# Accenture Previous Years Placement Questions 6

1. A teacher can divide her class into groups into groups of 5,13 and 17. What is the smallest possible strength of the class?
a.835
b.940
c.1105
d.1220
Explanation:
For smallest possible class strength, we consider LCM of the given numbers.  So LCM (5, 13, 17) = 1105.

2. If QUESTION = DOMESTIC what will b code for RESPONSE ?
1. OMESUCEM
2. OMESICSM
3. OMESICEM
4. OMESISCM
Explanation:
Q U E S T I O N
D O M E S T I C
We can see in the word ‘RESPONSE’ only R and P are the letters whose code is not given, by having a look on options we can say code for R is O and code for P is S (because common in all options).
Now we can easily code the word ‘RESPONSE’as ‘OMESICEM’.

3. A box contains 90 bolts each of 100 gm and 100 bolts each of 150 gm. If the entire box weighs 35.5 kg., then the weight of the empty box is :
A. 10 kg
B. 10.5 kg
C. 11 kg
D. 11.5 kg
E. None of the above
Explanation:
Let the weight of Empty box be x
90 × 0.100 + 100 × 0.150 + x = 35.5
Then x = 10.5 kg

4. The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
Explanation:
Let the average of all 10 members 4 years ago was x. After 4 years, 10 members age increases by 10 × 4 = 40.  But there is no change in the average as a person is replaced by an younger one. Which means, the younger one age is 40 years less than the old one.

5. If the radius of a circle is increased by 20% then the area is increased by:
a) 44%

b) 120%
c) 144%
d) 40%
e) None of the above
Explanation:
Let say radius = πr2πr2 = π102π102 = 100π100π
New radius = 10 × 120% =  12
New area = π122π122 = 144π144π
Increment = 144π−100π100π×100144π−100π100π×100 = 44%

6. When Raja was born, his father was 32 years older than his brother and his mother was 25 years older than his sister. If Raja’s brother is 6 years older than Raja and his mother is 3 years younger than his father, how old was Raja’s sister when Raja was born?
Explanation:
Raja’s brother 6 year old when Raja was born.
His father’s age = 32 + 6 = 38 year
His mother’s age = 35 year
His sister age  =35 – 25 = 10 years

7. A man purchased a watch for Rs.400 and sold it at a gain of 20% of the selling price. The selling price of the watch is:
Explanation:
Cost price = 400
Gain% = 20%
Gain  = 400 × 2010020100 = 80
Cost price = 400 + 80 = 480

8.If
1111 = r
2222 = t
3333 = e
4444 = n
5555 = ?
Explanation:
1 + 1 + 1 + 1 = 4
Four last letter is r
Similarly
5 + 5 + 5 + 5 = 20 i.e twenty so ans is Y.

9.  A new apartment complex purchased 60 toilets and 20 shower heads. If the price of a toilet is three times the price of a shower head, what percent of the total cost was the cost of all the shower heads?
a) 9%
b) 10%
c) 11%
d) 13%
e) 15%

Explanation:
Let the cost of shower head is x. Then the cost of the toilet = 3x.
Total cost of 60 toilets and 20 shower heads = 60 × 3x + 20 × x = 200x
So shower heads cost as a percentage of total cost = 20x20x20x20x × 100 = 10%

10. Find the smallest number which leaves 22,35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.

Explanation:

LCM of (26,39,52,65) = 780
Required number= 780 – 4 = 776

# Accenture Previous Years Placement Questions 5

1. If a certain computer is capable of printing 4900 monthly credit card bills per hour, while a new model is capable of printing at a rate of 6600 per hour, the old model will take approximately how much longer than the new model to print 10000 bills?
Explanation:
Old model is capable of printing at a rate of 4900 per hour
New model is capable of printing at a rate of 6600 per hour
Old model time taken to print 10000 cards = 10000/4900 = 100/49
New model time taken to print 10000 cards = 10000/6600 = 100/66
Old model – new model: 100/49 – 100/66 = 1700 /(49 × 66) = 850/(49 × 33) = 0.525 hrs => 31 mins
Therefore, the old model will take approximately 31 mins longer than the new model to print 10000 bills

2. 3 men and 8 women complete a task in same time as 6 men and 2 women do. How much fraction of work will be finished in same time if 4 men and 5 women will do that task.
1) 5/6
2) 6/13
3) 7/9
4) 11/12
5) 11/15
6) 13/14
Explanation:
3 m + 8 w = 6 m + 2 w
3 m = 6 w
1 m = 2 w
Therefore  3 m + 8 w = 14 w
4 m + 5 w =13 w

3. How can a cake(circular) be cut into 8 pieces by making just 3 cuts?
Explanation:
Cut the cake using three mutually perpendicular planes. It leaves 8 pieces, one each in each octant.

