Search found 9 matches
As the denominator value increases the value of the ratio decreases.So once should choose the value in the lower range.So the answer is 133.
- by avanishjoshi
Sun Jun 21, 2009 11:30 pm- Forum: Problem Solving
- Topic: Test Prep Ratios
- Replies: 6
- Views: 1875
As the denominator value increases the value of the ratio decreases.So once should choose the value in the lower range.So the answer is 133.
- by avanishjoshi
Sun Jun 21, 2009 11:27 pm- Forum: Problem Solving
- Topic: Test Prep Ratios
- Replies: 6
- Views: 1875
Let the capacity of machine A = A sprockets/hr Let the capacity of machine B = B sprockets/hr Let B work for X hours to produce 660 sprockets Then A(X + 10) = B(X) But B produces 10% more sprockets per hour than A .Therefore B = 1.1A. A(X + 10) = 1.1A(X) = > X+10 = 1.1X = > X = 100. [spoiler]A (100 ...
- by avanishjoshi
Thu Jun 18, 2009 11:01 pm- Forum: Problem Solving
- Topic: Work Rate Problem
- Replies: 4
- Views: 2953
Dividend = Divisor * Quotient + remainder Therefore X = YQ + 9 Now given that X/Y = 96.12 , that means the when X is divide by Y the quotient is 96.12 Therefore X = Y(96.12) + 9 But the product of divisor and the quotient should always be a integer. Thefore for Y(96.12) to be an integer, from the gi...
- by avanishjoshi
Tue Jun 16, 2009 1:13 am- Forum: Problem Solving
- Topic: OG PS #106 12th edition - Properties of numbers
- Replies: 5
- Views: 1425
Say 15 = YN + Remainder = YN + Y-3
18 = Y(N + 1).
Now Y will be a integer for N = 1,2,5,8 and 17
so the corresponding values of Y are 9,6,3,2 and 17 .
- by avanishjoshi
Thu Jun 04, 2009 4:02 am- Forum: Problem Solving
- Topic: remainders
- Replies: 4
- Views: 1270
Another approach is as follows: LCM * GCD = product of two numbers. If the two numbers are 3x and 4x then 120 * GCD = 12X^2. Now substitute the options. (A) 5 ie 120 * 5 = 12X^2 => X^2 = 50 which is not possible as square of a natural number cannot be 50 (B) 10 ie 120*10 = 12X^2 => X^2 = 100 which i...
- by avanishjoshi
Tue May 26, 2009 9:49 am- Forum: Problem Solving
- Topic: If the least common multiple
- Replies: 6
- Views: 5340
You need to look at the common solution that will satisfy the inequality. so if (Z-5)(Z+1) > 0,. then the area covering Z - 5 >0 and Z+1 > 0 or Z- 5 < 0 and Z+1<0 will satisfy the inequality. Now if we choose Z - 5 > 0 and Z + 1 > 0 then the common area that satisfies this equation is z> 5 as repres...
- by avanishjoshi
Mon May 25, 2009 7:30 am- Forum: Problem Solving
- Topic: z² - 4z > 5
- Replies: 7
- Views: 1882
I have solved the first one using the alligation.I donno how to solve the second one. Wine solution contains 3 wine and 1 water. We are adding water to make 1/2 wine. 3(parts of wine in first solutn) 0(parts of wine in 2nd soln ie water) 1/2 0- 1/2 1/2-3 0 minus 1/2 : 1/2 -3 gives the ratio of the w...
- by avanishjoshi
Mon May 25, 2009 7:18 am- Forum: Problem Solving
- Topic: Mixture & Percentage problem
- Replies: 4
- Views: 3340
Jack Jill At end of 1st game 600 200 At end of 2nd game 300 500 At end of 3rd game 550 250 At end of 4th game 275 525 Therefore jack has to pay 275 chips to jill.
- by avanishjoshi
Sun May 24, 2009 4:06 am- Forum: Problem Solving
- Topic: Pls help to sovle
- Replies: 4
- Views: 1209