TCS - 2015 - 2

TCS – 2015 - 2

1.       Rajiv can do a piece of work in 10 days , Venky in 12 days and Ravi in 15 days.  They all start the work together, but Rajiv leaves after 2 days and Venky leaves 3 days before the work is completed.  In how many days approximately the work is completed?
(a) 5
(b) 6
(c) 9
(d) 7
Ans: (d)

Rajiv in one day does 1/10 work.

Venky in one day does 1/12 work

Ravi in one day does  1/15 work.   All the three in one day does – 1/10 + 1/12 + 1/15 = ¼ work.

They work for 2 days and the work completed is ¼ * 2 = ½.   Remaining work is ½

Venky and Ravi in one day do – 1/12 + 1/15 = 3/20 work.

Hence to complete the remaining ½ work they would have taken -> ½ * 20/3 = 10/3 or 3 1/3 days. Venky after doing work for 1/3 day leaves.

Thus in 1/3 days the work done by both is 3/20 * 1/3 = 1/20 work.

Remaining work is ½ - 1/20 -> 9/20

To complete this Ravi will take ->  (9/20) / (1/15) -> 9/20 * 15 -> 6 ¾ days -> roughly 7 days.    

2.       On a 26 question test, five points were deducted for each wrong answer and eight points were added for each correct answer.  If all the questions were answered, how many were correct, if the score was zero?
(a) 10
(b) 12
(c) 11
(d) 13
Ans: (a)

Let the number of correctly answered questions be ‘x’. Then we have an equation:

8x – 5(26 – x) = 0.

Solving, we get the value of ‘x’ as 10.

3.        X = 101102103104105106107......146147148149150  (From numbers 101-150).  Find out the remainder when this number is divided by 9.

Ans: 2

                The rule of divisibility for 9 is the sum of the digits is to be divisible by 9.

To arrive at the sum of the digits we calculate the sum in the unit, tenth and the hundredth places separately.

(a)    Sum of the digits in the hundredth place is – 1 * 50 = 50

(b)   Sum of the digits in the tenths place is:

1*10 + 2*10 + 3*10 + 4*10 + 5*1 = 105. (Note the initial 0*9 will give only ‘0’ value)

(c)    Sum of the digits in the Unit’s place is:   (1 + 2 + 3 + 4 + …………+ 9) * 5 = 225.

Totalling (a), (b) and (c) we get …380. This when divided by 9 will leave a reminder 2.

4.       A number is 101102103104...150. As in the earlier question what is the reminder if

divided by 3?  

Ans: 2

The rule of divisibility for 3 is the same as that of 9 the reminder will remain the same as 2

5.       7^1+7^2+7^3+.......+7^205.

In the above series how many numbers have 3 in the unit place? 

Ans: 51

The unit digits for 7^1, 7^2, 7^3 and 7^4 are  7, 9, 3 and 1. These digits repeat again and again. In other words, the unit digits in 7 to the power repeat after every fourth power and each can be called one cycle.

Thus in 7^205 we have 205/4 -> 51 completed cycles.

Thus 51 powers of 7 will have unit digit as 3.

Hence the answer is 51.

6.       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?

Ans: Paper B

A simple question to answer.

Percentage of marks scored in Paper A is 18/70*100 = 25.7%

Percentage of marks scored in Paper B is 14/30*100 = 46.6%

In paper B he fared well.

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

Ans: 28.

The number of factors of a given number N is given by a^p*b^q*c^r ….

Where a, b, c,…. Are prime numbers and (p+1)*(q+1)+(r+1)*……. are factors.

Applying the above we have 1728 = 2^6*3^3

Hence the factors are (6+1) * (3+1) = 28

8.       The sum of two numbers is 45. Sum of their quotient and reciprocal is 2.05. Find the product of the numbers.

Ans: 500

Let x and y be the numbers.

We are given that x + y = 45 (i) and (x/y) + (y/x) = 2.05 (ii)

From (ii) we have (x² + y²) / xy = 2.05. expanding

[(x+y)² – 2xy] / xy = 2.05 -> (x+y)²  = 2.05xy + 2xy = 4.05xy

(x+y) = 45. We now have  ->   45*45/ 4.05 = xy

Thus xy = 500.

