Question

Bailey’s restaurant bought 6 1/3 kg of onions. The restaurant bought 7 times as much potatoes as onion. How many kg of potatoes did the restaurant buy? In a day the restaurant needs 2 1/3 kg of potatoes. For how many days will the total potatoes last?

Answer 1

Given :

The quantity of onions that restaurant bought = 6 $\dfrac{1}{3}$  = $\dfrac{19}{3}$ kg

The quantity of potatoes that restaurant bought = 7 × quantity of onions

The quantity of potatoes does restaurant needs in a day = 2 $\dfrac{1}{3}$  = $\dfrac{7}{3}$ kg

To Find :

( a ) How many kg of potatoes did the restaurant buy

( b ) For how many days will the total potatoes last

Solution :

( a )

Let The quantity of potatoes did the restaurant buy = x kg

∵ The quantity of potatoes that restaurant bought = 7 × quantity of onions

Or,                                                                          x   = 7 ×  $\dfrac{19}{3}$ kg

Or,                                                                          x    =  $\dfrac{133}{3}$ kg

The quantity of potatoes that restaurant bought =  x   =  $\dfrac{133}{3}$ kg

( b )

Let The number of days will the total potatoes last = n days

Applying unitary method

∵ For $\dfrac{7}{3}$ kg of potatoes , number of day required = 1

Or, For 1 kg of potatoes , number of day required = $\dfrac{1}{\dfrac{7}{3} }$ = $\dfrac{3}{7}$

∴  For $\dfrac{133}{3}$ kg of potatoes , number of day required = $\dfrac{3}{7}$ × $\dfrac{133}{3}$

Or,                                                                            n = $\dfrac{3 \times 133}{3 \times 7}$

i.e                                                                             n  = $\dfrac{133}{7}$

∴                                                                               n = 19 days

So, The number of days will the total potatoes last = n = 19 days

Hence,

( a ) The quantity of potatoes that restaurant bought is $\dfrac{133}{3}$ kg

( b ) The number of days will the total potatoes last is 19 days           . Answer