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 pointsIt was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Correct
Answer: Option 3
Explanation:
On 31^{st} December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
$\therefore$ On 31^{st} December 2009, it was Thursday.
Thus, on 1^{st} Jan, 2010 it is Friday.
Incorrect
Answer: Option 3
Explanation:
On 31^{st} December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
$\therefore$ On 31^{st} December 2009, it was Thursday.
Thus, on 1^{st} Jan, 2010 it is Friday.

Question 2 of 20
2. Question
1 pointsThe sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Correct
Answer: Option 1
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
$\Rightarrow$ 5x = 20
$\Rightarrow$ x = 4.
$\therefore$ Age of the youngest child = x = 4 years.
Incorrect
Answer: Option 1
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
$\Rightarrow$ 5x = 20
$\Rightarrow$ x = 4.
$\therefore$ Age of the youngest child = x = 4 years.

Question 3 of 20
3. Question
1 pointsA father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:
Correct
Answer: Option 1
Explanation:
Let the son’s present age be x years. Then, (38 – x) = x
$\Rightarrow$ 2x = 38.
$\Rightarrow$ x = 19.
$\therefore$ Son’s age 5 years back (19 – 5) = 14 years.
Incorrect
Answer: Option 1
Explanation:
Let the son’s present age be x years. Then, (38 – x) = x
$\Rightarrow$ 2x = 38.
$\Rightarrow$ x = 19.
$\therefore$ Son’s age 5 years back (19 – 5) = 14 years.

Question 4 of 20
4. Question
1 pointsWhat was the day of the week on 17^{th} June, 1998?
Correct
Answer: Option 3
Explanation:
17^{th} June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days
$\therefore$ 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.
Incorrect
Answer: Option 3
Explanation:
17^{th} June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days
$\therefore$ 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

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, ${\class{maths_class}{{{5x + 3}\over{4x + 3}} = {11 \over 9}}}$
$\Rightarrow$ 9(5x + 3) = 11(4x + 3)
$\Rightarrow$ 45x + 27 = 44x + 33
$\Rightarrow$ 45x – 44x = 33 – 27
$\Rightarrow$ x = 6.
$\therefore$ Anand’s present age = 4x = 24 years
Incorrect
Answer: Option 1
Explanation:
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, ${\class{maths_class}{{{5x + 3}\over{4x + 3}} = {11 \over 9}}}$
$\Rightarrow$ 9(5x + 3) = 11(4x + 3)
$\Rightarrow$ 45x + 27 = 44x + 33
$\Rightarrow$ 45x – 44x = 33 – 27
$\Rightarrow$ x = 6.
$\therefore$ Anand’s present age = 4x = 24 years

Question 6 of 20
6. Question
1 pointsExcluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Correct
Answer: Option 2
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = $\Bigg(\class{maths_class}{9\over54} \times 60 \Bigg)$ min = 10 min.
Incorrect
Answer: Option 2
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = $\Bigg(\class{maths_class}{9\over54} \times 60 \Bigg)$ min = 10 min.

Question 7 of 20
7. Question
1 pointsOn 8^{th} Dec, 2007 Saturday falls. What day of the week was it on 8^{th} Dec, 2006?
Correct
Answer: Option 4
Explanation:
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8^{th} Dec, 2007 will be 1 day beyond the day on 8^{th} Dec, 2006.
But, 8^{th} Dec, 2007 is Saturday.
$\therefore$ 8^{th} Dec, 2006 is Friday.
Incorrect
Answer: Option 4
Explanation:
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8^{th} Dec, 2007 will be 1 day beyond the day on 8^{th} Dec, 2006.
But, 8^{th} Dec, 2007 is Saturday.
$\therefore$ 8^{th} Dec, 2006 is Friday.

Question 8 of 20
8. Question
1 pointsA father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:
Correct
Answer: Option 1
Explanation:
Let the son’s present age be x years. Then, (38 – x) = x
$\Rightarrow$ 2x = 38.
$\Rightarrow$ x = 19.
$\therefore$ Son’s age 5 years back (19 – 5) = 14 years.
Incorrect
Answer: Option 1
Explanation:
Let the son’s present age be x years. Then, (38 – x) = x
$\Rightarrow$ 2x = 38.
$\Rightarrow$ x = 19.
$\therefore$ Son’s age 5 years back (19 – 5) = 14 years.