9.       A number when divided by 406 leaves remainder 115. What will be the reminder if the number is divided by 29?

Ans: 28

Let the number be N and the quotient X.

Then N = 406X + 115.  Since 406 is exactly divisible by 29, to arrive at the answer

We have to divide the reminder 115 alone by 29.

Thus 115/29 reminder is 28.

10.   The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
(a) 50
(b)100
(c) 150
(d) 216

Ans: (a)

We can write 10 as (1 + 2 + 3 + 4) four consecutive numbers.

Now adding 5 to each of these digits we get:

6 + 7 + 8 + 9 = 30 (again consecutive four digits.)

Again adding 5 to each of the above digits we get:

11 + 12 + 13 + 14 = 50

From the above we observe that 10, 30, 50, 70 ……that are multiple of 10 have consecutive four digits.

Thus we have 50 numbers that can be written in four consecutive digits between 1 and 1000.

Please note that 20, 40, 60 …… all cannot be written through four consecutive numbers.

TCS - 2015 Questions-1

TCS – 2015 Questions - 1

1.       Given that 0 < a < b < c < d, which of the following the largest?
(a) (c+d) / (a+b)
(b) (a+d) / (b+c)
(c)  b+c) / (a+d)
(d) (b+d) / (a+c)
Ans: (a)

These type of questions are best answered by apportioning values to the four alphabets.

Let ‘a’ be equal to 1, ‘b’ 2, ‘c’ 3 and ‘d’ 4. Apportioning these values we observe that option (a) is the largest.

2.        

           *       *     *     7     *       *      *      X     *      *      *       8      *       *

The fourteen digits of a credit card are to be written in the boxes shown above.  If the sum of every three consecutive digits is 18, then the value of x is :
(a) 3
(b) cannot be determined from the given information.
(c) 2
(d) 1
Ans: (a)

The sum of each three squares is 18. Starting from the left we have four sets of three squares.

Let the last two squares be A and B.

Now we have the total sum of all the squares as -> 18*4 + A + B -> 72 + A + B ………. (i)

Excluding the squares having digits 7, X and 8 we have three sets of three squares.

The total sum of these squares is -> 3*18 + 7 + X + 8 + A + B -> 69 + X + A + B ……….. (ii)

Solving (i) and (ii) we get the value of ‘X’ as 3.

3.       Jake can dig a well in 16 days.  Paul can dig the same well in 24 days.  Jake, Paul and Hari together dig the well in 8 days.  Hari alone can dig the well in
(a) 96 days
(b) 48 days
(c) 32 days
(d) 24 days
Ans: (b)

Jake in one day can do 1/16 work. In 8 days he can do 8/16 -> ½ work

Paul in one day can do 1/24 work. In 8 days he can do 8/24 -> 1/3 work.

Hari in one day can do 1/H work and in 8 days can do 8/H work.

Thus we have an equation :  ½ + 1/3 + 8/H = 1

8/H = 1 – (1/2 + 1/3).  8/H = 1/6  ->  H = 48. Thus Hari will take 48 days to complete the work.

4.       The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is

109.  If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order.  Find the number.
(a) 773
(b) 683
(c) 944
(d) 863
Ans: (d)

The two points to note here are (i) the sum of the squares is 109 and (ii) and after subtracting 495 the digits are the same in the reverse order. Hence, we can answer this question from the options given.

Only option (b) and (d) fulfil the first condition. But, option (ii) fulfils both the conditions.

Hence the answer is (d) 863. 

5.       Mark told John "If you give me half your money I will have Rs.75.  John said, "if you give me

one third of your money, I will have Rs.75/-  How much money did John have ?
(a) 45
(b) 60
(c) 48
(d)37.5
Ans: (b)

Let M and J be the initial money that the two have.

