Quantitative Aptitude - Arithmetic Ability Questions

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.

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

The square of a natural number never ends in 7.
42437 is not the square of a natural number

Let the Required cost be X paise.  Less Weight , Less cost (Direct Proportion)

Therefore, 250 : 200 :: 60 : x

=> (250 * x) = (200 * 60)

=> x= 48

Let l = 9 ft.

Then l + 2b = 37

=> 2b = 37 – l = 37 – 9 = 28

b = 28/2 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

nA=6, nB=5, nA∩B=1

∴Required probability = PA∪B

 = PA+PB-PA∩B

=  636+536-136 = 518

Let the number of failed students be x

=> Number of passed students = 4x 

So total number of students was 5x

From the given data,

If total number of students be 5x – 35

=> 4x-35-9/x+9 = 2/1

=> 4x – 44 = 2(x + 9)

=> 4x – 2x = 18 + 44

=> x = 31

Total number = 31×5

= 155