Aptitude Quiz
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This quiz consists series of questions from all the categories of Aptitude. Few points before you proceed
 There are total of 20 questions in this quiz.
 You will be given 20 minutes to finish this quiz.
 Each question carry 1 mark, and there is no negative marking.
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Question 1 of 20
1. Question
1 pointsThe greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
Correct
Answer: Option 3
Explanation:
Greatest number of 4digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Greatest number of 4digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
$\therefore$ Required number (9999 – 399) = 9600.
Incorrect
Answer: Option 3
Explanation:
Greatest number of 4digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Greatest number of 4digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
$\therefore$ Required number (9999 – 399) = 9600.

Question 2 of 20
2. Question
1 pointsIf the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
Correct
Answer: Option 3
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
$\therefore$ The required sum = $\class{maths_class}{{1 \over a}+{1\over b} = {{a+b}\over ab} = {55 \over 600} = {11 \over 120}}$
Incorrect
Answer: Option 3
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
$\therefore$ The required sum = $\class{maths_class}{{1 \over a}+{1\over b} = {{a+b}\over ab} = {55 \over 600} = {11 \over 120}}$

Question 3 of 20
3. Question
1 pointsThe product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:
Correct
Answer: Option 3
Explanation:
Let the numbers be x and y.
Then, xy = 9375 and $\class{maths_class}{x \over y} = 15$
$\class{maths_class}{xy \over (x/y)} = \class{maths_class}{9375 \over 15}$
$\Rightarrow y^2 = 625$
$\Rightarrow y = 25$
$\Rightarrow$ x = 15y = (15 x 25) = 375.
Incorrect
Answer: Option 3
Explanation:
Let the numbers be x and y.
Then, xy = 9375 and $\class{maths_class}{x \over y} = 15$
$\class{maths_class}{xy \over (x/y)} = \class{maths_class}{9375 \over 15}$
$\Rightarrow y^2 = 625$
$\Rightarrow y = 25$
$\Rightarrow$ x = 15y = (15 x 25) = 375.

Question 4 of 20
4. Question
1 pointsWhat is the twodigit number?
I. The difference between the twodigit number and the number formed by interchanging the digits is 27.
II. The difference between the two digits is 3.
III.The digit at unit’s place is less than that at ten’s place by 3
Correct
Answer: Option 5
Explanation:
Let the tens and units digit be x and y respectively.
I. (10x + y) – (10y + x) $\Leftrightarrow$ x – y = 3.
II. x – y = 3.
III. x – y = 3.
Thus, even all the given three statements together do not give the answer.
$\therefore$ Correct answer is (E).
Incorrect
Answer: Option 5
Explanation:
Let the tens and units digit be x and y respectively.
I. (10x + y) – (10y + x) $\Leftrightarrow$ x – y = 3.
II. x – y = 3.
III. x – y = 3.
Thus, even all the given three statements together do not give the answer.
$\therefore$ Correct answer is (E).

Question 5 of 20
5. Question
1 pointsPresent ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
Correct
Answer: Option 1
Explanation:
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, $\class{maths_class}{{{5x + 2}\over{4x + 3}} = {11 \over 9}}$
$\Rightarrow$ 9(5${x}$ + 3) = 11(4${x}$ + 3)
$\Rightarrow$ 45${x}$ + 27 = 44${x}$ + 33
$\Rightarrow$ 45${x}$ – 44${x}$ = 33 – 27
$\Rightarrow$ ${x}$ = 6.
$\therefore$ Anand’s present age = 4${x}$ = 24 years.
Incorrect
Answer: Option 1
Explanation:
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, $\class{maths_class}{{{5x + 2}\over{4x + 3}} = {11 \over 9}}$
$\Rightarrow$ 9(5${x}$ + 3) = 11(4${x}$ + 3)
$\Rightarrow$ 45${x}$ + 27 = 44${x}$ + 33
$\Rightarrow$ 45${x}$ – 44${x}$ = 33 – 27
$\Rightarrow$ ${x}$ = 6.
$\therefore$ Anand’s present age = 4${x}$ = 24 years.