We now have two equations:  

M + J/2 = 75 => 2M + J = 150 …….. (i)

M/3 + J = 75 => M + 3J = 225 …….. (ii)

Multiplying (ii) by 2 we get -> 2M + 6J = 450 …….. (iii)

Subtracting (i) from (iii) we get 5J = 300 and J = 60

6.       Eesha has a wheat business.  She purchases wheat from a local wholesaler of a particular

cost per pound.  The price of the wheat of her stores is $3 per kg.  Her faulty spring balance reads 0.9 kg for a KG.  Also in the festival season, she gives a 10% discount on the wheat.  She found that she made neither a profit nor a loss in the festival season.  At what price did Eesha purchase the wheat from the wholesaler?
(a) 3
(b) 2.5
(c) 2.43
(d) 2.7
Ans: (c)

Since her spring balance is faulty sale of 1 kg is shown as 0.9kg and thus she loses 10% on the sale price and gets only Rs 2.70 per kg.

Further, she allows additional 10% discount on this price and thus she gets only Rs 2.43 per kg.

At this price there is no gain or loss for her. Hence her purchase price is Rs 2.43 per kg.

7.       Raj goes to market to buy oranges.  If he can bargain and reduce the price per orange by

Rs.2, he can buy 30 oranges instead of 20 oranges with the money he has.  How much money does he have?
(a) Rs.100
(b) Rs.50
(c) Rs.150
(d) Rs.120
Ans: (d)

Let M be the money that Raj has. Then we have an equation

M/20 – M/30 = 2.  Solving we get the value of M as Rs 120.

8.       A city in the US has a basketball league with three basketball teams, the Aziecs, the Braves and the Celtics.  A sports writer notices that the tallest player of the Aziecs is shorter than the shortest player of the Braves.  The shortest of the Celtics is shorter than the shortest of the Aziecs, while the tallest of the Braves is shorter than the tallest of the Celtics.  The tallest of the Braves is taller than the tallest of the Aziecs.
Which of the following can be judged with certainty?
(X) Paul, a Brave is taller than David, an Aziec
(Y) David, a Celtic, is shorter than Edward, an Aziec
(a) Both X and Y
(b) X only
(c) Y only
(d) Neither X nor Y
Ans: (b)

Let us take the height of the shortest of the Braves as 4 ft. Then the tallest of the Aziecs is less than these heights say 3 ft. Now we can arrive at the range of heights of the three teams.

Aziecs  height range from 2 – 3

Braves height range from 4 – 6

Celtics height range from 1 – 7

We can now conclude that X is correct while Y cannot be determined.

9.       There are 3 classes having 20, 24 and 30 students respectively having average marks in
an examination as 20,25 and 30 respectively.  The three classes are represented by A, B and C and you have the following information about the three classes.
(i) In class A highest score is 22 and lowest score is 18
(ii) In class B highest score is 31 and lowest score is 23
(iii) In class C highest score is 33 and lowest score is 26.
If five students are transferred from A to B, what can be said about the average score of A; and what will happen to the average score of C in a transfer of 5 students from B to C ?
(a) definite decrease in both cases
(b) can't be determined in both cases
(c) definite increase in both cases
(d) will remain constant in both cases
Ans: (b)
Class A average is 20. And their range is 18 to 22
Class B average is 25. And their range is 23 to 31
Class A average is 30. And their range is 26 to 33
If 5 students transferred from A to B, A's average cannot be determined but B's average comes down as the highest score of A is less than lowest score of B.
If 5 students transferred from B to C, C's average cannot be determined the B's range of marks and C's range of marks are overlapping.

10.   The value of a scooter depreciates in such a way that its value of the end of each year is 3/4 of its value of the beginning of the same year.  If the initial value of the scooter is Rs.40,000, what is the value at the end of 3 years?
(a) Rs.13435
(b) Rs.23125
(c) Rs.19000
(d) Rs.16875
Ans: (d)
40000 * (3/4)³ = 16875.

Aptitude Questions - 4 (for Bank Exams)

Aptitude Questions – 4 (For Bank Exams)

1.                   Find the reminder when 333666777888999 divided by 3 or 9 or 11?

Ans: Reminder 5 when divided by 11.  

Rule of divisibility for 3 and 9 – I f the sum of the digits divisible by 3 or 9 then the whole number is divisible by 3 and 9.

Rule of divisibility for 11 --- If the difference between the sum of the odd placed numbers and even placed numbers is either zero or divisible by 11 then the whole number is divisible by 11.

In this case the difference is 5 and that’s the reminder.

