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Mathematics

The adjoining figure represents the preferences of the students during breakfast in a hostel mess. If the total number of students in the mess is 540, then with reference to the given figure, answer the following questions:

Here are two different factor trees of the number 90. Write the missing numbers: Fractions, ML Aggarwal Understanding Mathematics Solutions ICSE Class 6.

(i) What is the number of students who prefer coffee?

(ii) Whose number is greater, milk drinkers or orange juice drinkers and by what number?

(iii) What is the total number of students who drink mango shake or coffee? Is it equal to milk drinker?

(iv) Is the sum of all fractions in the given figure equal to 1?

Fractions

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Answer

Total number of students = 540.

From the figure, the fractions are:

Orange juice = 14\dfrac{1}{4}, Milk = 13\dfrac{1}{3}, Tea = 112\dfrac{1}{12}, Coffee = 16\dfrac{1}{6}, Mango shake = 16\dfrac{1}{6}.

(i) Number of students who prefer coffee = 16×540\dfrac{1}{6} \times 540

540690\Rightarrow \dfrac{540}{6}\\[1em] \Rightarrow 90

Hence, the number of students who prefer coffee = 90.

(ii) Number of milk drinkers = 13×540=5403=180\dfrac{1}{3} \times 540 = \dfrac{540}{3} = 180.

Number of orange juice drinkers = 14×540=5404=135\dfrac{1}{4} \times 540 = \dfrac{540}{4} = 135.

Since 180 > 135, milk drinkers are more than orange juice drinkers.

Difference = 180 - 135 = 45.

Hence, milk drinkers are greater than orange juice drinkers by 45.

(iii) Number of mango shake drinkers = 16×540=90\dfrac{1}{6} \times 540 = 90.

Number of coffee drinkers = 90.

Total = 90 + 90 = 180.

Number of milk drinkers = 180.

Yes, total of mango shake and coffee drinkers (180) is equal to milk drinkers (180).

Hence, the total number of students who drink mango shake or coffee = 180, which is equal to the number of milk drinkers.

(iv) Sum of all fractions = 14+13+112+16+16\dfrac{1}{4} + \dfrac{1}{3} + \dfrac{1}{12} + \dfrac{1}{6} + \dfrac{1}{6}

LCM of 4, 3, 12, 6 = 12.

1×34×3+1×43×4+112+1×26×2+1×26×2312+412+112+212+2123+4+1+2+21212121\Rightarrow \dfrac{1 \times 3}{4 \times 3} + \dfrac{1 \times 4}{3 \times 4} + \dfrac{1}{12} + \dfrac{1 \times 2}{6 \times 2} + \dfrac{1 \times 2}{6 \times 2}\\[1em] \Rightarrow \dfrac{3}{12} + \dfrac{4}{12} + \dfrac{1}{12} + \dfrac{2}{12} + \dfrac{2}{12}\\[1em] \Rightarrow \dfrac{3 + 4 + 1 + 2 + 2}{12}\\[1em] \Rightarrow \dfrac{12}{12}\\[1em] \Rightarrow 1

Hence, yes, the sum of all fractions in the given figure is equal to 1.

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