Q: A certain number of answer sheets can be checked in 12 days working 5 hours each day by 9 examiners. For how many hours a day should 4 examiners work to check twice the number of answer sheets in 30 days?
Solution: Answer sheets are work here. So M1*D1*H1*W2 = M2*D2*H2*W1 9*12*5*2 = 4*30*H2*1 Solve, H2 = 9
Q: If x3 + y3 = 9 and x + y = 3, then the value of x4+y4 is,
Solution: x3+y3 = (x + y) × (x2 − xy + y2)
Putting given values of x3+y3 and (x + y)
9 = 3 × ((x+y)2 − 3xy)
= 3 × (9 − 3xy)
= 27 − 9xy
9xy = 18
xy = 2
x4 + y4 = (x2 + y2)2 - 2x2y2
= (x2 + y2)2 - 2*4
[Putting value of xy]
= ((x + y)2 - 2xy)2 - 2*4
[Putting values of (x+y) and xy]
= (9 - 4)2 - 2*4
= 17Q: L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :
Solution: H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161. Let the numbers be a and b . Then , ab= 161. Now, co-primes with product 161 are (1, 161) and (7, 23). Since x and y are prime numbers and x >y , we have x=23 and y=7. Therefore, 3y-x = (3 x 7)-23 = -2
Q: The product of two numbers is 4107. If the H.C.F of these numbers is 37, then the greater number is
Solution: Let the numbers be 37a and 37b. Then , 37a x 37b =4107 => ab = 3. Now co-primes with product 3 are (1,3) So, the required numbers are (37 x 1, 37 x 3) i.e, (37,111). Therefore, Greater number = 111.
Q: The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is
Solution: Product of numbers = 11 x 385 = 4235 Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35 Now, co-primes with product 35 are (1,35) and (5,7) So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7) Since one number lies 75 and 125, the suitable pair is (55,77) Hence , required number = 77
Q: K can lay Highway road between two cities in 16 days. L can do the same job in 12 days. With the help of M, they completes the job in 4 days. How much days does it take for M alone to complete the work ?
Solution: Amount of work K can do in 1 day = 1/16 Amount of work L can do in 1 day = 1/12 Amount of work K, L and M can together do in 1 day = 1/4 Amount of work M can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48 => Hence M can do the job on 48/5 days = 9 (3/5) days
Q: If 10 men take 15 days to complete a work. In how many days will 37 men complete the work?
Solution: Given 10 men take 15 days to complete a work => Total mandays = 15 x 10 = 150 Let the work be 150 mandays. => Now 37 men can do 150 mandays in 150/37 =~ 4 days
Q: K, L and M can do a piece of work in 20, 30 and 60 days respectively. In how many days can K do the total work if he is assisted by L and M on every third day ?
Solution: In one day work done by K = 1/20 Work done by K in 2 days = 1/10 Work done by K,L,M on 3rd day =1/20 + 1/30 + 1/60 = 1/10 Therefore, workdone in 3 days is = 1/10 + 1/10 = 1/5 1/5th of the work will be completed at the end of 3 days. The remaining work = 1 - 1/5 = 4/5 1/5 of work is completed in 3 days Total work wiil be done in 3 x 5 = 15 days.
Q: Lasya alone can do a work in 16 days. Srimukhi’s efficiency is 20 % lesser than that of Laya. If Rashmi and Srimukhi together can do the same work in 12 days, then find the efficiency ratio of Rashmi to that of Lasya?
Solution: Given Lasya can do a work in 16 days. Now, time taken by Srimukhi alone to complete the work = 16 x 100/80 Time taken by Rashmi = n days => (12 x 20)/(20 - 12) = (12 x 20)/n => n= 30 days. Required ratio of efficiencies of Rashmi and Lasya = 1/30 :: 1/16 = 8 : 15.
Q: Kashundra Jones plans to make a lump sum deposit so that she can withdraw $3,000 at the end of each quarter for 10 years. Find the lump sum if the money earns 10% per year compounded quarterly
Solution: A=R[(1+i)^n-1]/i(1+i)^n
| You Have Score | /10 |