2.                   Which is the biggest perfect square amongst the following

(a) 15129,    (b) 12348,            (c) 23716,             (d) 20736

Ans: (c)

3.                   The greatest area of the following:

(a) The radius of circle is 4

(b) The square of diagonal is 4

(c) The square of side is 4

Ans: (a)

4.                   The area of the maximum size of the circle inscribed within a 10 inch square?

Ans: 78.57 sq. inch  ( The radius of the circle will be half the side of the square)

5.                    A gold density is 19 times greater than the water and for copper it is 9 times. At

what ratio you can mix gold and copper to get 15 times denser than water?

                       Ans: 3 : 2

                       Using the rule of alligation we have for Gold 15 – 9 = 6 parts and for Copper

                       19 – 15 = 4 parts. Thus the ratio is 3 : 2

6.                   Find the value of (1.99)2

Ans: 3.9601

7.                   There is a room with 6' x 8'. A one foot tile is fixed along the 4 walls in one row. How

many more one foot tiles are required  to finish the work.

Ans. 24 Tiles

Since all along the wall one foot tile is already fixed, the dimension of the room now is reduced to  -> 4 * 6 = 24 sq.ft.

8.                   Two persons can finish a job in 8 days. First person alone can finish the work in 24

days. How many days does the second person take to finish the job?

Ans: 12 days

Let the two persons be A and B.

A in one day does 1/24 job and B in one day 1/B job.

They both together finish the job in 8 days.

Hence the job done together in one day is 1/8. Thus we have an equation

1/24 + 1/B = 1/8 -> 1/B = 1/8 – 1/24 = 1/12.

1/B = 1/12. B will alone finish the job in 12 days.

9.                   A person's salary is getting reduced by 20%. What percentage should be added to

get back his original salary?

Ans: 25%

Use the formula ----  100R/100-R (where R is the percentage of reduction)

10.               Two persons start from the same point and walk in opposite directions with 5km/hr and

       5.5km/hr speed respectively. What is the distance separating them after 2 and half hrs?

       Ans: 26.25 km

       Relative speed between the two is ---- 5 + 5.50 = 10.50 km/hr. (They walk in opp.

                     direction.

      Thus the distance between them after 2.30 hours will be 10.50 * 2.30hrs = 26.25 km.    

11.               A person starts walking at a speed of 5km/hr through half the distance, rest of the distance he covers with a speed 4km/hr. Total time of travel is 9 hours. What is the maximum distance he can cover?

Ans: 40.50 km.

Average speed is ---(5 + 4)/2 = 4.50 km. ( Note: He covers the total distance in two equal half)

Distance/ Speed is = Time.  Let X be the distance.

Thus ------  x/4.5 = 9 hrs. solving we get … X = 40.50km

12.               Initially two cups of same volume are present with milk filled upto 3/5th and 4/5th

of their volumes. Water is then filled. Then two mixtures are mixed. Find the ratio of water to milk in the mixture.

Ans: 3 : 7

A simple question. Let the volume of the two cups be 5 each. Then total volume is 10. The total volume of Milk in these two cups initially is 3 + 4 = 7parts. The remaining is 3 parts now filled with water.

Thus the ratio of water and milk is ------- 3 : 7.

13.               16 grams of radioactive material decays into 8 grams in 10 years. How long will it

take to decay to 1 gram?

Ans: 70 years.

16 gm reduced to 8gm in 10yrs. So 8 gm get decayed.
For decaying 1 gm it takes 10/8 years. For decaying 8 gm it takes 10/8*8
Remaining 8 gm need to decay 7 to get 1 gm remaining.
For decaying 7 gm it will take
10/8*8*7 = 10*7 = 70 years
 

14.               In a rectangle the length is increased by of the original length. By what proportion

should the width be reduced so that the area will be the same?

Ans: 50%

Assume the original Length and Breadth as 5 and 4. The area is 5 * 4 = 20.

The new length is 5 + 5 = 10. To retain the same area the breadth should be 2.

Breadth reduced from 4 to 2 or 50%.

15.               A person’s salary is decreased by steps of 20%, 15% and 10%. What will be the percentage decrease, if the salary is decreased in a single shot?

Ans: 38.8%

Successive percentage change is given by the formula ->  x + y + xy/100 where x and y are the two successive percentages.

