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/a2-bc) + (1/b2-ca) + (1/c2-ab) = ?
(a) 0
(b) 1
(c) 2
(d) 4
Answer: (a) 0
ab+bc+ca = 0
⇒ ab+ca = -bc
⇒ a2 + ab + ca = a2 - bc
⇒ a (a+b+c) = a2 - bc
Similarly b2 - ca = b(a+b+c)
c2 - ab = c(a+b+c)
| Now | 1 | + | 1 | + | 1 | = | 1 | + | 1 | + | 1 |
| a2 - bc | b2 - ca | c2 - 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) |
Practice Test Exam