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

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Delhi University MCA Logical Ability And Logical Reasoning PYQ

Qus : 1 Delhi University MCA PYQ

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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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ

Solution

Y alone can finish the work in 9 days.

⇒ Y’s 1-day work $= \dfrac{1}{9}$

X works twice as fast as Y.

⇒ X’s 1-day work $= 2 \times \dfrac{1}{9} = \dfrac{2}{9}$

Together, their 1-day work $= \dfrac{2}{9} + \dfrac{1}{9} = \dfrac{3}{9} = \dfrac{1}{3}$

Time taken to complete the whole work $= \dfrac{1}{(1/3)} = 3 \text{ days}$

Answer: $\boxed{3\text{ days}}$ ✅

Qus : 2 Delhi University MCA PYQ

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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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ

Solution

Average of 10 numbers $= 25$

⇒ Sum of 10 numbers $= 10 \times 25 = 250$

New average $= 25 + 5 = 30$

⇒ New sum $= 10 \times 30 = 300$

Increase in sum $= 300 - 250 = 50$

Let the new number be $x$.

Then $x - 15 = 50$

⇒ $x = 65$

Answer: $\boxed{65}$ ✅

Qus : 3 Delhi University MCA PYQ

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

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ

Solution

Total solution $= 10 \text{ liters}$

Percentage of acid $= 60\%$

⇒ Amount of acid $= \dfrac{60}{100} \times 10 = 6 \text{ liters}$

Answer: $\boxed{6 \text{ liters}}$ ✅

Qus : 4 Delhi University MCA PYQ

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

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ

Solution

Given series: $2,\,7,\,14,\,23,\,34,\,\_\_$

Find the differences:

$7-2=5$, $14-7=7$, $23-14=9$, $34-23=11$

Difference pattern: $5,\,7,\,9,\,11$ (increases by 2 each time)

Next difference $= 11 + 2 = 13$

Next term $= 34 + 13 = 47$

Answer: $\boxed{47}$ ✅

Qus : 5 Delhi University MCA PYQ

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

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ

Solution

Logic: Each letter is shifted forward by 5 positions (D→I, O→T, G→L).

Apply to ITL: I→N, T→Y, L→Q

Answer: NYQ ✅

Qus : 6 Delhi University MCA PYQ

$\sqrt{\sqrt{1296+x^2}}=60%$ of 70. Then value of x is

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ

Solution

Qus : 7 Delhi University MCA PYQ

Given that $3^x$=656. The value of $5^{x-5}$ is ...

(a) 81

(b) 125

(d) 1/125

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ

Solution

Qus : 8 Delhi University MCA PYQ

What is the next term in the series 6, 20, 42, 72, 110, ... is

(a) 156

(b) 210

(d) 119

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ

Solution

Given series: $6,\,20,\,42,\,72,\,110,\,\_\_$

First differences: $14,\,22,\,30,\,38$

Second differences: $8,\,8,\,8$ (constant)

Next first difference $= 38 + 8 = 46$

Next term $= 110 + 46 = 156$

Answer: $\boxed{156}$ ✅

Qus : 9 Delhi University MCA PYQ

Next term in the series zaa, yeb, xic, wod, vue, ...

(a) uaf

(b) uef

(d) tag

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ

Solution

Reasoning: 1st letters go backward: z, y, x, w, v, u 2nd letters are vowels forward: a, e, i, o, u, then wrap to a 3rd letters go forward: a, b, c, d, e, f So the next term is uaf.

Qus : 10 Delhi University MCA PYQ

Age of a child is half of the age of father. Twenty years ago the age of father was 10 times the age of son. What is the age of father?

(a) 60 years

(b) 55 years

(d) 45 years

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ

Solution

Let father's age be $F$ and child's age be $C$.