Applying this after the second decrease, the percentage change is ->  -20 - 15 + 300/10 =  -32%

After the third decrease the change will be:

-32 – 10 + 320/100 = - 38.8%

Polaris - Aptitude - 1

Polaris – Aptitude - 1

1.       I have marbles in three different colours as follows:
Blue 60;  Green 156;  Red 204
I distribute them to children in such a way that each one of them get equal number of marbles and all in one colour.
How many children are there?
(a) 12                     (b) 35                    (c) 67                     (d) 31
Ans: (b)

Find the HCF of 60, 156 and 204. The HCF is 12. Total number of marbles is 420. Divide this by the HCF -> 420/12 = 35.

2.       A square of side 6cm has a circle inscribed in it. Calculate the approximate ratio of surface area of the circle to square.
(a) 11/14              (b) 13/14              (c) 12/14              (d) 14/18
Ans: (d)

The area of the Square is 6*6 = 36 sq.cm.  The radius of the circle if half the side of the square – 3 cm. The area of the circle is πr² -> 22/7 * 3 * 3 = 28 sq.cm (approx.)

Hence the ratio is 28 : 36 or 14 : 18

3.       There are 3 numbers whose ratio is as follows: first: second=2:3 and second : third= 5:8. There sum is 98, what is the second number?
(a) 20                     (b) 30                    (c) 40                     (d) 50
Ans: (b)

Let the three numbers be A, B and C. We now have two ratios

A : B :: 2 : 3  …….. (i)     and  B : C :: 5 : 8 ……….. (ii)

We have two different values for B in the two ratios. Let us make this common in both the ratios.

Multiply (i) by 5 and (ii) by 3.

We now have one common ratio ->    A : B : C :: 10 : 15 : 24

The sum of A + B + C = 98.  Hence the value of B is ->  98 * 15/49 = 30.

4.       A 30% saline solution is to be converted to a 10% saline solution. How much pure water needs to be added to 100 kg of 30% saline solution?
(a) 150 kg             (b) 100 kg            (c) 200 kg             (d) 250 kg
Ans: (c)

In 100 kg saline solution the salt is 30 kg. This quantum of salt will remain as such in the 10% saline solution also. Since the quantum of salt is 30 kg and its percentage is 10% only the total quantum of solution should be 300 kg.

Hence 200 kg of water needs to be added.

5.       In a given factory, average salary combined for the staff and supervisors is Rs 60 while that for only staff is Rs. 56. The average salary of supervisors is Rs. 400. There are a total of 12 supervisors. How many labourers are employed in this factory?
(a) 1020                (b) 1130                (c) 1500                (d) 1532
Ans: (a)

This question can best be answered through Rule of Alligation figure as under.

                          A (56)                                                          B (400)
                                  

                                                                X (60)           

                       B – X (340)                                                 X – A (4)

From the above we observe the ratio of staff and supervisors as 340 : 4

We are informed that there are 12 supervisors.

3 Parts of supervisors amount to 12.

Hence the number of staff is 340 * 3 = 1020.

6.       Alok completed a journey from his home to mall in 42 minutes at an average speed of 10kmph. He walked initially at 4kmph and then took an auto which moved at 25kmph. What part of his journey did he walk?
(a) 1km                 (b) 2km                                (c) 3km                 (d) 4km
Ans: (b)
The distance covered in 42 minutes t an average speed of 10 kmph is -> 10 * 42/60 = 7 km.

Let ‘x’ be the distance travelled by walking at a speed of 4 kmph. Then 7-x is the distance travelled by auto at a speed of 25 kmph.  We now have an equation:

x/4 + (7-x)/25 = 42/60. Solving we get the value of x as 2 km.

Hence, the distance covered by walking is 2km.

7.       Ashish can do 1/4 of his work in 10 days, Vijay 40% in 40 days and Deepak 1/3 in 13 days. Who finishes first?
(a) Ashish            (b) Deepak         (c) Vijay                                (d) Can’t say
Ans: (b)

Ashish ……. In 10 days does ¼th work. So, to complete the whole work will take 40 days.

Vijay in 40 days will complete 40% or 2/5 of the work. He will take 100 days to complete the work.

Deepak in 13 days completes 1/3^rd of the work and would take 39 days to complete the whole work.

Hence, Deepak will complete the work first.

