Quantitative Aptitude - Arithmetic Ability Questions

Let x and y be the respective km's travelled by man via taxi and by his own car.
Given x + y = 90 => x = 90 - y
But according to the question,
7x + 6y = 595
7(90-y) + 6y = 595
=> 630 - 7y + 6y = 595
=> y = 630 - 595 = 35
=> x = 90 - 35 = 55
Therefore, the distance travelled by taxi is 55 kms.

Let the number of sides be n. 

The number of diagonals is given by ⁿC₂ - n  

^{Therefore, n}C₂ - n = 44, n>0 

ⁿC₂ - n = 44

n² - 3n - 88 = 0 

n² -11n + 8n - 88 = 0  

As n>0, n will not be -8. Therefore, n=11.

$$\begin{gathered} ?2=400+16650x100 \\ ?= 8 \end{gathered}$$

From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of  interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.

when the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 20 liters of milk.

 Therefore, profit percentage = (5/20) x 100 =25 %       [ Profit %= (Profit/C.P)*100]

Since S.P. = 5 (M.P. - S.P.) => 5 M.P. = 6 S.P.
M.P. = (6/5) S.P.
Since the percentage discount = Percentage profit,
65S.P - S.P65S.P×100 = S.P -C.PC.P×100
Therefore, 1/6 = (S.P/C.P) – 1 => S.P = (7/6) C.P
M.P = 6/5 S.P = 6/5 x 7/6 C.P = 7/5 C.P
Therefore, Ratio of discount to C.P
= (M.P – S.P)/C.P = [{(7/5)C.P - (7/6)C.P} / C.P.]
= 7/30

Therefore, the ratio is 7 : 30

Assume A be the cost price.

Therefore,
35×A×10100 - 25×A×5100 = 1500

=>  A = Rs 37,500.