Aptitude Test 5
Questions Result & StatisticsQuizsummary
0 of 14 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
Information
Aptitude
Instruction:
 Total number of questions : 15
 Time allotted : 25 minutes.
 Each question carry 1 mark, no negative marks.
 Click the “Finish quiz” button given in bottom of this page to submit your answer.
 Test will be submitted automatically if the time expired.
 Don’t refresh the page.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
Marks : 0 of 14 questions answered correctly  
Total number of questions  :  14 
Number of currect answered  :  0 
Your time  : 

Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Not categorized 0%

Thank you for submitting online aptitude test!
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 Answered
 Review
 Question 1 of 14
1. Question
1 points.
A batsman makes a score of 87 runs in the 17th innings and thus increases his avg by 3. Find his average after 17th innings.
CorrectAnswer : A
Solution : Let the average after 17th inning = x.
then, average after 16th inning = (x – 3).
:. 16 (x – 3) + 87 = 17x or x = (87 – 48) = 39
IncorrectAnswer : A
Solution : Let the average after 17th inning = x.
then, average after 16th inning = (x – 3).
:. 16 (x – 3) + 87 = 17x or x = (87 – 48) = 39

Question 2 of 14
2. Question
1 points.
Rajeev’s age after 15 years will be 5 times his age 5 years back. What is the Present age of Rajeev?
CorrectAnswer : C
Solution : Sol. Let Rajeev’s present age be x years. Then,
Rajeev’s age after 15 years = (x + 15) years.
Rajeev’s age 5 years back = (x – 5) years.
:. X + 15 = 5 (x – 5) óx + 15 = 5x – 25 ó 4x = 40 ó x = 10.
hence, Rajeev’s present age = 10 years.
IncorrectAnswer : C
Solution : Sol. Let Rajeev’s present age be x years. Then,
Rajeev’s age after 15 years = (x + 15) years.
Rajeev’s age 5 years back = (x – 5) years.
:. X + 15 = 5 (x – 5) óx + 15 = 5x – 25 ó 4x = 40 ó x = 10.
hence, Rajeev’s present age = 10 years.

Question 3 of 14
3. Question
1 points.
Mr. Bhaskar is on tour and he has rs. 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by rs. 3. For how many days is mr. Bhaskar on tour?
CorrectAnswer : B
Solution : Suppose mr. Bhaskar is on tour for x days. Then,
360 – 360 = 3 ó 1 – 1 = 1 ó x(x+4) =4 x 120 =480
x x+4 x^{ }x+4 120
ó x^{2} +4x –480 = 0 ó (x+24) (x20) = 0 ó x =20.
Hence mr. Bhaskar is on tour for 20 days
IncorrectAnswer : B
Solution : Suppose mr. Bhaskar is on tour for x days. Then,
360 – 360 = 3 ó 1 – 1 = 1 ó x(x+4) =4 x 120 =480
x x+4 x^{ }x+4 120
ó x^{2} +4x –480 = 0 ó (x+24) (x20) = 0 ó x =20.
Hence mr. Bhaskar is on tour for 20 days

Question 4 of 14
4. Question
1 points.
If 1/8 of a pencil is black ½ of the remaining is white and the remaining 3 ½ is blue find the total length of the pencil?
CorrectAnswer : D
Solution : let the total length be xm
Then black part =x/8cm
The remaining part=(xx/8)cm=7x/8cm
White part=(1/2 *7x/8)=7x/16 cm
Remaining part=(7x/87x/16)=7x/16cm
7x/16=7/2
X=8cm
IncorrectAnswer : D
Solution : let the total length be xm
Then black part =x/8cm
The remaining part=(xx/8)cm=7x/8cm
White part=(1/2 *7x/8)=7x/16 cm
Remaining part=(7x/87x/16)=7x/16cm
7x/16=7/2
X=8cm

Question 5 of 14
5. Question
1 pointsCorrectAnswer : A
Solution : The given numbers are 1, 3, 5, 7, …, 99.
this is an a.p. with a = 1 and d = 2.
let it contain n terms. Then,
1 + (n – 1) x 2 = 99 or n = 50.
required sum = n (first term + last term)
2
= 50 (1 + 99) = 2500.
2
IncorrectAnswer : A
Solution : The given numbers are 1, 3, 5, 7, …, 99.
this is an a.p. with a = 1 and d = 2.
let it contain n terms. Then,
1 + (n – 1) x 2 = 99 or n = 50.
required sum = n (first term + last term)
2
= 50 (1 + 99) = 2500.
2

Question 6 of 14
6. Question
1 points.
A train 100 metres long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train. Find the speed of the train.?
CorrectAnswer : A
Solution : let the speed of the train be x kmph.
speed of the train relative to man = (x + 5) kmph = (x + 5) *5/18 m/sec.
therefore 100/((x+5)*5/18)=6 <=> 30 (x + 5) = 1800 <=> x = 55
speed of the train is 55 kmph.
IncorrectAnswer : A
Solution : let the speed of the train be x kmph.
speed of the train relative to man = (x + 5) kmph = (x + 5) *5/18 m/sec.
therefore 100/((x+5)*5/18)=6 <=> 30 (x + 5) = 1800 <=> x = 55
speed of the train is 55 kmph.

Question 7 of 14
7. Question
1 pointsCorrectAnswer : B
Solution : Let each side of the square be a. Then, area = a^{2}.
New side =(125a/100) =(5a/4). New area = (5a/4)^{ 2} =(25a^{2})/16.
Increase in area = ((25 a^{2})/16)a^{2} =(9a^{2})/16.
Increase% = [((9a^{2})/16)*(1/a^{2})*100] % = 56.25%.
IncorrectAnswer : B
Solution : Let each side of the square be a. Then, area = a^{2}.
New side =(125a/100) =(5a/4). New area = (5a/4)^{ 2} =(25a^{2})/16.
Increase in area = ((25 a^{2})/16)a^{2} =(9a^{2})/16.
Increase% = [((9a^{2})/16)*(1/a^{2})*100] % = 56.25%.

