Quantitative Aptitude for Competitive Exams
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000
Length of the rectangle is halved
=> New length = $\dfrac{\text{Original length}}{2}=\dfrac{100}{2}=50$
Breadth is tripled
=> New breadth= Original breadth × 3 = 100 × 3 = 300
New area = 50 × 300 = 15000
Increase in area = New Area - Original Area = 15000 - 10000= 5000
Percentage of Increase in area = $\dfrac{\text{Increase in Area}}{\text{Original Area}} \times 100$
$=\dfrac{5000}{10000} \times 100=50\% $
Total weight of students in division A = $ 36 \times 40$
Total weight of students in division B = $ 44 \times 35$
Total students = 36 + 44 = 80
Average weight of the whole class
$= \dfrac{\left(36 \times 40\right)+\left(44 \times 35\right)}{80} $
$= \dfrac{\left(9 \times 40\right)+\left(11\times 35\right)}{20} $
$= \dfrac{\left(9 \times 8\right)+\left(11\times 7\right)}{4} $
$= \dfrac{72+77}{4}\\$
$= \dfrac{149}{4}\\$
=$37.25$
Sum of the present ages of husband, wife and child = $\left(27 \times 3 + 3 \times 3\right)$ years = 90 years.
Sum of the present ages of wife and child = $\left(20 \times 2 + 5 \times 2\right)$ years = 50 years.
$\therefore$ Husbands present age = (90 - 50) years = 40 years.
| Total quantity of petrol consumed in 3 years | =$ \left(\dfrac{4000}{7.50} +\dfrac{4000}{8} +\dfrac{4000}{8.50} \right) $ litres |
| = 4000$ \left(\dfrac{2}{15} +\dfrac{1}{8} +\dfrac{2}{17} \right) $ litres | |
| =$ \left(\dfrac{76700}{51} \right) $ litres |
Total amount spent = Rs. $\left(3 \times 4000\right)$ = Rs. 12000.
| $\therefore$ Average cost = Rs.$ \left(\dfrac{12000 \times 51}{76700} \right) $= Rs.$ \dfrac{6120}{767} $ = Rs. 7.98 |
Let x is the length of the train and v is the speed
Time taken to move the post = 8 s
=> x/v = 8
=> x = 8v --- (1)
Time taken to cross the platform 264 m long = 20 s
(x+264)/v = 20
=> x + 264 = 20v ---(2)
Substituting equation 1 in equation 2, we get
8v +264 = 20v
=> v = 264/12 = 22 m/s
= 22×36/10 km/hr = 79.2 km/hr
Length of the train = distance covered in crossing the post = speed × time = speed × 18
Speed of the train = 300/18 m/s = 50/3 m/s
Time taken to cross the platform = 39 s
$\left(300+x\right)$/(50/3) = 39 s where x is the length of the platform
300+x = (39 × 50) / 3 = 650 meter
x = 650-300 = 350 meter
Required H.C.F. =$ \dfrac{H.C.F. of 9, 12, 18, 21}{L.C.M. of 10, 25, 35, 40} $=$ \dfrac{3}{1400} $
Let the numbers be $ a $ and $ b $.
Then, $ a $ + $ b $ = 55 and ab = 5 x 120 = 600.
$\therefore$ The required sum =$ \dfrac{1}{a} $+$ \dfrac{1}{b} $=$ \dfrac{a + b}{ab} $=$ \dfrac{55}{600} $=$ \dfrac{11}{120} $
Given that,
True discount = dollar 200
Simple interest = dollar 240
Time = 3 years
Consider,
Sum due=$\dfrac{(S.I.) \times (T.D.)}{(S.I.) - (T.D.)}$
=>Sum due=$\dfrac{(240) \times (200)}{(240) - (200)}$
⇒ Sum due = dollar. 1200
Now,
Rate = $\dfrac{(100) \times (S.I.)}{(Sum due ) \times (T)}$
⇒ Rate = $\dfrac{(100) \times (240)}{(1200 ) \times (3)}$
⇒ Rate = $\dfrac{24000}{3600}$
⇒ Rate = 6.66 %
Therefore, Sum due = dollar. 1200 and Rate = 6.66 %