TCS Quantitative Aptitude Questions

If 10 3D Printer Takes 10 minutes to print 10 models then,

100 models will be printed in 100*1/10= 10 Minutes

Since the individual time required to build the wall has increased due difficult terrain and slushy conditions at site.

The new time taken by three of them individually will be

Ravi can alone build the wall in = (120/100) * 5 = 6 days.

Sanjay can alone build the wall in = (125/100) * 8 = 10 days.

Mukund can alone build the wall in = (150/100) * 10 = 15 days.

So, in 1 day

Ravi can do = (1/6)th of work

Sanjay can do = (1/10)th of work

Mukund can do = (1/15)th of work

∴Total work done in 1 day if three of them work together = 1/6 + 1/10 + 1/15 = 10/30 = 1/3

By unitary method, we can get

If 1/3rd of whole work is done in 1 day.

Then,

The whole work will be done in = 1/(1/3) = 3 days.

∴Total time required by Ravi, Sanjay & Mukund to build the wall, if they work together = 3 days

youngest be x , then eldest = 3x.
3x = y + x + 2 => y = 2x -2
so ages are 3x , 2x-2 and x respectively.
also difference between two of the child's ages is 3
so 2x-2 -x = 3=> x = 5 , so 3x = 15

.6M*.7=.42 voted
.7F*.6=.42 voted
so, fraction voted =(.84/2)=.42

Let X be the number of 1 rupee and Y be the number of 2 rupees coin

X+Y= 100

Take x = 100, Y=0

Likewise,Y=1, X=98

It will go till Y=50, X=0

Therefore there are 51 ways

The car moves 1/2 mile in 1 minute.

Speed of the car = 1/2 mile for 1 min.

30 miles for 60 min(1 hr).

Therefore, speed of car is 30 mph.

Also, speed of the observant car is 55 mph.

And both are moving in the same direction, we must add the speeds.

So, speed of the car = (55 + 30) mph

= 85 mph.

Chocolate method : Take Lcm of george,paul and Hilllari we get LCM(8,10,12)=120
Let the no of chocolates are 120 in that
George can eat 15 chocolates/hr
paul can eat 12cho/hr
Hillari can eat 10cho/hr
total choclates they can eat in 1 hr =Paul + George+hillari=15+12+10=37choclates
for 9:00 to 11:00 difference -2hrs
so for 2 hrs they can eat 37*2=74
George left after 11 remaing choclates are 120-74=46
Paul and hillari can eat=12+10=32/hr
=46/32 =2 hrs approx so ans is 11:00 + 2hrs =1:00 pm

Let the money to be paid = 30 rupees.
So, George daily wage = 30/15 = 2
and Mark daily wage = 30/10 = 3.
If both are working,
then 5 rupees to be paid.
So given sum is sufficient for 30 / 5 = 6 days.

Consider the provided information.

Let us consider the marked price of the jacket is Rs. x.

It is given that the marked price of pants is Rs. 8500 less than the marked price of the jacket. Therefore the marked price of pants is Rs x - 8500.

If there is 30% off on marked price, that means someone needs to pay 70% of the marked price.

If the marked price of the jacket is Rs. x. Then, someone needs to pay 70% of x. Which can be written as:

70/100*x

=0.7 * x = 0.7x

If there is 40% off on marked price, that means someone needs to pay 60% of the marked price.

If the marked price of the pants is Rs. x - 8500 then, someone needs to pay 60% of x - 8500. Which can be written as:

60/100 * (x-8500)

0.6 * (x - 8500)

The total sale price of both items is Rs. 23500, which can be written as:

0.7x + 0.6 ( x- 8500) = 23500

0.7x + 0.6 x -5100 = 23500

1.3x = 23500 + 5100

1.3x = 28600

x = 22000

Therefore, the marked price of jacket is 22000.

Sona had three times the money than Tina had.

Things = Rs. 480

Tina's share = 65 %

Sona was left with 4 times the amount that Tina had.

To find:

The amount with Tina at the beginning.

Solution:

We know,

Sona = 3 Tina

Tina's share = 65 % of 480

65 / 100 * 480

Tina's share = 312 rupees.

Remaining will be of Sona's shar.