Question 9 of 20
9. Question
1 pointsIn a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
Correct
Answer: Option 1
Explanation:
Total number of balls = (8 + 7 + 6) = 21.
let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
$\therefore$ n(E) = 7.
$\therefore$ P(E) = $\class{maths_class}{{n(E) \over n(S)} = {7 \over 21} = {1 \over 3}}$
Incorrect
Answer: Option 1
Explanation:
Total number of balls = (8 + 7 + 6) = 21.
let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
$\therefore$ n(E) = 7.
$\therefore$ P(E) = $\class{maths_class}{{n(E) \over n(S)} = {7 \over 21} = {1 \over 3}}$

Question 10 of 20
10. Question
1 pointsThree unbiased coins are tossed. What is the probability of getting at most two heads?
Correct
Answer: Option 4
Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
$\therefore$ P(E) = $\class{maths_class}{{n(E) \over n(S)} = {7 \over 8}}$
Incorrect
Answer: Option 4
Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
$\therefore$ P(E) = $\class{maths_class}{{n(E) \over n(S)} = {7 \over 8}}$

Question 11 of 20
11. Question
1 pointsA rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
Correct
Answer: Option 2
Explanation:
Area of the park = (60 x 40) m^{2} = 2400 m^{2}.
Area of the lawn = 2109 m^{2}.
$\therefore$ Area of the crossroads = (2400 – 2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x – x^{2} = 291
$\Rightarrow$ x^{2} – 100x + 291 = 0
$\Rightarrow$ (x – 97)(x – 3) = 0
$\Rightarrow$ x = 3.
Incorrect
Answer: Option 2
Explanation:
Area of the park = (60 x 40) m^{2} = 2400 m^{2}.
Area of the lawn = 2109 m^{2}.
$\therefore$ Area of the crossroads = (2400 – 2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x – x^{2} = 291
$\Rightarrow$ x^{2} – 100x + 291 = 0
$\Rightarrow$ (x – 97)(x – 3) = 0
$\Rightarrow$ x = 3.

Question 12 of 20
12. 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 = 360º.
Angle traced by hour hand in 5 hrs 10 min. i.e., ${\class{maths_class}{31 \over 6}}$ hrs = $\Bigg(\class{maths_class}{{360\over12} \times {31\over6}}\Bigg)^\circ$ = 155º.
Incorrect
Answer: Option 3
Explanation:
Angle traced by hour hand in 12 hrs = 360º.
Angle traced by hour hand in 5 hrs 10 min. i.e., ${\class{maths_class}{31 \over 6}}$ hrs = $\Bigg(\class{maths_class}{{360\over12} \times {31\over6}}\Bigg)^\circ$ = 155º.

Question 13 of 20
13. Question
1 pointsAt 3:40, the hour hand and the minute hand of a clock form an angle of:
Correct
Answer: Option 3
Explanation:
Angle traced by hour hand in 12 hrs. = 360°.
Angle traced by it in $\class{maths_class}{11\over3}$ hrs = $\Bigg(\class{maths_class}{{360\over12} \times {11\over3}}\Bigg)$ = 110°
Angle traced by minute hand in 60 min. = 360°.
Angle traced by it in $\class{maths_class}{11\over3}$ hrs = $\Bigg(\class{maths_class}{360\over60} \times {40}\Bigg)$ = 110°
$\therefore$ Required angle (240 – 110)° = 130°
Incorrect
Answer: Option 3
Explanation:
Angle traced by hour hand in 12 hrs. = 360°.
Angle traced by it in $\class{maths_class}{11\over3}$ hrs = $\Bigg(\class{maths_class}{{360\over12} \times {11\over3}}\Bigg)$ = 110°
Angle traced by minute hand in 60 min. = 360°.
Angle traced by it in $\class{maths_class}{11\over3}$ hrs = $\Bigg(\class{maths_class}{360\over60} \times {40}\Bigg)$ = 110°
$\therefore$ Required angle (240 – 110)° = 130°