Question 6 of 20
6. Question
1 pointsWhat is Sonia’s present age?
I. Sonia’s present age is five times Deepak’s present age.
II. Five years ago her age was twentyfive times Deepak’s age at that time.
Correct
Answer: Option E
Explanation:
I. S = 5D $\rightarrow D = \class{maths_class}{S \over 5}$
II. S – 5 = 25 (D – 5) $\Leftrightarrow {S = 25D – 120 ….(ii)} $
Using (i) in (ii), we get $ S = \Bigg(25 \times \class{maths_class}{S \over 5}\Bigg) – 120 $
$\Rightarrow 4S = 120. $
$\Rightarrow S = 30. $
Thus, I and II both together give the answer. So, correct answer is (5).
Incorrect
Answer: Option E
Explanation:
I. S = 5D $\rightarrow D = \class{maths_class}{S \over 5}$
II. S – 5 = 25 (D – 5) $\Leftrightarrow {S = 25D – 120 ….(ii)} $
Using (i) in (ii), we get $ S = \Bigg(25 \times \class{maths_class}{S \over 5}\Bigg) – 120 $
$\Rightarrow 4S = 120. $
$\Rightarrow S = 30. $
Thus, I and II both together give the answer. So, correct answer is (5).

Question 7 of 20
7. Question
1 pointsIf 3^{(x – y)} = 27 and 3^{(x + y)} = 243, then x is equal to:
Correct
Answer: Option 3
Explanation:
3^{x – y} = 27 = 3^{3} $\Leftrightarrow$ x – y = 3 ….(i)
3^{x + y} = 243 = 3^{5} $\Leftrightarrow$ x + y = 5 ….(ii)
On solving (i) and (ii), we get x = 4.
Incorrect
Answer: Option 3
Explanation:
3^{x – y} = 27 = 3^{3} $\Leftrightarrow$ x – y = 3 ….(i)
3^{x + y} = 243 = 3^{5} $\Leftrightarrow$ x + y = 5 ….(ii)
On solving (i) and (ii), we get x = 4.

Question 8 of 20
8. Question
1 pointsSome articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
Correct
Answer: Option 4
Explanation:
Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.
C.P. of 30 articles = Rs. $\Bigg(\class{maths_class}{5\over6} \times 30\Bigg)$ = Rs. 25.
S.P. of 30 articles = Rs. $\Bigg(\class{maths_class}{6\over5} \times 30\Bigg)$ = Rs. 36.
$\therefore$ Gain % = $\Bigg(\class{maths_class}{11\over25} \times 100\Bigg)$ = 44%.
Incorrect
Answer: Option 4
Explanation:
Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.
C.P. of 30 articles = Rs. $\Bigg(\class{maths_class}{5\over6} \times 30\Bigg)$ = Rs. 25.
S.P. of 30 articles = Rs. $\Bigg(\class{maths_class}{6\over5} \times 30\Bigg)$ = Rs. 36.
$\therefore$ Gain % = $\Bigg(\class{maths_class}{11\over25} \times 100\Bigg)$ = 44%.

Question 9 of 20
9. Question
1 pointsHow many times in a day, are the hands of a clock in straight line but opposite in direction?
Correct
Answer: Option 2
Explanation:
The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clcok only).
So, in a day, the hands point in the opposite directions 22 times.
Incorrect
Answer: Option 2
Explanation:
The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clcok only).
So, in a day, the hands point in the opposite directions 22 times.

Question 10 of 20
10. Question
1 pointsThe true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is:
Correct
Answer: Option 2
Explanation:
Let P.W. be Rs. x.
Then, S.I. on Rs. x at 16% for 9 months = Rs. 189.
$\therefore {x \times 16 \times \class{maths_class}{9 \over 12} \times \class{maths_class}{1 \over 100} = 189 \hbox{ or } x = 1875}$
$\therefore$ P.W. = Rs. 1575.
$\therefore$ Sum due = P.W. + T.D. = Rs. (1575 + 189) = Rs. 1764.
Incorrect
Answer: Option 2
Explanation:
Let P.W. be Rs. x.
Then, S.I. on Rs. x at 16% for 9 months = Rs. 189.
$\therefore {x \times 16 \times \class{maths_class}{9 \over 12} \times \class{maths_class}{1 \over 100} = 189 \hbox{ or } x = 1875}$
$\therefore$ P.W. = Rs. 1575.
$\therefore$ Sum due = P.W. + T.D. = Rs. (1575 + 189) = Rs. 1764.

Question 11 of 20
11. Question
1 points41, 43, 47, 53, 61, 71, 73, 81
Correct
Answer: Option 4
Explanation:
Each of the numbers except 81 is a prime number.
Incorrect
Answer: Option 4
Explanation:
Each of the numbers except 81 is a prime number.

