Quantitative Aptitude - Arithmetic Ability Questions

Can be easily solved with following formula

a2-b2=a-b×a+b

= (25.732 - 15.732 ) x (25.732 + 15.732)

=414.64

Total numbers = 25

 Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9

 The probability (divisible by 4 or 7) = 9/25

Digits are 0,1,2,3,4,5,6,7,8,9. So no. of digits are 10

First all possible case => 9(0 excluded) x 10 x 10 = 900

Second no repetition allowed =>9 x 9 x 8 = 648

Third all digits are same => 9 (111,222,333,444,555,666,777,888,999)

Three digit numbers where two of the three digits are same = 900 - 648 - 9 = 243 ;

Given (3 Men + 4 Women + 6 Children) -----> 9 days

But W = 2M and C = M/2

Now, convert Men and Children into Women by

$$\begin{gathered} 3W2 + 4W + 6W4 \\ => 32 + 4 + 32 = 7 Women \end{gathered}$$ 

Therefore, 7 women alone can complete this work in 9 days.

Let us first calculate the cost of price of the mixture. the selling price of the mixture is given as Rs.96 and the profit is given as 20%. So, Cost price = 100120×96 = Rs. 80

Now, let us apply allegation method :

= 40 : 20 =  2 : 1

Ayush's present age = 10 years.
His mother's present age = (10 + 20) years = 30 years.
Ayush's father's present age = (30 + 5) years = 35 years.
Ayush's father's age at the time of Ayush's birth = (35 - 10) years = 25 years.

Therefore Ayush's father's age at the time of marriage = (25 - 2) years = 23 years.

Sum =Rs.(50 x 100/2x5)

= Rs. 500.

Amount

=[Rs.500x(1+5/100)²]

=Rs(500x21/20x21/20).

=Rs. 551.25

C.I

= Rs. (551.25 - 500)

= Rs. 51.25

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =122×129=1211