Enhanced Curriculum Support

SecA

1. Pick up the rational numbers from the following given numbers:

67, −12, 0, 10, 1000

2. Write two such rational numbers whose multiplicative inverse is same as they are.

3. The property represented by a + b = b + a is:

(a) closure property

(b) additive identity

(c) associative property

(d) commutative property

4. Which of the following is not true?

(a) Rational numbers are closed under multiplication

(b) Rational numbers are closed under division

(c) Rational numbers are closed under addition

(d) Rational numbers are closed under subtraction

5. What properties, the following expressions show?

(i) 23 + 45 = 45 + 23

(ii) 13 × 23 = 23 × 13

6. The property represented by a × (b + c) = a × b + a × c is:

(a) closure property

(b) distributive property

(c) associative property

(d) commutative property

7. The numerical expression: 38+ −57 = −1956 shows that

(a) Addition of rational numbers is not commutative

(b) Rational numbers are not closed under addition

(c) Rational numbers are closed under multiplication

(d) Rational numbers are closed under addition

8. Which of the following statements is correct:

(i) -5 + 3 ≠ 3 + (-5)

(ii) −812 = 10−15

(iii) 2 is not a natural number

(iv) 17 is not a prime number

(a) Option (iii)

(b) Option (ii)

(c) Option (iv)

(d) Option (i)

Sol

1. Solution: Since rational numbers are in the form of ab where b ≠ 0.

Only 67, −12 and 0 are the rational numbers.

2. Solution:

Reciprocal of 1 = 11 = 1

Reciprocal of -1 = 1−1 = -1

Hence, the required rational numbers are -1 and 1.

3. Solution: Option (d)

4. Solution: Option (b)

5. Solution:

(i) This shows the commutative property of addition of rational numbers.

(ii) This shows the commutative property of multiplication of rational numbers.

6. Solution: Option (b)

7. Solution: Option (d)

8. Solution: Option (a)

SecB

1. Simplify the following expression: 23× 94 + 56.

Sol

1. Solution:

23× 94 + 56 = 2×93×4 + 56 = 1812 + 56 = 32 + 56 = 3×3+56 = 9+56

= 146 = 73

SecC

1. If the cost of 4 12 litres of milk is ₹89 12, find the cost of 1 litre of milk.

2. If a car travels 710 kilometers in one minute, how many kilometers will it travel in 15 minutes?

Sol

1. Solution: Converting cost and amount into improper fractions:

4 12 L = 4×2+12 = 8+12 = 92 L

Rs. 89 12 = 89×2+12 = 178+12 = Rs. 1792

Cost of 1L milk = 179292 = Rs. 1799 = Rs. 19.89

2. Solution:

Distance traveled in one minute = 710 kilometers

Time = 15 minutes

Total distance = 710 kmmin × 15 = 7×1510 = 10510 = 10.5 kilometers

Therefore, the car will travel 10.5 kilometers in 15 minutes.

Problem 1

1. During a study group session, 6 students share 92chocolate bars equally. How much chocolate does each student get? Explain how sharing resources equitably can strengthen bonds among friends and peers.

Sol

Solution: To determine how much chocolate each student gets, we need to divide the total amount of chocolate by the number of students.

Amount of chocolate each student gets: 92 ÷ 6 = 92 × 16 = 912 = 34

Therefore, each student gets 34 of a chocolate bar.

Sharing resources equitably not only ensures that everyone's needs are met but also fosters a positive and supportive environment where strong, lasting relationships can thrive.

When resources are shared equally, individuals are more likely to cooperate and support each other. This cooperative spirit can lead to more effective and enjoyable group activities and projects.

Problem 2

2. A community service group collected 73 kilograms of recyclable materials from each member. If there are 9 members, how much material did they collect in total? Discuss the role of civic responsibility in community service and how individual efforts contribute to the greater good.

Sol

Solution: To find out the total amount of recyclable materials collected by the community service group, we need to multiply the amount collected by each member by the number of members.

