A thief escaped from police custody.Since he was a sprinter,he could run at a speed of 40km/hr.The police realized it after 3hr and started chasing him in the same direction at 50km/hr.The police had a dog which could run at 60km/hr.The dog would run to the thief and then return to the police and then would turn back towards the thief.It kept on doing so till the police caught the thief.Find the total distance travelled by the dog in the direction of the thief?
a)720km
b)600km
c)660km
pls do show detailed sol.
Rephrasing;
The grocer is selling 100g of mixture(butter+fat) for $100. This $100 includes a 25% profit because he would have only spent $80 for 80g of butter and 0$ for 20g fat.
*****************************************
Actually a better example would be;
Think that the grocer spent $100 for buying 100g pure butter. Now; in order to gain a profit of 25%, he will have to sell the same butter for $125.
How he would do that?
He would add 25g of fat(worth $0) to 100g of pure butter(worth $100) and sell 125g of mixture(butter+fat), saying that it is pure butter, for $125
Now you see;
125g of mixture is sold for $125 for which he had spent only $100. He made a profit of 25%.
125g of mixture contains 25g of fat; which is 25/125*100=20%
A thief escaped from police custody.Since he was a sprinter,he could run at a speed of 40km/hr.The police realized it after 3hr and started chasing him in the same direction at 50km/hr.The police had a dog which could run at 60km/hr.The dog would run to the thief and then return to the police and then would turn back towards the thief.It kept on doing so till the police caught the thief.Find the total distance travelled by the dog in the direction of the thief?
a)720km
b)600km
c)660km
pls do show detailed sol.
iam getting 600km.wats the oa?
A thief escaped from police custody.Since he was a sprinter,he could run at a speed of 40km/hr.The police realized it after 3hr and started chasing him in the same direction at 50km/hr.The police had a dog which could run at 60km/hr.The dog would run to the thief and then return to the police and then would turn back towards the thief.It kept on doing so till the police caught the thief.Find the total distance travelled by the dog in the direction of the thief?
a)720km
b)600km
c)660km
pls do show detailed sol.
since the police realized after 3 hrs he already ran 120kms so to catch him the time taken would be 120/10=12 hrs and the police would cover 600kms the dog covers 720(in both directions ) so 600kms in the forward direction and rem 120kms would be divided equally 60kms either side so total distance in the direction of the thief is 660kms .
hope it helps :):)
since the police realized after 3 hrs he already ran 120kms so to catch him the time taken would be 120/10=12 hrs and the police would cover 600kms the dog covers 720(in both directions ) so 600kms in the forward direction and rem 120kms would be divided equally 60kms either side so total distance in the direction of the thief is 660kms .
hope it helps :):)
ohhh 60 km dog aage bhi toh aayega peeche jane ke baad....my mistake 😃
i still couldn't get it.pls explain once more,"600kms in the forward direction and rem 120kms would be divided equally 60kms either side so total distance in the direction of the thief is 660kms"
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i still couldn't get it.pls explain once more,"600kms in the forward direction and rem 120kms would be divided equally 60kms either side so total distance in the direction of the thief is 660kms"
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The best way to solve these type of ques is to calculate the time for which the event takes place and then multiply by the speed.
So, in order to calculate the time.
The thief had been running for 3 hrs which created a distance of 3 * 40 = 120km. Now the relative speed of police wrt. to thief is 10km/hr (50 - 40)
So, to cover this distance of 120km, the police would have to run for a total of 12 hrs. Since the dog runs for an equal duration of time, the distance covered by the dog would be 12 * 60 = 720km
Similarly, the policeman covers a distance of 12 * 50 = 600km
Since there is an extra distance of 120km covered by the dog, this would be 60km in both the directions as he is running to and fro. To understand this, assume that he covers x km running backwards, then in the onward journey he'll have to cover these x km in addition to the distance that he would normally cover.
So, these 120km are equally distributed.
Hope that helps.
hi guyss..i need help on this question plzzz
In how many ways can a selection be made of 5 letters out of 5As,4Bs,3Cs,4Ds and 1E?
Ans is 71...can anyone plzz explain me howww??
hi guyss..i need help on this question plzzz
In how many ways can a selection be made of 5 letters out of 5As,4Bs,3Cs,4Ds and 1E?
