Explanation: Given that,

Total distance covered by man in 2 hours = 50km.

in first hour, he covers = 20km.

So,

Distance left to be cover = 50 - 20 = 30km.

Time left = 2 - 1 = 1 hour.

now, given that, he had to stop for 10 minutes for refuelling.

So,

Time left for travel = 1 hour - 10 min. = 60 - 10 = 50 minutes.

Therefore,

Distance to be covered now = 30km.

Time left = 50 minutes = (50/60) = (5/6) hours.

New speed = Distance/Time = 30/(5/6) = 30 * (6/5) = 36 km/h.

Hence,

 Speed in first hour = 20km/h .

 speed in second time = 36km/h.

 Increased speed in factor = (36 /20) = 1.8 times. (Ans.)

So, he has to increased his speed 1.8 times in order to complete the journey as per schedule time.

Explanation:

Explanation: we know that,

If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

A number is divisible by 8 if the last three digits are evenly divisible by 8.

If the number is separately divisible by 3 and 8 then the number is also divisible by 24.

So, we can conclude that, if given 6 digit number is divisible by 24, it must be divisible by 3 and 8.

checking by 8 first we get: -

352/8 = 44 quotient, 0 remainder.

So, it is divisible.

now, in order to divisible by 3, sum must be divisible by 3.

So,

(1 + n + n + 3 + 5 + 2) / 3 = 0 remainder.

(11 + 2n) / 3 = 0 remainder.

Putting values of n now, we get,

if n = 0 => 11 /3 = Remainder not equal to 0.

if n = 1 => (11 + 2)/3 = Remainder not equal to 0.

if n = 2 => (11 + 4)/3 = Remainder equal to 0.

if n = 3 => (11 + 6)/3 = Remainder not equal to 0.

if n = 4 => (11 + 8)/3 = Remainder not equal to 0.

if n = 5 => (11 + 10)/3 = Remainder equal to 0.

if n = 6 => (11 + 12)/3 = Remainder not equal to 0.

if n = 7 => (11 + 14)/3 = Remainder not equal to 0.

if n = 8 => (11 + 16)/3 = Remainder equal to 0.

if n = 9 => (11 + 18)/3 = Remainder not equal to 0.

Therefore,

→ sum of Possible values of n = 2 + 5 + 8 = 15 (Ans.)

Explanation: Let the MP of 2/7 and 32/343 be x

x = √(2/7)*(32/343)

= √(64/4901)

= (8/49)

Let the MP of 2 and 1/5000 be y

y = √(2)*(1/5000) = √1/2500 = 1/50

MP of x and y = √(8/49)*(1/50) = √8/2450

= √4/1225 = 2/35

Explanation: (0.025/100) * (240/100) * 1.5

= 1/4000 * 24/10 * 1.5

= 9/10000

Let x% of 0.9 be 9/10000

x% (0.9) = 9/10000

(x/100) (0.9) = 9/10000

x = (9/10000)*(100/0.9)

x = 0.1

A. 10.5%

B. 10%

C. 8%

D. 12.5%

Explanation: Let initial number of customer's are x .

So,

Initial cost of filling a gas tank = Rs.800

Initial number of customer's = x

then, Initial total revenue = 800 * x = Rs. 800x

Now,

New cost of filling a gas tank = 15% reduced = (85 * 800)/100 = Rs. 680

New number of customer's = 30% increases = (130 * x)/100 = 1.3x

then, New total revenue = 680 * 1.3x = Rs. 884x

therefore,

Revenue increased by = 884 - 800 = Rs. 84x

hence, Revenue increased in % = (84x * 100)/800x = 10.5% 

A. 20%

B. 40.9%

C. 130.9%

D. 44%

Explanation: Here it is given that Ankush bought (x+2) apples at the rate of Rs. 12 each.

So total cost price = Rs 12(x + 2)

Now one apple got rotten during transportation

So, number of apples sold = x + 2 - 1 = x + 1

Total selling price = Rs 300

Profit percentage = 25%

So, by the given condition,

12(x+2) * 125/100 = 300.

