Quantitative Aptitude - Arithmetic Ability Questions

By replacing the “?” with “x”

(12.5×14÷8.75) x (100/100) + 42 = 50 + X

(125*140)/875 + 42 = 50 + X

140/7 + 42 = 50 + X

20 + 42 = 50 + X

62 = 50 + X

X = 62 - 50

X = 12

Let the cost price of 1-liter pure milk be Re.1, then 

 6 liters (milk) → C.P = Rs. 62 liters (water) → C.P = Rs. 0→CP = Rs.6 only

8 litre mixture  =>

SP =>  8 x 2 = Rs. 16

Profit % = 16 - 66x100 = 10006 = 166.66%

Given number sequence is 28, 16, 28, 76, 274

Here the sequence follows a pattern that

28

28 x 0.5 + 2 = 14 + 2 = 16

16 x 1.5 + 4 = 24 + 4 = 28

28 x 2.5 + 6 = 70 + 6 = 76

76 x 3.5 + 8 = 266 + 8 = 274

274 x 4.5 + 10 = 1233 + 10 = 1243

Hence, the next number in the given number sequence is 1243.

Let there be x voters and k votes goes to loser then 

      0.8x - 120 = k + (k + 200)

         k+200 = 0.41x

 => k = 1440

and  (k + 200) =1640

Therefore  1440/3200 x 100 = 45%

Each of the integers in the first set could be combined with any from the second set, giving a total of 4 x 5 = 20 possible pairs.

Of these the combinations that could give a sum of 23 are (5 + 18), and (9 + 14)

This means that the probability of getting a sum of 23 is 2/20 = 1/10

Here the given series is 123 112 133 112 143

It follows a pattern that,

123

123 - 11 = 112

112 + 21 = 133

133 - 31 = 102 (not 122)

102 + 41 = 143

Hence, the odd man in the given number sequence is 122 instead of 102.

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.