Accenture
– Aptitude
1.
A train overtakes two persons walking along a railway track. The first one
walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and
8.5 seconds respectively to overtake them. What is the speed of the train if
both the persons are walking in the same direction as the train?
A.
66
km/hr B.
72 km/hr C.
78
km/hr D.
81 km/hr
Ans-D
Let the speed of the train be ‘x’
km/hr.
1st Person: Relative speed (x – 4.5) km/hr. The train takes 8.4 seconds
to cross the person. So, the distance travelled during this time is (x –
4.5)*8.4 = 8.4x – 37.8 ……… (1)
2nd Person: Relative
speed (x - 5.4) km/hr. The train takes
8.5 seconds to cross the person. So, the distance travelled during this time is
(x – 5.4)* 8.5 = 8.5x – 45.9 ………(2)
1 and 2 are equal and
represents the distance travelled by the train which incidentally is its own
length. Hence solving (1) and (2) we get the value of ‘x’
as 81 km/hr.
2.
Two stations A and B are 110 km apart on a straight line. One train starts from
A at 7 a.m. and travels towards B at 20 km/ph. Another train starts from B at 8
a.m. and travels towards A at a speed of 25 km/ph. At what time will they meet?
A.
9 a.m. B.
10 a.m. C. 10.30
a.m. D.
11 a.m.
Ans-B
The distance between the two
stations is 110 km.
The train from A starts at
7.00 a.m. and moves at 20 km/ph. Speed. Hence when the train from B starts at
8.00 a.m. the distance between the two trains is only 110 - 20 = 90 km.
The train from B moves at 25
km/ph. Speed and both the trains are moving towards each other. Hence the
relative speed between the trains is 20 + 25 = 45km/ph. Thus to cover the
distance of 90 km it would take two hours.
Hence the time at which both the trains would meet is 8.00 a.m. + 2 hours =
10.00 a.m.
3.
A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for
5 consecutive months. How much sale must he have in the sixth month so that he
gets an average sale of Rs. 6500?
A.
Rs.
4991 B.
Rs. 5991
C. Rs.
6001 D.
Rs. 6991
Ans-A
To get an average sale of Rs
6500 in six months the trader’s total sale should be 6500 * 6 = Rs 39,000.
The total sale for five months
achieved by him is 6435 + 6927 + 6855 + 7230 + 6562 = Rs 34,009.00. Hence the
trader should achieve a sale of 39000 – 34009 = Rs 4991.00 in the sixth month.
4.
The average weight of 8 person's increases by 2.5 kg when a new person comes in
place of one of them weighing 65 kg. What might be the weight of the new
person?
A.
76 kg B.
76.5 kg C.
85 kg D.
Data inadequate
Ans-C
Add the total increase in
weight after the new comer ie 2.5 * 8 = 20 kg to the earlier average of 65.
Hence the weight of the new comer is 65 + 20 = 85 kg.
5.
Complete the series 8, 7, 11, 12, 14, 17, 17, 22, (....)
Ans-20 ( 8 + 3 =11 + 3 = 14 + 3 =
17 + 3 = 20)
6.
Complete the series 1, 8, 27, 64, 125, 216, (....)
Ans-343 ( 13, 23,
33, 43, 53, 63, 73 )
7.
A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 meters and still beats
him by 8 seconds. The speed of B is:
A.
5.15
kmph B.
4.04 kmph C.
4.25 kmph D.
4.4 kmph
Ans-B
A runs at 5km/ph. Converting
his speed into per second A runs 5 * 5/18 = 25/18 meters / sec.
So for running 100 meters he
takes 100 /(2 5/18) = 72 seconds.
A beats B by 8 seconds. Ie the
difference in distance between them is 25/18 * 8 = 100 /9 = 11.11 meters. A
also has given B a start of 8 meters. Hence during the time of 72 seconds that
A took to cover the distance, B had actually run 100 – (8 + 11.11) = 80.89
meters only. For this the time taken by him was 72 seconds. Hence his
speed is (80.89/72) * (18/5) = 4.04 kmph.
8.
At a game of billiards, A can give B 15 points in 60 and A can give C to 20
points in 60. How many points can B give C in a game of 90?
