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 pointsA fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
Correct
Answer: Option 1
Explanation:
Let the total number of shots be x. Then,
Shots fired by A = $\class{maths_class}{5\over8}x$
Shots fired by B = $\class{maths_class}{3\over8}x$
Killing shots by A = $\class{maths_class}{1\over3} {\rm of} \class{maths_class}{5\over8}x = \class{maths_class}{5\over24}x$
Killing shots by B = $\class{maths_class}{1\over2} {\rm of} \class{maths_class}{3\over8}x = \class{maths_class}{3\over16}x$
$$ \therefore {3x\over24} = 27 {\rm or} x = \left({27 \times 16}\over3\right) = 144 $$
Incorrect
Answer: Option 1
Explanation:
Let the total number of shots be x. Then,
Shots fired by A = $\class{maths_class}{5\over8}x$
Shots fired by B = $\class{maths_class}{3\over8}x$
Killing shots by A = $\class{maths_class}{1\over3} {\rm of} \class{maths_class}{5\over8}x = \class{maths_class}{5\over24}x$
Killing shots by B = $\class{maths_class}{1\over2} {\rm of} \class{maths_class}{3\over8}x = \class{maths_class}{3\over16}x$
$$ \therefore {3x\over24} = 27 {\rm or} x = \left({27 \times 16}\over3\right) = 144 $$

Question 2 of 20
2. Question
1 pointsThe average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
Correct
Answer: Option 2
Explanation:
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 …. (i)
Q + R = (6250 x 2) = 12500 …. (ii)
P + R = (5200 x 2) = 10400 …. (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 …. (iv)
Subtracting (ii) from (iv), we get P = 4000.
$\therefore$ P’s monthly income = Rs. 4000.
Incorrect
Answer: Option 2
Explanation:
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 …. (i)
Q + R = (6250 x 2) = 12500 …. (ii)
P + R = (5200 x 2) = 10400 …. (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 …. (iv)
Subtracting (ii) from (iv), we get P = 4000.
$\therefore$ P’s monthly income = Rs. 4000.

Question 3 of 20
3. Question
1 pointsA library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
Correct
Answer: Option 4
Explanation:
Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average = $\Bigg(\class{maths_class}{{{510 \times 5} + {240 \times 25}}\over30}\Bigg)$
$ = \class{maths_class}{8550\over30}$
= ${285}$
Incorrect
Answer: Option 4
Explanation:
Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average = $\Bigg(\class{maths_class}{{{510 \times 5} + {240 \times 25}}\over30}\Bigg)$
$ = \class{maths_class}{8550\over30}$
= ${285}$

Question 4 of 20
4. Question
1 pointsThe sum of two number is 25 and their difference is 13. Find their product.
Correct
Answer: Option 2
Explanation:
Let the numbers be x and y.
Then, x + y = 25 and x – y = 13.
4xy = (x + y)^{2} – (x– y)^{2}
= (25)^{2} – (13)^{2}
= (625 – 169)
= 456
$\therefore$ xy = 114.
Incorrect
Answer: Option 2
Explanation:
Let the numbers be x and y.
Then, x + y = 25 and x – y = 13.
4xy = (x + y)^{2} – (x– y)^{2}
= (25)^{2} – (13)^{2}
= (625 – 169)
= 456
$\therefore$ xy = 114.

