Accenture Aptitude Questions

11. After 10 years A will be twice the age of B before 10 years and now if the difference is 9 years between them then what is the age of B after 10 years.

Let present ages of A and B be x and y respectively.

According to the question

x+10 = 2(y-10) after solving we get

x-2y = -30 eq. (1)

x-y = 9 eq. (2) after solving above equation we will get there present ages as such A = 48 and B = 39.

After 10 years B will be 49 yrs

8men and 10 womanmeans 8x and 10 yso 8x+10y=25 days.......(1eq)also10x+5y=25 days.......(2 eq)equating(subracting).. we get ..2x-5y=0 daysthat means....x/y=5/2x=5 and y=2so now ....we want to know the work done by 2 children and 3 menie.... 2x+3y=2*5+3*2=10+6=16so work done is 16 days..ANS: 16 days

3-)ans is 241 man 1 day work=1/441women 1day work =1/881 boy 1day work =1/132,so that 1 man 1 woman and 1 boy together can complete the work in 1/44 +1/88 +1/132on solving this we get 1/24,its the work in 1 day of all threeso reverse it we will get 24which is the required answers.

7) the propotion of the milk is 80%

5) i)there are 19 ways to open the lock if the last  digit is 9 (i.e) 9,19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99

ii)there is  one way to open the lock if the sum of the first teo digit is lesstha or equal to the last digit numbers

8) H+T+D= 15 (Hindu +  Times + Deccan)

    H+D=22

    D+T=23

    only Hindu =53-(15+22) =53-37 = 16

    only Times =46-(15+23) =46-38 =8

10) A beats c by 25 meters

plz explain how ?

1man or 2woman or 3 Boy ==44 days

so  1man=44 days

     1 wman=88 days

     1  boy=132 days

all together can do the work in 44+88+132=264 days

Answer: 264 days

Ans 6. 0

998(1-1)(2-1)...=998*0=0

->2ans  3

->3ans 24  (bcoz 1man's 1day work is 1/44 similarly 1 woman's  1 day work is1/88 and that of a child is 1/132 => 1/44(1+1/2+1/3)=1/24)

->5ans (i)90ways (if 0 is not allowed in hundreds place) bcoz as the last digit is 9,the digit in hundreds can b filled in 9c1ways(repetetion allowed) and that in tens place in 10c1 ways.therefore,it is 9c1*10c1=90

->6ans 0

->7ans 51.2%

->9ans 500

->10ans 49yrs

The final ratio will be definitely lesser than 3:5 coz initial percentage of milk is 37.5%(7.5 ltrs) now if we remove 4 litres of mixture then in total we have removed (3/8)*4 litres of milk that is the final concn of milk in the mixture is 7.5 - 1.5 litres = 6 ltrs and rest all is water ... so final ratio is 6:14 which simplifies to 3:7 !!!! so the final concentration of milk is 30% qed..

Question 7 ... last post was for quest 3/4

assume

last post was for quest 3/4

this is for ques 7..

Suppose initially we have 100 ltrs of milk

1) after removing 20% of pure milk we are left wid

       80ltr pure milk

       20ltr water

2) after again removing 20% or mixture we are left with

      80-(20% of 80) = 64 ltr

      36 ltr of water

3) next xchange

     64 - (20% of 64) = 51.2ltr

      rest all is water i.e. 48.8 ltr

11. A+10=2B.....EQ1

     A-B=9.......EQ2

A-2B=-10.........

-A+2B=+10.........

A-B=9.........

ON SOLVING THESE TWO EQNS WE GET B=19.

AND AGE OF B AFTER 10YRS IS B+10=29YRS

Hi

1 man or 2 woman or 3 boy=44 days

1 man's one day work is 1/44

2 womans 1 day work is 1/(44*2)=1/88

3 Boys 1 day work is 1/(44*3)=1/132

Altogether do a work means (1/44)+(1/88)+(1/132)

They complete the work in 24 Days

3

    20% of his loss = 20

    If loss = x;

    20% of x = 20

   x = 100

   that means his loss is 100

  Therefore price of  fan = 400 + 100 = 500

ANS2: 7

Ans for 9:>the original cost of fan is 475

This is for question no4

in 20 lts of mixture if 4 ltrs mixture is removed 16 lts of mixture is left ,so the intial proportion is 3:5.

That means 6 ltrs of milk 10 ltrs of water.

so when u add 4 lts of milk the ratio changes to 1:1(10:10)

This is for question no 7

lets assume that 100 lts of pure milk is there

>when replaced by 20 % with water means 80 lts pure milk and 20 lts water,so the ratio is 4:1.