4. A number of cats got together and decided to kill between them 999919 mice. Every cat killed an equal number of mice. Each cat killed more mice than there were cats. How many cats do you think there were?
Explanation: 991
999919 = 1000000 – 81 = 1000210002 – 9292 = (1000 + 9)(1000 – 9) = 1009 × 991.
Since there were more mice than there were cats, 991 cats killed 1009 mice each.

5. An  alloy of zinc and copper contains the metals in the ratio 5 : 3. The quantity of zinc to be added to 6 kg of the alloy so that the ratio of the metal may be 3 : 1 is:
A)3
B)6
C)8
D)9
Explanation:
In sixteen kg of alloy 10 kg of zinc and 6 kg copper is present
To make the ratio 3:1 we must add 8 kg of zinc to make it 18 kg of zinc and 6 kg copper

6. A class consists of 100 students, 25 of them are girls and 75 boys; 20 of them are rich and remaining poor; 40 of them are fair complexioned. The probability of selecting a fair complexioned rich girl is
Explanation:
The probability of selecting girl is: 25/100 = 1/4
The probability of selecting rich is: 20/100 = 1/5
The probability of selecting fair complexioned is: 40/100 = 2/5
Three are independent;probability of rich and fair complexioned girl is:
(1/4) ×(1/5) × (2/5) = 1/50

7. If one-third of one-fourth of a number is 15, then three-tenth of that number is:
A. 35
B. 36
C. 45
D. 54
Explanation:
The number is 1/3 of 1/4 is =15 then 1/3 × 1/4 = 15
Number is 180 then 180 × 3/10 = 54

8. 100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
A) 100/7% gain
B) 15% gain
C) 100/7% loss
D) 15 % loss
Explanation:
C.P. of 100 orange=350
S.P. of 12 is 48 i.e, 4 of each now of 100 will be 400
So profit = 400 – 350 = 50
profit =  profit×100%cpprofit×100%cp i.e 100/7% gain

9. If 3/p = 6 and 3/q = 15 then p – q = ?
A) 1/3
B) 2/5
C) 3/10
D) 5/6
E) None of the above
Explanation:
3p3p = 6  ,   p = 3636 = 1212and
3q3q = 15  ,   q = 315315 = 1515 then
p – q = 1/2 – 1/5 = 3/10

10. Two trains leaving from two station 50 miles away from each other with constant speed of 60 miles per hour, approaches towards each other on different tracks. if length of each train is 1/6 mile. when they meet How much time they need to pass each other totally ?
Explanation:
The trains are coming towards each other so their relative speed is 60 + 60 = 120 mph
In this case the distance would be addition of lengths of trains.i.e 1/6 + 1/6 = 1/3 m
T = dsds = 1/31201/3120 = 1136011360 h

# Accenture Previous Years Placement Questions 4

1. There are two bags A and B. A contains 6 red flowers and 3 pink flowers. where as bag B contains 2 red flowers and 7 pink flowers. One flower is chosen from a bag randomly. What is the probability that the flower chosen is pink?
a. 4/9
b. 1/3
c. 5/4
d. 5/9
Explanation:
Bag A 6 red + 3 pink
Bag B 2 red + 7 pink
Probability of picking 1 pink from bag A is 3/9
Probability of picking 1 pink from bag B is 7/9
Total = 3/9 + 7/9 = 10/9
And probability of selecting a bag is 1/2
So finally probability of choosing a pink flower = 1/2(10/9) = 5/9

2. The sum of a number and its square is 1406. What is the number?
1) 38
2) 39
3) 37
4) 29
Explanation:
By option Verification 37 + 37 × 37 = 1406

3. How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
A. 40
B. 400
C. 5040
D. 2520
Explanation:
The Word LOGARITHMS is contain 10 letters.
To find how many 4 letter word we can find from that = 10 × 9 × 8 × 7 = 5040.

