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Godrej Infotech Aptitude Paper 2011

Posted on :01-02-2016
Q1. Lucia is a wonderful grandmother. Her age is between 50 and 70. Each of her sons have as many sons as they have brothers. Their combined number gives Lucia

a) 55
b) 64
c) 73

ANS: 64


Q2. Complete the series:5, 20, 24, 6, 2, 8, ?

a) 32
b) 12
c) 16
d) 3

ANS: 12


Q3. A 1 k.m. long wire is held by n poles. If one pole is removed, the length of the gap becomes 12/3m. What is the number of poles initially?

a) n=196
b) n=200
c) n=251

ANS: 251


Q4. A software engineer returns from America. As he is fat he decided to have evening walk.....he started at 3pm. he walks along the road at 4km/hr for some time then he climbs a upward slope area at 3km/hr then downwards at the rate of 6km/hr. then back to the home through the road at 4km/hr. what is the distance he covered in one way if he reaches back home at 9pm.

a) 8 Km
b) 6 Km
c) 3 Km 
d) 9 Km

ANS: 6


Q5. In a race Andrew beats Jim. Jack is not the last. Dennis loses to both Jack and Lucia.Jim beats Jack. Who won the race?

ANS: Andrew

Explanation:

The possible positions in the race are,
Andrew Andrew
Jim Jim
Jack Lucia
Lucia Jack
Dennis Dennis
In either case Andrew is the winner.


Q6. Find the digits X,Y,Z

X X X X
Y Y Y Y +
Z Z Z Z
-----------------
 Y X X X Z
-----------------

ANS: X Y Z 
 9 1 8

Explanation:

x+y+z = z ==> x+y = 10
max value for y = 1
hence x = 1
also, (carry)1+x+y+z = x ==> 1+y+z = 10
 hence z=8
 X Y Z 
 9 1 8


Q7. A 1 k.m. long wire is held by n poles. If one pole is removed, the length of the gap becomes 12/3m. What is the number of poles initially?

ANS:  251

Explanation:

Let the no. of poles originally = n
After taking away one pole
(n-1)*12/3 = 1000
n = 251


Q8. A + B + C +D = D + E + F + G = G + H + I =17.IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH LETTER TAKEN ONLY ONE OF THE DIGIT FROM 1 TO 9. 

ANS: A = 4 ,B = 2, C =6, D = 5, E = 3, F = 8, G = 1,H = 7, I = 9.


Q9. Can you find out what day of the week was January 12, 1979?

ANS: Friday.


Q10. Find x+2y

(i). x+y=10
(ii). 2x+4y=20

ANS: (ii)


Q11. Is angle BAC is a right angle

(1) AB=2BC
(2) BC=1.5AC


Q12. Is x greater than y

(i) x=2k
(ii) k=2y


Q13. Find the antonym for gaurish.

a) Cheap
b) Flashy
c) Costly

ANS: a) Cheap


Q14. Choose an appropriate antonym for the word deliberate.

a) unintended
b) targeted
c) focussed

ANS: a) Unintended


Q15. Choose the antonym for Sorrow.

a) Joy
b) empath
c) sympathy

ANS: a) Joy


Q16. How many three digit numbers exist with the property that first digit of the 3-digit number is the product of last two digits?

a) 19 
b) 17
c) 12 
d) None of these

ANS: b) 17.

Explanation:

First digit (100s place) of a 3-digit number cannot be zero, let us start with 1.
If the first digit is 1, then the possible last two digits to get the product 1 is (1,1)
Therefore, the number is 111.
If the first digit is 2, then the possible last two digits to get the product 2 are (1,2) and (2,1).
Therefore, the numbers are 212 and 221.
Proceeding like this, we get,
First digit Corresponding Total number
 3-digit numbers of numbers
1 111 
2 212, 221 1+2 = 3
3 313, 331 3+2 = 5
4 414, 441 5+2 = 7
5 515, 551 7+2 = 9
6 616, 661 9+2 = 11
7 717, 771 11+2 = 13
8 818, 881 13+2 = 15
9 919, 991 15+2 = 17
Thus, the required number of such 3-digit numbers is 17.


Q17. How many 3-digit numbers satisfy the following property: Last digit of cube of the 3-digit number is 6 ?

a) 10 
b) 100 
c) 90 
d) none of these

ANS: a)10.

