Quantitative Aptitude - Arithmetic Ability Questions

10C5*28C1*10C6 = 7266

1/3 ---- 8

1 -------?
Hari can do total work in = 24 days

As satya is 60% efficient as Hari, then

Satya = 1/24 x 60/100 = 1/40

=> Satya can do total work in 40 days

1 ----- 40

2/3 ---- ? => 26 2/3 days.

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

  

 A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
 B = { 7, 14, 21, 28 }, n(B)= 4 

 

Since A and B are not mutually exclusive So :

Area of the park = (60 x 40) = 2400 sq.m  

Area of the lawn = 2109  sq.m 

Area of the crossroads = (2400 - 2109) = 291 sq.m   

Let the width of the road be x metres. Then,   

60x + 40x - (x * x) = 291  

=>(x * x) - 100x + 291 = 0    

=>(x - 97)(x - 3) = 0  

=>x=3

Suppose total earnings = Rs. 10000
10% saving = 10000 x (10/100) = Rs. 1000
This year 5% more income than prev year = Rs. 10500
12% saving = 10500 x (12/100) = Rs. 1260
Percentage of amount saved = (1260/1000) x 100 = 126%

Given Word is PROMISE.

Number of letters in the word PROMISE = 7

Number of ways 7 letters can be arranged = 7! ways

Number of Vowels in word PROMISE = 3 (O, I, E)

Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways

Now, the number of ways of arrangements so that three vowels should not come together

= 7! - (5! x 3!) ways = 5040 - 720 = 4320.

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

Let A be the event of driver A drive safely after consuming liquor. 

Let B be the event of driver B drive safely after consuming liquor. 

Let C be the event of driver C drive safely after consuming liquor. 

P(A)=2/5,P(B)=3/7,P(C)=3/4

The events A, B and C are independent . Therefore,

PA∩B∩C=P(A)P(B)P(C)=25*37*34=970

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70