# Verbal Reasoning - Arithmetic Reasoning - Discussion

Discussion Forum : Arithmetic Reasoning - Section 1 (Q.No. 5)
5.
A, B, C, D and E play a game of cards. A says to B, "If you give me three cards, you will have as many as E has and if I give you three cards, you will have as many as D has." A and B together have 10 cards more than what D and E together have. If B has two cards more than what C has and the total number of cards be 133, how many cards does B have ?
22
23
25
35
Explanation:

Clearly, we have :

B-3 = E ...(i)

B + 3 = D ...(ii)

A+B = D + E+10 ...(iii)

B = C + 2 ...(iv)

A+B + C + D + E= 133 ...(v)

From (i) and (ii), we have : 2 B = D + E ...(vi)

From (iii) and (vi), we have : A = B + 10 ...(vii)

Using (iv), (vi) and (vii) in (v), we get:

(B + 10) + B + (B - 2) + 2B = 133 5B = 125 B = 25.

Discussion:
18 comments Page 1 of 2.

Ashutosh Priyadarshan said:   2 years ago
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.

Sonia said:   5 years ago
Thanks @Gajjela.

Gajjela said:   5 years ago
@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].

Soma datta said:   5 years ago
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.

Siva said:   6 years ago
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.

I'm not getting this. Please explain me with simple steps.

Manimegalai said:   6 years ago
Not getting this, Please explain step by step clearly.

Pavan pj said:   6 years ago

SHUDHI said:   6 years ago
How A + B = D + E + 10?

It must be D + E = A + B + 10.
(1)

Lavanya said:   7 years ago