4. At 6′o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12′o clock
Explanation:
For ticking 6 times, there are 5 intervals.
Each interval has time duration of 30/5 = 6 secs
At 12 o’clock, there are 11 intervals,
So total time for 11 intervals = 11 × 6 = 66 secs.

5. A man travelled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village
Explanation:
Average speed = (2 ×a × b)/(a + b) here a = 25 b = 4
Average speed = 2 × 25 × 4/(25 + 4) = 200/29 kmph.
Distance covered in 5 hours 48 minutes = Speed × time
Distance = (200/29) × (29/5) = 40 kms
Distance covered in 5 hours 48 minutes = 40 kms
Distance of the post office from the village = (40/2) =20 km.

6. One year payment to the servant is Rs. 200 plus one shirt. The servant leaves after 9 months and receives Rs. 120 and a shirt.Then find the price of the shirt.
Explanation:
12 months  ==   Rs 200 + shirt –(1)
9 months    ==   Rs 120 + shirt –(2)
After subtracting equation (2) from (1) we get,(shirt is cancelled)
3 months = Rs 80
12 months = (80 × 12)/3= 320
Hence from equation (1) we get 320 = 200 + shirt
Therefore shirt = 120 Rs/-

7. In a class there are 45 pupil, out of them 12 are in debate only and 22 in singing only. Then how many in both?
Explanation:Total pupil = 45
Debate + Singing = 12 + 22 = 34
The intersection for two = 45 – 12 – 22 = 11 play both games.

8. Silu and Meenu were walking on the road.
Silu said, “I weigh 51 Kgs. How much do you weigh?”
Meenu replied that she wouldn’t reveal her weight directly as she is overweight.
But she said, “I weigh 29 Kgs plus half of my weight.”How much does Meenu weigh?
Explanation:
It is given that Meenu weighs 29 Kgs plus half of her own weight.
It means that 29 Kgs is the other half. So she weighs 58 Kgs.
Solving mathematically, let’s assume that her weight is A Kgs.
A = 29 + A/2
2 × A = 58 + A
A = 58 Kgs.

9.  What is the 56743 rd term in the series 1234567891011121314…….?
a) 1
b) 3
c) 7
d) 5
Explanation:
1 to 9 = 9 no.s 1 digit each no.
9 to 99 = 90 no.s 2 digit each,Total digit = 90 × 2 = 180 terms
99 to 999 = 900 no.s 3 digit each,total digit = 900 × 3 = 2700
999 to 9999 = 9000 no.s 4 digit each,total digit = 9000 × 4 = 36,000
Till Now 999,we have = 9 + 180 + 2700 = 2889 digits
Upto 9999 we have = 2889 + 36000 = 38889 digits
56743 – 38889 = 17854
After 9999 each no. has 5 digit
So 17584/5= 3570 with remainder 4.
3570th no after 9999 =9999+3570=13569
Next term=13570.

10.  A cycled from P to Q at 10 kmph and returned at the rate of 9 kmph. B cycled both ways at 12 kmph. In the whole journey B took 10 minutes less than A. Find the distance between P and Q.
Explanation:
Let the distance between P and Q = d km
A – time = d/10 + d/9 = 19 d/90 hours
B – time = 2 d/12 = d/6 hours
10 minutes = 1/6 hours
Thus 19 d/90 – d/6 = 1/6
(19 d – 15 d)/90 = 1/6
4 d/90 = 1/6
thus d = 15/4 km = 3.75 km

11. A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
A. 18
B. 24
C. 42
D. 81
Explanation:
Just go in reverse order.
Check the options with th subtracting 18 from each option for getting reverse number.
42 – 18 = 24 (which is in reverse).

