Quantitative Aptitude - Arithmetic Ability Questions

The given number series is 3, 11, 80, 291, 803

$$\begin{gathered} 3 \\ 3 + 23 = 3 + 8 = 11 \\ 11 + 43 = 11 + 64 = 75 (not 80) \\ 75 + 63 = 75 + 216 = 291 \\ 291 + 83 = 291 + 512 = 803 \end{gathered}$$

Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m

Cost Price = Rs.100125×8750 = Rs. 7000. Let the labelled price be Rs. x. 

Then, 70100*x=7000 ⇒x=Rs.7000*10070 

= Rs.10000

Here the given number series is 0, 3, 8, 15, 24, ?

The next number in the given series can be found such a way that the number series follows a pattern,

$$\begin{gathered} 22 - 4 = 0 \\ 32 - 6 = 3 \\ 42 - 8 = 8 \\ 52 - 10 = 15 \\ 62 - 12 = 24 \\ 72 - 14 = 35 \\ Hence, the next number in the iven series is 35. \end{gathered}$$

M x T / W = Constant
where, M= Men (no. of men)
T= Time taken
W= Work load
So, here we apply
M1 x T1/ W1 = M2 x T2 / W2
Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?
Note that here, W1 = W2 = 1 road, ie. equal work load.
Clearly, substituting in the above equation we get, M2 = 14 men.

Let C.P be Rs. x 

900 - x = 2(x - 450)     => x = Rs.600   

C.P = 600 gain required is 25% 

S.P = [(100+25) x 600] / 100 = Rs.750

I. gives, S.I for 3 years = Rs. 4500.

II. gives, Rate = 10% p.a.

III. gives, (C.I.) - (S.I.) = Rs. 465.

Clearly, using I and III we get C.I. = Rs. (465 + 4500).

Thus, II is redundant.

Also, from I and II, we get sum =

100 x 4500

= 15000.

10 x 3

Now C.I. on Rs. 15000 at 10% p.a. for 3 years may be obtained.

Thus, III is redundant.

Either II or III is redundant.