Directions(1-5): Two equations are given you have to solve both of them and compare,then choose a correct option.
- Quantity I: A train can cross a pole and a platform 440 m long in 12 seconds and 28 seconds respectively. Find the speed of the train.
Quantity II: Distance between Rajan’s college and house is 150 km.When Rajan decreases his original speed by 10 km/hr. he is late by 10 minutes. Find the original speed of the Rajan.Quantity I >Quantity IIQuantity I = Quantity IIQuantity I >=Quantity IIQuantity I =< Quantity IIQuantity I < Quantity IIOption E
From I: Length of the train be x.
Now, x/12 = (x+440)/28
=> x = 330
Speed of the train = 330/12 = 99 km/hr.
From II: Let the original speed of the Rajan be x km/hr.
150 = x(x-10)/10*10/60
=> x^2 – 10x – 9000 = 0
=> x = 100, – 90
Original speed of Rajan = 100 km/hr.
I <II - Quantity I: Pipes A and B alone can fill the empty tank in 20 hours and 25 hours resp. Pipe C alone can empty the full tank in 40 hours. Pipe A is opened at the start and after 5 hours pipe B is also opened. In how many hours the tank is filled completely?
Quantity II: A and B together can do a certain piece of work in 6 hours. A is 50% more efficient than B. In how many days A alone can complete the work?Quantity I = Quantity IIQuantity I >Quantity IIQuantity I >=Quantity IIQuantity I =< Quantity IIQuantity I < Quantity IIOption B
From I: LCM(20,25 and 40) = 200 litres
Quantity of water filled by A alone in 5 hours = 10*5 = 50 litres
Quantity of water filled by A and B together in 4 hours = (10+8)*4 = 72 litres
Quantity of remaining water to be filled by A,B and C together = 200 -50-72 = 78 litres
Time taken by these three pipes to fill the remaining 78 litres = 78/(10+8-5) = 78/13 = 6 hours
So, time taken to fill the empty tank = 5+4+6 = 15 hours
From II: Let time taken by B alone to complete the work be x hours.
So, time taken by A alone to complete the work = x/1.5 = 2x/3 hours
1/x+3/2x = 1/6
=> x = 15
Time taken by A alone to complete the work = 15/1.5 = 10 hours
I>II - Quantity I: A shopkeeper marked the article 50% above the cost price and sold it after two consecutive discounts of 10% and 20%. Find the profit percent in the transaction
Quantity II: A shopkeeper marked an article 25% above the cost price and sold it after a discount of x%. Find the value of x if in this transaction he had a profit of 15%.Quantity I = Quantity IIQuantity I >=Quantity IIQuantity I < Quantity IIQuantity I =< Quantity IIQuantity I >Quantity IIOption A
From I: Let the CP of the article be Rs.100.
MP = Rs.150 SP = 0.90*0.80*150 = Rs.108 Profit% = (108-100)/100*100 = 8%
From II: Let the CP be Rs.100. MP = Rs.125
SP = 125*(100-x)/100 = 115 125 – 1.25x = 115
=> x = 8
I = II
- Quantity I: x^2 + (6+4(2)^1/2)x + 24(2)^1/2 = 0
Quantity II: y^2 + 7(2)^1/2y + 24 = 0Quantity I =< Quantity IIQuantity I >=Quantity IIQuantity I < Quantity IIQuantity I >Quantity IIQuantity I = Quantity IIOption A
From I: x^2 + (6+4(2)^1/2)x + 24(2)^1/2 = 0
=> x(x+6)+4(2)^1/2 (x+6) = 0 => x = -6,-4(2)^1/2
From II: y^2 + 7(2)^1/2y + 24 = 0
=> y^2 + 4(2)^1/2y + 3(2)^1/2y +24 =0
=> (y+4(2)^1/2)(y+3(2)^1/2) = 0
=> y = -4(2)^1/2, – 3(2)^1/2
I= <II - Quantity I: 11x + 9y = 169
Quantity II: 13x – 5 y = 59Quantity I >Quantity IIQuantity I = Quantity IIQuantity I < Quantity IIQuantity I >=Quantity IIQuantity I =< Quantity IIOption B
On solving I and II,
we get Quantity I = Quantity II - Varun invested Rs. 27000 partially in scheme A offering 13% simple interest and rest in scheme B offering 10% compound interest compounded annually. The amount invested in scheme B is how much percent less/ more than the amount invested in scheme A if the total interest earned by Varun after three years is Rs. 9822?
30%25%20%15%10%Option C
Let amount invested by him Rs.x in scheme A. and in scheme B = 27000 – x
Interest earned by him from A = x*0.13*3 = Rs. 0.39x
Interest earned by him from B = (27000-x)*{(1+0.10)^3 – 1} = Rs. 8937 – 0.331x
Now, 0.39x + 8937 – 0.331x = 9822
=> x= 15000 Amounts invested by him in A and B resp. = Rs.15000 and Rs. 12000
Required% = (15000 – 12000)/15000*100 = 20% - Monthly income of A is 40% more than the monthly income of B. Monthly savings of A and B are in the ratio 2:1 resp. Monthly expenditure of A is equal to the monthly income of B. Find the average monthly savings of A and B if the monthly expenditure of B is Rs.3000 less than A.
Rs.2500Rs.4200Rs.3300Rs.4500Rs.4000Option D
Let the monthly income of B be x and A be 1.4x.
Monthly expenditure of A and B = Rs. x and Rs. x-3000 resp.
Now, (1.4x – x)/(x-x+3000) = 2/1
=> x = 15000
Monthly incomes of A and B are Rs.15000 and Rs. 21000 resp.
Monthly savings of A and B = Rs.6000 and Rs.3000 resp.
Average monthly savings = (6000+3000)/2 = Rs.4500 - Average age of a family of 5 members six years ago was 32 years. In between these 6 years a new baby was born. Now the average age of the family is same as it was 6 years ago. Find the present age of the baby.
3 years2 years5 years4 years6 yearsOption B
Sum of the present ages of the family members excluding the baby = (32+6)*5 = 190 years
Sum of the present ages of the family members including the baby = 32*6 = 192 years
Present age of the baby = 192 – 190 = 2 years - Vessel A contains 75 litre mixture of petrol and diesel mixed in the ratio 8:7 resp. Vessel B contains 125 litres mixture of petrol and kerosene mixed in the ratio 2:3 resp. All of the mixture present in vessels A and B are put in an empty vessel C. Now, what is the percentage of petrol in vessel C.
58%30%40%45%50%Option D
Quantity of petrol in A,B and C = 40 l, 50 l and 90 l resp.
Quantity of mixture in vessel C = 75+125 = 200 l
Required% = 90/200*100 = 45% - In an election between two candidates, the winning candidate got 57% of the valid votes and won by 1680 votes. Find the total number of votes cast if 20% of the cast votes are invalid.
2500020000150003000010000Option C
Let the number of votes cast = x
Number of valid votes = 0.8x
Number of votes received by the winning candidate = 0.8x*0.57 = 0.456x
Number of votes received by the losing candidate = 0.8x*0.43 = 0.344x
Therefore, 0.456x-0.344x = 1680
=> x = 15000