X = 18

Hence the number of apples sold by Ankush

= 18 + 1

= 19

A. 245

B. 243

C. 220.30

D. 244.80

Explanation: Let CP of article = Rs. 100x

MP = 36% above CP = (136 * 100x)/100 = Rs. 136x

Discount = 10%

SP = MP * (100 - D)/100 = (136x * 90)/100 = Rs. 122.4x

now,

Another discount = Rs.12.60

SP after another discount = Rs. (122.4x - 12.6)

According to the question,

Total Profit = 15.4%

122.4x - 12.6 = (100x * 115.4)/100

122.4x - 12.6 = 115.4x

7x = 12.6

x = 1.8

therefore,

MP of article = 136x = 136 * 1.8 = Rs. 244.80 

Explanation: 25% of x: 65% of y : 70% of z = 5 : 26 : 15

(25 * x)/100 : (65 * y)/100 : (70 * z)/100 = 5 : 26 : 15

5x:13y:15z = 5:26:15

5X = 5a

x = a

and 13Y = 26a

Y = 2a

and, 15Z = 15a

Z = a

also,

(1/16) (x + y + z) = 10

(x + y + z) = 160

a+2a+a = 160

a=40

therefore, x = 40, Y = 80 and Z = 40

(x - 2y + 3z) = 40 -160 +120 = 0

A. 9km

B. 7.5km

C. 8km

D. 4.5km

Explanation: Let saving = 40, expenditure = 100 hence income = 140

Income increases by 20% = 168, expenditure increases by 40% = 140, saving = 28

% decrease = 40-28 /40 * 100 = 30 %

Explanation: Profit = 20%

Hence 120% of cost price = 3240

Cost price = 2700

New selling price = 2781/- , profit = 81

% profit = 81 / 2700 * 100 = 3%

Explanation: Let there be 100 apples with a cost price of 100 and cost amount=100*100=10000

Sale amount at 92% profit over all=10000*192/100=19200

Number of rotten apples at 20%=20/100*100=20

Number of saleable apples =100-20=80

Sale price at 92% profit over all =19200/80=240

Marked price to allow discount at 20%=240/80*100=300

Mark up=mp-cp=300–100=200

Percentage of mark-up=200/100*100=200%

Explanation: 45% of 80 = 36

60% of 80 = 48

Total passed = 80 - 35% of 80 = 80 - 28 = 52

passed in both = 48+36-52= 32

Explanation: Total Days = 40

Days gone = 30

Days left = 40 - 30 = 10

number of hours per day = 10 hrs

as work is same

M1*H1*D1 = M2*H2*D2

50*8*30 = (50+X) * 10 * 10

X = 70

Explanation: Speed of bus = 2x km/hr

Speed of car = 5x km/hr

Distance = Speed * time

280 = 5x * T

distance covered by the bus = y

 y = 2x * T

280/y = 5/2

y = 280 * 2/ 5

y = 112

distance covered by the bus = 112  km

Explanation: Age of Ajay  + Sushma = 82

Shashank = 12 , hence Sushma = 28+12 = 40

Ajay = 82-40 = 42

Sashank after 4 years = 16

Ajay after 4 years = 46

Sum = 16+46 = 62 years

Explanation: From condition 1,

Explanation: We can solve this question easily using linear equations and the concept of average.

Let the number of students initially present in the class be = x

then the weight of all the students = 62x (? total = avg × no. of things)

when 8 students leave the class, the weight of the remaining

students = 62n - 8(55)

and when other 13 students join the class, the weight of all students =

62n - 8(55) + 13(65) = 63.9 (n - 8 + 13)

62n = 63.9(n+5) + 440 - 845

62n - 63.9n = 319.5 - 405

- 1.9n = -85.5

x = 45

∴ The total number of students in the class initially were 45.

Explanation: Given,

Ramesh completes work in 20 days

Let the Total work to be completed = 1 (1 unit)

So,

Ramesh can complete (1/20) part of work in 1 day

Given,

Mohan is 66.67% as efficient as Ramesh

i.e., If Ramesh completes X part of work in 1 day, then Mohan completes 2X/3 part of work in 1 day.

Here, Ramesh completes (1/20) part of work in 1 day

So, Mohan can complete (2/3)*(1/20) = (1/30) part of work in 1 day

Mohan can complete work in 30days

Let D be the days Ramesh has worked i.e., after D days Ramesh left the work.