Question 5 of 20
5. Question
1 pointsWhat is the number?
I. The sum of the two digits is 8. The ratio of the two digits is 1 : 3.
II. The product of the two digit of a number is 12. The quotient of two digits is 3.
Correct
Answer: Option 3
Explanation:
Let the tens and units digit be x and y respectively. Then,
I. ${x + y = 8}$ and $\class{maths_class}{{x \over y} = {1\over3}}$
$ \therefore $ I gives, $\class{maths_normal}{4y = 24}$ $\Leftrightarrow $ $\class{maths_normal}{y = 6}$
So, $\class{maths_normal}{x + 6 = 8} \Leftrightarrow \class{maths_normal}{x = 2.}$
II. $\class{maths_normal}{xy=12}$ and $\class{maths_class}{{x \over y} = {3 \over 1}}$
$\therefore$ II gives, $\class{maths_normal}{{x^2} = 36 \Leftrightarrow {x = 6}}$
So, $\class{maths_normal}{{3y = 6} \Leftrightarrow {y = 2}}$
Therefore, Either I or II alone sufficient to answer.
Incorrect
Answer: Option 3
Explanation:
Let the tens and units digit be x and y respectively. Then,
I. ${x + y = 8}$ and $\class{maths_class}{{x \over y} = {1\over3}}$
$ \therefore $ I gives, $\class{maths_normal}{4y = 24}$ $\Leftrightarrow $ $\class{maths_normal}{y = 6}$
So, $\class{maths_normal}{x + 6 = 8} \Leftrightarrow \class{maths_normal}{x = 2.}$
II. $\class{maths_normal}{xy=12}$ and $\class{maths_class}{{x \over y} = {3 \over 1}}$
$\therefore$ II gives, $\class{maths_normal}{{x^2} = 36 \Leftrightarrow {x = 6}}$
So, $\class{maths_normal}{{3y = 6} \Leftrightarrow {y = 2}}$
Therefore, Either I or II alone sufficient to answer.

Question 6 of 20
6. Question
1 pointsA is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
Correct
Answer: Option 4
Explanation:
Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.
$\therefore$ (2x + 2) + 2x + x = 27
$\Rightarrow$ 5x = 25
$\Rightarrow$ x = 5
Hence, B’s age = 2x = 10 years.
Incorrect
Answer: Option 4
Explanation:
Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.
$\therefore$ (2x + 2) + 2x + x = 27
$\Rightarrow$ 5x = 25
$\Rightarrow$ x = 5
Hence, B’s age = 2x = 10 years.

Question 7 of 20
7. Question
1 points$\class{maths_normal}{{(17)^{3.5}} \times {(17)^{?}}}$ = $17^8$
Correct
Answer: Option 4
Explanation:
Let $\class{maths_normal}{{(17)^{3.5}} \times {(17)^{?}}}$ = $17^8$
Then, $\class{maths_normal}{{(17)^{3.5+x}}}$ = $17^8$
$\therefore$ 3.5 + x = 8
$\Rightarrow$ x = (8 – 3.5)
$\Rightarrow$ x = 4.5
Incorrect
Answer: Option 4
Explanation:
Let $\class{maths_normal}{{(17)^{3.5}} \times {(17)^{?}}}$ = $17^8$
Then, $\class{maths_normal}{{(17)^{3.5+x}}}$ = $17^8$
$\therefore$ 3.5 + x = 8
$\Rightarrow$ x = (8 – 3.5)
$\Rightarrow$ x = 4.5

Question 8 of 20
8. Question
1 pointsSpeed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is:
Correct
Answer: Option 4
Explanation:
Speed upstream = 7.5 kmph.
Speed downstream = 10.5 kmph.
$\therefore$ Total time taken = $\Bigg(\class{maths_class}{{105 \over 7.5} + {105 \over 10.5}}\Bigg)$ hours = 24 hours.
Incorrect
Answer: Option 4
Explanation:
Speed upstream = 7.5 kmph.
Speed downstream = 10.5 kmph.
$\therefore$ Total time taken = $\Bigg(\class{maths_class}{{105 \over 7.5} + {105 \over 10.5}}\Bigg)$ hours = 24 hours.

