EduRev Aptitude Test- 2 - Computer Science Engineering (CSE) MCQ

Total female population = 150 + 200 + 350 + 450 + 500 = 1650 

Female population above 80 years age = 30 × 5 = 150 

Required average = 1650−150/ 5 = 300

let salary of Manoj be Rs 100x

The amount given to wife = 60/100 × 100x = Rs. 60x

ATQ, 60x × 50/100 = 18000

x = 600

Salary of Manoj = 100x = Rs 60000

I. 2x² - 7x – 60 = 0 

2x² - 15x + 8x – 60 = 0 

x (2x – 15 ) + 4 (2x – 15) = 0 (x + 4) (2x − 15) = 0 

x = −4, 15/2 

II. 3y² + 13y + 4 = 0 

3y² + 12y + y + 4 = 0 

3y (y + 4) + 1 (y + 4) = 0 

(3y + 1) (y + 4) = 0 y = − 1/3 , −4 

Hence, No relation between x and y

let cost price be Rs. 100x 

Marked price = 140 /100 × 100x = Rs 140x 

Selling price = Rs (140x − 224) 

Selling price after tax = 110/100 × (140x − 224) = Rs (154x − 246.4) 

ATQ, 100x + 158.6 = 154x − 246.4 

x = 7.5 

Cost price of article = 100x = Rs 750

From observing the data given, we find that it is a closed 3 set Venn diagram.

Let the three sports be F, T and C for Football, Tennis and Cricket respectively
n (F U T U C) = 256 , n(F) = 144, n(T) = 123, n(C) = 132, n(F ∩ T) = 58, n(C ∩T) = 25, n(F ∩ C) = 63

We know that (A U B U C) = n(A) + n(B) +n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩B ∩ C)
So, 256 = 144 + 123 + 132 - 58 - 25 - 63 + n (F ∩ T ∩ C)
n (F ∩ T ∩ C) = 256 - 144 + 123 +132 - 146
n (F ∩ T ∩ C) = 256 - 253 = 3
Now, it is easy to calculate the number of students who only play tennis using a Venn diagram.
n (Students who play only Tennis) = 123 - (55 + 3 + 22) = 123 - 80
n (Students who play only Tennis) = 43 students

Meena scores 40 % in an exam

After review, she scores 50 % more => Increase of 50 % from 40 % = 40% + 20% = 60% She fails by 35 marks, by scoring 60%

60% score = Pass mark - 35 ----- (1)

If her post review score is increased by 20%, she would have 7 more than the pass mark. 20% of 60% = 12 %

So, 60% + 12% = 72% of marks = Pass mark + 7 ------ (2)

So, 12% marks = 35 + 7 (5: 1 ratio)

So, similarly 12% can be re written as 10 % and 2 % (maintaining the 5:1 ratio)

Hence the pass percentage = 60 % + 10 % = 70%

(or)

Pass percentage = 72 % - 2 % = 70%

a + b + c = 10. a, b, c are whole numbers. Now this is similar to the previous question that we solved by placing 10 sticks and simplifying.

We cannot follow an exactly similar approach, as in this case a, b and c can be zero. Let us modify the approach a little bit. Let us see if we can remove the constraint that a, b, c can be zero.

If we give a minimum of 1 to a, b, c then the original approach can be used. And then we can finally remove 1 from each of a, b, c. So, let us distribute 13 sticks across a, b and c and finally remove one from each.

a + b + c = 13. Now, let us place ten sticks in a row

|       |       |       |       |       |       |       |       |       |       |       |       |

This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance,

|       |       |       |  +  |       |       |       |       |  +  |       |       |       |,

This would be the equivalent of 4 + 5 + 4. or, a = 4, b = 5, c = 4.

There are 12 slots between the sticks, out of which one has to select 2 for placing the '+'s. The number of ways of doing this is ¹²C₂. Hence, correct answer is 66.

 Exactly one of ab, bc and ca is odd
=> Two are odd and one is even.

abc is a multiple of 4
=> the even number is a multiple of 4.

