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il63 [147K]
3 years ago
5

Write a word problem that you can solve using multiples to estimate the quotient. Include a solution

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
4 0

Answer:

You have 21 candies and you want to share them among 4 friends. How many candies correspond to each one?

Step-by-step explanation:

Multiples of four are: 4 (= 4*1), 8 (= 4*2), 12 (= 4*3), 16 (= 4*4), 20 (= 4*5), 24 (= 4*6), ...

Then, 21/4 can be approximated as 20/4 = 5

hichkok12 [17]3 years ago
3 0

Let us consider,

The problem: 'Find two numbers between which the quotient of 50 ÷ 7 lies'.

Solution: We will solve this in two steps.

Step 1: Form a table of multiples of 7.

Counting number               Multiple of 7

          1                                       7

          2                                     14

          3                                     21

          4                                     28

          5                                     35

          6                                     42

          7                                     49

           8                                    56

Step 2: Use the table to find the multiples closest to 50.

Since, 49 = 7 × 7 and 56 = 7 × 8

Thus, 50 lies between 49 and 56.

Then, the quotient is either 7 or 8.

As 50 is closest to 49.

Hence, the quotient of 50 ÷ 7 is 7.

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. All the students in SS3 of a named school take either Mathematics (M), or Physics (P) or Chemistry (C). 40 take Mathematics, 4
aivan3 [116]

Answer:

(a) = 13

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(c) = 5

Step-by-step explanation:

Addition theorems on sets are

Theorem 1 :

n(AuB) = n(A) + n(B) - n(AnB)

Theorem 2 :

n(AuBuC) : = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

Total number of students in the school is not given

so let there are 60 students in the school

using theorem 2

n(AuBuC) : = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

let n(A) = Mathematics, n(B) =Physics and n(C) = Chemistry

so putting values,

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therefore, there are total 13 students who take all three subjects

Number of students who had taken only Mathematics  =
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Number of students who had taken only Physics  =

n(B) - n(BnA) - n(BnC) + n(AnBnC)

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2 years ago
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