Quantitative Aptitude - Arithmetic Ability Questions

Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4

Let A's age 10 years ago be x years 

Then B's age 10 years ago  be 2x yeats

 Total of their present ages =                                        

                                          = 

                                          = 35 years

Cash price = $21 000

Deposit = 10% × $21 000 = $2100 

Loan amount = $21000 − $2100 = $18900

I=p x r x t/100 

I=11340

Total amount = 18900 + 11340 = $30240

Regular payment = total amount /number of payments

Let X = Time taken for each of Type B Problems(100 Problems)
And 2X = Time taken for each of Type A problems(50 problems)

Total time period = 3hrs = 3 x 60min = 180 minutes

100X + 50(2X) = 180
100X + 100X = 180
200X = 180 min

X = 180/200
X = 0.90 min

By convertiing into seconds,

X = 0.90 x 60 seconds
X = 54 sec

So, time taken for Part A problems is = 54 x 2 x 50 = 5400 seconds
= 5400/60sec = 90 minutes.

When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.

When B runs 900 m, distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 m.

In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.

In a race of 600 m, the number of meters by which A beats C

= (600 x 212.5)/1000 = 127.5 m.

P(black ball)=3/12

P(red ball)=5/12

P(black or red)=3/12+5/12=2/3

Let the numbers be x, x+1, x+2 

Then x + x+1 + x+2 = 42

           =>  3x + 3 = 42

           => 3x = 39 

          =>  x = 13

The numbers would be 13, 14, 15 

Therefore, Sum of the digits of middle number = 1+4 = 5

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways