Online Aptitude Test - Aptitude Test 5
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
C.P. of 1 orange = Rs. | ![]() |
350 | ![]() |
= Rs. 3.50 |
100 |
S.P. of 1 orange = Rs. | ![]() |
48 | ![]() |
= Rs. 4 |
12 |
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0.50 | x 100 | ![]() |
= | 100 | % = 14 | 2 | % |
3.50 | 7 | 7 |
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
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140 | x 5x | ![]() |
, | ![]() |
150 | x 7x | ![]() |
and | ![]() |
175 | x 8x | ![]() |
100 | 100 | 100 |
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21x | and 14x. |
2 |
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21x | : 14x |
2 |
14x : 21x : 28x
2 : 3 : 4.
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
Three friends, P, Q and R started a partnership business investing money in the ratio of 5 : 4 : 2 respectively for a period of 3 years. What is the amount received by P as his share profit? | |
I. | Total amount invested in the business in Rs. 22,000. |
II. |
Profit earned at the end of 3 years is |
III. | The average amount of profit earned per year is Rs. 2750. |
I and II give, profit after 3 years = Rs. | ![]() |
3 | x 22000 | ![]() |
= Rs. 8250. |
8 |
From III also, profit after 3 years = Rs. (2750 x 3) = Rs. 8250.
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8250 x | 5 | ![]() |
= Rs. 3750. |
11 |
Thus, (either III is redundant [or] I and II are redundant).
Correct answer is (B).
Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
250 : 200 :: 60 : x
250 x x = (200 x 60)
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(200 x 60) |
250 |
x = 48.
P can complete the work in (12 x 8) hrs. = 96 hrs.
Q can complete the work in (8 x 10) hrs. = 80 hrs.
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1 | and Q's 1 hour's work = | 1 | . |
96 | 80 |
(P + Q)'s 1 hour's work = | ![]() |
1 | + | 1 | ![]() |
= | 11 | . |
96 | 80 | 480 |
So, both P and Q will finish the work in | ![]() |
480 | ![]() |
hrs. |
11 |
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480 | x | 1 | ![]() |
= | 60 | days = 5 | 5 | days. |
11 | 8 | 11 | 11 |
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x ![]() |
![]() |
4 x 20 | ![]() |
5 |
x = 16 days.
Hence, Tanya takes 16 days to complete the work.
Speed of the train relative to man = | ![]() |
125 | ![]() |
10 |
= | ![]() |
25 | ![]() |
2 |
= | ![]() |
25 | x | 18 | ![]() |
2 | 5 |
= 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45
x = 50 km/hr.
Let the speed of the slower train be x m/sec.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
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(100 + 100) | = 3x |
8 |
24x = 200
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25 | . |
3 |
So, speed of the faster train = | 50 | m/sec |
3 |
= | ![]() |
50 | x | 18 | ![]() |
3 | 5 |
= 60 km/hr.


Total distance covered |
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= 3 min. |
Let man's rate upstream be x kmph.
Then, his rate downstream = 2x kmph.
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![]() |
2x + x | ![]() |
: | ![]() |
2x - x | ![]() |
2 | 2 |
= | 3x | : | x |
2 | 2 |
= 3 : 1.
S.I. = Rs. (15500 - 12500) = Rs. 3000.
Rate = | ![]() |
100 x 3000 | ![]() |
= 6% |
12500 x 4 |
Video Explanation: https://youtu.be/SIcQu1HOmOk
Let the rate be R% p.a.
Then, 1200 x | ![]() |
1 + | R | ![]() |
2 | = 1348.32 |
100 |
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1 + | R | ![]() |
2 | = | 134832 | = | 11236 |
100 | 120000 | 10000 |
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1 + | R | ![]() |
2 | = | ![]() |
106 | ![]() |
2 |
100 | 100 |
![]() |
R | = | 106 |
100 | 100 |
R = 6%
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
What is the cost painting the two adjacent walls of a hall at Rs. 5 per m2 which has no windows or doors? | |
I. | The area of the hall is 24 sq. m. |
II. | The breadth, length and height of the hall are in the ratio of 4 : 6 : 5 respectively. |
III. | Area of one wall is 30 sq. m. |
From II, let l = 4x, b = 6x and h = 5x.
Then, area of the hall = (24x2) m2.
From I. Area of the hall = 24 m2.
From II and I, we get 24x2 = 24 x = 1.
l = 4 m, b = 6 and h = 5 m.
Thus, area of two adjacent walls = [(l x h) + (b x h)] m2 can be found out and so the cost of painting two adjacent walls may be found out.
Thus, III is redundant.
Correct answer is (C).
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
- Give answer (A) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
- Give answer (B) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
- Give answer (C) if the data either in Statement I or in Statement II alone are sufficient to answer the question.
- Give answer (D) if the data even in both Statements I and II together are not sufficient to answer the question.
- Give answer(E) if the data in both Statements I and II together are necessary to answer the question.
What is the height of a circular cone? | |
I. | The area of that cone is equal to the area of a rectangle whose length is 33 cm. |
II. | The area of the base of that cone is 154 sq. cm. |
II gives the value of r.
But, in I, the breadth of rectangle is not given.
So, we cannot find the surface area of the cone.
Hence, the height of the cone cannot be determined.
Correct answer is (D).
At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.
55 min. spaces are gained in 60 min.
40 min. spaces are gained in | ![]() |
60 | x 40 | ![]() |
= | 43 | 7 | min. |
55 | 11 |
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7 | min. past 5. |
11 |

Let investment in each case be Rs. (143 x 117).
Income in 1st case = Rs. | ![]() |
11 | x 143 x 117 | ![]() |
= Rs. 1287. |
143 |
Income in 2nd case = Rs. | ![]() |
39 | x 143 x 117 | ![]() |
= Rs. 1394.25 |
4 x 117 |
Clearly, 9 | 3 | % stock at 117 is better. |
4 |
T.D. = | B.G. x 100 | = Rs. | ![]() |
6 x 100 | ![]() |
= Rs. 50. |
R x T | 12 x 1 |
The numbers are squares of odd natural numbers, starting from 5 up to 15.
So, 36 is wrong.