50614.If 12% of x is equal to 6% of y, then 18% of x will be equal to how much % of y ?
7%
9%
11%
none of these
Explanation:
We have ,
12% of X = 6% of Y
=> 2% of X = 1% of Y
=>(2 x 9)% of X = ( 1 x 9)% of Y
Thus, 18% of X = 9% of Y.
50615.If a number is 20% more than the another, how much % is the smaller number less than the first ?
12(1/3)%
16(2/3)%
16(1/3)%
none of these
Explanation:
Take a number 100,
Then the other number is 120
% the smaller number is less than the first = [(20/(120)) x 100]% = 16(2/3)%.
50616.If the given two numbers are respectively 7% and 28% of a third number, then what percentage is the first of the second ?
20%
25%
18%
none of these
Explanation:
Here, l = 7 and m = 28
Therefore, first number = l/m x 100% of second number = 7/28 x 100% of second number = 25% of second number
50617.A man walks 6 km at a speed of 1 1/2 kmph, runs 8 km at a speed of 2 kmph and goes by bus another 32 km. Speed of the bus is 8 kmph. Find the average speed of the man.
4 (5/6) kmph
3 (5/6) kmph
5 (7/6) kmph
None of these
Explanation:
Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 32 km at a speed of 8kmph
time taken indivisually:
=> 6/1.5 = 4 m
=> 8/2 = 4 m
=> 32/8 = 4 m
Average speed of man= total distance/ total time
=> 46/12 = 3 (5/6)
50618.Two men start together to walk a certain distance, one at 4 kmph and another at 3 kmph.The former arrives half an hour before the latter. Find the distance.
6 km
16 km
9 km
8 km
Explanation:
Let the distance be x km. Then,
x/3 - x/4 =1/2
(4x - 3x)/12 = 1/2
x = 6 km
50619.One type of liquid contains 25 % of benzene, the other contains 30% of benzene. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the new mixture.
28 %
29 %
30 %
27 %
Explanation:
Let the percentage of benzene =X
(30 - X)/(X- 25) = 6/4 = 3/2
=> 5X = 135 or
X = 27 so,
required percentage of benzene = 27 %
50620.A jar contains a mixture of two liquids Acid( nitric acid) and Base(ammonium chloride) in the ratio 4 : 1. When 10 litres of the mixture is taken out and 10 litres of liquid Base is poured into the jar, the ratio becomes 2 : 3. How many litres of liquid Acid was contained in the jar?
14 lts
18 lts
20 lts
16 lts
Explanation:
%age of liquid Base in the original mixture
=1/5 x 100 = 20%
In the final mixture % of the liquid Base
= 3/5 x 100 = 60%
Now using the rule of allegation
Hence reduced quantity of the first mixture and the quantity of mixture B which is to be added are the same.
Total mixture = 10 + 10 = 20 liters and quantity of liquid A = 20/5 x 4 = 16 lts
50621.A factory employs skilled workers, unskilled workers and clerks in the proportion 8 : 5 : 1 and the wage of a skilled worker, an unskilled worker and a clerk are in the ratio 5 : 2 : 3. When 20 unskilled workers are employed, the total daily wages of all amount to Rs. 3180. Find the daily wages paid to each category of employees.
Rs. 2100, Rs. 800, Rs. 280.
Rs. 2400, Rs. 480, Rs. 300
Rs. 2400, Rs. 600, Rs. 180.
Rs. 2200, Rs. 560, Rs. 420.
Explanation:
Skilled workers, unskilled workers and clerks are in the proportion 8 : 5 : 1
Given 20 unskilled workers, So 5/14 x K = 20, K = 56,
Therefore there are 32 skilled workers, 20 unskilled workers and 4 clerks
Ratio of amount of 32 skilled workers, 20 unskilled workers and 4 clerks
= 5 x 32 : 2 x 20 : 3 x 4
= 160 : 40 : 12 or 40 : 10 : 3
Now, divide Rs 3,180 in the ratio 40 : 10 : 3
We get, Rs. 2400, Rs. 600, Rs. 180
50622.The soldiers in two armies when they met in a battle were in the ratio of 10 : 3. Their respective losses were as 20 : 3 and the survivors as 40 : 13. If the number of survivors in the larger army be 24,000, find the original number of soldiers in each army ?
28000, 8400
25000, 7500
29000, 2750
26000, 7800
Explanation:
Let the soldiers in the two armies be 10X, 3X and losses be 20Y, 3Y,
Then we have,
10X - 20Y = 24000 ...(i)
And 3X - 3Y = 24000 x 13/40 = 7800
Solving, we have 10X = 28000, 3X = 8400
