tuliram runs in a triathlon consisting of 3 phases in the following manner.running 12 km,cycling 24 kkm and swimming 5 km.his speeds in the 3 phases are in the ratio 2:6:1.he completes the race in n minutes.later he changes his stratgey so that the distances he covers in each phase are constant but his speeds are now in the ratio 3:8:1.the end result is that he completes the race taking 20 mins more than the earlier speed.it is also known that he has not changed his running speed when he changes his strategy.
1)what is his intial speed? ans:0.5km/min
2)if his speeds are in the ratio 1:3:1,with the running time remaining unchanged what is his finishing time ans:250/3 min
3)what is tuliram's original speed of running? 18kmph
pls explainin detail with steps..thanks in advance
A number is such that when divided by 4,5,6, or 7 it leaves the remainder 2,3,4, or 5 resp. Which is the largets number below 4000 that satisfies this property. 1. 3352 2. 3378 3. 3898 4. 2938
Ans : LCM of 4 5 6 7 = 420...SO largest multiple of 420 below 4000 = 420 x 9 =3780. Since the difference between all the divisors and remainders are same which is equal to 2 final answer is 3780 - 2 =3778. You sure about the options dude?....
Sorry, you're right. The second option is 3778 instead of 3378. Thanks. So, this is the way to solve these kind of questions. What if the difference of the divisors and remainders are different?
dude best possible way to do this kind of questions is by using options...they will give you sure shot answer and that also in quick time..... p.s.-condition apply
But still I wanted an approach for these kind of questions.
Sorry, you're right. The second option is 3778 instead of 3378. Thanks. So, this is the way to solve these kind of questions. What if the difference of the divisors and remainders are different?
When the difference is different you will have to work from the basics. Take an eg : say if a number leaves a remainder 5 on being divided by 7 and a remainder of 2 when divided by 5 then proceed as below... Let N be the number... then N = 7k + 5 also 7k + 5 leaves a remainder of 2 when divided by 5. So 7k + 5 - 2 = 7k + 3 is divisible by 5. On simple inspection u can get the lowest value of k which gives the lowest number which in this case is k = 1 . So the value is 10. But the real numbe will be 10 + 2 = 12 which on verification you can see that meet the criteria. Say what is the smallest such 3 digit number then the answer is 10 x 10 + 2 = 102 (verify) will meet the specifications. Hope this helps. 😃
at iim bangalore class 95, sonali , a first yr student has taken 10 courses, earning grades A (worth 4 points each), b (worth 3 points each), c (worth 2 points each). Her grade point avg is 3.2 and if the course in which she get c's were deleted, her GPA in the remaining courses would be 3.333 . how many a's b's and c's did she get?
i got two eqns but im unable to get the third one Let A's be a, B's be b & C's be c... 4a + 3b + 2c/10 = 3.2 a + b + c = 10
pls explain in detail with proceedings..thanks in advance
at iim bangalore class 95, sonali , a first yr student has taken 10 courses, earning grades A (worth 4 points each), b (worth 3 points each), c (worth 2 points each). Her grade point avg is 3.2 and if the course in which she get c's were deleted, her GPA in the remaining courses would be 3.333 . how many a's b's and c's did she get?
i got two eqns but im unable to get the third one Let A's be a, B's be b & C's be c... 4a + 3b + 2c/10 = 3.2 a + b + c = 10
pls explain in detail with proceedings..thanks in advance
Grades A B C Number x y z Points 4 3 2
Now (4x+3y+2z)/10 = 3.2 ie 4x+3y+2z = 32 ----> (1) also x+y+z = 10 (number of subjects) ----> (2) now since the course getting C are deleted GPA becomes 3.33 or 10/3 so (4x+3y)/(10-z) = 10/3 on rearranging we get 12x+9y+10z = 100 ---->(3) Multiplying (1) by 3 and subtracting from (3) we get z=1. so x+y=9 ----> (4) and from (2) 4x+3y=30 ---->(5) solving (4) and (5) we get x = 3 and y = 6.
SO number of As = 3 Bs = 6 Cs = 1 Correct me if I am wrong... 😃
Now (4x+3y+2z)/10 = 3.2 ie 4x+3y+2z = 32 ----> (1) also x+y+z = 10 (number of subjects) ----> (2) now since the course getting C are deleted GPA becomes 3.33 or 10/3 so (4x+3y)/(10-z) = 10/3 on rearranging we get 12x+9y+10z = 100 ---->(3) Multiplying (1) by 3 and subtracting from (3) we get z=1. so x+y=9 ----> (4) and from (2) 4x+3y=30 ---->(5) solving (4) and (5) we get x = 3 and y = 6.
