A trader was moving along a road selling eggs. An idler who didn't have much to do started to get the trader into a wordy duel. This grew into a fight; he pulled the basket with eggs and dashed it on the floor. The eggs broke. The trader requested the Panchayat to ask the idler to pay for the broken eggs. The Panchayat asked the trader how many eggs were broken. He gave the following response:
If counted in pairs one will remain ;
If counted in threes, two will remain;
If counted in fours, three will remain;
If counted in fives, four will remain;
If counted in sixes, five will remain;
If counted in sevens, nothing will remain;
My basket cannot accommodate more than150eggs.
So, how many eggs were there?

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115

116

118

119

answer is D.

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Detailed Solution

Suppose the number of eggs is ′m′.
The answer 'If you count in pairs, there is one left' means that if we divide all the eggs into groups of two, there will be one left at the end. We can also say that if we divide the total number of eggs by 2, we get 1 remaining unpaired egg.
⇒ The number is odd and can be represented as:

(i) Again, the response ‘If
counted in threes, two will remain’ means if the total number of eggs is divided into a group of 33 then 22 eggs will remain ungrouped.
⇒ Number of eggs can also be represented as:

(ii) Similarly, for each of the responses we have:
‘If counted in fours, three will remain’ means:

(iii)
‘If counted in fives, four will remain’ means:

(iv)
‘If counted in sixes, five will remain’ means:

(v)
‘If counted in sevens, nothing will remain’ means that the number of eggs is exactly divisible by the number seven, i.e.

(vi)
And also the total number of eggs is less than 150, so m<150
Now, because we know that all multiples of five always have either 0 or 50 or 5 in the ones place. So using relation (iv) we can say that instead of units ′m′ will be either

.
Also having four at its units place will make the number ′m′ an even number. But from relation (i), we know that ′m′ is an odd number.
Therefore, 99 will be at the unit place of the number ′m′ .
As we know that ′m′ is multiple of seven and hence lie between

.
So, from relation (vi) ⇒t belongs to interval [1,21]
But from relation (i) and (iv), we know that t≠2,4,6,8...... or t≠5,10,15,20,25......
Therefore, we are left with the values ⇒t=1,3,7,9,11,13,17,19,21
But also from relation (ii) we know ′m′ is not a multiple of 3 . So after eliminating the multiples of 3 from the set of values of ′t′ , we will see:
⇒t=1,7,11,13,17,19
For the above possible values of ′t′ , the possible values of ′m′ are:
⇒ For t=1,7,11,13,17,19 ; m=7,49,77,91,119,133
But since ′m′ can only have nine at its units place, i.e. possible values of ′m′ are m=49 or m=119
For m=49 , we can write it as:

which is not of the form

From relation (v), we can therefore conclude that m≠49
So we have the final value as: m=119
So there were a total of 119 eggs in the basket.
So option 4 is correct.

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