Question 14 of 20
14. Question
1 pointsThe salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
Correct
Answer: Option 3
Explanation:
Let A = 2k, B = 3k and C = 5k.
A’s new salary = $\class{maths_class}{115\over100} \hbox{ of } {2k} = \Bigg({\class{maths_class}{115\over100}\times 2k}\Bigg) = \class{maths_class}{23k\over10}$
B’s new salary = $\class{maths_class}{110\over100} \hbox{ of } {3k} = \Bigg({\class{maths_class}{110\over100}\times 3k}\Bigg) = \class{maths_class}{33k\over10}$
C’s new salary = $\class{maths_class}{120\over100} \hbox{ of } {5k} = \Bigg({\class{maths_class}{120\over100}\times 5k}\Bigg) = {6k}$
$\therefore$ New ratio $\Bigg(\class{maths_class}{23k\over10}:\class{maths_class}{33k\over10}:6k\Bigg)$ = 23 : 33 : 60
Incorrect
Answer: Option 3
Explanation:
Let A = 2k, B = 3k and C = 5k.
A’s new salary = $\class{maths_class}{115\over100} \hbox{ of } {2k} = \Bigg({\class{maths_class}{115\over100}\times 2k}\Bigg) = \class{maths_class}{23k\over10}$
B’s new salary = $\class{maths_class}{110\over100} \hbox{ of } {3k} = \Bigg({\class{maths_class}{110\over100}\times 3k}\Bigg) = \class{maths_class}{33k\over10}$
C’s new salary = $\class{maths_class}{120\over100} \hbox{ of } {5k} = \Bigg({\class{maths_class}{120\over100}\times 5k}\Bigg) = {6k}$
$\therefore$ New ratio $\Bigg(\class{maths_class}{23k\over10}:\class{maths_class}{33k\over10}:6k\Bigg)$ = 23 : 33 : 60

Question 15 of 20
15. Question
1 pointsAn error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
Correct
Answer: Option 4
Explanation:
100 cm is read as 102 cm.
$\therefore$ A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2} – A_{1}) = [(102)^{2} – (100)^{2}]
= (102 + 100) x (102 – 100)
= 404 cm^{2}.
$\therefore$ Percentage error = $\Bigg({\class{maths_class}{404\over{100\times100}}\times 100}\Bigg)%$ = 4.04%
Incorrect
Answer: Option 4
Explanation:
100 cm is read as 102 cm.
$\therefore$ A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2} – A_{1}) = [(102)^{2} – (100)^{2}]
= (102 + 100) x (102 – 100)
= 404 cm^{2}.
$\therefore$ Percentage error = $\Bigg({\class{maths_class}{404\over{100\times100}}\times 100}\Bigg)%$ = 4.04%

Question 16 of 20
16. Question
1 pointsWhat least value should be replaced by * in 223*431 so the number become divisible by 9?
Correct
Answer: Option 1
Explanation:
Trick: Number is divisible by 9, if sum of all digits is divisible by 9, so (2+2+3+*+4+3+1) = 15+* should be divisible by 9,
15+3 will be divisible by 9,
so that least number is 3.Incorrect
Answer: Option 1
Explanation:
Trick: Number is divisible by 9, if sum of all digits is divisible by 9, so (2+2+3+*+4+3+1) = 15+* should be divisible by 9,
15+3 will be divisible by 9,
so that least number is 3. 
Question 17 of 20
17. Question
1 pointsFor the integer n, if n*n*n is odd, then what is true
Correct
Incorrect

Question 18 of 20
18. Question
1 pointsThere are four prime numbers written in ascending order. The product of first three is 385 and that of last three is 1001. The last number is:
Correct
Answer: Option 2
Explanation:
$\class{maths_class}{{abc \over bcd} = {385\over1001} {\Rigtharrow} {a \over d} = {5\over13}}$
Incorrect
Answer: Option 2
Explanation:
$\class{maths_class}{{abc \over bcd} = {385\over1001} {\Rigtharrow} {a \over d} = {5\over13}}$

Question 19 of 20
19. Question
1 pointsA number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder. The number is
Correct
Answer: Option 2
Explanation:
(555 + 445) * 2 * 110 + 30 = 220000 + 30 = 220030
Incorrect
Answer: Option 2
Explanation:
(555 + 445) * 2 * 110 + 30 = 220000 + 30 = 220030

Question 20 of 20
20. Question
1 points6897 is divisible by
Correct
Incorrect