IIT JEE Practice Question and Answer

Q: Ravi and Ramu are partners in a business.Ravi contributes 1/6 of the capital for 16 months and Ramu received 2/3 of the profit. For how long Ramu's money was used ? 1885 0

  • 1
    11 months
    Correct
    Wrong
  • 2
    10 months
    Correct
    Wrong
  • 3
    9 months
    Correct
    Wrong
  • 4
    8 months
    Correct
    Wrong
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Answer : 4. "8 months"
Explanation :

Answer: D) 8 months Explanation: Let the total Profit be Z.Ramu's profit share is 2/3 of profit (i.e 2Z/3)Ravi's profit is (Z-2Z/3) =Z/3Hence the profit ratio is , Ravi : Ramu = Z/3 : 2Z/3 = 1:2 Let total Capital be Rs.X and Ramu has contributed for Y months.Since Ramu's profit share is 2/3, his invest share will be 2/3 in capital. Ravi's invest for 16 months / Ramu's invest for Y months = Ravi's profit share / Ramu's profit sharei.e. (X/6 x 16) / (2X/3 x Y) = 1/2 Solving the above equation, we get Y = 8.So Ramu's money has been used for 8 months.

Q: In how many ways can the letters of the word 'LEADER' be arranged ? 1054 0

  • 1
    360
    Correct
    Wrong
  • 2
    420
    Correct
    Wrong
  • 3
    576
    Correct
    Wrong
  • 4
    220
    Correct
    Wrong
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Answer : 1. "360"
Explanation :

Answer: A) 360 Explanation: No. of letters in the word = 6No. of 'E' repeated = 2Total No. of arrangement = 6!/2! = 360

Q: A boat travels 72 km downstream in 8 hours and 84 km upstream in 12 hours. Find the speed of the boat in still water and the speed of the water current ? 1572 0

  • 1
    9 and 3 kmph
    Correct
    Wrong
  • 2
    6 and 7 kmph
    Correct
    Wrong
  • 3
    8 and 1 kmph
    Correct
    Wrong
  • 4
    7 and 2 kmph
    Correct
    Wrong
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Answer : 3. "8 and 1 kmph"
Explanation :

Answer: C) 8 and 1 kmph Explanation: Downstream speed = 72km/8hrs  = 9 kmphupstream speed = 84km/12hrs = 7 kmphspeed of boat = avg of downstream and upstream speedsspeed of boat = (9+7)/2kmph = 8 kmph.current speed = half of the difference of downstream and upstream speedscurrend speed = (9-7)/2kmph = 1 kmph

Q: A train moving at 2/3 rd of its normal speed reaches its destination 20 minutes late. Find the normal time taken ? 2965 0

  • 1
    4/3 hrs
    Correct
    Wrong
  • 2
    2/3 hrs
    Correct
    Wrong
  • 3
    3/2 hrs
    Correct
    Wrong
  • 4
    1/4 hrs
    Correct
    Wrong
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Answer : 2. "2/3 hrs"
Explanation :

Answer: B) 2/3 hrs Explanation: Let the original speed and time is S and Tthen distance = S x TNow the speed changes to 2/3S and T is T+20As the distance is sameS x T = 2/3Sx(T+20)solving this we get t = 40 minutes =40/60 = 2/3 hour

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Answer : 3. "48"
Explanation :

Answer: C) 48 Explanation: No. of students will be 98 x 12 = 1176let us consider n circles then 1 + 2 + 3 + ....+ n = 1176n(n+1)/2 = 1176n = 48 circles.

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Answer : 1. "48"
Explanation :

Answer: A) 48 Explanation: 3X + 4y = 240by substitute through options 48 is correct

Q: Find by how much percentage the average of first 10 odd numbers (starting from 1) is greater than the last term ? 1415 0

  • 1
    991/16 %
    Correct
    Wrong
  • 2
    900/19 %
    Correct
    Wrong
  • 3
    874/13 %
    Correct
    Wrong
  • 4
    719/17 %
    Correct
    Wrong
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Answer : 2. "900/19 %"
Explanation :

Answer: B) 900/19 % Explanation: First ten odd numbers form an arithmetic progression of the form1,3,5,7,9......here a = first term = 1d = common difference = 2Average of first n numbers = (2a + (n - 1)d)/2n th term of the AP = a + (n - 1)dSubstituting a = 1, d = 2 and n = 10 in the above formulas average of first 10 numbers = 1010 th term of the AP = 19Therefore average of first 10 terms is 19 - 10 = 9 greater thanthe last term. Hence the average is greater than the sum by9/19 X 100 = 900/19 %

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Answer : 3. "80%"
Explanation :

Answer: C) 80% Explanation: circumference of A = 2πr = 8 => so r = 4πvolume = π×4π2×10 = 160πcircumference of B = 2πr   = 10 => so r = 5πvolume = π×5π2×8 =  200πso ratio of capacities = 160/200 = 0.8 so capacity of A will be 80% of the capacity of B.

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