Given $C = \dfrac{F}{2}$

Twenty years ago: $F - 20 = 10(C - 20)$

Substitute $C = \dfrac{F}{2}$:

$F - 20 = 10\left(\dfrac{F}{2} - 20\right)$

$F - 20 = 5F - 200$

$180 = 4F \Rightarrow F = 45$

Answer: $\boxed{45\text{ years}}$ ✅ (Option d)

Qus : 11 Delhi University MCA PYQ

In a race of 1 Kilometer, athlete A beats athlete B by 50 meters, in another race of 500 meters, athlete B beats athlete C by 50 meters. By how many meters (Approximately) can athlete A beat athlete C in a race of 400 meters?

(a) 50 meters

(b) 58 meters

(d) 72 meters

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ

Solution

$\dfrac{v_A}{v_B}=\dfrac{1000}{950}=\dfrac{20}{19}$, $\dfrac{v_B}{v_C}=\dfrac{500}{450}=\dfrac{10}{9}$

$\Rightarrow \dfrac{v_A}{v_C}=\dfrac{200}{171}$

When A runs $400$ m, C runs $400\times\dfrac{171}{200}=342$ m.

Difference $=400-342=58$ m.

Answer: 58 meters ✅

Qus : 12 Delhi University MCA PYQ

Persons A, B, and C can finish a work in 15, 10, and 12 days respectively. Person B and person C start the work together but are asked to leave after 4 days. In how many days will the remaining work be done by person A?

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ

Solution

Rates: $A=\frac{1}{15},\; B=\frac{1}{10},\; C=\frac{1}{12}$ (work/day)

Work done by $B+C$ in 4 days: $4\!\left(\frac{1}{10}+\frac{1}{12}\right)=4\cdot\frac{11}{60}=\frac{11}{15}$

Remaining work: $1-\frac{11}{15}=\frac{4}{15}$

Time for $A$: $\dfrac{\frac{4}{15}}{\frac{1}{15}}=4$ days

Answer: $\boxed{4\text{ days}}$ ✅

Qus : 13 Delhi University MCA PYQ

Find the odd man out:

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

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ

Solution

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

Odd Term 216.

Qus : 14 Delhi University MCA PYQ

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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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ

Solution

Total distance = 6 km, Total time = 90 min = 1.5 hr

Distance covered = $\frac{2}{3}\times6=4$ km, Time taken = $\frac{2}{3}\times90=60$ min = 1 hr

Speed in first part = $\dfrac{4}{1}=4$ km/hr

Remaining distance = 2 km, Remaining time = 0.5 hr

Required speed = $\dfrac{2}{0.5}=4$ km/hr

Answer: $\boxed{4\text{ km/hr}}$ ✅

Qus : 15 Delhi University MCA PYQ

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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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ

Solution

Let rates be $a,b,c$. Given: $a+b=\tfrac{1}{15}$, $b+c=\tfrac{1}{20}$, $a+b+c=\tfrac{1}{10}$.

Using $(a+b)+(a+c)+(b+c)=2(a+b+c)$:

$\tfrac{1}{15}+(a+c)+\tfrac{1}{20}=\tfrac{1}{5}\ \Rightarrow\ a+c=\tfrac{1}{5}-\left(\tfrac{1}{15}+\tfrac{1}{20}\right)=\tfrac{1}{12}$

Time by $A+C = \dfrac{1}{1/12}=12$ days.

Answer: $\boxed{12\text{ days}}$ ✅

Qus : 16 Delhi University MCA PYQ

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

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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ

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 Delhi University MCA PYQ

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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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ

Solution

Speed of faster train $= 50$ km/h $= 13.89$ m/s

Speed of slower train $= 40$ km/h $= 11.11$ m/s

Relative speed (opposite directions) $= 13.89 + 11.11 = 25$ m/s

Distance to be covered $= 500$ m

Time $= \dfrac{500}{25} = 20$ seconds

Answer: $\boxed{20\text{ seconds}}$ ✅

Qus : 18 Delhi University MCA PYQ

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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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ

Solution

Speed in still water $= 12$ km/hr

Rate of current $= 3$ km/hr

Downstream speed $= 12 + 3 = 15$ km/hr

Time $= 30$ min $= \dfrac{1}{2}$ hr

Distance $= 15 \times \dfrac{1}{2} = 7.5$ km

Answer: $\boxed{7.5\text{ km}}$ ✅

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