Algebra
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Q.

$A, B, C$ and $D$ play a game of cards. $A$ says to $B$ "If I give you 8 cards, you will have as many as $C$ has and I shall have 3 less than what $C$ has. Also if I take 6 cards from $C$, I shall have twice as many as $D$ has". If $B$ and $D$ together have 50 cards, how many cards have $A$ got?

 A.

40

 B.

37

 C.

23

 D.

27

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Solution:
Option(A) is correct

$B+8=C$

$A−8=C–3$

$A+6=2D$

$B+D=50$

solving these we get $A =40$

Edit: Based on Akanksha and Ravish's input changed the final answer from 27(option D) to 40(option A)

Edit2: For explanation on how to get $A=40$, check Shireen's detailed explanation.


(7) Comment(s)


Doncon
 ()

$A$ says to $B$ "if you give $8$ cards......"

So, $B$ only giving $8$ cards to $A$ ...then how $ B+8$,

it should be $B-8$ and $A+8$.

Am I correct?



Poonam
 ()

I am not getting the answer 40?


Shireen
 ()

$B+8=C$ -------- (1)

$A−8=C–3$ -------- (2)

$A+6=2D$ -------- (3)

$B+D=50$ -------- (4)

Putting value of $C$ from (1) into (2),

$A−8=C–3=(B+8)-3$

$\Rightarrow A-B-13=0$ --------(A)

Similarly, putting value of $D$ from (4) into (3),

$A+6=2D=2(50-B)$

$\Rightarrow A+2B-94=0$ --------(B)

Now we have 2 equations, eq(A) and eq(B) in terms of variables $A$ and $B$. These equations can now be used to find the value of $A$ or $B$.

In order to find $A$, we multiply eq(A) by $2$ and add to eq(B).

$\Rightarrow \text{eq(B)} +2\times \text{eq(A)}$

$(A+2B-94)+2(A-B-13)=0$

$\Rightarrow 3A-94-26=0$

$\Rightarrow 3A=120$

$\Rightarrow A=40$


Bhai
 ()

i am nt getting how ans 40


Shireen
 ()

Kindly see my answer in reply to Poonam.


Ravish
 ()

i marked the right answer as 40..solution also shows 40 and answer showng as wrong ..y???



Akanksha
 ()

answer is 40 rather than the marked answer.