Question 9 of 20
9. Question
1 pointsIn how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Correct
Answer: Option 1
Explanation:
Required number of ways = $({^7C_5} \times {^3C_2}) = ({^7C_2} \times {}^3C_1) = \Bigg(\class{maths_class}{{7 \times 6} \over {2 \times 1}}\times 3\Bigg) = 63.$
Incorrect
Answer: Option 1
Explanation:
Required number of ways = $({^7C_5} \times {^3C_2}) = ({^7C_2} \times {}^3C_1) = \Bigg(\class{maths_class}{{7 \times 6} \over {2 \times 1}}\times 3\Bigg) = 63.$

Question 10 of 20
10. Question
1 points8, 13, 21, 32, 47, 63, 83
Correct
Answer: Option 1
Explanation:
Go on adding 5, 8, 11, 14, 17, 20.
So, the number 47 is wrong and must be replaced by 46.
Incorrect
Answer: Option 1
Explanation:
Go on adding 5, 8, 11, 14, 17, 20.
So, the number 47 is wrong and must be replaced by 46.

Question 11 of 20
11. Question
1 points1, 4, 9, 16, 25, 36, 49, (….)
Correct
Answer: Option 3
Explanation:
Numbers are ${1^2}, {2^2}, {3^2}, {4^2}, {5^2}, {6^2}, {7^2}.$
So, the next number is ${8^2}$
Incorrect
Answer: Option 3
Explanation:
Numbers are ${1^2}, {2^2}, {3^2}, {4^2}, {5^2}, {6^2}, {7^2}.$
So, the next number is ${8^2}$

Question 12 of 20
12. Question
1 pointsQ is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?
Correct
Answer: Option 4
Explanation:
Given that:
1. The difference of age b/w R and Q = The difference of age b/w Q and T.
2. Sum of age of R and T is 50 i.e. (R + T) = 50.
Question: R – Q = ?.
Explanation:
R – Q = Q – T
(R + T) = 2Q
Now given that, (R + T) = 50
So, 50 = 2Q and therefore Q = 25.
Question is (R – Q) = ?
Here we know the value(age) of Q (25), but we don’t know the age of R.
Therefore, (RQ) cannot be determined.
Incorrect
Answer: Option 4
Explanation:
Given that:
1. The difference of age b/w R and Q = The difference of age b/w Q and T.
2. Sum of age of R and T is 50 i.e. (R + T) = 50.
Question: R – Q = ?.
Explanation:
R – Q = Q – T
(R + T) = 2Q
Now given that, (R + T) = 50
So, 50 = 2Q and therefore Q = 25.
Question is (R – Q) = ?
Here we know the value(age) of Q (25), but we don’t know the age of R.
Therefore, (RQ) cannot be determined.

Question 13 of 20
13. Question
1 pointsSimran started a software business by investing Rs. 50,000. After six months, Nanda joined her with a capital of Rs. 80,000. After 3 years, they earned a profit of Rs. 24,500. What was Simran’s share in the profit?
Correct
Answer: Option D
Explanation:
Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.
$\therefore $ Simran’s share = Rs. $\Bigg({24500 \times \class{maths_class}{3 \over 7}}\Bigg)$ = Rs. 10,500
Incorrect
Answer: Option D
Explanation:
Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.
$\therefore $ Simran’s share = Rs. $\Bigg({24500 \times \class{maths_class}{3 \over 7}}\Bigg)$ = Rs. 10,500

Question 14 of 20
14. Question
1 pointsIf 6^{th} March, 2005 is Monday, what was the day of the week on 6^{th} March, 2004?
Correct
Answer: Option A
Explanation:
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
$\therefore$ The day on 6^{th} March, 2005 will be 1 day beyond the day on 6^{th} March, 2004.
Given that, 6^{th} March, 2005 is Monday.
$\therefore$ 6^{th} March, 2004 is Sunday (1 day before to 6^{th} March, 2005).
Incorrect
Answer: Option A
Explanation:
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
$\therefore$ The day on 6^{th} March, 2005 will be 1 day beyond the day on 6^{th} March, 2004.
Given that, 6^{th} March, 2005 is Monday.
$\therefore$ 6^{th} March, 2004 is Sunday (1 day before to 6^{th} March, 2005).

