from a pack of cards king queen jack and ace are removed then the algebraic sum of the rest of

total no. of cards in a pack =52
total no of face cards=4*4=16
remaining cards=52-16=36
algebraic sum of rest cards=(2+3+4+5+6+7+8+9+10)*4
=54*4=216
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Ans is 216
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52 cards are divided into 4 suits there four each suit contains 13 cards including king, queen, jack and ace. the values of king , queen and jack are more than 10 so leave them and also leave 1 because ace value is 1. Then make sum of 2+3+4+5+6+7+8+9+10=54.Sum of each suit=54, totally we have 4 suit therefore 4*54=216. Option c is the correct answer.
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ans is 216
4(2+3+4+5+6+7+8+9+10)
216
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(2+3+4+5+6+7+8+9+10)*4
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4*(2+10)*9/2
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54*4=216
2+3+4+5+6+7+8+9+10=54
since we have 4 types in card so 54*4=216
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There are total 16 cards of queen,king,jack and ace in a pack of 52 cards.
The remaining cards are 2,3,4,5,6,7,8,9,10.
There are total 4 pairs . So add them all and multiply by 4.
(2+3+4+5+6+7+8+9+10)*4=54*4=216
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total number of cards=52
total number of symbol is 4
total number of card in each symbol=13
we have to remove Ace,Queen,king,Jack=4
so, 13-4 from each.
so remaining card numbers sum=(2+3+4+5+6+7+8+9+10)=54
then 54*4=216.
Answer=216.
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from a pack of cards king, queen, jack and ace are removed. then the algebraic sum of the rest of the cards are?
a)53 b)314 c)216 d)65