APTITUDE TEST 2

From 6¹⁰ × 7¹⁷ × 55²⁷, we have to write each base in terms of multiplication of its prime factors.

That is,=(2X3)¹⁰×(7)¹⁷×(5X11)²⁷

=2¹⁰X3¹⁰X7¹⁷X5²⁷X11²⁷

The no. of prime factors = sum of the exponents
= 10+10+17+27+27
= 91

Hence, the number of prime factors is 91.

When the two train running in opposite direction, relative speed = 60+48 = 108 kmph = 108X5/18 m/sec = 30 m/sec

Sum of the lengths of the two trains = sum of the distances covered by the two trains in the above relative speed

Sum of the lengths of the two trains = 30X15 = 450 m

When the two trains running in the same direction, relative speed = 60-48 =12 kmph = 12X5/18 = 10/3 m/sec

When the two trains running in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. The distance he covered in 36 seconds in the relative speed is equal to the length of the slower train.

Length of the slower train = 36X10/3 = 120 m

Length of the faster train = 450-120 = 330 m

Hence, the length of the two trains are 330m and 120m

We can apply L.C.M method to solve this problem

Total work = 90 units (L.C.M of 15,30,18,9)
(A+B) can complete 6 units/day (90/15 = 6)
(B+C) can complete 3 units/day (90/30 = 3)
(A+C) can complete 5 units/day (90/18 = 5)
By adding, we get 2(A+B+C) = 14 units/day
(A+B+C) = 14/2 = 7 units/day

work done by A = (A+B+C)-(B+C)=7-3=4 units/day
work done by B = (A+B+C)-(A+C)=7-5=2 units/day
work done by C = (A+B+C)-(A+B)=7-6=1 unit/day
work done by (A+B+C)in 9 days = 9X7 = 63 units.
Balance work = 90-63 = 27 units

This 27 units of work to be completed B & C. Because A left after 9 days of work.

No. of days taken by B & C to complete the balance 27 units of work = 27/3 = 9 days. (Because B&C can complete 3 units of work per day)

Average speed = 2pq/(p+q), here p=25 q=4
= (2X25X4)/(25+4)
Therefore, average speed = 200/29 km/hr

And 5 hour 48 min = 5⁴⁸⁄₆₀hrs = 29/5 hours
Distance = Speed X Time

Distance = (200/29)X(29/5) = 40 km
Distance covered in 5 hrs 48 min = 40 km

Hence,distance of the post office from the village = 40/2 = 20km

Let "x" be the present age of the son.Then the present age of father is (3x+3)

3 years hence, father's age will be 10 years more than twice the age of the son
(3x+3)+3 = 2(x+3)+10
3x+6 = 2x+16
x = 10
To find the present age of the father, plug x = 10 in (3x+3)

Present age of the father = 3(10)+3 = 33 years

If "x' be the first even number, then the four consecutive even numbers are x, x+2, x+4, x+6

Average of the four consecutive even numbers = 27

(x+x+2+x+4+x+6)/4 = 27
(4x+12) = 108
4x = 96 ===> x = 24

Hence the largest even number = x+6 = 30

From the given information, we have
Speed downstream = (14+4) = 18 kmph
Speed upstream = (14-4) = 10 kmph

Let "x" be the distance between A and B
x/18 + (x/2)/10 =19 (Hint: Time = Distance/Speed)
x/18 + x/20 =19
(10x + 9x)/180 = 19 (L.C.M of 18,20 is 180)
19x/180 = 19
x = 180

Hence the distance between A and B is 180 km.

Original weight of John = 56.7 kg (given)

He is going to reduce his weight in the ratio 7:6

[Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a']

His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.

Let the cost price of 1 ltr pure milk be $1

Now we take some quantity of milk (less than 1 ltr),add some water and make it to be 1 ltr mixture

Let "x" be the money we invest for milk in 1 ltr of milk-water mixture

Since the gain is 20%, selling price = x + 20%of x
Selling price = 120% of x = (120/100)x = (6/5)x

But the mixture is sold at the cost price of pure milk
So, we have (6/5)x = 1
x = 5/6
Therefore cost price of the milk in the mixture = $(5/6

Cost price of the water = $0

Rule to find the ratio for producing mixture = (d-m):(m-c)
(d-m):(m-c) = 1-5/6:5/6-0 = 1/6:5/6 = 1:5

So, water and milk to be mixed in the ratio to gain 20% is 1:5

Let A,B and C be the incomes of A,B and C respectively

From the given information, we have C = $1600
10% of B = 30% of C
(10/100)B = (30/100)X1600
B = $ 4800

15% of A = 25% of B
(15/100)A = (25/100)X4800
A = $8000

A+B+C = 8000+4800+1600 = 14400

Hence, the total income of A,B and C is $14400