Question 15 of 20
15. Question
1 pointsRobert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
Correct
Answer: Option 3
Explanation:
Let the distance travelled by x km.
Then, $\Bigg(\class{maths_class}{{x \over 10}{x \over 15}}\Bigg)$ = 2
$\Rightarrow$ 3x – 2x = 60
$\Rightarrow$ x = 60 km.
Time taken to travel 60 km at 10 km/hr = $\Bigg(\class{maths_class}{6 \over 10}\Bigg)$ hrs = 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
$\therefore$ Required speed = $\Bigg(\class{maths_class}{60 \over 5}\Bigg)kmph$ = 12kmph
Incorrect
Answer: Option 3
Explanation:
Let the distance travelled by x km.
Then, $\Bigg(\class{maths_class}{{x \over 10}{x \over 15}}\Bigg)$ = 2
$\Rightarrow$ 3x – 2x = 60
$\Rightarrow$ x = 60 km.
Time taken to travel 60 km at 10 km/hr = $\Bigg(\class{maths_class}{6 \over 10}\Bigg)$ hrs = 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
$\therefore$ Required speed = $\Bigg(\class{maths_class}{60 \over 5}\Bigg)kmph$ = 12kmph

Question 16 of 20
16. Question
1 pointsA clock is started at noon. By 10 minutes past 5, the hour hand has turned through:
Correct
Answer: Option 3
Explanation:
Angle traced by hour hand in 12 hrs = ${360^\circ}$
Angle traced by hour hand in 5 hrs 10 min. i.e., $\class{maths_class}{31 \over 6} {\rm hrs} = {\Bigg(\class{maths_class}{{360 \over 12} \times {31 \over 6}}\Bigg)}^\circ = {155^\circ}$
Incorrect
Answer: Option 3
Explanation:
Angle traced by hour hand in 12 hrs = ${360^\circ}$
Angle traced by hour hand in 5 hrs 10 min. i.e., $\class{maths_class}{31 \over 6} {\rm hrs} = {\Bigg(\class{maths_class}{{360 \over 12} \times {31 \over 6}}\Bigg)}^\circ = {155^\circ}$

Question 17 of 20
17. Question
1 pointsOne card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
Correct
Answer: Option 2
Explanation:
Clearly, there are 52 cards, out of which there are 12 face cards.
$\therefore$ P (getting a face card) = $\class{maths_class}{{12 \over 52} = {3 \over 13}}$
Incorrect
Answer: Option 2
Explanation:
Clearly, there are 52 cards, out of which there are 12 face cards.
$\therefore$ P (getting a face card) = $\class{maths_class}{{12 \over 52} = {3 \over 13}}$

Question 18 of 20
18. Question
1 pointsThe least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Correct
Answer: Option 4
Explanation:
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.
$\therefore$ Required number = (90 x 4) + 4 = 364.
Incorrect
Answer: Option 4
Explanation:
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.
$\therefore$ Required number = (90 x 4) + 4 = 364.

Question 19 of 20
19. Question
1 pointsThe greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Correct
Answer: Option 2
Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.
Incorrect
Answer: Option 2
Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.

Question 20 of 20
20. Question
1 pointsIf a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?
Correct
Answer: Option 1
Explanation:
Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
$\therefore$ 250 : 200 :: 60 : x $\Leftrightarrow {250 \times x}$ = (200 x 60)
$\Rightarrow x = \class{maths_class}{{200 \times 60} \over 250}$
$\Rightarrow x = 48$
Incorrect
Answer: Option 1
Explanation:
Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
$\therefore$ 250 : 200 :: 60 : x $\Leftrightarrow {250 \times x}$ = (200 x 60)
$\Rightarrow x = \class{maths_class}{{200 \times 60} \over 250}$
$\Rightarrow x = 48$