If A men working A hours a day can do A units of a work in A days, then B men working B hours per day would be able to complete, how many units of work?
More men, more units (Direct)
More working hours, more units (Direct)
More days, more units (Direct)
Men ------------ A:B
Working hrs ---- A:B --- :: A:P
Days ----------- A:B
A banker paid Rs.5767.50 for a bill of Rs.5840, drawn on April 4, at 6 months. On what day was the bill discounted, the rate of interest being 7%?
OPtion
1) 2 August
2) 9 August
3) 4 August
4) 8 August
5) 6 August
6) 7 August
7) 1 August
8) 3 August
9) 5 August
10)None of these
Solution
B.D. = Rs.(5840 - 5767.20) = Rs.72.80
=> Rs.72.80 is S.I. on Rs.5840 at 7%
So unexpired time = 100*72.80 / 7*5840 years = 13/73 = 65 days
Now, date of draw of bill = April, 4 at 6 months
Nominally due date = October, 4
Legally due date = October, 7
So, we must go back 65 days from October, 7
Oct. Sept. Aug.
7+ 30+ 28
i.e., The bill was discounted on 3rd August.
A boat covers 24 km upstream and 36 km downstream in 6 hours, while it covers 36 km upstream and 24 km downstream is 13/2 hours. The velocity of the current (in km/hr) is.
Let speed upstream & speed downstream be x km/hr and y km/hr respectively. Then
24/x + 36/y = 6 and 36/x + 24/y = 13/2
=> 24u + 36v = 6 and 36u + 24v = 13/2 (u = 1/x and v = 1/y)
Adding we get
u = 1/8 and v = 1/2
=> x = 8 and y = 12
Velocity of current = 1/2 (12-8) km/hr = 2 km/hr
The area of a rectangle is thrice that of a square. Length of the rectangle is 40 cm and the breadth of the rectangle is 3/2 times that of the side of the square. The side of the square in cm is
A lump of two metals weighing 18 gms. is worth Rs.87 but if their weight be interchanged, it would be worth Rs.78.60. If the price of one metal be Rs.6.70 per gm., find the weight of the other metal in the mixture.
If one lump is mixed with another lump with the quantities of metals interchanged then the mixture of the two lumps would contain 18 gm. of first metal and 18 gm. of second metal and the price of the mixture would be Rs. (87+78.60) or Rs. 165.60
=> Cost of (18 gm of 1st metal + 18 gm of 2nd metal)
= Rs. 168.60
So cost of (1 gm of 1st metal + 1 gm of 2nd metal)
= Rs. 165.60/18 = Rs. 9.20
Now mean price of lump = Rs. (87/18) per gm = Rs. (29/6)
By alligation rule
Quantity of 1st metal/Quantity of 2nd metal = 14/6 : 56/30 = 5:4
In 9 gm of mix, 2nd metal = 4 gm
In 18 gm of mix 2nd metal = 4/9 *18 = 8 gm
Option 3)
A bill falls due in 9 months. The creditor agree to accept immediate payment of half and to defer the payment of the other half for 18 months. He finds that by this arrangement he gains Rs.4.50. What is the amount of the bill, if money be worth 4%.
Let the amount of bill be Rs.200. Then, according to the agreement, the creditor agree to pay after 9 month, an amount which is the sum of the amount of Rs. 100 for 9 months and P.W. of Rs. 100 due 9 months hence.
Now, S.I. = Rs.(100*3*4 / 4*100) = Rs.3
=> Amount = Rs.(100+3) = Rs.103
Also, P.W. = Rs.(100*100 / (100+(4* 3/4) = Rs.(10000/103)
Total amount after 9 months = Rs.(103 + 10000/103) = Rs. 20609/103
Gain = Rs.(20609/103 - 200) = Rs.(9/103)
If gain is Rs.(9/103), sum due = Rs.200
If gain is Rs.9/2, Sum due = Rs.(200*103*9 / 9*2) = Rs.10300
Two metallic right circular cones having their height 4.1 cm and 4.3 cm and the radii of their bases 2.1 cm each, have been melted together and recast into a sphere. The diameter of the sphere is
OPtion
1) 4.2 cm
2) 4.8 cm
3) 2.5 cm
4) 1.4 cm
5) 1.8 cm
6) 3.9 cm
7) 3.5 cm
8) 5.8 cm
9) 6.3 cm
10)None of these
Solution
Sum of the volume of the cones = volume of the sphere
1/3*Pi*2.1^2*4.1 + 1/3*Pi*2.1^2*4.3 = 4/3*Pi*R^3
Solve
Diameter = 4.2 cm
Let length = l and breadth = b.
Let, new breadth = x
Then, (110% of l)*x = lb
=> 110/100 lx = lb
x = 10/11 b
Decrease in breadth = [(b - 10/11 b)*1/b*100]% = 100/11 %
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