# Accenture Previous Years Placement Questions 3

1. A man purchased a watch for Rs. 400 and sold it at a gain of 20% of the selling price. The selling price of the watch is ?
Explanation:
C.p. = Rs 400
Gain = 20%
S.P = C.P × 100+p%100100+p%100 = 480

2.The cost of two varieties of paint is Rs. 3969 per 2 kg and Rs. 1369 per 2 kg respectively. After how many years will the value of both paint be the same, if variety 1 appreciates at 26% per annum and variety 2 depreciates at 26% per annum ?

Explanation:
Simply appreciates variety 1 by 26% and depreciates variety 2 by 26% as:
3969(1−126)n=1369(1+126)n3969(1−126)n=1369(1+126)n
For n = 2 we get both values equal.
Variety 2      Variety1
3969.00       1369                Initially
2937.06       1724.94          after I year
2173.42       2173.42         after II year
So the price become same after 2 years.

3. An exhibition was conducted for 4 weeks. The number of tickets sold in 2nd week was increased by 20% and increased by 16% in the 3rd week but decreased by 20% in the 4th week. Find the number of tickets sold in the beginning, if 1392 tickets were sold in the last week ?

Explanation:

Let initially A tickets have been sold.
So now in 2nd week 20% increases so
A × 120100120100
In 3rd week 16% increases so
A × 120100120100 × 116100116100
In 4th week 20% decrease so
A × 120100120100 × 116100116100 × 120100120100 = 1392
A = 1250

4. Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?

Explanation:
Let two numbers be ah and bh.
As h is 13, we get the numbers as 13a, 13b.
LCM = 13ab.
So 13ab = 273
⇒ab = 21.
So a = 7 or 3.
One of this number is 39 or 91.  Given that the number is greater than 60, we take 91 as the required number.

5. In an exam, Ajith, Sachu, Karna, Saheep and Ramesh scored an average of 39 marks. Saheep scored 7 marks more than Ramesh. Ramesh scored 9 fewer than Ajith. Sachu scored as many as Saheep and Ramesh scored. Sachu and Karna scored 110 marks between them. If Ajith scores 32 marks then how many marks did Karna score?

Explanation:

Let marks of Ajith = a ; Sachu = sc ; Karna = k ; Saheep = sh and Ramesh = r  then
sh – r = 7        – – – – (i)
a – r = 9          – – – – (ii)
sc = sh + r      – – – – (iii)
sc + k = 110   – – – – (iv)
Also given a = 32
So from (ii) r = 23  and from (i) sh = 30 , from (iii) sc = 53, from (iv) k = 57.

6. The average number of visitors of a library in the first 4 days of a week was 58. The average for the 2nd, 3rd, 4th and 5th days was 60. If the number of visitors on the 1st and 5th days were in the ratio 7:8 then what is the number of visitors on the 5th day of the library?

Explanation:

If number of visitors on 1st, 2nd, 3rd, 4th and 5th day are a, b, c, d and e respectively then
a + b + c + d = 58 × 4 = 232   – – – (i)
b + c + d + e = 60 × 4 = 240   – – – (ii)
Subtracting (i) from (ii), e – a = 8  – – – (iii)
Given that e : a = 8 : 7
Let e = 8x and a = 7x.
Given, 8x  – 7x = 8 ⇒ x = 8  – – – (iv)
So  a = 56 and e = 64.

7. A man said to a lady, “Your mother’s husband’s sister is my aunt”. How is the lady related to the man.
Explanation:
Mother’s husband is nothing but Father of the lady.  So fathers sister will be Aunt to the Lady.  But in question, they gave that, mother’s husband’s sister is Man’s Aunt.  So she is aunt to both the Lady and Man.  So they should be brother and sister.

8. If a man reduces the selling price of a fan from 400 to 380 his loss increases by 20% .What is the cost price of fan ?
Explanation:
Let the cost price be x. Then initial loss = x – 400
Given that 20% ( x – 400) = 20
⇒ x – 400 = 100
⇒ x = 500

9. 260 can be represented as:
A) @****@**
B) @@*@@@@@@
C) @@*@@@@**
D) @*****@**
Explanation:
260 can be written in binary format as (100000100)2(100000100)2
Replacing 1 with @ and 0 with *, we get option d.