A.
30 points B.
20 points C.
10
points D.
12 points
Ans-C
A give 15 points to B in a
game of 60 points. In other words B scores 45 points only when A scores 60
points.
Similarly When A scores 60
points C scores only 40 points and A gives C 20 points.
Now considering B and C, When
B scores 45 points C scores only 40 points. Thus in a game of 90 when B scores
90 points, C would have scored only 80 points. Hence, B can
give C 10 points in a game of 90.
9.
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the
same remainder in each case. Then sum of the digits in N is:
A.
4 B.
5 C.
6 D.
8
Ans-A
Since the divider leaves the
same reminder in all the three cases, we shall find the HCF of the difference
between the numbers.
The difference between 4665
and 1305 is 3360 and the difference between 6905 and 4665 is 2240.
The HCF of these two differences
is 1120. Thus 1120 is the divider that leaves the same reminder in all the
three cases. Hence the sum of the digits in the divider is 1+1+2+0= 4.
10.
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
A.
9000 B.
9400 C.
9600 D.
9800
Ans-C ( Can easily answer from the
choices )
11.
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a.
in 5 years. What is the sum?
A.
Rs. 4462.50 B.
Rs. 8032.50 C. Rs.
8900 D.
Rs. 8925
Ans-D
Simple interest per annum is
9% and for five years it would amount to 9 * 5 = 45%. We are informed the total
simple interest over 5 years is Rs 4016.25. Hence the principle which is 100%
is
(4016.25 * 100) / 45 = Rs 8925.00
12.
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A
and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the
total amount of simple interest earned in 2 years be Rs. 3508, what was the
amount invested in Scheme B?
A.
Rs.
6400 B.
Rs. 6500 C.
Rs.
7200 D.
Rs. 7500
Ans-A
Let ‘x’ be the amount invested
in scheme A at 14% p.a. So in 2 years it would have given him a total of 28%
interest.
The amount invested in scheme
B at 11% p.a. is (13,900 – x). This would have given him a total interest in 2
years of 22%. We note the total interest
in both schemes earned over a period of 2 years is Rs 3,508.
Thus we have an equation
(28 * x)/ 100 + {22 ( 13900 – x)}/ 100 = 3,508. Solving this
we get the value of ‘x’ as Rs 7500. This is the amount invested in scheme A.
Hence, the amount invested in scheme B is 13,900 – 7500 =
6,400.00
13.
Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 :
8. There is a proposal to increase these seats by 40%, 50% and 75%
respectively. What will be the ratio of increased seats?
A.
2 : 3 :
4 B.
6 : 7 : 8 C.
6 : 8 :
9 D.
None of these
Ans-A
The current ratio of Mathematics,
Physics and Biology is 5 : 7 : 8.
The proposed increases are Mathematics
– 40%, Physics – 50% and Biology – 75%.
Apply these increased
percentages to the original ratios. We now have
Mathematics 5 + 40% = 7. Physics 7 + 50% = 10.5 Biology 8 + 75% = 14.
Thus the new ratios are 7 : 10.5 : 14 or reduced to 2 : 3 : 4.
14.
In a mixture 60 liters, the ratio of milk and water 2 : 1. If the this ratio is
to be 1 : 2, then the quantity of water to be further added is:
A.
20
litres B.
30 litres C.
40
litres D.
60 litres
Ans-D
The quantity of Milk and Water
in 60 liters at present is in the ratio 2 : 1. Thus the quantity of Milk is 40
liters and Water 20 liters.
Now the ratio is to be changed
to 1 : 2 where the quantity of water will be twice that of Milk. The existing
quantity of Milk of 40 liters remains as such. Two parts of this quantity should
be Water in the new ratio.
Hence we should have 80 liters
of water in the mixture. We already have 20 liters as per the original ratio. Hence
additionally further 60 liters of water is to be added.
15.
Two numbers are respectively 20% and 50% more than a third number. The ratio of
the two numbers is:
A.
2 : 5 B.
3 : 5 C. 4 :
5 D.
6 : 7
Ans-C
Very simple. Assume the third
number as 100. Then the two numbers will be 120 and 150 and the ratios of them
will be 4 : 5.
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