Reasoning Aptitude
ARITHMETIC REASONING MCQs
Arithmetical Reasoning
Let number of horses = number of men = x.
Then, number of legs = 4x + 2 x (x/2) = 5x.
So, 5X = 70 or x = 14.
We have : A = 3B ...(i) and
C - 4 = 2 (A - 4) ...(ii)
Also, A + 4 = 31 or A= 31-4 = 27.
Putting A = 27 in (i), we get: B = 9.
Putting A = 27 in (ii), we get C = 50.
Let the father's age be x and the son's age be y.
Then, x - y = y or x = 2y
Now, x = 36. So. 2y = 36 or y = 18.
Son's present age = 18 years.
So. son's age 5 years ago = 13 years.
Let the number of correct answers be x.
Number of incorrect answers = (75 - x).
4x - (75 - x) = 125 or 5x = 200 or x = 40.
Let number of keepers be x. Then,
Total number of feet = 2 x 50 + 4 x 45 + 4 x 8 + 2x = 2x + 312.
Total number of heads = 50 + 45 + 8 + x= 103 + x.
Therefore (2x + 312) = (103 + x) + 224 or x = 15.
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 wil l have ai
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 numbei
of cards be 133, how many cards does B have ?
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 or 5B = 125 or B = 25.
Clearly, we have :
J - 40 = `1/2`A ...(i)
A - 40 = J ...(ii)
A - 40 = R + 40 ...(iii)
Solving (i) and.(ii) simultaneously, we get : J = 120 and A= 160.
Putting A 160 in (iii), we get R = 80.
Total money = R + J + A= Rs.(80 + 120 + 160) = Rs.360.
A, B, C, D and E play a game of cards. A says to E, "If you give me 3 cards,
you will have as many as I have at this moment while if D takes 5 cards from
you, he will have as many as E has.'' A and C together have twice as many
cards as E has. B and D together also have the same number of cards as A and
C taken together. If together they have 150 cards, how many cards has C got ?
Clearly, we have :
A = B - 3 ...(i)
D + 5 = E ...(ii)
A+C = 2E ...(iii)
B + D = A+C = 2E ...(iv)
A+B + C + D + E=150 ...(v)
From (iii), (iv) and (v), we get: 5E = 150 or E = 30.
Putting E = 30 in (ii), we get: D = 25.
Putting E = 30 and D = 25 in (iv), we get: B = 35.
Putting B = 35 in (i), we get: A = 32.
Putting A = 32 and E = 30 in (iii), we get: C = 28.
Clearly, the required number would be such that it leaves a remainder of 1 when
divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.