Aptitude Test 4
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 Question 1 of 15
1. Question
1 pointsCorrectAnswer: Option B
Solution: Clearly, unit’s digit in the given product = unit’s digit in 7^{153} x 1^{72}.
now, 74 gives unit digit 1.
7^{152} gives unit digit 1,
∴ 7^{153} gives unit digit (l x 7) = 7. Also, 1^{72} gives unit digit 1.
hence, unit’s digit in the product = (7 x 1) = 7.
IncorrectAnswer: Option B
Solution: Clearly, unit’s digit in the given product = unit’s digit in 7^{153} x 1^{72}.
now, 74 gives unit digit 1.
7^{152} gives unit digit 1,
∴ 7^{153} gives unit digit (l x 7) = 7. Also, 1^{72} gives unit digit 1.
hence, unit’s digit in the product = (7 x 1) = 7.

Question 2 of 15
2. Question
1 points.
Kiran had 85 currency notes in all , some of which were of Rs.100 denomination and the remaining of Rs.50 denomination the total amount of all these currency note was Rs.5000. How much amount did she have in the denomination of Rs.50?
CorrectAnswer: Option D
Solution: Let the no of fifty rupee notes be x
Then, no of 100 rupee notes = (85 – x)
50x + 100(85 – x) = 5000
x + 2(85 – x) = 100
x = 70
So, required amount = Rs.(50*70)= Rs.3500
IncorrectAnswer: Option D
Solution: Let the no of fifty rupee notes be x
Then, no of 100 rupee notes = (85 – x)
50x + 100(85 – x) = 5000
x + 2(85 – x) = 100
x = 70
So, required amount = Rs.(50*70)= Rs.3500

Question 3 of 15
3. Question
1 points.
There were 35 students in a hostel. Due to the admission of 7 new students, the expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs 1. What was the original expenditure of the mess?
CorrectAnswer: C
Solution: Let the original average expenditure be Rs.x. Then,
42 (x – 1) – 35x = 42 ==> 7x = 84 ==> x = 12.
Original expenditure = Rs. (35 x 12) = Rs. 420.
IncorrectAnswer: C
Solution: Let the original average expenditure be Rs.x. Then,
42 (x – 1) – 35x = 42 ==> 7x = 84 ==> x = 12.
Original expenditure = Rs. (35 x 12) = Rs. 420.

Question 4 of 15
4. Question
1 points.
Distance between two stations Haldia and Varanasi is 778 km. A train covers the journey from Haldia to Varanasi at 84 km per hour and returns back to Haldia with a uniform speed of 56 km per hour. Find the average speed of the train during the whole journey.
CorrectAnswer: Option A
Solution: Required average speed = ((2xy)/(x+y)) km / hr
=(2 x 84 x 56)/(84+56) km/hr
= (2*84*56)/140 km/hr
= 67.2 km/hr.
IncorrectAnswer: Option A
Solution: Required average speed = ((2xy)/(x+y)) km / hr
=(2 x 84 x 56)/(84+56) km/hr
= (2*84*56)/140 km/hr
= 67.2 km/hr.

Question 5 of 15
5. Question
1 points.
10% of the inhabitants of village having died of cholera. a panic set in, during which 25% of the remaining inhabitants left the village. The population is then reduced to 4050. Find the number of original inhabitants.
CorrectAnswer: C
Solution: Let the total number of original inhabitants be x.
((75/100))*(90/100)*x) = 4050 ==> (27/40)*x = 4050
==> x=((4050*40)/27) = 6000.
IncorrectAnswer: C
Solution: Let the total number of original inhabitants be x.
((75/100))*(90/100)*x) = 4050 ==> (27/40)*x = 4050
==> x=((4050*40)/27) = 6000.

Question 6 of 15
6. Question
1 points.
By selling 33 meters of cloth, one gains the selling price of 11 meters. Find the gain percent.
CorrectAnswer: B
Solution: Let sp = selling price, cp = cloth price
(sp of 33m) – (cp of 33m) = gain = sp of 11m
sp of 22m = cp of 33m
Let cp of each meter be Rs.1, then, cp of 22m = Rs.22, sp of 22m = Rs.33
Gain% = [(11/22)*100]% = 50%
IncorrectAnswer: B
Solution: Let sp = selling price, cp = cloth price
(sp of 33m) – (cp of 33m) = gain = sp of 11m
sp of 22m = cp of 33m
Let cp of each meter be Rs.1, then, cp of 22m = Rs.22, sp of 22m = Rs.33
Gain% = [(11/22)*100]% = 50%

