HackerRank Averages Quiz -1

 Here ,  Total Distance = (28 + 35)km 

                                    =  63 km

 Total Time = (2 + 3)hrs = 5 hrs

Therefore, Average Speed = (63 / 5 ) km/hrs

                                            = 12.6 km/hrs

Let the present age of Soham’s mother, Soham and Akshay be f, a and s respectively.

a + 7 = (s + 5) (1.25)

5s – 4a = 3 ...(i)

Also,

(f – 5) + (a – 5)

= 23 2

f + a = 56 ........(ii)

Soham was 4 years old when Akshay was 1 year old. 

Difference = 4 – 1 = 3 years

a – s = 3 ........(iii)

Solving (i) and (iii)

s = 15 years

a = 18 years

f = 38 years

Akshay was 10 years old, 5 years ago

So, Soham’s mother age 5 years ago was = 38 – 5 = 33 years

The total students whose number was multiple of 2 = 25

The total students whose number was multiple of 5 = 10

The total students whose number was multiple of 5 and 2 both i.e. of 10 = 5

The total students whose number was either multiple of 2 or multiple of 5 = (25 + 10 – 5) = 30

It means, 30 students were selected and 20 students were not selected

The sum of the weight of all the students = (56 \times 20 + 58 \times 30) kg = 2860 kg

The average weight of all the students = \frac{2860}{50} = 57.2 kg

Average Time= T1+T2+T3/3

T1=50/40=4/5

T2=20/60=1/3

T3=40/30=4/3

Average Time=37/15=2.5 hours (approx.)

Let the numbers be “ab” and “xy”

After they are written in reverse numbers will be “ba” and “yx”

As there is a increase of 27 marks in average the total increase in the total marks = 5 \times 27 = 135

As rest of the numbers remain same the increase is only in the two numbers

So “ba” + “yx ” – “ab” – “xy” = 135

10b + a + 10y + x – (10a + b + 10x + y) = 135

9 (b – a) + 9 (y – x) = 135

(b – a) + (y – x) = 15

Hence the sum of difference of the digits of the two numbers is 15

Ratio of incorrect number = 13 : 10

Multiplying both 13 and 10 by number like 5,6,7,8 etc we will get

13 × 7, 10 × 7 = 91, 70.

91, 70 as only combination where the sum of difference of digits is 15 (9 – 1 + 7 – 0)

The original numbers are 19 and 07 and their sum = 19 + 7 = 26

Sum of weight of A, B, C and D = 67 \times 4 = 268 kg

Sum of weight of A, C, D and E = 63 × 4 = 252 kg

Let, weight of B = x kg

Then, weight of E will be (x – 16) kg

Weight of A = (x – 16 – 18) = (x – 34) kg

So, x – 16 + x – 34 = 58 × 2

2x = 116 + 50

x = 83

Therefore, weight of E = (83 – 16) = 67 kg

Reqd. average = \frac{(83 + 67)}{2}= 75 kg

Let the total number of students sent for the global quiz competition be x.

Given that,

48x = 5 + 7 + 9 + 11 + (x – 4)50

2x = 168

x = 84

According to the above formula , 

a = 60 km/hr

b = 76 km/hr

Therefore,

Average speed = (\frac{60+76}{2})

                                     = 68 km/hr

Let the volume of each bottle = 100k, volume of four glasses = 4 \times 100k = 400k

The volume of sharbat in four bottles =
{(30% + 35% + 88% + 99%)* 100k= 252k}

All bottles are emptied and refilled to a volume of 60%, the total volume of four bottles = 60% * 400K= 240k

Remaining volume of sharbat = (252k – 240k) = 12k

12k = 60ml → k = 5ml

Total volume of sharbat in bottles = 240 \times 5ml = 1200ml

Height of 3 boys Ranbeer, Samuel and Arijit is

\frac{208}{3} \times 3 = 208 inches.

Height of Ranbeer, Shan and Amitabh is

\frac{203}{3} \times 3 = 203 inches.

With the help of this information, the height of 5 men cannot be determined.