SO number of As = 3 Bs = 6 Cs = 1 Correct me if I am wrong... :)
A precious stone weighing 35 grams worth Rs. 12250 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5. If price varies as square of the weight find the loss incured. a)5750 b)6000 c)5500 d)5000 the answer is d) i proceeded till here 12250=k(35)^2 k=10. wat next?? pls explain in detail
A precious stone weighing 35 grams worth Rs. 12250 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5. If price varies as square of the weight find the loss incured. a)5750 b)6000 c)5500 d)5000 the answer is d) i proceeded till here 12250=k(35)^2 k=10. wat next?? pls explain in detail
You got k = 10. Correct. Now it gets broken into weights of ratio 2:5. ie weights are 10 and 25gms. So price from 10gm = 10 x 10^2 = 1000 and the same for 25 gm = 10 x 25^2 = 6250 Total price now = 7250. Initial price = 12250 So loss = 5000...Choice (d)
A mother divided an amount of Rs. 61000 between her two daughters aged 18 years and 16 years respectively and deposited their shares in a bond. If the interest rate is 20% compounded annually and if each received the same amount as the other when she attained the age of 20 years, their shares are
Anshu gave Bobby and Chandana s many pens as each one of them already had.The Chandana gave Anshu and Bobby as many pens as each already had.Now each had an equal number of pens.The total number of pens is 72. (a) How many pens did Bobby have initially (b) How many pens did Chandana have initially (c) How many pens did Anshu have initially
A mother divided an amount of Rs. 61000 between her two daughters aged 18 years and 16 years respectively and deposited their shares in a bond. If the interest rate is 20% compounded annually and if each received the same amount as the other when she attained the age of 20 years, their shares are pls expain in detail with steps
16 waali ka share x 18 waali ka share 61000-x x*(1.2)^4 = (61000-x)*(1.2)^2 1.44 x = 61000 - x x = 25000
The volume of a pyramid varies jointly as its height and the area of its base; when the area of the base is 60 square dm and the height 14 dm the volume is 280 cubic dm.what is the area of base of a pyramid whose volume is 390 cubic dm and whose height is 26 dm? a)40 b)45 c)50 d)none of these the answer is 45 pls explain the proceeding thanks in advance
The volume of a pyramid varies jointly as its height and the area of its base; when the area of the base is 60 square dm and the height 14 dm the volume is 280 cubic dm.what is the area of base of a pyramid whose volume is 390 cubic dm and whose height is 26 dm? a)40 b)45 c)50 d)none of these the answer is 45 pls explain the proceeding thanks in advance
Volume = k * (height) * (Area of base) where k is any constant
Given, 280 = k * 60 * 14...(1) Also, 390 = k * 26 * A ....(2) where A is the area asked for.
A five digit number is formed using digits 1,3,5,7,9 without repeating any of them. what is the sum of all such possible numbers? a)6666600 b)6666660 c)6666666 d)none
If harmonic mean between two positive numbers is t the inverse of their geometric mean as 12:13; then the numbers could be in ratio a)12:13 B)13:12 c)4:9 d)2:3
could nt get the answer in the second one w/o using options..solution??
A five digit number is formed using digits 1,3,5,7,9 without repeating any of them. what is the sum of all such possible numbers? a)6666600 b)6666660 c)6666666 d)none
If harmonic mean between two positive numbers is t the inverse of their geometric mean as 12:13; then the numbers could be in ratio a)12:13 B)13:12 c)4:9 d)2:3
A five digit number is formed using digits 1,3,5,7,9 without repeating any of them. what is the sum of all such possible numbers? a)6666600 b)6666660 c)6666666 d)none
If harmonic mean between two positive numbers is t the inverse of their geometric mean as 12:13; then the numbers could be in ratio a)12:13 B)13:12 c)4:9 d)2:3
could nt get the answer in the second one w/o using options..solution??
Ans
1) for this type of questions always remember the formula : (n-1)! x (sum of the given digits) x 1111..(n times) ---> works with distinct digits only. So here n = 5. Sum of digits = 25 So sum = 4! x 25 x 11111 = 6666600 ---> option (A)
2) I didnt quite get the question... can you please rephrase it if you dont mind?
If harmonic mean between two positive numbers is to the inverse of their geometric mean as 12:13; then the numbers could be in ratio a)12:13 B)13:12 c)4:9 d)2:3 corrected the question
and suja bhai 1+4/7+(9/7^2)+(16/7^3)...... ye series kaise karte the?? 1 month mein sab bhul gaya hu 😞
A can do a piece of work in 20 days.he works at it for 5 days and then B finishes it in 10 more days.In how many days will A and B together finish this work.
A's 5 days work=5/20=1/4 So 3/4th of work left 3/4*1/10=3/40?? wat next..im unable to proceed Pls explain in detail with steps thanks in advance
Since there are total 120 digits in the number. hence applying the AP because it is caused by successively placing consecutive natural nos. till 1-9= 9x1= 9 digits from 10-99= 45x2=90 digits ( using 10+ (2n-1)x1=99 where n is the nth digit of the progression) now 9+90=99 digits are covered.. remaining 120-99= 21 digits can similarly be expressed as 21= x*3 where the last natural no is at xth place from 100. since x comes out to be 7 hence the no. in the last is 107. now when divided by 8.. possible remainders are 1,3,5,7.. since only 7 is vailable in the options hence answer must be 7. Sorry evn i can make out how the last no. is given to be 646.
Hope this helps! Cheers!!
1-9 => 9 numbers with single digit each => 9 digit and 9th digit will be 9 10-59 => 50 numbers with two digit each => 50*2=100 digit ie. (9+100 => 109th digit as 9 (because number 59's unit place will be 109th digit) Now, 60-64 => 5 numbers with two digit each => 5*2= 10 digit ie. (9+100+10 => 119th digit as 4 (because number 64's unit place will be 119th digit and 6 at 118th digit) at 120th digit we will get number 6 (because next consecutive number we will write in series is 65..so 6 will be at 120th digit and 5 will be at 121th digit) but we have to consider only upto 120th digit Hence we know 118th digit= 6 119th digit= 4 120th digit= 6 So remainder 646/8 => 6 so answer is 6