8.       You can complete a carpet weaving in 10 days while I can do it in 15 days. We work together but I leave in 2 days. How many days will you now require to finish the work?
(a) 7                       (b) 8                       (c) 19/3                 (d) 20/3
Ans: (d)

In one day I will do 1/10 work and my friend will do 1/15 work. Together we both in one day will do 1/10 + 1/15 = 5/30 -> 1/6 work. In two day we both will do 1/6 * 2 = 1/3 work.

The remaining work now is 1 – 1/3 = 2/3 which I alone will have to do.

The time taken by me to complete this work is ----- (2/3) / 1/10 = 20/3 days.

9.       Increase in population for a given country in three consecutive years is 10%, 25% and 10%. What is the combined percentage increase for the three years?
(a) 45%                 (b) 52%                 (c) 51.25%           (d) 45.25%

Ans: (c)

Successive percentage increase for two years is given by the formula -> x + y + xy/100 where x and y are the percentage variation of two successive periods.

Applying this, we get the percentage variation at the end of 2^nd year as -> 10 + 25 + 250/100 = 37.50%

Now the variance between the second and third year is ->  37.50 + 10 + (37.50*10)/100

= 51.25% 

10.   Sheila and Sunita start a business with investment of Rs8000 and Rs5000 resp. Sunita is an active partner and so she gets 10% of profit separately for supervision. If the total profit is Rs 3240/-, then what is the share of profit for Sunita?
(a) 1445                (b) 1334                (c) 1500                (d) 324
Ans: (a)

Ratio of Shiela and Sunita’s investment are :  8 : 5

10% of total profit of Rs 3240/- is Rs 324/- and this Sunita gets separately.

Remaining profit is --- 3240 – 324 = 2916.

Sunita’s share in this profit is ->     2916 * 5/13 =1121.

Hence, Sunita’s share of profit in the business is ----  1121 + 324 = 1445.

Accenture - Aptitude - 2

Accenture – Aptitude – 2

1.       A car is filled with four and half gallons of oil for a full round trip. The car however consumes quarter gallon more of oil during its up trip than its down trip. What is the quantum of oil consumed by the car during its down trip?

(a)    2 gallons       (b) 3 gallons        (c) 4 gallons        (d) 5 gallons.

Ans: (a)

Assuming the car consumes equal quantity of oil for the whole trip then each of the up and down trip would have consumed to and a quarter gallons of oil since the total oil consumption is given as four and half gallons.

However, for the up trip the car consumes quarter gallon oil more -> 2.25 + 0.25 = 2.50 gallons.

Since the total oil consumed is only four and half gallons, the oil consumed for the down trip is -> 4.5 – 2.5 = 2.00 gallons.

2.       Two men can complete a work in 24 minutes. One of them can complete the work alone in 40 minutes. How much time the other man will take to complete the work alone?

(a)    35 min           (b) 50 min            (c) 60 min            (d) 24 min

Ans: (c)

Let the two men be A and B. Together they complete the work in 24 minutes.

Let us assume A alone completes the work in 40 minutes.

Now in one minute A and B will do 1/24 work.

In one minute A alone will do 1/40 work. B alone will do 1/x work where ‘x’ is the minutes taken by him to do the work.

Now we have an equation ……….. 1/A + 1/B = 1/24 substituting the values

We have ….. 1/40 + 1/x = 1/24

1/x = 1/24 – 1/40 = 1/60. Hence B will take 60 minutes to complete the work.

3.       In a theatre the parking slot allows the parking of four wheeler and two wheelers in the ratio 1 : 8. One day the total number of wheels in the parking slot was observed as 100. How many four wheelers and two wheelers were parked there?

(a)     4 : 24             (b) 5 : 40               (c) 5 : 20               (d) 10 : 30

Ans: (b)

This is a simple but tricky question. If we go by the number of wheels both (b) and (d) would give the total number as 100. But, the ratio of parking slot is given as 1 : 8. On this count option (b) alone will fit and is the answer.

Similar questions were asked in TCS papers also.

4.       Two oranges, three bananas and four apples cost Rs 15.00. Three oranges, two bananas and one apple cost Rs 10.00. What is the cost of three oranges, three bananas and three apples?