Given:

Each member collected 73 kilograms of recyclable materials. There are 9 members. Total material collected = 73 × 9 which gives us 21 kg

Therefore, the total amount of recyclable materials collected by the group is 21 kilograms.

Civic responsibility is crucial in community service as it fosters engagement, social cohesion, and the addressing of community needs. Individual efforts, while seemingly small, collectively contribute significantly to the greater good, creating a positive and lasting impact on society.

Problem 3

3. Rohan and Meera were given an assignment to simplify rational numbers. Rohan finds the solution to be 58 while Meera finds it to be 1016. Meera argues that her answer is correct because she worked hard on it. How should they resolve their difference of opinion while upholding the values of honesty and integrity?

Sol

Solution:

First verify both solutions mathematically.

We see that Meera's Solution needs to be simplified. Further simplify 1016.

Find the Greatest Common Divisor (GCD) of 10 and 16. The GCD is 2. We can simplify it by dividing both the numerator and the denominator by 2:

10÷216÷2 = 58

​Now, compare the simplified results, where we see that both the obtained solutions are identical.

Now, discuss the process openly. Acknowledge that both Rohan and Meera arrived at the correct answer, but in different forms initially. Emphasize the importance of simplifying rational numbers to their lowest terms. Appreciate their efforts.

Learn from the Experience. Both students should understand that honesty involves verifying facts (mathematical correctness in this case) and integrity involves acknowledging the correctness irrespective of who provided the initial answer.

Q1

1.A group of friends shared a pizza. Two of the friends each ate 38 of the pizza, and the third friend ate 14of the pizza. How much of the pizza was left uneaten? Explain how you calculated your answer.

Sol

Solution:

Since two person ate 38 of the pizza and one person had 14 of the pizza:

Consumption of two persons = 2 × 38 = 68

Consumption of third person = 1 × 14 = 14

​ Total consumed = 68 + 14 = 6+28 = 1

Remaining Pizza = 1 - 1 = 0

There is no pizza remaining at the end of the meal.

Q2

2. In a marathon race, a runner consumes 34 of a energy bar every 10 kilometers. If the race is 42 kilometers long, how many energy bars does the runner need to complete the race? Round your answer to the nearest whole number.

Sol

Solution: To find out how many energy bars the runner needs to complete the 42-kilometer race, given that they consume 34 of an energy bar every 10 kilometers, we'll proceed with the following steps:

Calculate the total number of 10-kilometer segments in the race: 4210 = 4.2

This means the race consists of 4 full 10-kilometer segments, with 2 kilometers left over.

Calculate the total number of energy bars needed for the full segments:

Each 10-kilometer segment requires 34 of an energy bar. For 4 segments: 4 × 34 = 3

So, for the full segments, the runner needs 3 energy bars.

Consider the remaining 2 kilometers:

Since the runner consumes 34 of an energy bar every 10 kilometers, they will need to consume of an energy bar for the remaining 2 kilometers as well: 34 × 210 = 320

Total energy bars needed: 3 + 320 = 3.15

Round to the nearest whole number:

Since we need to round to the nearest whole number:3

Therefore, the runner needs approximately 3 energy bars to complete the 42-kilometer race, rounding to the nearest whole number. This calculation accounts for both the full 10-kilometer segments and the extra kilometers left over.

Questions

1.Which of the following is not true?

(a) 23 + 54 = 54 + 23

(b) 23 − 54 = 54 − 23

(c) 23 × 54 = 54 × 23

(d) 23 ÷ 54 = 23 × 45

2.Multiplicative inverse of 01 is:

(a) 1

(b) −1

(c) 0

(d) not defined

3.Three rational numbers lying between −34 and 12 are :

(a) −12, 0 , 34

(b) −14, 14, 34

(c) −14, 0, 14

(d) −54, 0, 14

Sol

1. Solution: Subtraction of rational numbers is not commutative.

Thus, 23 − 54 = 54 − 23

Therefore, (b) is the correct answer.

2. Solution: Reciprocal of 01 is 10 , which is not defined.

Thus, Multiplicative inverse of 01 is not defined.

