Quantitative Aptitude - Arithmetic Ability Questions

Let the number of boys = b

Let the number of girls = g

From the given data,

81b + 83g = 81.8(b + g)

81.8b - 81b = 83g - 81.8g

0.8b = 1.2g

b/g = 1.2/0.8 = 12/8 = 3/2

=> g : b = 2 : 3

Hence, ratio between the number of girls to the number of boys = 2 : 3.

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

Watson bought the book for Rs. 240 and sold to Johny at a profit of 50%.

S.P = C.P(1 + P%/100)

=> S.P for Watson = C.P for Johny = 240(1 + 50/100) = 240 x 1.5 = Rs. 360

Let Johny quoted the marked price of the book as Rs. M

We know, SP = M.P(1 - Discount(%)/100)

Here discount = 10% to Shekar,

S.P for Johny = M(1 - 10/100) = 0.9M

But Johny want to earn 25% profit,

=> S.P = C.P(1 + P%/100)

=> 0.9M = 360(1 + 25/100)

=> M = (360x1.25)/0.9

=> M = Rs. 500

Therefore, Johny should quote Rs. 500 as the marked price of the book to get 25% profit and allowing 10% discount to Shekar.

Using Pythagoras theorem we can see that 12² + 15² is not equal to 18²
Neither is 10² + 24² = 25²

SP of first article = 1000
Profit = 30%
CP = (SP)x[100/(100+P)] = 10000/13

SP of second article = 1000
Loss = 20%
CP = (SP)x[100/(100-L)] = 5000/4 = 1250

Total SP = 2000

Total CP = 10000/13 + 1250 = 2019.23

CP is more than SP, he makes a loss.

Loss = CP-SP = 2019.23 - 2000 = 19.23
Loss Percent = [(19.23)/(2019.23)]x100 = 0.95 %.

Reduction in consumption
= P100+P X 100%
= 25100+25 X 100%
= 20%

i=j/m

PV=
FV(1+  i)^-n

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.