Aptitude - Average - Discussion

10. In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with address of the first element X (1, 1) is 3000; find the address of X (8, 5).

11. One boy can eat 100 chocolates in half a minute, and another can eat half as many in twice the length of time. How many chocolates can both boys eat in 15 seconds?

12. Potatoes are made up of 99% water and 1% "potato matter." Jack bought 100 pounds of potatoes and left them outside in the sun for a while. When he returned, he discovered that the potatoes had dehydrated and were now only made up of 98% water. How much did the potatoes now weight?

13. You own a pet store. If you put in one canary per cage, you have one canary too many. If you put in two canaries per cage, you have one cage too many. How many canaries and cages do you have?

14. Find the value of @@+25-++@16, where @ denotes "Square" and + denotes "square root".

15. A power unit is there by the bank of the river of 750 meters width. A cable is made from power unit to power plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable Rs.12/- per meter. Find the total of laying the cable.

16. In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time.Find how much temperature more or less in 4pm to 9pm.

1. A bus is started from the bustand at 8am and after staying 30min at a destination return back to the busstand.The destination is 27 miles from the busstand.The speed of the bus is 18mph.In return journey the bus travels with 50% fast speed.At what time it is return to the bus stand.

a 11am b 1pm c 11pm d 10am

2. Person who decided to go to the weekend trip should not exceed 8hrs driving a day.Average speed of forward is 40mph.Due to the traffic in sunday in return journey average speed is 30mph.How far he can select a picnic spot?

a 120 b 240 c 360 d 40

3. Find the value of @@+25-++@16,where @ denotes square and + denotes "square root"?

4. In a two dimensional array,X(9,7) with each element occupy 4 bytes of memory, with the address of the first element x(1,1) us 3000.find the address of x(8,5)

Can anyone help me in finding the solution for this problem as soon as possible?

@Swathi

Answer for the !st problem.

Given :
1) Going to destination:
Distance = 27 miles.
Speed = 18mph.

2)Return journey:-
Distance= 27 miles
Speed=50% faster than 18mph(given).
So, Speed = 1/2*18 + 18 = 27.
Speed =27mph.
Halt of bus=30 minutes.

Answer:-

Let's first find the time to reach the destination:-
Speed = Distance/time
So, time = Distance/Speed
Time =1.5 hr.

Let's find time to return
Time = Distance/Speed
=27/27.
= 1 hr.

So total time taken to go and return
= going time + 1/2 hr (halt) + return time.
= 1.5 + 0.5 + 1 hr.
= 3hr.

So as journey began at 8 am
So Bus will return at 8am+ 3hr=11am.
Final Answer : a) 11am.

@Swathi

Answer for the 1st problem.

Given :
1) Going to destination:
Distance = 27 miles.
Speed = 18mph.

2)Return journey:-
Distance= 27 miles
Speed=50% faster than 18mph(given).
So, Speed = 1/2*18 + 18 = 27.
Speed =27mph.
Halt of bus=30 minutes.

Answer:-

Let's first find the time to reach the destination:-
Speed = Distance/time
So, time = Distance/Speed
Time =1.5 hr.

Let's find time to return
Time = Distance/Speed
=27/27.
= 1 hr.

So total time taken to go and return
= going time + 1/2 hr (halt) + return time.
= 1.5 + 0.5 + 1 hr.
= 3hr.

So as journey began at 8 am
So Bus will return at 8am+ 3hr=11am.
Final Answer : a) 11am.

I solved in this method.. Is this right?
Let x,y,z be the age of husband,wife and child respectively..
for the first condition.. we can write 27=((x+y+z)/3)-3
27=(x+y+z-9)/3
81=x+y+z-9
90=x+y+z....equation 1
for the second condition..we can write 20=((y+z)/2)-5
20=(y+z-10)/2
40=(y+z-10)
50=y+z........equation 2
equation1 - equation2 gives x+y+z-y-z=90-50
x=40
i.e husband's age is 40years..

@priya:
Hi priya. Look into your equation, send 3 to right side which gives the equation, h,w,c are current ages of husband, wife and child respectively....
The average of their ages three years back is as follows.
((h-3)+(w-3)+(c-3))/3=81
==> h+w+c=81+9
==> h+w+c=90 ----(1)(this is the sum of ages of husband, wife and child at present)
And further it is given that, five years back, average of wife and child is 20 years.
i.e. ((w-5)+(c-5))/2=20
==> (w-5)+(c-5)=40
==> w+c=50----(2)

Now, on substituting equation (2) in (1)..
we will get h=90-50 = 40 years.
Hence, the current age of husband is 40 years.
Hope you would get it now.

3 years ago, average age of husband, wife and their child = 27,
=> Sum of age of husband, wife and their child before 3 years = 3 * 27 = 81,
=> (h-3) + (w-3) + (c-3) = 81,
=> h + w + c = 81 + 9 = 90 --- equation(1).

5 years ago, average age of wife and child = 20,
=> Sum of age of wife and child before 5 years = 2 * 20 = 40,
=> (w-5) + (c-5) = 40,
=> w + c = 40 + 10 = 50 --- equation(2).

Substituting equation(2) in equation(1)
=> h + 50 = 90,
=> h = 90 - 50 = 40.

i.e., Present age of the husband = 40.

First separate the question:

In the first part, The average age of husband, wife and their 3 years ago was 27 years, so;

Lets consider Husdand age = x-3(-3 because it was 3 years ago from the present age)

Wife's age= y-3 and son's age = z-3,
Average= x-3+y-3+z-3/3=27,
x+y+z=81 ------> (1)

Second part of the question.
That of wife and the child 5 years ago was 20 years, so;
Wife age years ago is y-5,
Sons age is z-5,
Avg= y-5+z-5=20,
y+z=30---> (2)

Equating (1) and (2)
Then x=40.

This is ramya from vizag,can i solve the problem as below as per question,
present ages of hus=h;wife=w;child=c;
---> 3 years ago it was said the avg ages of the hus,wife,child was 27 years.so,
((h-3)+(w-3)+(c-3))/3=27;
so solving it i got h+w+c=90;--->(a)
and given 5 years ago wife and childs age avg=20years so,
i took((w-5)+(c-5))/2=20;solving it i got w+c=50;-->(B)
on solving (a)&(b) i got h age as 40years,can this procedure be correct??
please do reply me?

Let age of husband, wife and child is x, y, z.

Then 3 year ago age of husband, wife and child was x-3, y-3, z-z.

Given that 3 year ago average of their age was 27.

So (x-3+y-3+z-3) /3 = 27.

=> x+y+z-9 = 81.

=> x+y+z = 90____ (1).

5 year ago age of wife and child was y-5 and z-5.

Given that 5 year ago average of their age was 20.

So (y-5+z-5)/2 = 20.

=> y+z-10 = 40.

=> y+z = 50_______ (2).

Now, (1) - (2).

=> x = 40.

So age of husband is 40.