Quantitative Aptitude Prep Tips for MBA

The bag has 25-paise, 50-paise and 1 -rupee coins in the ratio 8 : 2 : 7. Now this is the ratio of the number of coins while the total value of the coins has been given. So we first need to convert coins into value.

Let the coins be 8x, 2x and 7x respectively.

8x coins of 1/4 rupee each = value is Rs. 2x.

2x coins of 1/2 rupee each = value is Rs. x.

7x coins of 1 rupee each = value is Rs. 7x.

Now

2x + x + 7x = 770

10x = 770 x = 77

Number of 25 paise and 50 coins = 10 x = 770 coins

Let daily working hours & wages per hour before increase be H and W respectively

So his Salary/day = HW

Working hours after increase = 0.82H, wages per hour = 1.45W

So his daily salary = $\frac{82}{100}H*\frac{29}{20}W=\frac{1189}{1000}HW$ $\frac{}{}\frac{}{}\frac{}{}$ . Thus an Increase of 18.9%

A + B + C = 18500…. (1)

A = B + 25% of   $B=\frac{5B}{4}$ $\frac{}{}$ …. (2)

B = C + 20% of $C=\frac{6C}{5}$ $\frac{}{}$ …. (3)

Equation (1) gives,

$\frac{5B}{4}+\frac{6C}{5}+C=18500$

$\Rightarrow \frac{3C}{2}+\frac{6C}{5}+C=18500$

37C = 185000 C = 5000

B = 6000, A = 7500.

A + B = 13500

Let the total work by LCM of 20, 24 and 30 i.e. 120 units

Work done per day by A and B = a + b = $\frac{120}{20}=6$ $\frac{}{}$

Work done per day by B and C = b + c = $\frac{120}{24}=5$ $\frac{}{}$

Work done per day by A and C = a + c = $\frac{120}{30}=4$ $\frac{}{}$

2 (a + b + c) = 15

So, a + b + c = 7.5

Required number of days = $\frac{120}{7.5}=16\text{?}days$

Let the numbers be x & y

(mean proportion)2 = xy.

So, xy = 784

x = $\frac{784}{y}$ $\frac{}{}$ .… (1)

x, y and 224 are in continued proportion

y2 = 224 x .… (2)

y2 = 224 * $\frac{784}{y}$ $\frac{}{}$

y3 = 14 * 16 * 28 * 28

y = 56 and x = 14

Alternate Method

Use options and check.

Let the speeds be 2Aand A kmph respectively

$\frac{216}{3A}=6$

A = 12 and the speed of B = 3 kmph

$\frac{\left(28-25\right)}{\left(45-28\right)}=\frac{3}{17}$

Rahul can occupy any one of the two corner positions and therefore, Rahul can occupy a position in two ways.

The other four people can seat themselves in 4! ways, that is, 24 ways.

Total ways = 2 * 24 = 48 ways.

As per the problem, we have Length of the first train = 120 m

Length of the second train = 180 m

As per the problem,

$\frac{(120+180)}{(8+s)}=5$

8 + s = 60

s = 52 m/sec