20 men in 6 days can build 112 meters
25 men in 30 days can build=112*(25/20)*(3/6)
= 70 meters
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.
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
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 the annual instalment be Rs. 100. 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 400. The total loan that can be discharged is Rs. 400 + 60 = Rs. 460. 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.
X/5 + 5 = x/4 - 5
⇒ x/5 - x/4 = 10
X/20 = 10
⇒ x = 200
Let x=speed of boat and y=speed of current
=30/ (x+y)=18/(x-y)=5 by solving y=1.2 km/hr
Distance=125 meter speed=50-5=45km/hr=>45*5/18=12.5 m/s
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
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)
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
So we have C.P. = 29.50
S.P. = 31.10
Gain = 31.10 - 29.50 = Rs. 1.6
Gain %=( Gain/Cost*100)%
The letters Increase by 1; the numbers are duplicated by 2.
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
The number nearest to 100 which is more noteworthy than 100 and divisible by 14 is 112, which is the principal term of the arrangement which must be summed. The number nearest to 1000 which is under 1000 and distinct by 14 is 994, which is the last term of the arrangement. 112 + 126 + .... + 994 = 14(8+9+ ... + 71) = 35392
Question 16. Gavaskar's Average In His Initial 50 Innings Was 50. After The 51st Innings, His Average Was 51. What Number Of Runs Did He Score In His 51st Inning? (assuming That He Lost His Wicket In His 51st Innings)
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.
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
Speed downstream = (13 + 4) km/hr = 17 km/hr.
Time taken to movement 68 km downstream = (68/17) hrs = 4 hrs
Question 19. A, B, C Are The Partner In A Business. During A Specific Year. A Got 33% Of The Benefit. B Got One-fourth Of The Benefit And C Got The Rest Of The Rs. 5000. What Amount Of Measure Of Cash Did A Get?
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
3/4 of his offer = 75000
So his offer = 100000.
2/3 of business esteem = 100000
So add up to esteem = 150000
Question 21. Nirmal And Kapil Began A Business Contributing Rs. 9000 And Rs. 12000 Separately. Following A Half Year, Kapil Pulled Back Portion Of His Speculation. In The Event That Following A Year, The Aggregate Benefit Was Rs. 4600, What Was Kapil's Share Initially ?
Nirmal: Kapil = 9000*12:(12000*6+6000*6) = 1:1
Kapil share = Rs. [4600 *(1/2)) = Rs. 2300
Question 22. Anirudh, Harish, And Sahil Put A Sum Of Rs.1, 35,000 In The Proportion 5:6:4 Anirudh Contributed Has The Capital For 8 Months. Harish Contributed For A Half Year And Sahil Contributed For 4 Months. On The Off Chance That They Acquire A Benefit Of Rs.75, 900, Then What Is The Offer Of Sahil In The Profit ?
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
So sahil's profit= (4/23)*75900 = 13200
Question 23. A Begins Riding His Bicycle At 10 Am With A Speed Of 20kmph And B Likewise Begins At 10 Am With A Speed Of 40kmph From A Similar Point In A Similar Way. Returns South At 12 O'clock And B Turns North At 11 Am. What Will Be The Distance Between A And B At 2 Pm ?
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
Question 24. 60 Liters Of Diesel Is Required To Movement 600 Km Utilizing An 800 Cc Motor. In The Event That The Volume Of Diesel Required To Cover A Separation Changes Specifically As The Limit Of The Motor, At That Point What Number Of Liters Of Diesel Is Required To Movement 800 Km Utilizing 1200 Cc Motor ?
Let keep 800 cc steady and compute the measure of diesel for 800 km
Presently, ascertain diesel required for new separation i.e. 800 km,
Let the real separation voyaged be x km.
At that point, x/10 = (x + 20)/14.
=> 14x = 10x + 200
=> 4x = 200.
=> x = 50 km.
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