13 + [(-12) + (-7)]**

**(v) (-4) + [15 + (-3)] = [-4 + 15] + …**

**Solution:**

(i) -5 + (-8) – (-8) + (-5)

(ii) -53 + 0 = -53

(iii) 17 + (-17) = 0

(iv) [13 + (-12)] + (-7) – 13 + [(-12) + (-7)]

(v) (-4) + [15 + (-3)] – [-4 + 15] + (-3)

**5.An elevator descends into a nine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350 m.**

**Solution:**

The present position of the elevator is at 10 in above the ground level.

Distance moved by the elevator below the ground level = 350 m

∴ Total distance moved by the elevator = 350 m + 10 m = 360 m

Rate of descent = 6 m/min.

Total time taken by the elevator =360m6m/min

= 60 minutes = 1 hour

Hence, the required time = 1 hour.

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## Ncert Exemplary

**1.Madhre is standing in the middle of a bridge which is 20 m above the water level of a river. If a 35 m deep river is flowing under the bridge, then the vertical distance between the foot of Madhre and bottom level of the river is.**

(a) 55 m (b) 35 m (c) 20 m (d) 15 m

The correct answer is

[Vertical distance = 20 m + 35 m = 55 m]

**2.[(– 10) × (+ 9)] + ( – 10) is equal to**

(a) 100 (b) –100 (c) – 80 (d) 80

Solution: Correct answer is

**3.–16 ÷ [8 ÷ (–2)] is equal to**

(a) –1 (b) 1 (c) 4 (d) –4

Solution: Correct answer is

Fill in the blanks to make the statement True.

**4.(– 25) × 30 = – 30 × .**

**5.75 ÷ = -75.**

**6.(–5) × (–7) is same as (–7) × (–5) .**

**7.(– 80) ÷ (4) is not same as 80 ÷ (–4) .**

**8.Find the odd one out* of the four options in the following:**

(a) (–2, 24) (b) (–3, 10) (c) (–4, 12) (d) (–6,8)

Here – 2 × 24 = – 48, = – 4 × 12 = – 48 and = – 6 × 8 = – 48

All the pairs i.e. (–2, 24); (–4, 12); (–6, 8) give same answer on multiplication, whereas –3 × 10 = –30, gives a different answer. So, odd one is

**9.Find the odd one out of the four options given below.**

(a) (–3, –6) (b) (+1, –10) (c) (–2, –7) (d) (–4, –9)

Solution: Here –3 + (–6) = –9,

+1 + (–10) = –9 and

–2 + (–7) = –9

All the above pairs i.e. (–3, –6); (+1, –10); (–2, –7) give same answer on adding, whereas – 4 + (–9) = –13, gives a different answer. So, odd one out is (d).

**10.Match the integer in Column I to an integer in Column II so that the sum is between –11 and – 4.**

Column I | Column II |
---|---|

(a)-6 | (i)-11 |

(b)+1 | (ii)-5 |

(c)+7 | (iii)+1 |

(d)-2 | (iv)-13 |

**11.If a is an integer other than 1 and –1, match the following.**

Column I | Column II |
---|---|

(a)a ÷ (–1) | (i)a |

(b)1 ÷ (a) | (ii)1 |

(c)(–a) ÷ (–a) | (iii)Not an Integer |

(d)a ÷ (+1) | (iv)-a |

**12.Write a pair of integers whose sum is zero (0) but difference is 10.**

Since sum of two integers is zero, one integer is the additive inverse of other integer, like – 3, 3; – 4, 4 etc.But the difference has to be 10. So, the integers are 5 and – 5 as 5–(–5) is

**13. Social Studies Application: Remembering that 1AD came immediately after 1BC, while solving these problems take 1BC as –1 and 1AD as +1.**

**(a) The Greeco-Roman era, when Greece and Rome ruled Egypt started in the year 330 BC and ended in the year 395 AD. How long did this era last?**

**(b) Bhaskaracharya was born in the year 1114 AD and died in the year 1185 AD. What was his age when he died?**

**(c) Turks ruled Egypt in the year 1517 AD and Queen Nefertis ruled Egypt about 2900 years before the Turks ruled. In what year did she rule?**

**(d) Greek mathematician Archimedes lived between 287 BC and 212 BC and Aristotle lived between 380 BC and 322 BC. Who lived during an earlier period?**

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**14.Write a pair of integers whose product is –12 and there lies seven integers between them (excluding the given integers).**

**15.Match the following.**

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## Competency Based

**1.Rahul and his friends went on a mountain hiking trip. The hike involved various ascents and descents. Below is a record of their movements during the hike.**

**1.Monday: They started from the base camp at an elevation of 0 meters. They ascended 150 meters.**

**2.Tuesday: They descended 80 meters.**

**3.Wednesday: They ascended 120 meters.**

**4.Thursday: They descended 100 meters.**

**5.Friday: They ascended 200 meters.**

**Questions :**

**1.What is their elevation at the end of Friday?**

**2.On which day did they have the highest elevation, and what was it?**

**3.If they wanted to return to the base camp (0 meters) from their elevation at the end of Friday, how many meters would they need to descend?**

**4.If they ascended 50 meters every day after Friday for the next three days, what would their final elevation be?**

**5.Compare the elevation change on Tuesday and Thursday. On which day did they descend more, and by how much?**

**1.Solution:**

Let's calculate the elevation for each day:

Monday: 0 + 150 = 150 meters

Tuesday: 150 - 80 = 70 meters

Wednesday: 70 + 120 = 190 meters

Thursday: 190 - 100 = 90 meters

Friday: 90 + 200 = 290 meters

Final elevation at the end of Friday: 290 meters.

**2.Solution:**

By checking the calculations for each day:

Monday: 150 meters

Tuesday: 70 meters

Wednesday: 190 meters

Thursday: 90 meters

Friday: 290 meters

Highest elevation: 290 meters on Friday.

**3.Solution:**

Current elevation at the end of Friday: 290 meters.

To return to 0 meters, they need to descend: 290 meters.

**4.Solution:**

Current elevation at the end of Friday: 290 meters.

Ascending 50 meters each day for three days:

Day 1 after Friday: 290 + 50 = 340 meters

Day 2 after Friday: 340 + 50 = 390 meters

Day 3 after Friday: 390 + 50 = 440 meters

Final elevation after three more days: 440 meters.

**5.Solution:**

Tuesday descent: 80 meters

Thursday descent: 100 meters

Comparing the descents:

Thursday: 100 meters

Tuesday: 80 meters

Difference: 100 - 80 = 20 meters

They descended more on Thursday by 20 meters.