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Linear Equations

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CAT 2025 Lesson : Linear Equations - Items Measured in Groups of 2 or more

4.8 Items measured in groups of 2 or more

The following can be ascertained in these questions

1

) Note that the highest and lowest groups are that of the heaviest possible or lightest possible items.

2

) The second highest group will have just one change compared to highest, i.e. replacing the lightest in the group with the next lightest person.

3

) Likewise, the second lightest group will have just one change compared to lightest, i.e. replacing the heaviest in the group with the next heaviest person.

4

) Let's say from

n

items, the weights are measured

r

at a time. Then each item is grouped with

r - 1

other out of a total of

n - 1

items. Therefore, number of times an item is measured

=

Number of ways in which other items can be selected

= ^{n-1} \text{C}_{r-1}

5

) The average is the sum total of all these weight combinations divided by the number of times each item was weighed.

Example 17

4

girls had different sums of money, which were all in integers and in rupees. The total money with all possible combination of

3

of the

4

girls were noted to be Rs.

44

, Rs.

47

, Rs.

49

and Rs.

52

. What was the individual sums of money in ascending order?

Solution

Let the sum of money with the

4

girls in ascending order be

a, b, c

and

d

respectively.

Lightest:

a + b + c = 44

-----(

1

)

2^{\text{nd}}

Lightest:

a + b + d = 47

-----(

2

)
Heaviest:

b + c + d = 52

-----(

3

)

2^{\text{nd}}

Heaviest:

a + c + d = 49

-----(

4

)

Adding all of these equations,

3 (a + b + c + d) = 192

a + b + c + d = 64

-----(

5

)

Eq(

5

)

-

Eq(

1

) ⇒

d = 20

Eq(

5

)

-

Eq(

2

) ⇒

c = 17

Eq(

5

)

-

Eq(

3

) ⇒

a = 12

Eq(

5

)

-

Eq(

4

) ⇒

b = 15

The sums of money in ascending order are

12, 15, 17, 20

.

Answer:

12, 15, 17, 20

Example 18

A class teacher had to find the weight of

5

of her students. As

25

kg was the minimum weight for the weighing machine and all her students weighed less than it, she decided to make

2

students stand at a time. She did this for all possible combinations of

2

students. The weights she noted were

30

kg,

32

kg,

34

kg,

35

kg,

36

kg,

37

kg,

38

kg,

39

kg,

40

kg and

43

kg.

(I) What was the average weight of the class?
(II) Arrange the weights of the

5

students in ascending order.

Solution

2

out of

5

students are measured each time. With each student,

4

other students can be selected at a time.
Number of times each student is weighed

= ^{5-1}C_{2-1} = ^{4}C_{1} = 4

Therefore, when we add the

10

weights given we get

4(a + b + c + d + e) = 364

⇒

a + b + c + d + e = 91

-----(

1

)

(I) Average weight of the class

= \dfrac{\text{Sum of weights}}{\text{Number of Students}} = \dfrac{91}{5} = 18.2

kg

(II) Heaviest and Lowest weights

a + b = 30

-----(

2

)

d + e = 43

-----(

3

)
Adding these ⇒

a + b + d + e = 73

-----(

4

)

Eq(

4

)

-

Eq(

1

) ⇒

\bm{c = 18}

Second heaviest will be

c + e = 40

⇒

\bm{e = 22}

Second lightest will be

a + c = 32

⇒

\bm{a = 14}

Substituting in Eq(

2

) and Eq(

3

), we get

\bm{b = 16}

and

\bm{d = 21}

Answer: (I)

18.2

kg; (II)

14

kg,

16

kg,

18

kg,

21

kg,

22

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