Aptitude Test 5
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 Total number of questions : 15
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 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.