# Splification - Quantitative Aptitude

Q1. A boy read 3/8 th of a book on one day and 4/5 th of the remainder on another day. If there were 60 pages unread, how many pages did the book contain ?

(a) 452

(b) 466

(c) 480

(d) 495

**Answer:** (c) 480

Let, the total number of pages be x

1st day read = x X (3/8)

Remainder = (x - (3x/8)) = 5x/8

Read next day = ( | 4 | X | 5x | ) = | x |

5 | 8 | 2 |

A/Q (3/8)x + (x/2) + 60 = x

⇒ 3x + 4x + 480 = 8x

⇒ x = 480

Q2. In a class, there are two sections A and B. if 10 students of section B shift over to section A, the strength of A becomes three times the strength of B. But if 10 students shift over from A to B, both A and B are equal in strength. How many students are there in sections A and B ?

(a) 35, 21

(b) 39, 25

(c) 44, 29

(d) 50, 30

**Answer:** (d) 50, 30

Let, in section A total number of student be x and in section B total number of student be y

∴ x + 10 = 3(y - 10)

⇒ 3y - x = 40 ------- (1)

x - 10 = y + 10

x - y = 20 ------- (2)

(1) + (2) ⇒ 2y = 60 ⇒ y = 30

(2) ⇒ x = 20 + 30 = 50

∴ Answer is 50, 30

Q3. Students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2 rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is

(a) 77

(b) 79

(c) 85

(d) 96

**Answer:** (d) 96

Let, the number of rows be x and the number of student be y

∴ Total number of students = xy

Now (y+4)(x-2) = xy

⇒ 4x - 2y = 8

⇒ 2x - y = 4 ------- (1)

And (y-4)(x+4) = xy

⇒ -4x + 4y = 16

-x + y = 4 ------(2)

On solving (1) and (2) we get

x = 8 y = 12

∴ Number of student in the class = 8 X 12 = 96

Q4. On the Children's Day, sweet were distributed equally among 540 children. But, on that day 120 children were absent. Hence each child got 4 more sweets. How many sweets were to be distributed to each child originally ?

(a) 12

(b) 14

(c) 16

(d) 18

**Answer:** (b) 14

Let, required number of sweets be x

∴ 540x = (540-120)(x+4)

⇒ 540x = 420 (x+4)

⇒ 120x = 420 X 4

⇒ x = (420 X 4)/120 = 14

Q5. The difefrence of 19/16 and its reciprocal is

(a) 99/177

(b) 101/299

(c) 105/304

(d) 108/311

**Answer:** (c) 105/304

Difference = | 19 | - | 16 | = | 361 - 256 | = | 105 |

16 | 19 | 304 | 304 |

Q6. A sink contains exactly 12 liters of water. If water is drained from the sink until it holds 6 liters of water less than the quantity drained away, then how many liters of water were drained away ?

(a) 9 liters

(b) 11 liters

(c) 13 liters

(d) 16 liters

**Answer:** (a) 9 liters

Let x liters of water drained away

∴ (12 - x) = (x - 6)

⇒ 2x = 18

⇒ x = 9 liters

Q7. A total of 324 coins of 20 paise and 25 paise make a sum of Rs. 65. The number of 25 paise coins is

(a) 4

(b) 12

(c) 17

(d) 23

**Answer:** (a) 4

Let, No. of 25 paise coins be x

∴ No of 20 paise coins be (324 - x)

A/Q | 25 | x + | 20 | (324-x) = 65 |

100 | 100 |

⇒ | x | + | 324 - x | = 65 |

4 | 5 |

⇒ 5x + 1296 - 4x = 65 X 20

⇒ x = 1300 - 1296

⇒ x = 4

Q8. How many digits are required for numbering the pages of a book having 200 pages ?

(a) 457

(b) 492

(c) 521

(d) 556

**Answer:** (b) 492

One digit number = 9

Two digit numbers = 90

Three digit numbers = 101

∴ Total numbers of digits = 1 X 9 + 2 X 90 + 3 X 101 = 9 + 180 + 303 = 492

Q9. Hari went to a shop to buy 50 Kg of rice. He buys two varities of rice which cost him Rs. 4.50 per Kg and Rs. 5 per Kg. He spends a total of Rs. 240. What was the quantity of cheaper rice bought by him ?

(a) 17 Kg

(b) 20 Kg

(c) 23 Kg

(d) 25 Kg

**Answer:** (b) 20 Kg

Let the required quantity of rice be x Kg.

∴ Quantity of costlier rice bought = 50 - x

A/Q (9/2)x + 5 (50-x) = 240

⇒ 9x + 10 (50-x) = 480

⇒ x = 20 Kg

Q10. If (ab + bc + ca) = 0 then (1/a^{2}-bc) + (1/b^{2}-ca) + (1/c^{2}-ab) = ?

(a) 0

(b) 1

(c) 2

(d) 4

**Answer:** (a) 0

ab+bc+ca = 0

⇒ ab+ca = -bc

⇒ a^{2} + ab + ca = a^{2} - bc

⇒ a (a+b+c) = a^{2} - bc

Similarly b^{2} - ca = b(a+b+c)

c^{2} - ab = c(a+b+c)

Now | 1 | + | 1 | + | 1 | = | 1 | + | 1 | + | 1 |

a^{2} - bc |
b^{2} - ca |
c^{2} - ab |
a(a+b+c) | b(a+b+c) | c(a+b+c) |

= | bc + ca + ab | = | 0 | = 0 |

abc(a+b+c) | abc(a+b+c) |

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