Arithmetic Reasoning

1.The total of the ages of Amar, Akbar and Anthony is 80 years. What was the total of their ages three years ago ?

**Answer:**Option** 1**

Solution:

Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.
2.Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A ?

**Answer:**Option** 1**

Solution:

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

3.An institute organised a fete and 1/5 of the girls and 1/8 of the boys participated in the same. What fraction of the total number of students took part in the fete ?

4.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 ?

**Answer:**Option** 1**

Solution:

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.

5.A pineapple costs Rs. 7 each. A watermelon costs Rs. 5 each. X spends Rs. 38 on these fruits. The number of pineapples purchased is

**Answer:**Option** 1**

Solution:

6.A woman says, "If you reverse my own age, the figures represent my husband's age. He is, of course, senior to me and the difference between our ages is one-eleventh of their sum." The woman's age is

**Answer:**Option** 1**

Solution:

Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.Then, woman's age = (10X + y) years; husband's age = (10y + x) years.

Therefore (10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)

⇔ (9y-9x) = (1/11)(11y + 11x) = (x + y) ⇔ 10x = 8y ⇔ x = (4/5)y

Clearly, y should be a single-digit multiple of 5, which is 5.

So, x = 4, y = 5.

Hence, woman's age = 10x + y = 45 years.

7.A girl counted in the following way on the fingers of her left hand : She started by calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5 and then reversed direction calling the ring finger 6, middle finger 7 and so on. She counted upto 1994. She ended counting on which finger ?

**Answer:**Option** 1**

Solution:

Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....i.e. numbers of the form (8n + 1).

Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.

8.What is the product of all the numbers in the dial of a telephone ?

**Answer:**Option** 1**

Solution:

Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.
9.A is 3 years older to B and 3 years younger to C, while B and D are twins. How many years older is C to D?

**Answer:**Option** 1**

Solution:

Since B and D are twins, so B = D.Now, A = B + 3 and A = C - 3.

Thus, B + 3 = C - 3 ⇔ D + 3 = C-3 ⇔ C - D = 6.

10.The 30 members of a club decided to play a badminton singles tournament. Every time a member loses a game he is out of the tournament. There are no ties. What is the minimum number of matches that must be played to determine the winner ?

**Answer:**Option** 1**

Solution:

Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.