Ans is 71...can anyone plzz explain me howww??
when all are same = 1
when all are different = 1
when 4 are same and one different = 3C1*4C1 = 12
when 3 are same and others are different = 4C1*4C2 = 24
when 3 are same and rest of 2 are also same = 4C1*3C1 = 12
when 2 are same and other 2 are also same but one different = 4C2*3C1 = 18
when 2 are same and rest of 3 are different = 4C1*4C3 = 16
total = 84
when all are different = 1
when 4 are same and one different = 3C1*4C1 = 12
when 3 are same and others are different = 4C1*4C2 = 24
when 3 are same and rest of 2 are also same = 4C1*3C1 = 12
when 2 are same and other 2 are also same but one different = 4C1*3C1*3C1 = 36
when 2 are same and rest of 3 are different = 4C1*4C3 = 16
total = 101
shldn't we consider the case when all 5 are same..?
Is there any solution for the Additional Problems given at the end of each block... i am almost done with all chapters.. looking fwd to start the excersises at the end of each block..
guys please solve my doubts..( 2 questions)
M children (1 to M) are standing in a circle facing each other wearing different caps .In the 1st round each passes his cap to child on his left. In the next round each one passes the cap he now has to the second child on his left. In the next round each one passes the cap he now has to second child on his left. In the next round each one passes the cap he has to third child on his left and so on. They stop when the first child on his left and so on.They stop when the first child gets his original cap back for the first time. a) if M= 44 then after hw many rounds will dey stop..?
if it is true that mod of ( x^2 - 9) is less than K for all real nos X such that mod of ( x-3)
Guys please say the logic,
From AP,
*Three times the first of three consecutive odd integers is 3 more than twice the third . wat is the third integer?
a. 15
b. 9
c. 11
d. 5
Guys please say the logic,
From AP,
*Three times the first of three consecutive odd integers is 3 more than twice the third . wat is the third integer?
a. 15
b. 9
c. 11
d. 5
2a+1,2a+3 and 2a+5
6a+3 = 4a+10+3
2a =10
a=5
so the third integer 2*5+5=15:)
correct me if am wrong........
Guys please say the logic,
From AP,
*Three times the first of three consecutive odd integers is 3 more than twice the third . wat is the third integer?
a. 15
b. 9
c. 11
d. 5
Should be 15.
Let the 3 consecutive odd integers be -> a+1, a+3 and a+5 resp.
=> 3(a+1) = 3 + 2(a+5).
=> 3a + 3 = 3 + 2a + 10.
=> a = 10.
Hence, 3rd term (a+5) = 10+5 = 15.
2a+1,2a+3 and 2a+5
6a+3 = 4a+10+3
2a =10
a=5
so the third integer 2*5+5=15:)
correct me if am wrong........
@rakesh:
Answer is correct, but i din't get the logic how u choose 2a+1 instead of a+1, So plz explain 1st and 2nd step
@rakesh:
Answer is correct, but i din't get the logic how u choose 2a+1 instead of a+1, So plz explain 1st and 2nd step
i chose it because we generally represent any odd no. like that.
2a+1.........try putting any value of a u will always get an odd number.
for 2nd step....
i just follwed the question
3*(2a+1) = 2*(2a+5)+3..
hope it helps.:)
@rakesh:
Answer is correct, but i din't get the logic how u choose 2a+1 instead of a+1, So plz explain 1st and 2nd step
Actually it would be 2a only. I took a coz i knew that the 2 wld get canceled out in the end. The reason for taking 2a is that when we multiply a number by 2, the product is always even and when we add 1 to that product it becomes odd.
Hope it helps.
Should be 15.
Let the 3 consecutive odd integers be -> a+1, a+3 and a+5 resp.
=> 3(a+1) = 3 + 2(a+5).
=> 3a + 3 = 3 + 2a + 10.
=> a = 10.
Hence, 3rd term (a+5) = 10+5 = 15.
Ya, guys i get cleared. i interpreted question in another way.
thanks..
*In an AP Sp=q, and Sq=p (Sn being the sum of the first n terms of the progression). find Sp+q.