Quantitative Aptitude - Arithmetic Ability Questions

In this type of problems the formuae is
(5*x+ or - t)*12/11
Here x is replaced by the first interval of given time. Here x is 5.
t is spaces apart

  Case 1 : (5*x + t) * 12/11

 (5*5 + 3) * 12/11

 28 * 12/11 = 336/11= 31511 min

therefore the hands will be 3 min apart at 31 5/11 min past 5.

Case 2 : (5*x - t) * 12/11

(5*5 -3 ) * 12/11

 22 *12/11 = 24 min 

 therefore the hands will be 3 min apart at 24 min past 5

Let there be x pupils in the class.

Total increase in marks = (x * 1/2) = x/2.

x/2 = (83 - 63) => x/2 = 20 => x = 40.

(2000 + 3)(2000 + 4) - (2000 + 1)(2000 + 2) = ?
Since (2000 x 2000) - (2000 x 2000) is equal to zero. ?
= (8000 + 6000 + 12) - (4000 + 2000 + 2)
=> 14012 - 6002 = 8010.

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 

Jan   Feb   Mar   Apr   May   Jun   Jul  

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 

Total number of odd days = (0 + 2) = 2.  

Required day was 'Tuesday'.

Step (i): Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100 -------------- (i)

Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x + 2.40y = 312 -------------- (ii)

Step (iii):
Solving (i) and (ii)
3.60x + 3.60y = 360 --------- Multiply (i) by 3.60
3.60x + 2.40y = 312 --------- (ii)
1.20y = 48
y = 48 / 1.20
= 40

 Number of girls = 40

From the given data,

$$\begin{gathered} 40100 x 75 + 80100 x 25 = K100 x 250 \\ 30 + 20 = 5K2 \\ 50 x 2 = 5K \\ K = 1005 = 20 \end{gathered}$$

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind 

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.

 Hence, price of the third variety = Rs.175.50 per kg.

By the rule of alligation:

Profit of first part                         Profit of second part

So, ratio of 1st and 2nd parts = 4 : 6 = 2 : 3. 

Therefore, Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.