Explanation:

Note that,When three numbers are multiplied, the units digit of the product will be the last digit of the product of the last digit of each of the numbers.
If last digit of a 3-digit number is 1 then last digit of cube of the number will be 1.Similarly, if last digit of the cube of a 3-digit number is 6 then last digit of its corresponding 3-digit number must be 6.Therefore, possible such three digits numbers are,
106, 116, 126,....,196 -> 10 numbers.
206, 216, 226,....,296 -> 10 numbers.
Proceeding like this, we have,
806, 816, 826,....,896 -> 10 numbers.
906, 916, 926,....,996 -> 10 numbers.
Thus, number of such numbers = 10+10+...+10 (9 terms) = 90.
Hence the answer is 90.


Q18. Consider a 4-digit number 5611 where last two digits are the sum of first two digits (5 + 6 = 11) and the unit digit is 1. How many such 4-digit numbers are exist?

a) 9 
b) 10
c) 51 
d) none of these

ANS: a) 9

Explanation:

Since last digit of required 4-digit numbers is 1 then the possible last two digits of such numbers are 01, 
11, 21, 31,41,...,91.
Note that, the maximum possibility of sum of two single digits is 18. (9 + 9 = 18).
Therefore, 21, 31,...91 are not possible.
Then only possible last two digits are 01 and 11.
If last two digits is 01 then the number is 1001.
And if the last two digit is 11 then the numbers are
211, 8311, 7411, 6511,5611, 4711, 3811 and 2911.
Hence, the total number of numbers is 1 + 8 = 9.


Q19. A man lent out a certain amount of money to A on simple interest (S.I) at the rate 20% per annum for 2 years and the same amount of money at same rate to B on compound interest (C.I) which is compounded half yearly for 2 years. At the end of 2 years, he got Rs.1282 as the difference amount of S.I and C.I then what was the amount he lent out?

a) Rs.20,000 
b) Rs.21,000 
c) Rs.22,000 
d) Rs.23,000

ANS:  a) Rs.20,000

Explanation:

Let the required amount be Rs. P.
S.I on P for 2 years at 20% per annum = P x R x T /100 = P x 2 x 20 / 100 = 2P/5.
C.I on P for 2 years at 20% (compounded half yearly) = P {1 + (R/2)/100}2n
= P {1 + (20/2)/100}2(2)
= P {1 + 10/100}4
= P {1 + 1/10}4
= P {11/10}4
Given that, the difference is Rs.1282.
i.e., P{11/10}4
= P {11/10}4
= P{14641/10000 7/5} = Rs.1282
= P{14641- 14000/10000 } = Rs.1282
= P{641/10,000} = Rs.1282
= P = Rs. 20,000.
Hence, the answer is Rs.20,000.


Q20. A man invested Rs. A in bank X which offers simple interest at 10% per annum and he invested Rs. B in  bank Y which offers compound interest at 6% per annum. If the total amount of interest (together bank X and Y) in 2 years was Rs.8017 and the total amount invested by him was Rs. 52500 then which of the following is equal to A?

a) Rs.20,000 
b) Rs.21,500 
c) Rs.22,000 
d) Rs.23,500

ANS: a) Rs.20,000.

Explanation:

We have to find Rs. A
Total amount invested by him is Rs. 52500
i.e., Rs. (A + B) = Rs. 52,500
Then B = Rs.52500 - A
He invests Rs. Rs.52500 - A in bank Y on C.I at 6% per annum for 2 years.
Total S.I from bank X = PRT/100 = A x 10 x 2 / 100 = Rs. A / 5 .....(1)
Total C.I from bank Y = P {1 + R/100}n
= (52500 - A)[{1+ 6/100}2
= (52500 - A)[{1+ 3/50}2
= (52500 - A)[{53/50}2
= (52500 - A)[{2809/2500} -1 ]
= (52500 - A){309/2500} ......(2)
Total interest amount of S.I and C.I = Rs.8017.
From (1) and (2), A/5 + (52500  A)(309 / 2500) = 8017
A/5 + 52500 x (309/2500)  A (309/2500) = 8017
191A/2500 + 21x309 = 8017
191A/2500 = 8017   6489
191A/2500 = 1528
A = 8 x 2500 = 20,000
Hence the required answer is Rs.20,000.


