Speed of the bus excluding stoppages = 54 kmph.

Speed of the bus including stoppages = 45 kmph.

Loss in speed when including stoppages = 54 - 45 = 9kmph.

=> In 1 hour, bus covers 9 km less due to stoppages

Hence, time that the bus stop per hour = time taken to cover 9 km.

=distance / speed= 9 / 54 hour = 1/6 hour = 60/6 min=10 min.

Work done by A in 1 day = 1/16

Work done by B in 1 day = 1/12

Let C takes x days to lay railway track between two given stations.

So work done by C in 1 day = 1/x

A, B and C together take 4 days to lay the railway track.

So, in 1 day they complete 14 of the total work together

So, 1/16 + 1/12 + 1/x = 1/4

=> 1/x = 1/4 - 1/16 - 1/12

=> 1/x = 5/48

=> x = 48/5.

A completes the work in 6 days.

So, Work done by A in 1 day = 1/6.

=> A's 3-day work = 3/6 = 1/2.

B completes the work in 8 days

So, Work done by B in 1 day = 1/8

=> B's 3-day work = 3/8

The remaining work is done by C.

So the part of work done by C = 1 - (1/2+ 3/8) = 1/8.

So, C's share is 18 of 3200 = 3200 * 1/8 = 400.

N X 50 = (325000 - 300000) = 25000

N = 25000 / 50

= 500.

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back.

From this, we can understand that time needed for riding one way = time needed for waking one way - 2 hours.

Given that time taken in walking one way and riding back = 5 hours 45 min

Hence the time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min.

Let the capital be Rs. x.

Then, = (x × 8×1/100) - (x × 31/4×1/100)

= 61.50.

‹=›32x - 31x =6150×4

‹=›x= 24600.

Cost price = Rs. 56000

Profit % = 25 %

Selling price = [(100 + Gain %)/100] * Cost price

Selling price = (125 /100) x 56000

Selling price = 70,000.

Other number = (11×7700 / 275)

= 308.

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

‹=› 10_{p4}

‹=› (10 x 9 x 8 x 7)

‹=› 5040.

Let the number of students appeared be x.

Then, 65% of x = 455

‹=›65 / 100 x = 455

‹=› x= [455 x 100 / 65]

= 700.

Let the numbers be x and y.

Then, x + y = 20 and x - y = 8.

Therefore, x2 - y2 = (x + y)(x - y)

‹=› 20 x 8

‹=› 160.

Let the C.P. of the plot = Rs. x.

Given that x - 15% of x = 18700.

=> (85/100) * x = 18700.

=> x = 22000.

So the cost price (C.P.) of the plot = Rs. 22000.

S.P. of plot for 15% profit = 22000 + 15% of 22000 = 22000*(115/100) = 25300.

Let the C.P. be x and the S.P. be y

So the profit is (y - x) ---------------------------- (1)

Now, the S.P. is doubled. So the new S.P. = 2y

The new profit = (2y - x)

Given that when S.P. is doubled, profit increases 3 times

=> New profit = 3 * old profit

=> (2y - x) = 3(y - x)

=> y = 2x

So, the profit = (y - x) = (2x - x) = x

% profit = (x / x) * 100 % = 100%.

Let the numbers be x and (22 - x).

Then, 5x = 6(22 - x)

‹=› 11 x = 132

x = 12.

So, the numbers are 12 and 10.

Required ratio = (6x1x1 / 6x5x5)

‹=›1/ 25

‹=›1: 125.

Speed of the train = (30 x 5/18) m/sec

= (25 / 3) m/sec.

Distance moved in passing the standing man = 100 m.

Required time taken = 100 / (25 / 3)

= (100 x 3 / 25) sec

= 12 sec.

Surface area = (34398 / 13)

‹=›2646cm3

‹=›6a2= 2646

‹=›a2= 441

‹=›a = 21.

So, volume = (21x21x21) cm3= 9261cm3.

