Wednesday 27 July 2016

Tata Digital Health Recent Aptitude Answers July 2016

Answer 1:

g = grandson's age
s = son's age
m = my age
:
"My grandson is about as many days as my son is weeks,"
s = 7g
:
my grandson is as many months as I am in years.
m = 12g
:
g + s + m = 120
Substitute for m and s
g + 7g + 12g = 120
20g = 120
g = 120%2F20
g = 6 yrs is grandson's age

then
s = 7(6) = 42 yrs is son's age
and
m = 12(6) = 72 yrs is my age
;
:
See if we can prove this:
"My grandson is about as many days as my son is weeks," a
g: 6*365 = 2190 days
s: 42*52 = 2184 days, that's why they said "about"
:
"my grandson is as many months as I am in years."
g: 12*6 = 72 months

Answer 2:

The way I did it was to simply add up the uninjured people from each injury and subtract it from the total. A lot easier in my mind.

So: 25 + 30 + 15 + 20 = 90


Total soldiers - ones lacking at least one injury = 100 - 90 = 10.

Answer 3:

Raj, Rajesh, Rahul, Ravi, Rohit, Rakesh

Answer 4:

1 sheep - 1 dog = 1 goat + 1 dog
1 sheep - 1 goat = 2 dogs
A sheep is 10 dinars, but what does a goat cost?
The money they received for their camels is a square. And the digit in the tens-place of this square is odd (total money / 10 dinars), because they could not divide the sheep equally.
Now the interesting thing is that all squares with an odd tens-place end with a 6. (16^2 = 256, 24^2 = 576). So a goat must cost 6 dinars.
10 - 6 = 2 dogs

So a dog costs 2 dinars.

Answer 5:

65292 

As per given conditions, there are three possible 
combinations for 2nd, 3rd and 4th digits. They are (3, 0, 
7) or (4, 1, 8) or (5, 2, 9) 

It is given that there are 3 pairs whose sum is 11. All 
possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now 
required number is 5 digit number and it contains 3 pairs 
of 11. So it must not be having 0 and 1 in it. Hence, the 
only possible combination for 2nd, 3rd and 4th digits is 
(5, 2, 9) 

Also, 1st digit is thrice the last digit. The possible 
combinations are (3, 1), (6, 2) and (9, 3), out of which 
only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the 
answer is 65292.

Answer 6:

Let the total distance travelled downhill, on the level, and uphill, on the outbound journey, be x, y, and z, respectively.
The time taken to travel a distance s at speed v is s/v.

Hence, for the outbound journey

x/72 + y/63 + z/56 = 4

While for the return journey, which we assume to be along the same roads

x/56 + y/63 + z/72 = 14/3

It may at first seem that we have too little information to solve the puzzle.  After all, two equations in three unknowns do not have a unique solution.  However, we are not asked for the values of x, y, and z, individually; but for the value of x + y + z.

Multiplying both equations by the least common multiple of denominators 56, 63, and 72, we obtain

7x + 8y + 9z = 4 · 7 · 8 · 9
9x + 8y + 7z = (14/3) · 7 · 8 · 9

Now it is clear that we should add the equations, yielding

16(x + y + z) = (26/3) · 7 · 8 · 9

Therefore x + y + z = 273; the distance between the two towns is 273 miles.

Answer 7:


Sunday

Answer 8:

Thursday

Answer 9:

let a = original amt to be divided among his children:
:
1sh. 1000 + .1(a-1000) = (1000 + .1a - 100)
(.1a + 900) = 1st child amt
Subtract that from original amt:
a - (900+.1a) = (.9a - 900)
:
2sh. 2000 + .1[(.9a-900)-2000] = 2000 + .1(.9a-2900) = (2000 + .09a - 290)
(.09a + 1710) = 2nd child amt
;
It says the shares are equal, therefore we can find a:
1st child sh = 2nd child sh

.1a + 900 = .09a + 1710
.1a - .09a = 1710 - 900
.01a = 810
a = 810%2F.01
a = $81,000; original amt to be divided
:
Find the share given to the 1st child using the equation
1sh = .1a + 900
1sh = .1(81000) + 900
1sh = 8100 + 900
1sh = $9000 amt to 1st child
:
Check to see if the 2nd child equation yields the same amt
2sh = .09a + 1710
2sh = .09(81000) + 1710
2sh = 7290 + 1710 
2sh = $9000 amt to 2nd child, so we are on the right track
:
It asks how many children are there 81000/9000 = = 9 children each getting 9000

Answer 10:

Let A = number of minutes the accurate clock moves. 
Let S = number of minutes the slower clock moves.

We know that: S =5/6A

When the slower clock shows 3 PM, it has moved 180 minutes: S = 180

Hence:180 = 5/6A   =>   A = 216

The accurate clock has moved 216 minutes.
The time will be 3:36 PM.




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