The arithmetic mean of a and b is an integer
=> a and b are odd.

and so is the arithmetic mean of a, b and c.
=> a + b + c is a multiple of 3.

c can be 4 or 8.

c = 4; a, b can be 3, 5 or 5, 9

c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible.

The question is "How many such triplets are possible (unordered triplets)?"

Hence the answer is "4"

Choice D is the correct answer.

Given that, log₅(x + y) + log₅(x - y) = 3
We know log A + log B = log (A x B)
log₅(x + y) + log₅(x - y) = log₅(x² - y²)
log₅(x² - y²) = 3
x² - y² = 5³
x² - y² = 125
Similarly, log₂y - log₂x = 1 - log₂3
log₂ y/x = log₂2 - log₂3
log₂ y/x = log₂ 2/3
3y = 2x
(3/2 y)² - y² = 125
9/2 y² - y² = 125
5/4 y² = 125
y² = 100
y = 10
x = 15
So, xy = 15 × 10 = 150

Sum = (1950 × 100) /(2 × 15) =Rs 6500 

CI in 2 years at 10% per annum= 10 + 10 + (10 × 10)/ 100 = 21% 

ATQ (6500 + x) × 21/100 = 1680
⇒ (6500 + x) = 8000 x = Rs 1500

Ram’s cost price = M.R. P.× 80/100 

Ramesh C. P. = M. R. P.× 80/100 × 90/100 

Ranjan C. P. = M.R. P.× 80/100 × 90/100 × 120/100 = 1,29,600 

⇒ M.R.P. = Rs. 1,50,000

Let, Males and females who use their coupons in Haircutting be 13x and 7x respectively.
⇒ Males who use their coupons in Pedicure = 7x + 72
Then Females who use their coupons in Pedicure = 450 − 13x − 7x − 7x − 72 = 378 − 27x 

ATQ,
7x + 72 + 13x − (7x + 378 − 27x) = 174
40x − 306 = 174
40x = 480 x = 12 

Required Ratio = 156 + 54/156 + 84 

= 210/240 = 7/ 8

Males who use their coupons in Spa = 156 × 5 /4 = 195 

Females who use their coupons in Spa = 84 × 11 /6 = 154 

Total number of people who use their coupon in Spa = 195 + 154 = 349

ATQ, 

⇒ x = 8

∴ Sum of red and blue ball = 8 + 6 = 14

Total numbers of ways → 7! 

Favorable numbers of ways → 5! ×3! 

Probability → =1/7

Solve using options. If we try 500 (option b) for convenience, we can see that the difference between the two is 64 (as the SI would amount to 300 and CI would amount to 100 + 120 + 144 = 364).
Since, we need a difference of only 48 we can realize that the value should be 3/4th of 500. Hence, 375 is correct

We need the sum of the series 2000 + 2400 + 2800 to cross 100000.
Trying out the options, we can see that in 20 years the sum of his savings would be: 2000 + 2400 + 2800 +…+ 9600.
The sum of the series would be 20 × 5800 = 116000.
If we remove the 20th year we will get the saving for 19 years.
The series would be 2000 + 2400 + 2800 + … + 9200. Sum of the series would be 116000 − 9600 = 106400.
If we remove the 19th year’s savings the savings would be 106400 − 9200 which would go below 100000.
Thus, after 19 years his savings would cross 100000. Option (a) is correct

Total rainfall for the week = 3.7 × 7 = 25.9 cm ...(i)
Rainfall for Monday, Tuesday and Wednesday = 4.3 × 3 = 12.9 cm ...(ii)
Rainfall for Friday, Saturday and Sunday = 3.9 × 3 = 11.7 cm ...(iii)
Rainfall for Thursday = Equation (i) – Equation (ii) – Equation (iii) = 25.9– 12.9 – 11.7 = 1.3 cm

The average weight per box is asked.
Hence, the container does not have to be counted as the 5th item.
Also, since the average for 23 boxes goes up by 1 kg, the total weight must have gone up by 23 kgs.
That weight is the actual weight of the container.
Hence, option (d) is correct. 

n(n + 1)/2 should be a perfect square. The first value of n when this occurs would be for n = 8.
Thus, on the 8th of March the required condition would come true

P (Amit) = 1/3 P (vikas) = 2/7 P (vivek) = 1/8. Required Probability = 1/3 + 2/7 + 1/8 = 125/168.

