HackerRank Quantitative Aptitude Test

A number is divisible by 9 if the sum of its digits is completely divisible by 9.

Here, 9+3+5+8+1+9+1=36,which is completely divisible by 9.

Hence,9358191 is also completely divisible by 9

By actual division, we find that 999999 is exactly divisible by 13. The quotient 76923 is the required number.

We need to find LCM of 3, 5 and 7

3,5 and 7 are prime numbers

So LCM = 3 × 5 × 7

= 105

So he write exams for all three courses in the same week after 105 weeks

The differences between numbers are

401 - 293 = 108

563 - 401 = 162

563 - 293 = 270

Required number=HCF of 108, 162 and 270

Prime factorization of the numbers:

108 = 2 × 2 × 3 × 3 × 3

162 = 2 × 3 × 3 × 3 × 3

270 = 2 × 3 × 3 × 3 × 5

HCF (108, 162, 270)

= 2 × 3 × 3 × 3

= 54

8m−2 =  8m82 \frac{ 8^{m}}{8^{2}} 828m​

= 198464 \frac{1984}{64} 641984​

=31

3 frogs died but they are still present in the well.

So, there are still 10 frogs in the well.

Profit percentage = ProfitRevenue×100\frac{Profit}{Revenue} \times 100RevenueProfit​×100

= 1.20200×100\frac{1.20}{200} \times 1002001.20​×100

= 0.6%

Let the names of Rashika, Manisha and Risha be assumed as x, y and z for our convenience.

The probabilities of the three friends to finish the race are 1/3,

1/5 and 1/4 respectively. It may be noted that one reaching the finishing point is independent of other

reaching.

If P(x), P(y) and P(z) denote the probabilities.

The probability of at least one of them reaching the finishing point = 1 – P (none of them finishing the race)

= 1−(23)(45 )(34 )=351 - \left ( \frac{2}{3} \right )\left (\frac{4}{5}  \right )\left (\frac{3}{4}  \right )= \frac{3}{5}1−(32​)(54​ )(43​ )=53​

Total distance travelled in 12 hours = (27+29+31+.....upto 12 terms)

This is an A.P with first term (a)=27,  number of terms (n)= 12 and d=2.

∴ Required distance    = 12/2 × [2×27 + (12-1) × 2]

=6×(54+22)

= 456 kms

Sum of all the 15 numbers = 13 * 65 = 845

Sum of the first 8 of them = 8 * 59 = 472

Sum of the last 7 of them = 7 * 62 = 434

So, the 8th number = 434 +472 - 840 = 66.

Paddy Considine’s age 4 years hence is (x + 4),

so his present age is x years.

Kerry Washington’s age 6 years ago is (y –6),

so her present age is y years.

y = 2x –10 .......(i)

x + 5 = y –5 .....(ii)

Solving both the equations,x = 20 and y = 30 years

Required Ratio=2030=2:3 \frac{20}{30} =2:3 3020​=2:3

Timothy Hutton got married 14 years ago,

Present age of Timothy Hutton= (7/5)*Timothy Hutton’s age at the time of marriage

Present age of Timothy Hutton: Timothy Hutton’s age at the time of his marriage = 7: 5

= > 2’s = 14

= > 1’s = 7

Present age of Timothy Hutton= 7’s = 49

Timothy Hutton’s brother = Timothy Hutton– 10 = 49 – 10 = 39 years

The age of Timothy Hutton’s brother at the time of his marriage = 39 – 14= 25 years

Let Imran rate of work be 100/2 = 50 work units per day.

Yousuf would do 100/3 = 33.33 work units per day and Pathan does 133.33/5 = 26.66 units of work per day.

Their 1 days work = 50 +33.33 + 26.66 = 110 units of work per day.

In 20 days, the total work done would be 2200 units of work

Hence for Yousuf to do it alone it would take: 2200/33.33 = 66 days to complete the samework.

Let the quantity of mixture X initially be ‘m’ litres

0.8m + 713×78\frac{7}{13} \times 78137​×78 = 92

0.8m+ 42 = 92

m = 5008\frac{500}{8}8500​ = 62.5 litres

Number of people who can drive = 2.

∴ Number of ways of selecting driver = 2C1 ^{2}\textrm{C}_{1} 2C1​

The other person who knows driving can be seated only in the rear three seats in 3 ways

Total number of ways of seating the two persons = 2C1×3 ^{2}\textrm{C}_{1}\times 3 2C1​×3

∴ Number of ways of seating remaining = 3!

Total number of all five can be seated = 2C1×3×3! ^{2}\textrm{C}_{1}\times 3\times 3! 2C1​×3×3! = 36