Delhi University MCA Logical Ability and Logical Reasoning Previous Year Question Paper and Solution with PDF

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Previous Year Questions

Delhi University MCA
Previous Year Questions
Delhi University MCA

Delhi University MCA Previous Year Questions (PYQs)

Delhi University MCA Logical Ability And Logical Reasoning PYQ

Qus : 1

X works twice as fast as Y . Y alone can finish the work in nine days . X and Y together can finish the work in _____ days.

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Qus : 2

Average of ten numbers in a list is 25.If one of the numbers in the list is exchanged with another number the average of the new list increases by 5. What is the new number included in the list , if the original number was 15?

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Qus : 3

How much of acid is in the 10 liter of a 60% solution, of acid and water solution?

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Qus : 4

What is the next term in the series? 2, 7, 14, 23, 34, ______

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Qus : 5

The code of DOG is ITL , what is the code of ITL?

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Qus : 6

Delhi University MCA Previous Year Paper

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Then value of x is

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Qus : 7

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Qus : 8

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Qus : 9

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Qus : 10

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Qus : 11

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Qus : 12

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Solution

A-------15

B-------10

C-------12

Total work (LCM of A,B,C) = 60

Work done by B and C in a day = (6+5)=11

B and C word done in 4 days = 44

Remaining work for A=16

A completes 4 work in a day, so A take 4 more days to complete 16 works.

Qus : 13

Find the odd man out:

1, 4, 27, 16, 125, 36, 216, 64, 729, 100

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Solution

The pattern of the series is 1³, 2², 3³, 4², 5³, 6², 7³. .....

Odd Term 216.

Qus : 14

An athlete has to cover a distance of 6 Kms in 90 Minutes. He covers two-third of the distance in two-thirds of the total time. To cover the remaining distance in the remaining time, his speed should be____ Km/Hr. ?

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Solution

An athlete needs to cover 6 km in 90 minutes.

Given that he covers 2/3 of the distance in 2/3 of the total time

⇒ he covers 2/3 of 6 km in two-thirds of 90 minutes

⇒ He covers 4 km in 60 minutes.

Now he needs to cover the remaining 2 km in remaining 30 minutes

Distance = 2 km

Time =30 minutes =1/2 hour

Required Speed =2/(1/2)km/hr = 4km/hr

Qus : 15

A and B can complete a work in 15 days. B and C can complete the same work in 20 days. If A, B and C together can finish it in 10 days, then A and C can complete the same work in____days.

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Solution

One Day work of A+B = 1/15

One Day work of B+C = 1/20

Lwt One Day work of A+C = 1/x

One Day work of A+B+C = 1/10

2.One Day work of A+B+C =(One Day work of A+B)+(One Day work of B+C)+(One Day work of A+C)

$\frac{2}{10}=\frac{1}{15}+\frac{1}{10}+\frac{1}{x}$

x=12

A+C can complete the same work in 12 days.

Qus : 16

If 2^x = (1024)^1/5, what is the value of x ?

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Solution

$2^x=1024^\frac{1}{5}$

$2^x=(2^{10})^{1/5}$

$2^x=2^{10/5}$

$2^x=2^2$

$x=2$

Qus : 17

Two trains each 500 metres long, are running in opposite directions on parallel tracks. If their speeds are 50 km/hr and 40 km/hr respectively, the time taken by the slower train to pass the driver of the faster one is _______seconds

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Solution

Relative speed = (50 + 40) km/hr

$90\times\frac{5}{18}$ )m/sec

=25 m/sec.

We have to find the time taken by the slower train to pass the Driver of the faster train and not the complete train.

So, distance covered = Length of the slower train.

Therefore, Distance covered = 500 m.

∴ Required time (

$500×\frac{1}{25}$ )=20sec.

Qus : 18

The speed of a boat in still water is 12 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 30 minutes is

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Solution

Speed downstream = (12 + 3) kmph = 15 kmph.

Distance travelled =

$15 \times \frac{30}{60}$ km = 7.5km.

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