Quantitative Aptitude - Arithmetic Ability Questions

To solve a pair of simultaneous equations such as those given we can add or subtract them.

Adding we get 4x + 4y = 20

Therefore 2x + 2y = 10

K : L : M = 100 : 45 : 30
              = 20 : 9 : 6

Given L share is 27

=> 9  ------  27
      35 ------  ?

      => 105

Let the total length of the pencil = x cms

From the given data,

'x/8' is black

Remaining part = x - x/8 = 7x/8

1/2 of 7x/8 is white

=> 7x/16 is white (Remaining part is 7x/16)

And given that the remaining part is blue and is 3 1/2 cm = 7/2 = 3.5 cm

That remaining part = 7x/16

Therefore,

7x/16 = 7/2

x = 16/2 = 8

Hence, the total length of the pencil = x = 8 cm.

Let the price of each note book be Rs.x.

Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.

Hence xy = 300

=> y = 300/x

(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy

=>5(300/x) - 10x - 50 = 0 => -150+x2+5x

multiplying both sides by -1/10x

=> x2+15x-10x-150=0

=> x(x + 15) - 10(x + 15) = 0

=> x = 10 or -15

As x>0, x = 10.

   I. gives, Rate = 5% p.a.

 II. gives, S.I. for 1 year = Rs. 600.

III. gives, sum = 10 x (S.I. for 2 years).

One hour = 60 min

18/60 = 3/10

Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15

Therefore, 15 sweets were each student originally supposed to get.

Distance = 70 x 1 ½ = 105 km

Relative Speed = 85 – 70 = 15

Time = 105/15 = 7 hrs

4:30 + 7 hrs = 11.30 p.m.