Quantitative Aptitude - Arithmetic Ability Questions

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.

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

A's 1 days work = 1/12 

B's 1 days work =1/6

Therefore, (A+B)'s 1 day's work= 1/12 + 1/6 = 1/4       

Therefore, Time taken by both to finish the whole work = 114 = 4 days

Alternatively :

Efficiency of A =100/12 =8.33%

Efficiency of B =100/6=16.66%

Combined efficiency of A and B both = 8.33+16.66=25%

Therefore, Time taken by both to finish the work (working together) =100/25= 4days

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

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.

Average speed=total distance/total time
Let distance to store be  K
then, total time =(K/20)+(K/30)=K/12
and, total time =(2K)
so average speed= 2K / (K/12) = 24kmph.