10. A piece of ribbon 4 yards long is used to make bows requiring 15 inches of ribbon for each. What is the maximum number of bows that can be made?
A. 8
B. 9
C. 10
D. 11
E. 12
Explanation:
1 yard = 3 feet = 3× 12 = 36 inches
4 yard = 4 × 36 = 144 inches
Number of maximum bows that can be made = 1441514415 = 9.6
Option B is correct.

# Accenture Previous Years Placement Questions 2

ACCENTURE PREVIOUS YEARS PLACEMENT QUESTIONS 2

1. One dog tells the other that there are two dogs in front of me.The other one also shouts that he too had two behind him. How many are they?

Explanation:

Dog

Dog

Dog

So there are 3 dogs.

They are in circle.2. In the following questions, the following letters indicate mathematical operations as indicated below:

A: Addition; V: Equal to;  S: Subtraction;  W:Greater than;  M: Multiplication;  X: Less than;  D: Division

Find Out of the four alternatives given in these questions, only one is correct according to the above letter symbols. Identify the correct one.

See the options given below

A) 6 S 7 A 2 M 3 W 0 D 7

B) 6 A 7 S 2 M 3 W 0 A 7

C) 6 S 7 M 2 S 3 W 0 M 7

D) 6 M 7 S 2 A 3 X 0 D

Explanation:By BODMAS rule

6 – 7 + 3 × 2 > 0/7

3. What will be the output of the following code statements?

Integer a = 10, b = 35, c = 5 print (a × b / c) – c

Explanation:

Apply BODMAS rule

10 × 35 = 35053505 =70 –5 =65

4. Usually the room tariff in this hotel is higher. At present, it is low because of the ______ season.

a. peak

b. off

c. down

d. slow

e. full

Explanation:

Because of the Off season.

5. The average number of visitors of a library in the first 4 days of a week was 58. The average for the 2nd, 3rd, 4th and 5th days was 60.If the number of visitors on the 1st and 5th days were in the ratio 7:8 then what is the number of visitors on the 5th day of the library?

Explanation

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then

a + b + c + d = 58 × 4 = 232     —-(i) &

b + c + d + e = 60 × 4 = 240     —-(ii)

Subtracting (i) from (ii), e – a = 8 —(iii)

Given

a/e=7/8 —(iv)

So from (iii) & (iv) a=56, e=64

6. An exhibition was conducted for 4 weeks. The number of tickets sold in 2nd work week was increased by 20% and increased by 16% in the 3rd work week but decreased by 20% in the 4th work week. Find the number of tickets sold in the beginning, if 1392 tickets were sold in the last week

Explanation:

let initially A ticket has been sold.

So now in 2nd week 20% increases so

A × 120100120100

In 3rd week 16% increases so

A × 120100120100 × 116100116100

In 4th week 20% decrease so

A × 120100120100 × 116100116100 × 120100120100 = 1392

A = 12507. Limited resolution of early microscopes was one of the reasons of _________ understanding of cells.

1. aided

2. discredited

3. increased

5. restricted

Explanation:

it should be restricted because of limited resolution

8. A constructor estimates that 3 people can paint Mr khans house in 4 days. If he uses 4 people instead of 3,how long will they take to complete the job?

a.4

b.2

c.3

d.5

Explanation:

Use formula For a work  Members × days = constant

3 × 4 = 4 × a

a  = 3

9. The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is:

A. 12%

B. 40/3%

C. 15%

D. 14%

Explanation:

Ans. C. 15%

The true discount on Rs. 2562 due 4 months hence is Rs. 122.

Therefore Present Worth is = 2562 – 122 = 2440.

It means that Rs. 122 is interest on Rs. 2440 for 4 months.

Therefore rate percent is = (122 × 100 × 12)/(2440 × 4) = 15%

10. Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?

A. 1 year

B. 2 years

C. 25 years

E. None of these

Explanation:

Given

R – Q = Q – T and R + T = 50 which gives Q = 25

As the difference between R & Q and Q & T is same

# Accenture Previous Years Placement Questions  1

1. What is the next number of the following sequence
21,77,165,285,……
Explanation:
Look series
3 × 7 = 21
7 × 11 = 77
11 × 15 = 165
15 × 19 = 285
19 × 23 = 437 ans
T(n) = (4n – 1) × (4n + 3)
Put value of n find any term.