Question 12 of 20
12. Question
1 points3, 5, 11, 14, 17, 21
Correct
Answer: Option 3
Explanation:
Each of the numbers except 14 is an odd number.
The number ’14’ is the only EVEN number.
Incorrect
Answer: Option 3
Explanation:
Each of the numbers except 14 is an odd number.
The number ’14’ is the only EVEN number.

Question 13 of 20
13. Question
1 points8, 27, 64, 100, 125, 216, 343
Correct
Answer: Option 2
Explanation:
The pattern is 2^{3}, 3^{3}, 4^{3}, 5^{3}, 6^{3}, 7^{3}. But, 100 is not a perfect cube.
Incorrect
Answer: Option 2
Explanation:
The pattern is 2^{3}, 3^{3}, 4^{3}, 5^{3}, 6^{3}, 7^{3}. But, 100 is not a perfect cube.

Question 14 of 20
14. Question
1 points2, 5, 10, 17, 26, 37, 50, 64
Correct
Answer: Option 4
Explanation:
(1*1)+1 , (2*2)+1 , (3*3)+1 , (4*4)+1 , (5*5)+1 , (6*6)+1 , (7*7)+1 , (8*8)+1
But, 64 is out of pattern.
Incorrect
Answer: Option 4
Explanation:
(1*1)+1 , (2*2)+1 , (3*3)+1 , (4*4)+1 , (5*5)+1 , (6*6)+1 , (7*7)+1 , (8*8)+1
But, 64 is out of pattern.

Question 15 of 20
15. Question
1 points835, 734, 642, 751, 853, 981, 532
Correct
Answer: Option 1
Explanation:
In each number except 751, the difference of third and first digit is the middle one.
Incorrect
Answer: Option 1
Explanation:
In each number except 751, the difference of third and first digit is the middle one.

Question 16 of 20
16. Question
1 pointsThe banker’s discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is:
Correct
Answer: Option 2
Explanation:
S.I. on Rs. 1600 = T.D. on Rs. 1680.
$\therefore$ Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
$\therefore$ Time = ${\Bigg(\class{maths_class}{{100 \times 80} \over {1600 \times 15}}\Bigg)}$ year = $\class{maths_class}{1 \over 3}$ year = 4 months.
Incorrect
Answer: Option 2
Explanation:
S.I. on Rs. 1600 = T.D. on Rs. 1680.
$\therefore$ Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
$\therefore$ Time = ${\Bigg(\class{maths_class}{{100 \times 80} \over {1600 \times 15}}\Bigg)}$ year = $\class{maths_class}{1 \over 3}$ year = 4 months.

Question 17 of 20
17. Question
1 pointsLet N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
Correct
Answer: Option 1
Explanation:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Incorrect
Answer: Option 1
Explanation:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Question 18 of 20
18. Question
1 pointsThe average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
Correct
Answer: Option 3
Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Incorrect
Answer: Option 3
Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.

Question 19 of 20
19. Question
1 pointsThe difference between a twodigit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
Correct
Answer: Option 2
Explanation:
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let ten’s and unit’s digits be 2x and x respectively.
Then, (10 x 2x + x) – (10x + 2x) = 36
$\Rightarrow$ 9x = 36
$\Rightarrow$ x = 4.
$\therefore$ Required difference = (2x + x) – (2x – x) = 2x = 8.
Incorrect
Answer: Option 2
Explanation:
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let ten’s and unit’s digits be 2x and x respectively.
Then, (10 x 2x + x) – (10x + 2x) = 36
$\Rightarrow$ 9x = 36
$\Rightarrow$ x = 4.
$\therefore$ Required difference = (2x + x) – (2x – x) = 2x = 8.

Question 20 of 20
20. Question
1 pointsThe sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
Correct
Answer: Option 4
Explanation:
Let the present ages of son and father be x and (60 –x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
$\Rightarrow$ 54 – x = 5x – 30
$\Rightarrow$ 6x = 84
$\Rightarrow$ x = 14.
$\therefore$ Son’s age after 6 years = (x+ 6) = 20 years..
Incorrect
Answer: Option 4
Explanation:
Let the present ages of son and father be x and (60 –x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
$\Rightarrow$ 54 – x = 5x – 30
$\Rightarrow$ 6x = 84
$\Rightarrow$ x = 14.
$\therefore$ Son’s age after 6 years = (x+ 6) = 20 years..