Question 8 of 14
8. Question
1 points.
The dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is 2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.
CorrectAnswer : C
Solution : Volume of the metal used in the box = external volume – internal volume
= [(50 * 40 * 23) – (44 * 34 * 20)]cm^{3}
= 16080 cm^{3}
weight of the metal =((16080*0.5)/1000) kg = 8.04 kg.
IncorrectAnswer : C
Solution : Volume of the metal used in the box = external volume – internal volume
= [(50 * 40 * 23) – (44 * 34 * 20)]cm^{3}
= 16080 cm^{3}
weight of the metal =((16080*0.5)/1000) kg = 8.04 kg.

Question 9 of 14
9. Question
1 pointsCorrectAnswer : B
Solution : The word bihar contains 5 different letters.
Required number of words = ^{5}p_{5} = 5! = (5x4x3x2x1) = 120.
IncorrectAnswer : B
Solution : The word bihar contains 5 different letters.
Required number of words = ^{5}p_{5} = 5! = (5x4x3x2x1) = 120.

Question 10 of 14
10. Question
1 pointsCorrectAnswer : B
Solution : S={HH,HT,TH,TT}
Let EE=event of getting one head
E={tt,ht,th}
P(e)=n(e)/n(s)=3/4
IncorrectAnswer : B
Solution : S={HH,HT,TH,TT}
Let EE=event of getting one head
E={tt,ht,th}
P(e)=n(e)/n(s)=3/4

Question 11 of 14
11. Question
1 pointsCorrectAnswer : A
Solution : Required unit’s digit = unit’s digit in (4)^{102} + (4)^{103.}
now, 4^{2} gives unit digit 6.
(4)^{102 }gives unjt digit 6.
(4)103 gives unit digit of the product (6 x 4) i.e., 4.
hence, unit’s digit in (264)m + (264)103 = unit’s digit in (6 + 4) = 0
IncorrectAnswer : A
Solution : Required unit’s digit = unit’s digit in (4)^{102} + (4)^{103.}
now, 4^{2} gives unit digit 6.
(4)^{102 }gives unjt digit 6.
(4)103 gives unit digit of the product (6 x 4) i.e., 4.
hence, unit’s digit in (264)m + (264)103 = unit’s digit in (6 + 4) = 0

Question 12 of 14
12. Question
1 points.
From a group of boys and girls,15 girls leave.there are then left 2 boys for each girl.after this,45 boys leave.there are then 5 girls for each boy. Find the number of girls in the beginning?
CorrectAnswer : D
Solution : At present there be x boys.
Then,no of girls at present=5x
Before the boys had left:no of boys=x+45
And no of girls=5x
X+45=2*5x
9x=45
X=5
No of girls in the beginning=25+15=40
IncorrectAnswer : D
Solution : At present there be x boys.
Then,no of girls at present=5x
Before the boys had left:no of boys=x+45
And no of girls=5x
X+45=2*5x
9x=45
X=5
No of girls in the beginning=25+15=40

Question 13 of 14
13. Question
1 pointsCorrectAnswer : C
Solution : Let the numbers be x and (15 – x).
Then, x^{2} + (15 – x)^{2} = 113 => x^{2} + 225 + x^{2} – 30x = 113
=> 2x^{2} – 30x + 112 = 0 => x^{2} – 15x + 56 = 0
=> (x – 7) (x – 8) = 0 => x = 7 or x = 8.
So, the numbers are 7 and 8.
IncorrectAnswer : C
Solution : Let the numbers be x and (15 – x).
Then, x^{2} + (15 – x)^{2} = 113 => x^{2} + 225 + x^{2} – 30x = 113
=> 2x^{2} – 30x + 112 = 0 => x^{2} – 15x + 56 = 0
=> (x – 7) (x – 8) = 0 => x = 7 or x = 8.
So, the numbers are 7 and 8.

Question 14 of 14
14. Question
1 points.
One year ago, the ratio of gaurav’s and sachin’s age was 6: 7 respectively.Four years hence, this ratio would become 7: 8. How old is sachin ?
CorrectAnswer : A
Solution : Let gaurav’s and sachin’s ages one year ago be 6x and 7x years respectively.
Then, gaurav’s age 4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin’s age 4 years hence = (7x + 1) + 4 = (7x + 5) years.
(6x+5)/ (7x+5) = 7/8 ó 8(6x+5) = 7 (7x + 5) ó 48x + 40 = 49x + 35 ó x = 5.
Hence, sachin’s present age = (7x + 1) = 36 years.
IncorrectAnswer : A
Solution : Let gaurav’s and sachin’s ages one year ago be 6x and 7x years respectively.
Then, gaurav’s age 4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin’s age 4 years hence = (7x + 1) + 4 = (7x + 5) years.
(6x+5)/ (7x+5) = 7/8 ó 8(6x+5) = 7 (7x + 5) ó 48x + 40 = 49x + 35 ó x = 5.
Hence, sachin’s present age = (7x + 1) = 36 years.
Share this:
 Click to share on Facebook (Opens in new window)
 Click to share on WhatsApp (Opens in new window)
 Click to share on Telegram (Opens in new window)
 Click to share on LinkedIn (Opens in new window)
 Click to share on Pinterest (Opens in new window)
 Click to share on Reddit (Opens in new window)
 Click to share on Tumblr (Opens in new window)