480-312

Sona's share = 168 rupees

We get,

Sona - 168 = 4 ( Tina - 312 )

Substituting the values,

3 Tina - 168 = 4 ( Tina - 312 )

3 Tina - 168 = 4 Tina - 1248

We get,

Tina = 1080 rupees

Hence, Tina had 1080 rupees in the beginning

as there are total 5 families having number of children in arthmetic progression.

the number of children in 5 families are-

2, 3, 4, 5, 6

each children have pets to look after =3 times the no. of children in family

therefore,

for a single child in first family the number of pets to look after = 3 * 2

similarly,

3(2*2 +3*3 +4*4 +5*5 +6*6)=270

[ 2- letters of Alphabets from (a-z)] and 2 digits from (1-9). ****No repetitions are allowed in any case]

Finding no. of ways of selecting 2 letters from a-z without repetition

Select any alphabet first.
26C1 = 26
now 25 Alphabets left,
(As another alphabet should not be equal to the one that is selected)
taking one from these 25
25C1=25

Finding no. of ways of selecting 2 digits from (0-9) without repetition:

Total 10 no.s are there in 0-9
First taking one of them,
10C1=10
now 9 no.s left
(The other should not be equal to this one)
9C1= 9

As all are ''and''ed multiply all,
I mean selecting one from the alphabets AND another one from alphabets left AND selecting one digit from 0-9 AND other digits from remaining digits....that is
26*25*10*9=58500

Let the first place team got Rs. a

Since, award money increases by the same amount for successive finishing places.

∴ Series formed in A.P.

Let the constant amount be Rs. d.

Now, l = 275 , n = 16 and S₁₆ = 8000

∴ l = a + ( n - 1)d

→ 275 = a + 15d ----------------------------------- (1)

and S₁₆ = 16/2 [ 2a + (16 -1)(d)]

→ 8000 = 8[ 2a + 15d]

→ 1000 = 2a + 15d ------------------------------------- (2)

On subtracting (1) from (2) we get,

( 2a + 15d) - (a + 15d) = 1000 - 275

→ 2a + 15d - a - 15d = 725

→ a = 725

Hence, Rs. 725 is received by first place team.

Lets analyze the individual ages.

Shashank = 13 years old

esha is 3 years older than Shashank hence she is :

13 + 3 = 16 years old.

The father was 34 years when Ayesha was born.

The father is thus :

34 + 16 = 50 years old now

The mother was 29 years when Shashank was born hence :

The mother is :

29 + 13 = 42 years old.

The sum of their parent's ages now is :

42 + 50 = 92 years

Let the amount be divided into three parts X, Y, and Z.

X + Y + Z = 2387

X/5 = Y/2 = Z/4 = K

X = 5K

Y = 2K

Z = 4K

Hence, 5K + 2K + 4K = 2387

11K = 2387

K = 217

5 times of 1st part + 3 times of 2nd part + 4 times of 3rd part = 5X + 3Y + 4Z

= 5(5K) + 3(2K) + 4(4K) = 5(5*217) + 3(2*217) + 4(4*217)

= 5425 + 1302 + 3472 = 10199

30L + 3Q = 1167 - (1)

30L + 6Q = 1284 -(2)

Subtracting (2) from (1),

30L + 6Q = 1284

-30L + 3Q = 1167

Hence 3Q = 117

Hence, Q=117/3

Q=39

30L + 6*39=1284

30L= 1284-234

L=1050/30

L=35

First 10 kg = 35*10

Hence, Rs.350 for the first 10kgs of Apples.

A and B are traveling in the same direction.

B is traveling at a constant speed of 55kmph.

The distance between A and B is 1.5 km

A crosses B within 1 minute

Let's find out the relative speed between A and B.

Relative speed between A and B is 1.5 * 60 = 90 kmph

Since A and B are traveling in the same direction , Relative Speed = Speed of A - Speed of B

So, the speed of A = 55 + 90 = 145kmph

Given,

Total question = 26,

Let x be the number of questions which are correct,

⇒ The number of questions that are incorrect = 26 - x,

Now, five points were deducted for each wrong answer and eight points were added for every correct answer,

Thus, the total score = 8x - 5(26-x)

According to the question,

8x - 5(26-x) = 0

⇒ 8x - 130 + 5x = 0

⇒ 13x = 130

⇒ x = 10

(mon+tue+wed)/3=27;

(mon+tue+wed)=81;------(1)

(tue+wed+thu)/3=24;

(tue+wed+thu)=72;----(2)

from 1 & 2 the avg value of tue and wed should not be greater than 54 and it cannot be 48 . so the value of the temperature in thu is 18

time right now =3:12
the bus arrived 1 min ago so 3:12-0.01=03:11
late 2 mins so right time should be 3:11-0.02=3:09
next bus timing is 3:18
interval timing 3:18- bus expected time 3:09= 0.09 min
so next bus will be at = 3:18+0.09= 3:27 pm