Q21. A man borrowed a certain amount on S.I and C.I at same rate of interest. At the end of two years, he paid back Rs.1600 and Rs.1664 as S.I and C.I respectively. If he took 3 years to return the amount then what was the difference of C.I and S.I ?

a) Rs.197.12
b) Rs.128.12 
c) Rs.293.12 
d) Rs.179.12

ANS: a) Rs.197.12

Explanation:

Let the amount he borrowed be Rs.P
Let C.I on P for 2 years be X and S.I on P for 2 years be Y.
S.I on P for 1st year = Y/2 .....(1)
C.I for 2nd year = Fixed Interest + Interest on interest
Therefore, X = Y + S.I on Y/2 for 1st year (by eqn(1))
X - Y = S.I on Y/2 for 1st year ....(2)
He paid S.I at the end of 2nd year (Y) = Rs.1600 then S.I for 1st year (Y/2) = Rs.800
S.I for 3rd year = S.I for 2nd years + S.I for 1st year = Rs.1600 + Rs.800 = Rs.2400 ....(3)
He paid C.I at the end of 2nd year = Rs. 1664
Difference S.I  C.I = (X - Y) = Rs.1664 1600 = Rs. 64.
sub. X - Y in eqn(2), we get
Rs.64 = S.I on Y/2 for 1st year.
S.I on Rs.800 for 1st year = Rs.64
Therefore, principal P1 = Rs.800, S.I = 64, T = 1 year.
Then R = 100 x S.I / P x T
= 100 x 64 / 800 x 1 = 8%
Now, R = 8%, t = 2 years and S.I = 1600
Then, principal P = 100 x S.I / R x T
= 100 x 1600 / 8 x 2 = Rs.10,000
Now, we have to calculate the C.I on Rs.10,000 for 3 years at 8% per annum.
C.I = P [{1 + R/100}n
= 10000[{1+ 8/100}3
= 10000[{27/25}3
= 10000[{273
= Rs.2597.12
C.I for 3 years = Rs.2597.12
Required difference = C.I for 3 years S.I for 3 years = Rs.( 2597.12  2400) = Rs.197.12


Q22. It costs Rs.P to print each first 500 copies of a document and Rs.Q to print each subsequent copy. Which of the following expression is equal to the total cost to print R number of copies of the document if R is greater than 500?

a) Rs. R(P - Q) + 500RQ 
b) Rs. 500(P - Q) + RQ 
c) Rs. 500(P + Q) - RQ 
d) Rs. R(P + Q) - 500RQ

ANS: b) Rs. 500(P - Q) + RQ.

Explanation:

Cost to print one copy of first 500 copies = Rs.P
Total cost to print first 500 copies = Rs.(500 x P).
We have to print R number of copies, and given that R > 500.
For more than 500 copies the cost to print is Rs.Q.
Total cost to print remaining (R-500) copies of R = Rs. (R-500) x Q.
Required cost to print total R copies = Rs. [500P + (R-500)Q] = 500P + RQ 500Q
= Rs. 500(P - Q) + RQ.
Hence, the answer is option b.


Q23. There were certain number of students in a seminar hall before the seminar started. 1 hour later, the number of students in the seminar hall is increased by n% and after another one hour it was decreased by n%. At the end of the seminar, there were only 100 students. Find the initial number of students.

ANS: 10,00,000 / (1002)

Explanation:

Let the number of students be X.
After 1 hour, number of students was increased by n%.
That is, after 1 hour, number of students = X + nX/100
After 2 hours, decreased by n%.
That is, after 2 hours, number of students = [X + nX/100]  n% of [X + nX/100]
= [X + nX/100] (n/100)[ X + nX/100]
=(X / 100) (100 + n) (n / 100) (X / 100) (100 + n)
=(X / 100) (100 + n){1 (n / 100)}
=(X / 100) (100 + n){100 n}(1 / 100)
= (X/10000)(1002)
There were only 100 students at the end of the seminar.
Therefore, from (1), 100 = (X/10000)(1002)
X = 10,00,000 / (1002)
Hence, the answer is 10,00,000 / (1002)


Q24. There are 6 members in a family and their average weight is A. If B is the average weight of three of them and C is the average weight of remaining three then which of the following is true?

a) A = 2(B + C) 
b) 2A = (B + C) 
c) A = 3(C + B)
d) none of these

ANS: b) 2A = (B + C)

Explanation:

Average weight of 6 members = A
Total weight of 6 members = 6A
Average weight of 3 of them = B
Total weight of this 3 members = 3B
Average weight of remaining 3 = C
Total weight of remaining 3 = 3C.
Therefore, 6A = 3B + 3C
i.e., 2A = B+C.
Hence the answer is option b.


  
   






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