Quant Mania : 10 Important Questions

A dishonest dealer professes to sell his goods at cost price, but he uses a weight of 960 gm for the kg weight. Find his gain per cent.

Solution: Suppose goods cost the dealer Re 1 per kg. He sells for Rs 1 what cost him Rs. 0.96.

∴ Gain on Re 0.96 = Re 1 – Re 0.96 = Re 0.04

∴ Gain on Re 100 = 0.04/0.96  × 100=4  1/6

∴ Gain % = 4 1/6 %

If a man purchase 11 oranges for Rs. 10 and sells 10 oranges for Rs 11. How much profit or loss does he make?

Solution:   Suppose that the person bought 11 x 10 = 110 oranges.

CP of 110 oranges = 10/11 x 110 = Rs 100

                SP of 110 oranges = 11/10 x 110 = Rs 121

∴ Profit = Rs 121 – Rs 100 = Rs 21

And % profit =Profit/CP  x 100 = 21/100  x 100 = 12%

A train 75 metres long overtook a person who was walking at the rate of 6 km an hour, and passed him in 71/2 seconds. Subsequently it overtook a second person, and passed him in 63/4 seconds. At what rate was the second travelling?

Relative Speed of train and the first person = (75 )/((15 )/2) = 10 m/s 

= 10 × 18/5 = 36 km/hr.

∴   Speed of train = 36 + 6 = 42 km /hr

Now, relative speed of train and 2nd person = 75/27 × 4 m/s = 300/27 × 18/5 = 40 km/hr

∴   Speed of 2nd person = 42 - 40 = 2 km /hr.

Two women, Ganga and Saraswati working separately can mow a field in 8 and 12 hrs respectively. If they work in stretches of one hour alternately, Ganga beginning 9 a.m., when will the mowing be finished?

Solution: In the first hour Ganga mows  1/8   of the field.

In the second hour Saraswati mows  1/12 of the field.

In the first 2 hrs ( 1/8+ 1/12= 5/24) of the field is mown.

In 8 hrs 5/24  ×4=  5/6 of the field is mown.

Now, (1 –  5/6  )=  1/6  of the field remains to be mown in the 9th hour 

Ganga mows  1/8   of the field.

Saraswati will finish the mowing of (  1/6- 1/8  )=  1/24 of the field in (1/( 24) ÷ 1/12  ) 

Or  1/2  of an hour.

The total time required is (8 + 1 + 1/2) or 9  1/2  hrs.

Thus, the work will be finished at 9 + 9  1/2 = 18  1/2 or 6  1/2 p. m.

A boy goes to school at a speed of 3 km/hr and returns to the village at a of 2 km/hr. if he takes 5 hrs in all, what is the distance between the village and the school?

Solution: Let the required distance be x km.

Then time taken during the first journey = x/3 hr.

                And time taken during the second journey = x/2 hr.

              ∴   x/3+ x/2=5 or,  (2x+3x)/6 = 5         or,      

           5x = 30.

∴ X = 6 ∴  the required distance = 6 km.

A positive number is by mistake multiplied by 5 instead of being divided by 5. By what percent more or less than the correct answer is the result obtained?

Solution :

Let the number be 1

Then the correct answer= 1/5

The incorrect answer that was obtained = 5

The result is more than the correct answer= 5-1/5=24/5

The required percent= (24/5)/(1/5)×100=2400%

Two pipes A and B can fill a tank in 16 minutes and 24 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that that the tank is full in 12 minutes?

Solution:

Let B be closed after x minutes. 

part filled by (A + B) in x min.+ part filled by A in (12 – x) min.= 1

∴  x[1/16+ 1/24] + ( 12-x)×  1/16 = 1

Or, 5x/48 + (12-x)/16 = 1 or, 5x + 3 (12– x) = 48

Or, 2x = 12 ∴   x = 6.

So, B should be closed after 6 min.

The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 yrs, then what was the average of the family at the birth of the youngest member?

Solution: Total ages of all the members = 6 × 22 = 132 yrs

7 yrs ago, total sum of ages = 132 – (6 × 7) = 90 yrs.

But at the time there 5 members in the family

∴    Average at that time = 90 ÷ 5 = 18 yrs.

A batsman in his 17th innings makes a score of 85, and thereby increases his average by 3. What is his average after 17 innings?

Solution: Let the average after 16th innings be x, then 16x + 85

= 17 (x +3) = Total score after 17th innings.

∴   X = 85 – 51 = 34

∴   Average after 17th innings = x + 3 = 34 + 3 = 37

A man can row 6 km/hr in still water. It takes him twice as long to row up as to row down the river. Find the rate of the stream. 

Solution:

Let man’s rate upstream = x km/hr

Then, man’s rate downstream = 2x km/hr

∴   Man’s rate in still water = 1/2 (x + 2x) km/hr

            ∴   3x/2 = 6 or x = 4 km/hr

Thus, man’s rate upstream = 4 km/hr

Man’s rate downstream = 8 km/hr

∴ Rate of stream = 1/2 (8 – 4 ) = 2 km/hr

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