Both Ramesh and Mohan can complete (1/20+1/30) part of work in 1 day.

In D days Ramesh and Mohan can complete

D*(1/20+1/30) part of work

In 10 days Mohan can complete

10*(1/30) part of work

We have,

D*(1/20+1/30) + 10*(1/30) = 1

L.C.M of (20,30) = 60

D*(3+2) + 10*2 = 60

5D + 20 = 60

5D = 40

D=8

Therefore, after 8 days Ramesh left the work

Explanation:

x  =    9A + 6

x =    21B  + 12

9A + 6 = 21B + 12

9A = 21B + 6

3A = 7B  + 2

B = 1, A = 3          => x = 33

B =4, A  = 10     => x =  96

B = 7, A = 17       => x =  159

and so on x = 222, 285, 348, 411

x between 250 and 400

So, x can be 285, 348

Sum = 285 + 348 = 633

sum of all possible values of x is 633.

Explanation: Yellow balls are = x + 5

Blue balls are = 2x + 1

Red balls =?

Total balls = 4(x + 2) = 4x + 8, So red balls are = x + 2.

Find value of x, to answer how many red balls are there.

Given, two balls were selected and the probability of getting a red and blue is 1/6.

From this,

{(x+2)C1 * (2x+1)C1}/(4x+8)C2 = 1/6

Solving above equation gives x = 2.

So total red balls are = x +2 = 2+2 = 4. Option B

Explanation: Let, fixed charged be = Rs. X

And charge per km = Rs. Y

1^st journey, X + 72y = 1107

2^nd journey, X + 55y = 898.

Solving both Y = Rs. 12.3.

Substituting Y in any one of the equations gives, X = Rs. 221

Then for 3^rd journey, X + 45y = 221 + 45*12.3 = 774

Explanation: C can complete the work alone in 60 days.

A can complete work in 60 + (40/100)60  = 84 days

B can complete work in 60 + (75/100)60  = 105 days

Work Done in 3 Days =

3 * (1/84) + 1/105 + 1/60 = (15 + 4 + 7)/420 =26/420

= 13/210

Work done in 16 * 3 = 48 days  

= 16 * 13/210 = 208/210

Work left = 2/210 = 1/105

1/105 Work done by A in (1/105)/(1/84) = 4/5  day

Explanation: Distance = speed × time

total distance = 50km×3hr= 150km

speed of bus is

50 - 40% of 50 = 30km/hr

time taken by bus without stopping

150/30 =5hr

since bus stops for 10 min at each 10km

150km/10km = 15 stops

but last stop we are not considering

because its the destination.

14 stops ×10 min =140 min

so total time spent will be 5hr + 140min i.e.,

Explanation: Selling price = 900 + C.P

Mark price = 2400 + C.p

Discount = M.P – S.P = 1500

Profit = S.P – C.P = 900

X% of C.p = 900

X% of MP = 1500

X% of ( 2400 + CP ) = 1500

X% of 2400 + x% of cp = 1500

X% of 2400 + 900 = 1500

X% of 2400 = 600

X = 25

Value of 4x = 100

Explanation: Let numbers are X, X+2, X+4, X+6, X+8

Average = sum /5

Sum = 5X + 20

5X +20 = 27*5

X = 23

Largest number = X+8 = 23+8 = 31, average = 27 sum = 31+27 = 58

Explanation: Let distance = X km

And speed = Y km/hr

A Car covers the distance between two points in 45minutes

X/Y = 45/60 hrs

x/y = ¾

4x = 3y

Y = 4x/3

If the speed of the car is reduced by 5km/hr, the time taken to cover the distance increases to 48minutes

X / (y-5) = 48/60

X / (y-5 ) = 4/5

5x = 4y – 20

Sub Y = 4x/3 value of X = 60km

Explanation: C.P = 1200

C.P : M.P = 3:5

Hence M.P = 2000

After 1^st discount of 30% on 2000 = 600 will be reduce

M.P = 1400 2^nd discount of 18% of 1400 = 252 will be reduced

S. P = 1148

LOSS = 1200 – 1148 = 52

% loss = 52 / 1200 * 100 = 4(1/3)loss %