Speed of the car in covering the remaining distance= 276/6= 46 kmph
Original speed of the car= 46*100/92= 50kmph
Distance travelled by the car in first 5 hours= 50*5= 250 km
Difference between distance travelled= 276-20= 26
Required percentage= 26*100/276= 9.42%

Let the let the money invested by B be x Money invested by A= x*120/100= 6x/5 Money invested by C= x*140/100= 7x/5 So, 7x/5 – 6x/5= 4000 x/ 5=4000 x= Rs 20000

So, Money invested by A= 20000*120/100= Rs 24000

No. of families who owns an SUV = 350
No. of families who owns both SUV and car = 50% of 350 = 175
It is given that 75% of the families own at least one vehicle
So, 250 families do not own any vehicle
No. of families who own a car and not an SUV = 750 – 350 = 400
So, no of families who own at most one vehicle
= 250 + (350 – 175) + (750 – 350) = 825
Hence, 825 is the correct answer

Assume the cost of material = Re 1/unit and normal marked price = 1.2/unit 

Consider the amount of money spent on the given day = Rs. 100 

Then the shopkeeper will get 110 units of material.
The marked price of 110 units = 110*1.2 

After giving 25% discount, new selling price = 110*1.2*0.75 = 99
Hence on the whole transaction he makes 100 - 99 = 1% loss

Let the CP of milk be Rs. 100/litre
As he is claiming 10% profit, SP of the total mixture = Rs. [(15 + 3) * 110] = Rs. 1980
Actual profit = 20%
Let actual CP be Rs. x.
Then, x + 20% of x = Rs. 1980
Solving for x, we get x = Rs. 1650
So, Actual CP = Total CP of milk + Total CP of water
Or, Total CP of water = Rs. 1650 – (15 * 100) = Rs. 150
Or, CP of water = Rs. 50/litre
Required ratio = 100:50 = 2:1
Hence, option C is correct.

Let us assume the no. of days as ‘3d’ and the output per day as ‘x’

Then, 3d*x = 1800 … (1) 

For the first ‘d’ days, the output was (x - 20).

For the next ‘2d - 1’ days, the output was (x + 20). 

=> d(x-20) + (2d - 1) (x + 20) = 1800

=> dx – 20d + 2dx + 40d – x – 20 = 3dx {Replacing 1800 with 3dx from equation (1)}

=> 20d = x + 20

=> d = (x + 20)/20 … (2) 

=> 3 [(x + 20)/20] x = 1800

=> x² + 20x = (1800/3) *20

=> x² + 20x – 12000 = 0

=> (x + 120) (x - 100) = 0

=> x = -120 or 100 

Since x is the output it cannot be negative.

So, the initial planned output is 100 tonnes. Thus, Option B

Speed of Garibrath express = 72 km/hr = 20m/s

Speed of Durunto express = 54 km/hr = 15m/s

When the trains start to cross each other, Ram is at front end of Garibrath express and Shyam is at rear end of Durunto express. So the initial distance between them is equal to the length of Durunto express.

Hence initial distance between them = 180m

Shyam is moving in the same direction as the train so his effective speed is 15+4 = 19 m/s

Ram is moving in the direction opposite to the train, so his effective speed is 20-3 =17 m/s.

Hence with reference to the train Ram and Shyam are moving in the same direction but with reference to ground they are moving in the opposite direction with relative speed of 19+17 = 36 m/s

Total distance to be covered = 180 m

Hence required time = 180/36 = 5 seconds