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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