2. A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
A. 18
B. 24
C. 42
D. 81
Explanation:
Just go in reverse order.
Check the options with th subtracting 18 from each option for getting reverse number.
42 – 18 = 24 (which is in reverse).

3. Two guys work at some speed.  After some time one guy realises he has done only half of the other guy completed which is equal to half of what is left So how much faster than the other is this guy supposed to do to finish with the first.
Explanation:
If x is the part of task this is completed then,
1st has done work = (1 – x)/4 and 2nd has done work = (1– x)/2
So 2nd will have to increase his speed by 2 times.

4. We are given 100 pieces of a puzzle. If fixing two components together is counted as 1 move ( a component can be one piece or an already fixed set of pieces), how many moves do we need to fix the entire puzzle.
Explanation:
Fixing 2 components together = 1 move
98 components are left ,to again fix one of these to the other two = 1 move
97 comp left,so 97 more moves
Total = 1 + 1 + 97 = 99.

5.  The ime a passenger train takes to cross another freight train is twice when the passenger train crosses the freight train running in opposite directions. What is the ratio of their speeds?
Explanation:
Speed of freight train = x
Speed of passenger train = y
Sum of their length = s
So time = s/(x – y)when both are in same direction
Time = s/(x + y) when opposite direction
From question s/(x – y) = 2.s/(x + y) so solving this we get, x/y = 3/1.
So x : y = 3 : 1

6. At a recent birthday party there were four mothers and their children. Aged 1,2,3 and 4. from the clues below can you work out whose child is whose and their relevant ages?

• It was jane’s child’s birthday party.
• Brian is not the oldest child.
• Sarah had Anne just over a year ago.
• Laura’s Child will be next birthday.
• Daniel is older than Charlie is.
• Teresa’s child is the oldest.
• Charlie is older than Laura’s child.

Explanation:
According to given points.
Anne is one yr old her mother is Sarah
Brian is 2 yr old his mom is Laura
Charlie is 3 yr old his mom is Jane
Daniel is 4 yr old his mom is Teresa.
And its charlie’s birthday party.

7. Ajith was driving down the country side when he saw a farmer tending his pigs and ducks in his yard. Ajith asked the farmer how many of each he had.  The farmer replied that there were 60 eyes and 86 feet between them.  How many ducks and how many pigs were there ?
Explanation:
Let the no of ducks be x and no of pigs be y.
Then,since there are 60 eyes in total and both ducks and pigs have 2 eyes we have  2x + 2y = 60 – – – (1)
And total number of legs are 86, ducks have 2 legs while a pig has four so 2x + 4y = 86 – – – (2)
Subtracting (1) from (2) ,we get
2y = 26. ie. y = 13
Putting value in – – – (1),
We get x = 17

8. A supportive young hare and tortoise raced in opposite directions around a circular track that was 100 yards in diameter. They started at the same spot, but the hare did not move until the tortoise had a start of one eighth of the distance ( that is, the circumference of the circle). The hare held such a poor opinion of the other’s racing ability that he sauntered along, nibbling the grass until he met the tortoise. At this point the hare had gone one sixth of the distance. How many times faster than he went before must the hare now run in order to win the race ?
Explanation:
The hare and the tortoise are at the same point when hare have to cover 5/6 of the distance and tortoise have to cover 1/6 of the distance to complete the race.
If x is the speed of tortoise then it’ll take 1/(6 x) time to finish.
So, hare will have to run more then 5 times the speed of tortoise to win the race.

9. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
A. 9
B. 11
C. 13
D. 15
Explanation:
Let numbers are n,n + 2,n + 4
Now 3 n = 2(n + 4) + 3
3 n – 2 n = 11
So
N = 11 then consecutive odd numbers are 11, 13, 15

10. A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
A. 12 days
B. 15 days
C. 16 days
D. 18 days
Explanation:
Work done in 3 days = (3/20) + (1/30) + (1/60) = 1/5
Remaining work = 4/5
If 1/5 work is done in 3 days then,
4/5 work is done in (3 × 5 × 4)/5 = 12 days
Total no. of days will become 12+3=15 days for the complete work to be done