Work done by the leak in 1 hour = (1/2 - 3/7)

‹=›1/14.

Leak will empty the tank in 14 hours.

Let the speed of the train be x km/hr.

Given, Speed of the man = 5 km/hr.

Since both the train and man are moving in the same direction so the speed of the train (relative to man) would be (x-5) km/hr.

Length of the train = 125 m = 125/1000 km = 18 km

Time taken to cross the man = 10 sec = 10/3600 hrs. = 1/360 hrs.

So speed =distance/time.

=> x-5 =360/ 8.

=> 8x - 40 = 360.

=> 8x = 400.

=> x = 50 km/hr.

Q20. If A Number X Is 10% Less Than Another Number Y And Y Is 10% More Than 125, Then X Is Equal To?

y = 125 + 10% of 125

= 125 + 12.50

= 137.50.

x = 137.50 - 10% of 137.50

= 137.50 - 13.75

= 123.75.

Clearly, 9261 is a perfect cube satisfying the given property.

(.000216)^{1/3} = (216 / 10^{6})^{1/3}

= (6 x 6 x / 10^{2} x 10^{2} x 10^{2})^{1/3}

= 6 / 10^{2}

= 6 / 100

= .06

At the time of meeting, let the distance traveled by the first train be x km.

Then, distance covered by the second train is (x + 70) km.

=> x/60 = (x + 70)/75.

=> 75x = 60x + 60*70.

=> 15x = 4200 = x = 280 km.

So, distance between A and B = (x + x + 70)

= 280 + 280 + 70

= 630 km.

Let B takes x days to complete the job.

Since A is 3 times better than B, A takes one third of the time taken by B.

So, A can finish the job in x/3 days.

Given that time taken by A to complete the work is 60 days less than that of B

So, x/3 = x - 60.

=> x = 90.

So, B does the job in 90 days and A does in (90-60) days = 30 days.

Fraction of job done by A in 1 day = 1/30 and fraction of job done by B in 1 day = 1/90.

When working together, work done by A and B in 1 day = 1/30 + 1/90 = 2/45.

So, total time taken by A and B to complete the work together = 45/2.

Let the number of boys = x.

Then, number of girls = x.

Now, 2(x - 8) = x.

x = 16.

Total number of students = 2x = 2 x16 = 32.

Let the rate be R% p.a.

Then, (5000xRx2/100) + (3000xRx4/100)

‹=›100R+120R= 2200

‹=›R= (2200/220)

Rate ‹=›10%.

P (getting a prize) = 10 / (10+25)

‹=› 10 / 35

‹=› 2 / 7.

Let the cost price of 1 article be z.

So, the cost price of 20 article = 20z ------------------------- (1)

Selling Price of 20 articles = 20z + 25% of 20z = 25z

=> Selling Price of 1 article = 25z/20 = (5/4) *z

=> Selling Price of x articles = (5/4) *z*x ------------------------- (2)

Given that Selling Price (S.P.) of x articles = Cost Price (C.P.) of 20 articles

=> (5/4) *z*x = 20z

=> x = 16.

Required numbers = (L.C.M of 12, 16, 18, 21, 28) + 7

‹=›1008 + 7

= 1015.

Part filled by (A+B+C) in 1 hour = (1/5 + 1/6 + 1/30)

‹=› 1/3.

All the three pipes together will fill the tank in 3 hours.

Relative Speed = (280 / 9) m/sec

= (280/9 x 18/5)

= 112 kmph.

Speed of the train = (112 - 50) kmph

= 62 kmph.

Let the speeds of the two trains be a m/sec and b m/sec respectively.

Then, length of the first train = 27a meters,

and length of the second train = 17b meters.

so time = distance/ speed.

=> 23 = (27a+17b) / (a+b).

=> 23a + 23b = 27a +17b

=> 23a - 27a = 17b - 23b

=> -4 a = -6 b

=> 2 a = 3 b

=> a/ b= 3/2.