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TCS assessment test questions

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Choose the correct option.

(p/q-q/p)=21/10. Then find 4p/q+4q/p?


A58/3

B58/5

C68/5

DNone of these

Answer: Option B

Explanation:

let us take p/q=x and q/p=1/x
x-1/x=21/10
then by solving we get a quadratic equation like 10x^2-10-21x=0
10x^2-25x+4x-10=0
then by taking factors we get two values for x,x=-2/5 and x=5/2
by substituting x=5/2
4(5/2)+4(2/5)=58/5

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NA
SHSTTON
27
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22
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Choose the correct option.

4 men can check exam papers in 8 days working 5 hours regularly. What is the total hours when 2 men will check the double of the papers in 20 days?


A8 hrs.

B9 hrs.

C4 hrs.

DNone of these

Answer: Option A

Explanation:

Let a man can do 1 unit of work in 1 hour.
Total units of work = 4 * 8 * 5 = 160 units.
Now work = 2 * 160 = 320 units.
Now 2 men work for 20 days. Let in x hours they have to work per day.
Now total work = 2*x*20 = 40 x
40x = 320 So x = 320/40 = 8 hours.

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NA
SHSTTON
10
Solv. Corr.
17
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27
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Choose the correct option.

7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?


A51

B49

C65

D45

Answer: Option A

Explanation:

Units digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats. So for every 4 terms we get one term with 3 in its unit digit. So there are total of 205/4 = 51 sets and each set contains one terms with 3 in its unit digit.

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NA
SHSTTON
18
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11
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Choose the correct option.

In paper A, one student got 18 out of 70 and in paper B he got 14 out of 30. In which paper he did fare well?


A46.6

B46

C45

D4.6

Answer: Option A

Explanation:

Find the percentages. Paper A = 18/70 * 100 = 25.7
Paper B = 14/30 * 100 = 46.6

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SHSTTON
38
Solv. Corr.
52
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Choose the correct option.

Find the total no of divisors of 1728 (including 1 and 1728)


A25

B22

C28

DNone of these

Answer: Option C

Explanation:

1728= 2^6 * 3^3
Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
Imp: if a number represented in standard form (a^m *b^n) , then the number of factors Is given by (m+1)(n+1).
=> (6+1)(3+1) = 28

ShortCut By :: chetan raj how did that formula (a^m*b^n) then number of factors given by (m+1)( n+1) come from?

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NA
SHSTTON
3
Solv. Corr.
6
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9
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Choose the correct option.

A certain function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x. The value of f(1) is


A1

B2

C3

DCannot be determined

Answer: Option C

Explanation:

Put x =1 => f(1)+2*f(6-1) = 1 => f(1) + 2*f(5) = 1
Put x = 5 => f(5)+2*f(6-5) = 5 => f(5) + 2*f(1) = 5
Put f(5) = 5 - 2*f(1) in the first equation
=> f(1) + 2*(5 - 2*f(1)) = 1
=> f(1) + 10 - 4f(1) = 1
=> f(1) = 3

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NA
SHSTTON
1
Solv. Corr.
1
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The rupee/coin changing machine at a bank has a flaw. It gives 10 ten rupee notes if you put a 100 rupee note and 10 one rupee coins if you insert a 10 rupee note but gives 10 hundred rupee notes when you put a one rupee coin! Sivaji, after being ruined by his rivals in business is left with a one rupee coin and discovers the flaw in the machine by accident. By using the machine repeatedly, which of the following amounts is a valid amount that Sivaji can have when he gets tired and stops at some stage (assume that the machine has an infinite supply of notes and coins):


A26975

B53947

C18980

D33966

 View Answer |  Submit Your Solution | Topic: Uncategorized| Asked In TCS |

Answer: Option B

Explanation:

The process works like this:

Rs.1 Coin  => 10 × 100 = Rs.1000

Rs.100  => 10 × 10

Rs.10  => 1 × 10

Sivaji gets more money when he inserts a rupee coin only.  For each rupee coin, he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000.  So he has 1999 with him.  Now if he inserts another coin, he has 1998 + 1000 = 2998.

Now each of these numbers is in the form of 999n + 1.  So option B can be written as 54 * 999 + 1.

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SHSTTON
11
Solv. Corr.
22
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Choose the correct option.

Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?


A59

B61

C49

D56

Answer: Option D

Explanation:

Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.
Solving the above we get N = 181
When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
=> 15x + 1 = 7c + 6 => c = (15x?5)/7 => c = 2x+(x?5)/7
Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
Solving we get M = 25.
(125 also satisfies but this is next number)
Now N - M = 56

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SHSTTON
3
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Choose the correct option.

The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573. What is the largest of the numbers A, B, C, D?


A1712

B1650

C1164

D1211

Answer: Option A

Explanation:

a+b+c=4024
b+c+d= 4087
a+c+d=4524
a+b+d=4573
Combining all we get 3(a+b+c+d) = 17208
=> a + b + c +d = 3736
Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

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NA
SHSTTON
19
Solv. Corr.
25
Solv. In. Corr.
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Anand packs 304 marbles into packets of 9 or 11 so that no marble is left. Anand wants to maximize the number of bags with 9 marbles. How many bags does he need if there should be atleast one bag with 11 marbles


A33

B32

C31

D30

Answer: Option B

Explanation:

Given 9x + 11y = 304.
x = (304?11y)/9 = 33 + (7?2y)/9 + y
So y = - 1 satisfies. Now x = 35,
Now other solutions of this equation will be like this. Increase or decrease x by 11, decrease or increase y by 9. So we have to maximize x. next solution is x = 24 and y = 8. So bags required are 32.

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