Quantitative Aptitude - Arithmetic Ability Questions

I. 2x2 + 19x + 42 = 0

2x2 + 12x + 7x + 42 = 0

2x (x+6) + 7(x +6) = 0

(2x+7) (x+6) = 0 x = -7/2, -6 I

I. 4y2 + 43y + 30 = 0

4y2 +40y + 3y + 30 = 0

4y(y + 10) + 3(y + 10) (4y + 3)(y + 10) = 0

y = -3/4, -10

So answer is no relation.

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Total volume of water displaced =(4 x 50) cu.m = 200 cu.m

Rise in water level = 20040×20=0.25m = 25cm

Let the original number be x.

Final number obtained = 110% of (90% of x) =(110/100 * 90/100 * x) = (99/100)x. 

x-(99/100)x=10

=> x =1000

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.

Let the required length be x meters

 More men, More length built     (Direct proportion) 

Less days, Less length built      (Direct Proportion)

Men20:35Days6 : 3⋮⋮ 56 :x 

=> (20 x 6 x X)=(35 x 3 x 56) 

=> x = 49 

Hence, the required length is 49 m.

Given numbers are 1.08 , 0.36 and 0.90

 H.C.F of 108, 36 and 90 is 18                  [ ∵ G.C.D is nothing but H.C.F]

 Therefore, H.C.F of given numbers = 0.18                 

Explanation:

Let rate = R% and time = R years.

Then, 1200*R*R100=432

12R2=432

=>R2=36⇔R=6