(a)    Rs 10              (b)  Rs 20              (c) Rs 15               (d) Rs 25

Ans: (c)

2O + 3B + 4A = 15

3O + 2B + 1A = 10   (adding both)

5O + 5B + 5A = 25 so, O + B + A = 5 and

3O + 3B + 3A = 5 x 3 = 15.

5.       In 8*8 chess board John discovered 204 squares. In 3.*3 tik tac toe board how many squares he will find?

(a)    15                   (b)  14                   (c)  12                    (d)  10

Ans: (b)

If the number of squares on each side is ‘n’ then the total number of squares in the board is given as follows:

N² + (n-1)² + (n-2)² + (n-3)² + ………..  Substituting 8 for ‘n’ we get total number of squares as 204.

Similarly substituting 3 for ‘n’ we get the value 14.

You can also use the formula for finding the sum of the first ‘n’ squares

{n x (n+1) x (2n+1)} / 6

6.       For typing a research report of 70 pages the amount payable is Rs 20.00. A types 42 pages and B types 28 pages respectively. How much money should be paid to A?

(a)    Rs 10              (b) Rs 12               (c) Rs 14               (d) Rs 15

Ans: (b)

The ratio of work done by A and B is 42 : 28 -> 3 : 2. Hence the amount also should be shared in this ratio.

Hence A will get ……. 20 x 3/5 -> Rs 12.

7.       A rectangular plate with length 11 inches, breadth 8 inches and thickness 2 inches is melted and drawn into a circular rod of diameter 8 inches. What is the length of the rod?

(a)    3 inches        (b)  7 inches        (c)  3.5 inches     (d)  4.5 inches

Ans: (c)

Volume of the rectangular plate is 11 * 8 * 2 -> 176 cubic inches.

The volume of the steel rod should equal this volume.     

Volume of steel rod is given by πr²l (where ‘r’ is the diameter and ‘l’ the length)

Thus π x (4)² x l = 176   l = 176 / π x 16 -> 3.5 inches.

8.       A tank gets filled in 6 minutes at the rate of 5 cu.ft/min. The length of the tank is 4 feet and the width half that of the length. What is the depth of the tank?

(a)    3.75ft            (b)  4.8ft               (c) 3.7ft                (d) 4.7ft

Ans: (a)

The tank gets filled in 6 minutes at the rate of 5 cu.ft/mt.

Hence the volume of the tank is 6 x 5 = 30 cubic feet.

The length of the tank is 4 feet and the width (half of length) 2 feet. If the depth is ‘l’

Then the volume of 4 x 2 x l = 8 l.

Thus  8l = 30 cu.ft and l = 30/8 = 3.75 ft   

9.       A man goes on his motor bike to the airport to collect a mail. The plane had arrived earlier and the mail was sent through mail van. The man meets the mail van after half an hour of starting collects his mail and returns home. In the process he saves 20 minutes of time. How early the plane had arrived?

(a)    10 min           (b) 15 min            (c) 20 min            (d) 25 min

Ans: (a)

The man collects the mail from the van and returns home in one hour or sixty minutes. In the process he saved 20 minutes. Thus it is evident he would have taken totally 60 + 20 = 80 minutes to go to the airport and return if the plane had not come early. In other words his travel time to airport will be 40 minutes and return time 40 minutes.

But, he collects the mail after travelling 30 minutes only and the difference of 10 minutes the van had covered.

But, the man had saved 20 minutes of which the van had taken 10 minutes. Hence, the plane had arrived 10 minutes earlier than the scheduled time

(NOTE: This puzzle originally appeared in a different format in Shakuntala Devi’s book of puzzles long ago and gets repeated in various placement papers over the years.)

10.   There is a six digit code number. The first two digits when multiplied by 3 give all ones. The second two digits when multiplied by 6 give all twos.  The remaining two digits when multiplied by 9 give all threes. What is the code number?

(a)    332211          (b) 272727           (c) 373737            (d) 333

Ans: (c)

We can answer this question from the choices.

Rule 1. The first two digits when multiplied by 3 give all ones. Hence, it implies the first of the two digits should be 7. Both (b) and (c) fulfils this condition.

But, only (c) gives all ones and option (b)fails on this score.

Option (c) fulfils the other two conditions also.