Therefore, (d) is the correct answer.

3. Solution: Three rational numbers lying between −34 and 12 are: −14, 0 , 14

Therefore, the correct answer is (c).

Questions

4. The product of a non-zero rational number and its reciprocal is .

5. If x = 13 and y = 67 then xy − yx = ?

Sol

4. Solution: Let x be the non-zero rational number.

Then, its reciprocal will be 1x.

Now, their product is x × 1x = 1.

The product of a non-zero rational number and its reciprocal is one.

5. Solution: Given that, x = 13 and y = 67

Then, xy − yx = 13 × 67 − (67/13)

Multiply and divide the numbers, xy − yx = 621 − 187

Take LCM and simplify, xy − yx = −4821 = −167

Therefore, if x = 13 and y = 67 then xy − yx = −167

Questions

6.True/False: Every rational number has a reciprocal.

7.True/False: −45 is larger than −54.

Sol

6. Solution: We know that The product of a non-zero rational number and its reciprocal is one.

But, there is no rational number which when multiplied with zero, gives one.

So, the rational number 0 has no reciprocal or multiplicative inverse.

Therefore, the given statement is false.

7. Solution: Make the denominators same by multiplying and dividing −45 by 4, −45 × 44 = −1620

Multiply and divide −54 by 5, −54 × 55 = −2520

Here, −1620 is larger than −2520.

Thus, −45 is larger than −54.

Therefore, the given statement is true.

Questions

8.Find 47 × 143 ÷ 23

9.Using appropriate properties, find 23 × −57 + 73+ 23 × −27

10.The product of two rational numbers is –7. If one of the number is –10, find the other.

Sol

8. Solution:

47 × 143 ÷ 23 = 47 × (143 × 32)

= 47 × 7 = 4

9. Solution: 23 × −57 + 73+ 23 × −27 = 2×−53×7 + 73+ 2×−23×7

= −1021 + 73 + −421 = −1421 + 73 = −14+7×721 = 49−1421 = 3521 = 53

10. Solution: Let the other rational number be 𝑥.

Given that the product of two rational numbers is −7, and one of the numbers is −10, we can set up the equation:

−10×x=−7

Now, solve for 𝑥: −10x = −7

Divide both sides by −10:

x = −7−10

Simplify the fraction: x = 710

Therefore, the other rational number is 710.

Questions

11.Let O, P and Z represent the numbers 0, 3 and -5 respectively on the number line. Points Q, R and S are between O and P such that OQ = QR = RS = SP. What are the rational numbers represented by the points Q, R and S. Next choose a point T between Z and O so that ZT = TO. Which rational number does T represent?

12.A farmer has a field of area of 49 45 ha. He wants to divide it equally among his one son and two daughters. Find the area of each one’s share. (ha means hectare; 1 hectare = 10,000 m2)

13.Why is 45 considered a rational number but 2 is not? Explain in terms of the definitions.

Sol

11. Solution: As OQ = QR = RS = SP and OQ + QR + RS + SP = OP therefore Q, R and S divide OP into four equal parts.

So, R is the mid-point of OP, i.e. R = 0+32 = 32

Q is the mid-point of OR, i.e. Q = 120+32 = 34

and S is the mid-point of RP, i.e. S = 123+32 = 94

Therefore, Q = 34 , R = 32 and S = 94

Also ZO = OT. So, T is the mid-point of OZ, i.e T = 0+−52 = −52

Thus, T represents −52

12. Solution: 49 45 ha = 2495 ha

Each Share = 13 × 2495 = 835 ha = 16 35 ha

Thus, each share is equal to 16 35 hectares.

13. Solution:

45 is written explicitly as a ratio of two integers, 4 and 5, with 5 not being zero. Its decimal form, 0.8, terminates after one digit.

Meanwhile, 2 cannot be expressed as a ratio of two integers. Its decimal form is non-terminating and non-repeating.

