Online Aptitude Test - Aptitude Test - Random
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- Total number of questions: 20.
- Time allotted: 30 minutes.
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Marks : 2/20
Test Review : View answers and explanation for this test.
x = 13p + 11 and x = 17q + 9
13p + 11 = 17q + 9
17q - 13p = 2
q = | 2 + 13p |
17 |
The least value of p for which q = | 2 + 13p | is a whole number is p = 26 |
17 |
x = (13 x 26 + 11)
= (338 + 11)
= 349
Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.
x = 5k + 3
x^{2} = (5k + 3)^{2}
= (25k^{2} + 30k + 9)
= 5(5k^{2} + 6k + 1) + 4
On dividing x^{2} by 5, we get 4 as remainder.
x + 3699 + 1985 - 2047 = 31111
x + 3699 + 1985 = 31111 + 2047
x + 5684 = 33158
x = 33158 - 5684 = 27474.
(112 x 5^{4}) = 112 x | 10 | 4 | = | 112 x 10^{4} | = | 1120000 | = 70000 | ||
2 | 2^{4} | 16 |
106 x 106 - 94 x 94 | = (106)^{2} - (94)^{2} |
= (106 + 94)(106 - 94) [Ref: (a^{2} - b^{2}) = (a + b)(a - b)] | |
= (200 x 12) | |
= 2400. |
Given Exp. = | (12)^{3} x 6^{4} | = | (12)^{3} x 6^{4} | = (12)^{2} x 6^{2} = (72)^{2} = 5184 |
432 | 12 x 6^{2} |
Let 3889 + 12.952 - x = 3854.002.
Then x = (3889 + 12.952) - 3854.002
= 3901.952 - 3854.002
= 47.95.
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
How many marks did Tarun secure in English? | |
I. | The average mark obtained by Tarun in four subjects including English is 60. |
II. | The total marks obtained by him in English and Mathematics together are 170. |
III. | The total marks obtained by him in Mathematics and Science together are 180. |
I gives, total marks in 4 subjects = (60 x 4) = 240.
II gives, E + M = 170
III gives, M + S = 180.
Thus, none of (A), (B), (C), (D) is true.
Correct answer is (E).
A shopkeeper sells some articles at the profit of 25% on the original price. What is the exact amount of profit? To find the answer, which of the following information given in Statements I and II is/are necessary? | |
I. | Sale price of the article |
II. | Number of articles sold |
Gain = 25% of C.P.
In order to find gain, we must know the sale price of each article and the number of articles sold.
Correct answer is (D).
Aman : Rakhi : Sagar = (70,000 x 36) : (1,05,000 x 30) : (1,40,000 x 24) = 12 : 15 : 16.
A's 1 day's work = | 1 | ; |
15 |
B's 1 day's work = | 1 | ; |
20 |
(A + B)'s 1 day's work = | 1 | + | 1 | = | 7 | . | ||
15 | 20 | 60 |
(A + B)'s 4 day's work = | 7 | x 4 | = | 7 | . | ||
60 | 15 |
Therefore, Remaining work = | 1 - | 7 | = | 8 | . | ||
15 | 15 |
Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.
= | 36 x | 5 | m/sec | |
18 |
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Time taken = | 360 | sec | = 36 sec. | |
10 |
S.I. = Rs. (15500 - 12500) = Rs. 3000.
Rate = | 100 x 3000 | % | = 6% | |
12500 x 4 |
Video Explanation: https://youtu.be/SIcQu1HOmOk
S.I. = Rs | 1200 x 10 x 1 | = Rs. 120. | ||
100 |
C.I. = Rs. | 1200 x | 1 + | 5 | 2 | - 1200 | = Rs. 123. | ||||
100 |
Difference = Rs. (123 - 120) = Rs. 3.
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What is the area of a right-angled triangle? | |
I. | The perimeter of the triangle is 30 cm. |
II. | The ratio between the base and the height of the triangle is 5 : 12. |
III. | The area of the triangle is equal to the area of a rectangle of length 10 cm. |
From II, base : height = 5 : 12.
Let base = 5x and height = 12x.
Then, hypotenuse = (5x)^{2} + (12x)^{2} = 13x.
From I, perimeter of the triangle = 30 cm.
5x + 12x + 13x = 30 x = 1.
So, base = 5x = 5 cm, height = 12x = 12 cm.
Area = | 1 | x 5 x 12 | cm^{2} | = 30 cm^{2}. | |
2 |
Thus, I and II together give the answer.
Clearly III is redundant, since the breadth of the rectangle is not given.
Correct answer is (A).
2(15 + 12) x h = 2(15 x 12)
h = | 180 | m = | 20 | m. |
27 | 3 |
Volume = | 15 x 12 x | 20 | m^{3} | = 1200 m^{3}. | |
3 |
Video Explanation: https://youtu.be/V8EQ1YIaH74
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = | 8! | = 10080. |
(2!)(2!) |
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = | 4! | = 12. |
2! |
Required number of words = (10080 x 12) = 120960.
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) = | n(E) | = | 4 | = | 1 | . |
n(S) | 36 | 9 |