>now when again 20 lts of mixture is taken out 80 lts of mixture is left that means apply the ratio of 4:1 which says that 64 lts of milk and 16lts of water.

if we replace it with 20 lts of water  the ratio cahenges to 64:36 i.e 16:9

>now again when 20 lts is removed 80 lts of mixture contains 51.2lts of milk and 28.8lts of water. that means when u add 20 lts of water there will be 48.8lts of water and 51.2 lts of milk will be left in the mixture.

Q4- 3:7

Q5- 100

Q6- 0

Q8-30

Q9-280

Q10-23.5

Q11-39

Answer to Q.11)Given: after 10 yrs A will be twice the age of B before 10 yrsthat means: A+10=2(B-10)2B-A=30 .... eqn1Given: Now the difference is 10that means:A~B=9let A-B=9 .... eqn2and B-A=9 .... eqn3 (Since only the difference is mentioned, and not the order of persons)Asked: B's age after 10 yrsSolving eqn1 & eqn2, A=48 yrs; B=39yrs So, ans is (B+10)=49yrsSolving eqn1 & eqn3, A=12 yrs; B=21 yrs So, ans is (B+10)=31yrs

if a single person is doing the work he takes longer time when compared to three people(man,woman and child) doing the work together....it's a indirect proportion....so it cannot be 246 days...the correct solution is 24 days...

ans: (3)

1 man do work in 44 days

1manday work=1/44

lly 1 womanday work=1/(2*44)

1 boyday work=1/(3*44)

if work together=1/((1/44)+(1/88)+(1/132))

answer is 24 days.

Ans (5)

teh combi is like this 009,019,029....099=10 combi

                              109,119..........189=9 combi

                              209,219..........279=8 combi

                              ..........................

                              909=1 combi

therefore no of combi are 1+2+3.......+10=55ans

Ans 9. Is 500 Rs.

It is not the correct one.

hi all

3 ans: is 90 and not 24 y means

first we have to find days so less people more days required

1:1

1:2      ::45:?(X)   

1:3

solving tis ratio we get 90 i.e.,

1*45+2*45+3*45=3X

270=3x

X=90. required answer

 

A+10=2B  {After 10 years, age of A will be equal to twice the age of B before 10 years}

A-B=9  {Now their diffference in age}

Solving these two eqtns. we get B=19 and A=28 (A-B=9)

Substituting this in eq1

28+10=2*19 ie;38=38 this is the age of B before 10 years

After 10 years B will be 48 years.

A  beats c by 25 meters.
A<---------------B<--------------------C
         10m                       15m
<---------------------------------------->
                         25m

Hi everyone!

Q-5 ii)
ans
let consider the case where 3 rd digit = 1 st digit + 2nd digit
for this the no . of cases will be

1+2+3+4..+10=55

this gives us an insight into the problem

now now for the first case where we had taken 000
here is only this case into consideration but taking the given ques where 3 rd digit is less than or equal to 1st and 2nd digit then this case has no other cases but see another case of (i ii 2)
here for equality 2 cases were there now there is increment of one more case which is actually from the earlier case because that is also applicable here .

seeing this the new series is:
1+(2+1)+(3+2+1)+(4+3+2+1)................+(10+9.....+1)
=
(1*10+2*9+3*8+4*7+5*6)*2=2*110=220

ans=220

8 men and 10 woman means 8x and 10y

so 8x+10y=25 days.......(1eq)

also 10x+5y=25 days.......(2eq)

equating (subracting) we get..

2x-5y=0 days

that means....x/y=5/2x=5 and y=2

so now....

we want to know the work done by 2 children and 3 men

ie.... 2x+3y=2*5+3*2=10+6=16

so work done is 16 days..

ANS: 16 days

How can you form second equation? if you know explain it briefly

In ques 2

T -> 6
H->14

The implementation is not correct as 46 - (15 + 23) , T and D considered but T and H is not.

34 locks haveto be opened
(1,1,9)
(2,1,9)
(1,2,9)
(2,2,9)
(3,2,9)
(2,3,9)
(4,1,9)
(1,4.,9)
......
......
......
......
......(6,3,9),(3,6,9),(4,5,9),(5,4,9)

In how many ways can a lock be opened if that lock has three digit number lock if
i. The last digit is 9
ii. Sum of the first two digits is less than or equal to the last digit numbers are from 0-9
since its a 3 digit number it starts from 009 and end at 909
so the formula that we can use is n(n+1)/2 where n=10 the number of ways is 55.

Hi guys i think for (i)  there are 100 ways and for (ii) 55 ways.

4) ans is   3:7
   

.8*.8*.8=.512, or 51.2% whole milk

55

5 a.) Since the lock has three digits and it should end with 9, we have 100 ways to unlock the lock, i.e., 009, 019, 029, 039, 049, 059, 069, 079, 089, 099, 109, 119, 129, 139, 149...989, 999. Count the number of values, you will get 100.