Questions

14. Explain why −78 is a rational number. Convert it to its equivalent positive fraction form.

15. Determine which is larger: 57or 68. Show your working.

16. Identify the rational number which is different from the other three : 2/3, −4/5, 1/2, 1/3. Explain your reasoning.

Sol

14. Solution:

−78 can be written explicitly as a ratio of two integers, say pq where q ≠ 0. Thus, it is a rational number.

Both forms, −78 and 7−8 are equivalent to −78. The negative sign can be placed in front of the entire fraction, in the numerator, or in the denominator without changing the value of the fraction.

15. Solution:

Find the least common multiple (LCM) of denominator (of 7 and 8): 7 × 8 = 56

Convert each fraction to have the denominator of 56:

57 = 5×87×8 = 4056

68 = 6×78×7 = 4256

​Compare the numerators: 40 < 42.

Thus, 4256 > 4056 i.e. 68 > 57

16. Solution: −45 is the rational number which is different from the other three, as it lies on the left side of zero while others lie on the right side of zero on the number line.

Questions

17. Find the rational numbers between −25 and 12.

18. Find multiplicative inverse of: 16+49×43

19. If 23 of the students in a class have brown eyes, and there are 30 students in total, predict how many students you would expect to have brown eyes. Explain whether this fraction would apply to any number of students and why.

Sol

17. Solution:

Let us make the denominators the same, say 50.

−25 = −2×105×10 = −2050

12 = 1×252×25 = 2550

Thus, ten rational numbers between −25 and 12 = ten rational numbers between −2050 and 2550

Therefore, ten rational numbers between -20/50 and 25/50 are -18/50, -15/50, -5/50, -2/50, 4/50, 5/50, 8/50, 12/50, 15/50, 20/50.

18. Solution:

Solving, we get: The least common denominator (LCD) of 6 and 9 is 18.

16+49×43 = 318+818×43 = 1118×43 = 4454 = 2227

The multiplicative inverse of a number ab is ba.

Thus, the required multiplicative inverse is 2722.

19. Solution:

Total number of students in the class: 30

Fraction of students with brown eyes: 23

Calculate the number of students with brown eyes:

Multiply the total number of students by the fraction that have brown eyes:

Number of students with brown eyes= 2×303 = 603​ = 20

In conclusion, based on the given fraction and total number of students, you would predict that approximately 20 students in the class have brown eyes. The fraction 23 applies universally because it represents a constant ratio of students with brown eyes to the total number of students in the class.

Q1

Sanmesh who is working in a multinational company earns Rs. 150000 per month. Out of his earnings he spend 110 th on food items, 14th on shopping with family, 15th of remaining on education of his two kids and rest of money he puts on his savings.

Based on the above information, answer the following questions:

(i) How much money did he spend on the food items?

(ii) How much money did he spend on the shopping?

(iii) Calculate the amount spend by Sanmesh on education of children.

Sol

Solution:

Sanmesh's monthly earnings: Rs. 150,000

(i) Expenditure on Food Items:

Amount spent on food items = 110 × 150,000 = 15,000

(ii) Expenditure on Shopping:

Amount spent on shopping = 14 × 150,000 = 37,500

(iii) Expenditure on Education of Children:

Remaining amount = Total earnings - (Amount spent on food items + Amount spent on shopping) = 150,000 − (15,000 + 37,500) = 150,000 − 52,500 = 97,500

Now, calculate how much Sanmesh spent on education of his children, which is 15th of the remaining amount:

Amount spent on education = 15 × 97,500 = 19,500

So,

(i) Sanmesh spent Rs. 15,000 on food items.

(ii) Sanmesh spent Rs. 37,500 on shopping.

(iii) Sanmesh spent Rs. 19,500 on the education of his children.

Q2

A school receives a donation of Rs. 15,000 to purchase classroom supplies. The funds are allocated as follows: 13 on stationery items, 15 on educational games, 14 of the remaining on art supplies. The remainder is used for purchasing books.

(a) How much money is spent on stationery items ?

(b) What is the amount allocated for educational games ?

(c) Calculate the expenditure on art supplies.

