Verbal Reasoning - Arithmetic Reasoning - Discussion

Let's start solving the problem by using the given information:

1. A says to B, "If you give me three cards, you will have as many as E has"
This means B has three cards more than E.

2. A says to B, "If I give you three cards, you will have as many as D has."
This means A has three cards less than D.

3. A and B together have 10 cards more than what D and E together have.
This can be represented as (A + B) = (D + E) + 10.

4. B has two cards more than C has.
This can be represented as B = C + 2.

5. The total number of cards is 133.
This can be represented as A + B + C + D + E = 133.

Using equation (1), we can write E = B - 3.
Using equation (2), we can write D = A + 3.
Substituting these values in equation (3), we get (A + B) = (A + B + 10) - 6.
Simplifying, we get A = 8.

Using equation (4), we get B = C + 2.
Using equation (5), we get B + C + D + E = 125.

Substituting the values of B, C, D, and E, we get:
B + (B - 2) + (A + 3) + (B - 3) = 125.
Simplifying, we get B = 23.

Therefore, B has 23 cards.

@All.

1. A says to B "If you give me 3 cards you will have as many as E has means, B has more cards than E. If you add 3 cards to E, then he equals to B.
So, B=E+3.

2. A says to B "If I give you 3 cards you will have as many as D has means, D has more cards than B.
So, D=B+3.

3. From 1 and 2, you will get ( D+E)=2B.

4. A and B together have 10 cards more than what D and E have (A+B) have more cards than (D+E).
So,( A+B)=(D+E)+10=2B+10 (from point 3).

5. B has two cards more than what C has.
So, B=C+2 => C=B-2.

6. Total no.of cards be 133.
then,
(A+B)+(C)+(D+E)=133,
(2B+10)+(B-2)+(2B)=133,
(5B)=125,
B=25.

So, the Answer is option [C].

Total no of cards=133.

i.e A+B+C+D+E=133 ---> (1)
Then B-3=E ---> (2)
B+3=D ---> (3)
B=2+C ie C=B-2 ---> (4)
A+B=10+D+E ---> (5)

Now, we know the values of E and D ie eq(2) and eq(3)
replace those values in eqn(4)
A+B = 10+B+3+B-3,
A+B = 10+2B,
A = 10+2B-B,
A = B+10.

Now we know all values except B.
Replace the all Known values in eq(1).
then

B+10+B+B-2+B+3+B-3=133
5B+8 =133,
5B = 133-8,
5B = 125,
B = 125/5,
B = 25.

Total no of cards=133.

i.e A+B+C+D+E=133 ---> (1)
Then B-3=E ---> (2)
B+3=D ---> (3)
B=2+C ie C=B-2 ---> (4)
A+B=10+D+E ---> (5)

Now, we know the values of E and D ie eq(2) and eq(3)
replace those values in eqn(4)
A+B = 10+B+3+B-3,
A+B = 10+2B,
A = 10+2B-B,
A = B+10.

Now we know all values except B.
Replace the all Known values in eq(1).
then

B+10+B+B-2+B+3+B-3=133,
5B+8 =133,
5B = 133-8,
5B = 125,
B = 125/5,
B = 25.

We have,

B - 3 = E .
B + 3 = D.
B=C+2=>C=B-2.

Then,
A+B = D + E+10.
Which we can be done into,

A+B = B-3 + B+3 + 10.
A+B= 2B + 10.

Then lets add all.

A+B+C+D+E=133.
2B+10+B-2+B+3+B-3=133.
5B=133-8.
B=125/5.
B=25 -> we get the answer.

A+B+C+D+E = 133.

B+3 = D(1).
B-3 = E(2).

B = C+2, C = B-2(3).

A+B = D+E+10 = (B+3)+(B-3)+10(4).

From 1, 2, 3 & 4.

We get...

B+3+B-3+10+B-2+B+3+B-3 = 133.

5B+8 = 133.
5B = 125.
B = 25.

B = x.
A + B = 2x + 10,
A = x + 10.
E = x - 3.
D = x + 3.
C = x - 2.
So, x + 10 + x + x - 2 + x + 3 + x - 3 = 133.
=> 5x = 125.
=> x = 25.

B=C+2=>C=B-2
2B=D+E=>D=2B-E
(B+10)+B+(B-2)+2B-E+E=133
B+10+B+B-2+2B=133
5B+8=133
5B=133-8
B=125/5
B=25

n = 96.

=> 4n-4 = 96.

=> 4(n-1) = 96.

=> n-1 = 96/4.

=> n-1 = 24.

=> n = 24+1.

=> n = 25.

A+B+C+D+E=133
(B+10)+B+(B-2)+2B=133

5B+8(i.e 10-2)=133
5B=133-8
5B=125
B=125/5
B=25.