When the selling price comes down from 400 to 380 the loss is increased by 20%
means the the reduction of Rs.20 in selling price leads to 20% increase in loss.
That implies the actual loss at Rs.400 was 20/100 * 20= 100
The cost price = 400 + 100 = 500

1) As there are two entities, men and women, the work is equally shared by 8 men and 10 women. so 8 men=10 women, 20 child=10 women from second eqn. therefore 8 men=20 child. 16 men=20 women=40 child= 25 days.
1 men=25*16=400 days... since it is an indirect variation......similarly,
3 men alone do it in (400/3) days, and 2 children alone do it in 500 days.so in one day the contribution of 3 men is (3/400) of the work and the contribution of 2 children is(2/1000) of the work. (3/400)+(1/500)=19/2000... so in one day 19/2000 of the work is done...hence the remaining work1981/2000 is done in 104.26 days ....so totally they take approximately 105.26 days to complete the work

ANS.105.26 days.....

As there r 2 entities d work s equally shared by 8 men n 10 women, 8 men = 10 women, from 2nd eqn. 20 child = 10 women, so 8 men = 20 child.  Therefore 16 men = 20 women = 40 child = 25 days. Since it is an indirect variation 1 men alone takes 25 * 16 = 400 days to complete the work. Similarly, 1 child alone takes 25 * 40 = 1000 days to complete the work.  The contribution of 1 man in 1 day is 1/400 of the work. 3 men contribution is 3/400 of the work in one day... contribution of 1 child in 1 day is 1/1000 of the work... 2 child contribution is 2/1000 of the work in one day...hence in 1 day (3/400) + (2/1000) = 19/ 2000 of the work... the remaining work is 1981/2000... The remaining work is done in 104.26 days...so totally it takes 105.26 days to complete the entire work.

 

ANS. 105.26 DAYS

for 7)

Milk :water is 64:61

consider pure milk is 10litres
20% of 10lit=2lit
adding 2lit water, the ration of M:W = 4:1
now taking 2lit again(20%) ,
total parts = 4+1(M+W)=5 parts , total quantity taken is 2lit
thus, 1part =2/5=0.4lit
thus, milk removed=4*0.4=1.6lit
water=1*0.4=0.4lit
thus now milk= 6.4lit and water=1.6lit(6.4+1.6=8lit)
add 2 lit of water , total water =1.6+2=3.6lit
thus now ratio is (6.4:3.6==16:9)

now remove 2 lit again
total part =16+9=25
1part=1/25lit=0.08lit

Thus milk removed =0.08*16=1.28lit
water removed =0.08*9=0.72lit

thus milk remaining =6.4-1.28=5.12lit
water remainin=3.6-0.72=2.88lit

20% water added=2lit adder=2.88+2=4.88lit

thus ratio now is (5.12:4.88)

solving we get ratio as 64:61

6.4days i.e 7 days

this is for question 1: answer is 7 days..

Q5) answer is: 10+9+8+7+6....+1 = 55

its a multiplicative series. value increases from 1 to n> 1000. so in between these value there should be the number 998. so
(998-1)(998-2)...................((998-998)=0)..(998-n)
any number multiplied by 0 is zero. so the whole term wil become 0.
So the ans:0

q(9) dont care about the fixed price 400, the selling price is 380,so 20% loss ,so the amount is 380 of80%,so the 20% is 380/4 is 95 so 380+95=475 is the ans"475"

Ans: If pure milk is 100%.
20% replaced by water i.e. 100-20=80% (Pure Milk)
again 20% replaced by water i.e. 80-20=60% (Pure Milk)
again 20% replaced by water i.e. 60-20=40% (Pure Milk)
Finally the proportion of milk in that mixture is 40%

51.2% percent

Data insufficient if L is length of race THEN answer is 25 - (150/L)

10 >ans
data is insufficient if L is length of race answer is 25 - (150/L) and it is explained below

let L be the length of race and T be the time taken by A to complete race then

As speed = L/T
Bs speed=(L-10)/T

time taken by B to complete race =L/((L-10)/T))
=LT/(L-10)

then cs speed = (L-15)/(time taken by B to complete race)

=((L-15)(L-10))/(LT)
distance traveled by C in T seconds = ((L-15)(L-10))/L

therefore A beats C by L- (distance traveled by C in T seconds)

= 25 -(150/L)
therefore answer is dependent on length of race
hence data is insufficient

Let us take pure milk contains 100 litre
20% replaced by water =>100*20/100=20
100-20=80
now 80 litre milk is available

Again 20% replaced by water =>80*20/100=16
80-16=64
now 64 litre milk is available

Again 20% replaced by water =>64*20/100=12.8
64-12.8=51.2
now 51.2 litre milk is available