3/4 of his offer = 75000

So his offer = 100000.

2/3 of business esteem = 100000

So add up to esteem = 150000

Let keep 800 cc steady and compute the measure of diesel for 800 km

800*60/600=80 liters.

Presently, ascertain diesel required for new separation i.e. 800 km,

80*1200/800=120 liters.

At 12 O'clock, A cover 40km and on the opposite side B at 11 o clock cover 40km, again they went towards each other (which is really the separation between them), that is A needs to make a trip 2hr (From 12 to 2 at 20km/hr.) i.e. 2*20=40km and opposite side B needs to Travelled out 3hr (From 11 to 2 at 40km/hr.) i.e. 3*40=120Km.

At that point Then the total distance traveled by them is the Actual distance between them i.e. 40+120= 160Km

Let the annual instalment be Rs. @The first instalment will be paid one year from now i.e. 3 years before it is actually due. The second instalment will be paid two years from now i.e. 2 years before it is actually due.

The third instalment will be paid 1 year before it is actually due.

The fourth instalment will be paid on the day the amount is actually due.

On the first instalment the interest will be paid for 3 years, on the second for 2 years, on the third for 1 year, on the fourth for 0 year. In total an interest for 6 years will be paid (3 + 2 + 1 + 0) on Rs. 100 @ 10%. Interest = (100 × 6 × 10)/100 = Rs. 60 and the principal is Rs 100 × 4 = Rs 4@The total loan that can be discharged is Rs. 400 + 60 = Rs. 46@Here the technique of Chain Rule will be applied. I.e. for Rs. 460 the instalment required is Rs. 100, for Rs. 4600 the instalment required is 4600 × 100/460 = Rs. 1000.

B is thrice as fast as C

C covered in 42 minutes

B covered in 42/3=14 min

A is twice as fast as B

A covers in 14*(1/2) = 7 min

Let the real separation voyaged be x km.

At that point, x/10 = (x + 20)/14.

=> 14x = 10x + 200

=> 4x = 200.

=> x = 50 km.

Add up to score after 50 innings = 50*50 = 2500

Total score after 51 innings = 51*51 = 2601.

So, runs made in the 51st innings = 2601-2500 = 101

If he had not lost his wicket in his 51st innings, he would have scored an unbeaten 50 in his 51st innings.

Let x=speed of boat and y=speed of current

=30/ (x+y)=18/(x-y)=5 by solving y=1.2 km/hr

Anirudh contribute for 8 months, Harish contributed for 6 and

sahil for 4 months in the proportion of 5:6:4

so proportion = 5*8 : 6*6 : 4*4

=> 40:36:16

=> 10:9:4

So sahil's profit= (4/23)*75900 = 13200

Lets expect Total benefit x

x * (1-1/3-1/4) = 5000

=> x*(12-4-3)/12 = 5000

x = 5000*12/5 = Rs. 12000

so An's offer = Rs. (1/3*12000) = Rs. 4000

A runs B runs C runs

600 metres race 600m 540 m

500 metres race 500 m 450m

Combing ratio A runs B runs C runs

300metres - 2700meters - 2430metres

Unitary A runs B runs C runs

Method 400mtres - 360 metres - 324 metres

∴ A beats C by 400-324 = 76 metres.

So we have C.P. = 29.50

S.P. = 31.10

Gain = 31.10 - 29.50 = Rs. 1.6

Gain %=( Gain/Cost*100)%

= (1.6/29.50*100)%=5.4%

Nirmal: Kapil = 9000*12:(12000*6+6000*6) = 1:1

Kapil share = Rs. [4600 *(1/2)) = Rs. 2300

Speed downstream = (13 + 4) km/hr = 17 km/hr.

Time taken to movement 68 km downstream = (68/17) hrs = 4 hrs

The letters Increase by 1; the numbers are duplicated by 2.

**37 ½ km ph Solution:** Time required for the initial 60 km = 120 min.

The Time required for the second 60 km = 72 min.

Add up to time required = 192 min

Average speed = (60*120)/192 = 37 1/2

On the off chance that Rani Age is x, at that point Teena age is x-6,

So (x-6)/x = 6/8

=> 8x-48 = 6x

=> 2x = 48

=> x = 24

So, Teena age is 24-6 = 18 years

Let P = Principal

A - Amount

We have a = P (1 + R/100)3 and CI = A - P

ATQ 993 = P (1 + R/100)3 - P

∴ P = Rs 3000/ -

Presently SI @ 10% on Rs 3000/ - for 3 yrs = (3000 x 10 x 3)/100

= Rs 900/ -

Let's assume A finishes the 600 m race in 60 sec, then

600/60 = 10 m/sec is his speed

B traveled (600-60 = 540 m in 60 sec, therefore

540/60 = 9 m/sec is B's speed

"in a race of 500 metres, B can beat C by 50 metres."

500/9 = 55.56 sec is B's time to finish a 500 m race

C traveled 500-50 = 450 m in 55.56 sec, therefore

450/55.56 = 8.1 m/sec is C's speed

By how many will A beat C in a race of 400 metres?

400/10 = 40 sec for A to run a 400 m race

C will travel 8.1*40 = 324 m in 40 sec therefore

C will be 400-324 = 76 m behind when A crosses the finish line

Distance=125 meter speed=50-5=45km/hr=>45*5/18=12.5 m/s

Time=125/12.5=10sec

20 men in 6 days can build 112 meters

25 men in 30 days can build=112*(25/20)*(3/6)

= 70 meters

X/5 + 5 = x/4 - 5

⇒ x/5 - x/4 = 10

X/20 = 10

⇒ x = 200

A's 1day's work = 1/40

B's 1day's work = 1/28

They can cooperate in = 1/40 + 1/28 = 16 days (estimate)

The quick typist's work done in 1 hr = 1/2

The moderate typist's work done in 1 hr = 1/3

If they work to join, work is done in 1 hr = 1/2+1/3 = 5/6 So,

the work will be finished in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min