Quantitative Aptitude - Arithmetic Ability Questions

(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys

 =>5 men= 10 boys    => 1man= 2 boys

 Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys

( 8 men + 6 boys) = (8 x 2 +6) boys = 22 boys.

 Let the required  number of days be x.

 More boys , Less days     (Indirect proportion)

 More work , More days    (Direct proportion)

 Boys22:11Work1 : 3⋮⋮ 14:x

 Therefore, (22 * 1 * x) = (11 * 3 * 14)

=> x = 21 

Hence, the required number of days = 21

Total number of students = 1100 + 700 = 1800.

Number of students passed = (42% of 1100 + 30% of 700) - (462 + 210) = 672.

Number of failues = 1800-672 = 1128.

Percentage failure = (1128/1800 * 100 )% = 62 * 2/3 %. 

 Lot the total number of workers be v Then,

8OOOv = (12000 * 7) + 6000 (v - 7) <=>  2000v = 42000  <=>  v = 21.

AC-AB<BC Or AB+BC>AC

BC-AC<AB Or AB+AC>BC

AB-BC<AC Or AC+BC>AB

Sum of 2 sides of triangle is always greater than the third side

So all three statements are true

Area of 4 walls = 2(l+b)h
=2(10+7) x 5 = 170 sq m
Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m

Cost of painting = Rs. 158 x 3 = Rs. 474

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

= 6*6*6*6=64

n(S) = 64

Let X be the event that all dice show the same face. 

X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}

n(X) = 6

Hence required probability = n(X)n(S)=664 =1216