Question 7 of 15
7. Question
1 points.
A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours each. How long will they take to do it, working together 8 hours a day?
CorrectAnswer: D
Solution: A can complete the work in (7 x 9) = 63 hours.
B can complete the work in (6 x 7) = 42 hours.
A’s 1 hour’s work = (1/63) and B‘s 1 hour’s work = (1/42)
(A + B)’s 1 hour’s work = (1/63)+(1/42) = (5/126)
both will finish the work in (126/5) hrs.
Number of days. of (42/5) hrs each = (126 x 5)/(5 x 42) = 3 days
IncorrectAnswer: D
Solution: A can complete the work in (7 x 9) = 63 hours.
B can complete the work in (6 x 7) = 42 hours.
A’s 1 hour’s work = (1/63) and B‘s 1 hour’s work = (1/42)
(A + B)’s 1 hour’s work = (1/63)+(1/42) = (5/126)
both will finish the work in (126/5) hrs.
Number of days. of (42/5) hrs each = (126 x 5)/(5 x 42) = 3 days

Question 8 of 15
8. Question
1 points.
In a stream running at 2kmph, a Motorboat goes 6km upstream and back again to the starting point in 33 minutes. Find the speed of the Motorboat in still water.
CorrectAnswer: Option B
Solution: Let the speed of the motorboat in still water be x kmph. then,
6/x+2 + 6/x2 = 33/60
11x^{2}240x44 = 0
11x^{2}242x+2x44=0
(x22)(11x+2) = 0
x = 22kmph
IncorrectAnswer: Option B
Solution: Let the speed of the motorboat in still water be x kmph. then,
6/x+2 + 6/x2 = 33/60
11x^{2}240x44 = 0
11x^{2}242x+2x44=0
(x22)(11x+2) = 0
x = 22kmph

Question 9 of 15
9. Question
1 points.
The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.
CorrectAnswer: Option C
Solution: Side of first square = (40/4) = 10 cm;
Side of second square = (32/4)cm = 8 cm.
Area of third square = [(10)^{ 2} – (8)^{ 2}] cm^{2} = (100 – 64) cm^{2} = 36 cm^{2}.
Side of third square = (36)^{(1/2)} cm = 6 cm.
Required perimeter = (6 x 4) cm = 24 cm.
IncorrectAnswer: Option C
Solution: Side of first square = (40/4) = 10 cm;
Side of second square = (32/4)cm = 8 cm.
Area of third square = [(10)^{ 2} – (8)^{ 2}] cm^{2} = (100 – 64) cm^{2} = 36 cm^{2}.
Side of third square = (36)^{(1/2)} cm = 6 cm.
Required perimeter = (6 x 4) cm = 24 cm.

Question 10 of 15
10. Question
1 points.
How many words can be formed from the letters of the word ‘director’, So that the vowels are always together?
CorrectAnswer: Option A
Solution: In the given word, we treat the vowels ieo as one letter.
thus, we have drctr (ieo).
this group has 6 letters of which r occurs 2 times and others are different.
number of ways of arranging these letters = 6!/2! = 360.
now 3 vowels can be arranged among themselves in 3! = 6 ways.
required number of ways = (360×6) = 2160
IncorrectAnswer: Option A
Solution: In the given word, we treat the vowels ieo as one letter.
thus, we have drctr (ieo).
this group has 6 letters of which r occurs 2 times and others are different.
number of ways of arranging these letters = 6!/2! = 360.
now 3 vowels can be arranged among themselves in 3! = 6 ways.
required number of ways = (360×6) = 2160

Question 11 of 15
11. Question
1 pointsCorrectAnswer: Option D
Solution: All 12 months have 28 days
IncorrectAnswer: Option D
Solution: All 12 months have 28 days

Question 12 of 15
12. Question
1 pointsCorrectAnswer: Option B
Solution: Angle traced by hour hand in 12 hrs = 360º.
Angle traced by hour hand in 5 hrs 10 min. i.e., 31 hrs = [(360/12) x (31/6)]º = 155º
IncorrectAnswer: Option B
Solution: Angle traced by hour hand in 12 hrs = 360º.
Angle traced by hour hand in 5 hrs 10 min. i.e., 31 hrs = [(360/12) x (31/6)]º = 155º

Question 13 of 15
13. Question
1 points.
Find out the wrong number in the given sequence of numbers.
22, 33, 66, 99, 121, 289, 594
CorrectAnswer: Option C
Solution: Each of the number except 289 is a multiple of 11.
IncorrectAnswer: Option C
Solution: Each of the number except 289 is a multiple of 11.

Question 14 of 15
14. Question
1 points.
A and B are working on an assignment. A takes 6 hours to type 32 pages on a computer, while B takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
CorrectAnswer: Option D
Solution: Number of pages typed by A in one hour=32/6=16/3
Number of pages typed by B in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours
IncorrectAnswer: Option D
Solution: Number of pages typed by A in one hour=32/6=16/3
Number of pages typed by B in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours

Question 15 of 15
15. Question
1 points.
A man covers a distance on scooter. Had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower, he would have taken 40min more. The distance is.
CorrectAnswer: Option B
Solution: Let distance = x m; Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x —————————————(1)
x/y2 – x/y = 40/60 hr y(y2) = 3x ———–(2)
divide 1 & 2 equations
by solving we get x = 40 kmIncorrectAnswer: Option B
Solution: Let distance = x m; Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x —————————————(1)
x/y2 – x/y = 40/60 hr y(y2) = 3x ———–(2)
divide 1 & 2 equations
by solving we get x = 40 km
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