The base of a triangular
field is three times its altitude. If the cost of cultivating the field at Rs.
24.68 per hectare be Rs. 333.18, find its base and height.
Sol: Area of the field = Total cost/Rate = (333.18/24.68) hectares =13.5
hectares.
= (13.5*10000) m^2 =135000m^2.
Let altitude = x meters and base = 3x meters.
Then, ½ *3x* x= 135000 or x^2 = 9000 or x= 300.
Therefore, base =900 m & altitude = 300m.
Find the area of a rhombus
one side of which measures 20cm and one diagonal
24cm.
Sol: Let, other diagonal = 2x cm,
Since halves of diagonals and one side of rhombus form a right angled triangle
with side as hypotenuse, we have:
(20)^2 =(12)^2+x^2 or x=Ö(20)^2-(12)^2 =Ö256=16 cm.
Therefore, other diagonal = 32 cm.
X alone can do a piece of
work in 15 days and Y alone can do it in 10 days. X and Y undertook to do it for
Rs. 720. With the help of Z they finished it in 5 days. How much is paid to Z?
Sol. In one day X can finish 1/15th of the work.
In one day Y can finish 1/10th of the work.
Let us say that in one day Z can finish 1/Zth of the work.
When all the three work together in one day they can finish 1/15 + 1/10 + 1/Z =
1/5th of the work.
Therefore, 1/Z = 1/30.
Ratio of their efficiencies = 1/15: 1/10: 1/30 = 2: 3: 1.Therefore Z receives
1/6th of the total money.
According to their efficiencies money is divided as 240: 360: 120.
Hence, the share of Z = Rs. 120.
How many number of times
will the digit ‘7' be written when listing the integers from 1 to 1000?
Sol:7 does not occur in 1000. So we have to count the number of times it appears
between 1 and 999. Any number between 1 and 999 can be expressed in the form of
xyz where 0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any
of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the
second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit,
2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits (
0 to 9 with the exception of 7). There will be 9 such numbers. However, this
digit which is not 7 can appear in the first or second or the third place. So
there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is
written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in
it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300
P can give Q a start of 20
seconds in a kilometer race. P can give R a start of 200 meters in the same
kilometer race. And Q can give R a start of 20 seconds in the same kilometer
race. How long does P take to run the kilometer?
Solution:
P can give Q a start of 20 seconds in a kilometer race. So, if Q takes 'x'
seconds to run a kilometer, then P will take x – 20 seconds to run the
kilometer.
Q can give R a start of 20 seconds in a kilometer race. So, if R takes 'y'
seconds to run a kilometer, then Q will take y – 20 seconds to run the
kilometer.
We know Q takes x seconds to run a kilometer
Therefore, x = y – 20
Therefore, P will take x – 20 = y – 20 – 20 = y – 40 seconds to run a kilometer.
i.e. P can give R a start of 40 seconds in a kilometer race, as R takes y
seconds to run a kilometer and P takes only y – 40 seconds to run the kilometer.
We also know that P can give R a start 200 meters in a km race.
This essentially means that R runs 200 meters in 40 seconds.
Therefore, R will take 200 seconds to run a km.
If R takes 200 seconds to run a km, then P will take 200 – 40 = 160 seconds to
run a km.
A and B enter in to a
partnership and A invests Rs. 10,000 in the partnership. At the end of 4 months
he withdraws Rs.2000. At the end of another 5 months, he withdraws another
Rs.3000. If B receives Rs.9600 as his share of the total profit of Rs.19,100 for
the year, how much did B invest in the company?
Solution:
The total profit for the year is 19100. Of this B gets Rs.9600. Therefore, A
would
get (19100 – 9600) = Rs.9500.
The partners split their profits in the ratio of their investments.
Therefore, the ratio of the investments of A : B = 9500 : 9600 = 95 : 96.
A invested Rs.10000 initially for a period of 4 months. Then, he withdrew
Rs.2000.
Hence, his investment has reduced to Rs.8000 (for the next 5 months).
Then he withdraws another Rs.3000. Hence, his investment will stand reduced to
Rs.5000 during the last three months.
So, the amount of money that he had invested in the company on a money-month
basis
will be = 4 * 10000 + 5 * 8000 + 3 * 5000 = 40000 + 40000 + 15000 = 95000
If A had 95000 money months invested in the company, B would have had 96,000
money months invested in the company (as the ratio of their investments is 95 :
96).
If B had 96,000 money-months invested in the company, he has essentially
invested
96000/12 = Rs.8000
A 20 litre mixture of milk
and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture
is removed and replaced with pure milk and the operation is repeated once more.
At the end of the two removal and replacement, what is the ratio of milk and
water in the resultant mixture?
Solution:
The 20 litre mixture contains milk and water in the ratio of 3 : 2. Therefore,
there will be 12 litres of milk in the mixture and 8 litres of water in the
mixture.
Step 1. When 10 litres of the mixture is removed, 6 litres of
milk is removed and 4 litres of water is removed. Therefore, there will be 6
litres of milk and 4 litres of water left in the container. It is then replaced
with pure milk of 10 litres. Now the container will have 16 litres of milk and 4
litres of water.
Step 2. When 10 litres of the new mixture is removed, 8 litres
of milk and 2 litres of water is removed. The container will have 8 litres of
milk and 2 litres of water in it. Now 10 litres of pure milk is added.
Therefore, the container will have 18 litres of milk and 2 litres of water in it
at the end of the second step.
Therefore, the ratio of milk and water is 18 : 2 or 9 : 1.
A zookeeper counted the
heads of the animals in a zoo and found it to be 80. When he counted the legs of
the animals he found it to be 260. If the zoo had either pigeons or horses, how
many horses were there in the zoo?
Solution:
Let the number of horses = x
Then the number of pigeons = 80 – x.
Each pigeon has 2 legs and each horse has 4 legs.
Therefore, total number of legs = 4x + 2(80-x) = 260
=>4x + 160 – 2x = 260
=>2x = 100
=>x = 50.
A group of workers can do a
piece of work in 24 days. However as 7 of them were absent it took 30 days to
complete the work. How many people actually worked on the job to complete it?
Solution:
Let the original number of workers in the group be 'x'
Therefore, actual number of workers = x-7.
We know that the number of manhours required to do the job is the same in both
the cases.
Therefore, x (24) = (x-7).30
24x = 30x - 210
6x = 210
x = 35.
Therfore, the actual number of workers who worked to complete the job = x - 7 =
35 -7 = 28.
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