(d) How much money is spent on purchasing books ?

Sol

Solution: Given:

Total donation received = Rs. 15,000

(a)

Allocation for stationery items = 13 of Rs. 15,000 Amount spent on stationery items = 13 × 15000 = Rs. 5,000

So, Rs. 5,000 is spent on stationery items.

(b)

Remaining after stationery items = Rs. 15,000 - Rs. 5,000 = Rs. 10,000

Allocation for educational games = 15 of Rs. 10,000 = 15 × 10000 = Rs. 2,000

So, Rs. 2,000 is allocated for educational games.

(c)

Remaining after educational games = Rs. 10,000 - Rs. 2,000 = Rs. 8,000

Allocation for art supplies = 14 of Rs. 8,000 = 14 × 8000 = Rs. 2,000

So, Rs. 2,000 is spent on art supplies.

(d)

Remaining after art supplies = Rs. 8,000 - Rs. 2,000 = Rs. 6,000

Therefore, Rs. 6,000 is spent on purchasing books.

Q3

Three friends Rajeev, Sarita and Rahul go to purchase some sweets, namkin and cold drinks for a party. The following chart shows the price and available stock of sweets and namkin in the shop.

(a) After purchasing 500gm of sweet laddu, jalebi and barfi each, Rajeev had Rs. 150 left with him. How much money does Rajeev had before the purchase?

(b) Rajeev wants to purchase one packet of Mixture and two packets of potato chips with the remaining Rs. 150. Explain whether he can purchase it or not.

(c) Rahul had Rs. 200 and wants to purchase one packet of Mixture, one packet of potato chips, 250gm Sweet laddu and one bottle of cold drink. But due to insufficient money he had to reduce the quantity of one of the item. Find out the name of that item along with reason.

(d) Sarita had Rs. 250 and wants to purchase those items which were not purchased by her friends. Choose the correct list of items she will purchase.

(i) Jalebi

(ii) (Roasted) Dry fruits

(iii) Barfi

(iv) Potato chips

Sol

Solution:

(a) Rajeev spent on sweets as follows:

Laddu: 500g at Rs. 350 per kg

Jalebi: 500g at Rs. 400 per kg

Barfi: 500g at Rs. 400 per kg

Total cost = (500g × 3501000) + (500g × 4001000) + (500g × 4001000) = Rs. (175 + 200 + 200) = Rs. 575

Given that Rajeev had Rs. 150 left after these purchases, his initial amount of money was:

Initial amount = Total cost + Remaining money = Rs. 575 + Rs. 150 = Rs. 725

So, Rajeev initially had Rs. 725.

(b) The cost for one packet of mixture and two packets of potato chips:

Mixture: Rs. 40 per packet

Potato chips: Rs. 60 per packet

Total cost = Rs. 40 + 2 × Rs. 60 = 40 + 120 = Rs. 160

Since Rajeev only has Rs. 150 left, he cannot afford these items.

(c) Rahul wants to buy:

Mixture: 1 packet (Rs. 40)

Potato chips: 1 packet Rs. 60

Laddu: 250g at Rs. 350 per kg = Rs 3502 = Rs. 175

Cold drink: 1 bottle (Rs. 35)

Total cost = 40 + 60 + 175 + 35 = Rs. 310 

Since Rahul has only Rs. 200, he needs to reduce the quantity of one item. The item he would reduce is the Sweet Laddu because it costs Rs. 175, which is higher than the cost of other items.

(d) Sarita has Rs. 250 and wants to buy items not purchased by her friends:

Items not purchased: Jalebi, Roasted Dry Fruits, Barfi

Comparing the prices:

Jalebi: Rs. 400 per kg

Roasted Dry Fruits: Rs. 980 per kg

Barfi: Rs. 400 per kg

Considering affordability with Rs. 250:

Jalebi (250g): 250g at Rs. 400 per kg = Rs. 4002 = Rs. 200

So, the correct list of items Sarita will purchase is: (i) Jalebi

Thus, Sarita will purchase Jalebi.