The number of ordered pairs of positive integers (a, b) satisfying the

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The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 is:

A. 13
B. 14
C. 15
D. 16
E. 17

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GMATinsight · Answer

2a + 3b = 100

First solution is (a, b) = (50, 0) NOT ACCEPTABLE

OTHER SOLUTION WILL BE (47, 2)
OTHER SOLUTION WILL BE (44, 4)
And so on

Now, the value of a must differ by co-efficient of y which is 3

i.e. Total values of a = 50/3 = 16 points

But (50,0) is not acceptable hence

Total acceptable points = 16-1 = 15

Answer: OPTION C

exc4libur · Answer

2a+3b=100, 3b=100-2a
b=100-2a/3=integer>0
100-2a=multiple(3)
100-2a=99, 2a=1, a=1/2≠integer
100-2a=96, 2a=4, a=2
100-2a=93, 2a=7, a≠integer
100-2a={even multiples of 3}
multiples of 3 from 1 to 100: {99-3}/3+1=33
even multiples: 16

Ans (D)

madgmat2019 · Answer

2a + 3b = 100

All the even no's below 34 will be acceptable for b

b=2,4,6,8,10......32

Total 16 pairs of a,b..are available for above equation

OA:D

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freedom128 · Answer

2a + 3b = 100 ---> Ordered pairs of positive integers (a, b) include (2,32), (5,30), (8,28), ..., (47,2)

Number of ordered pairs (u1=2, un=47, k=3): 
un= u1+(n-1)*k
47= 2+(n-1)*3
45= (n-1)*3
n=16

FINAL ANSWER IS (D)

unraveled · Answer

The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 is:

A. 13
B. 14
C. 15
D. 16
E. 17

Since both a and b are positive integers, starting with least values of both till maximum value helps find the range respectively.
Here, value of a would increase in the number that is equal to the value of coefficient that b has i.e. 3 and vice versa 
For a = 2, b = 32
a = 5, b = 30
.....

a = 47, b = 2

Total values =  \frac{47 - 2}{3} + 1  OR =  \frac{32 - 2}{3} + 1  = 16

Answer D.

ArunSharma12 · Answer

The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 is:

A. 13
B. 14
C. 15
D. 16
E. 17

Constraint: a,b > 0

Maximum value of b could be 33 when a is 0. so b has to be < 33.
b has to be even so that rest of sum (100 - 3b = even) could be assigned in terms of 2a.

values of b: 2,4,6...,32; Total 16 values
Ans: D

eakabuah · Answer

We are to find the number of ordered integer pairs of solutions for 2a+3b=100
2a+3b=100
Let there be at least n number of solutions, then n(2*3)<100 n<50/3 implying there are at least 16 solutions.
 2a+3b=100
a=50-3b/2
when b=2, a=47
when b=32, 
a=50-48=2
Note that the values of the variable a increase by the coefficient of the variable b.
So, the number of ordered positive integer pairs of solutions = (47-2)/3 + 1 = 15+1=16

The answer is therefore D.

Jawad001 · Answer

The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 is:

A. 13
B. 14
C. 15
D. 16
E. 17

Given:
2a + 3b = 100
Required:
2a + 3b = 100
2a = 100-3b
a = 50-(3/2)*b……………(1)
The value of a to be positive on condition that b must be multiple of 2.

When, b = 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32 in the equation (1) 
a = 50-(3/2)*b = 50-(3/2)*2 = 47 
In this way there are 16 ordered pairs of positive integers.
Answer: D

lacktutor · Answer

The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 is:

—> a = 50 —\frac{3b}{2} 
 a >0  —>  50 —\frac{3b}{2}> 0 
—>  b < 33.(3) 
( b must be between 0 and 33, inclusive)
 a = 50 —\frac{3b}{2} 
—>  b_1= 2 , —>  a= 47 
—>  b_2= 4  —>  a = 44 
....
—>  b_n = 32  —>  a = 2

b_n = b_1 +(n—1)d 
 32 = 2 + (n—1)2 
 15 = n—1 
 n= 16

The answer is D.

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Kinshook · Answer

Asked: The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 is:

(a,b) = (20,20) is a solution
a will change by 3 and b will change in opposite direction by 2 for each possible solution.

(a,b) = {(20,20), (23,18),(26,16),(29,14),(32,12),(35,10),(38,8),(41,6),(44,4),(47,2),(17,22),(14,24),(11,26),(8,28),(5,30),(2,32)}

The number of ordered pairs of positive integers (a, b) satisfying the equation 2a + 3b = 100 = 16

IMO D

NihaarB · Answer

2a+3b=100
for a and b belonging to integers, 2a will be even, thus 3b must also be even.
The greatest number that is divisible by 3 and less than 100 is 96.
The number of values satisfying the equation for b (as this is eq is linear, the no of ordered pairs will be the same as the solutions of b) be n.
32 = 2 +2(n-1)
(we have to start from 2 as b cannot be 0)
32 = 2n
n = 16.
Therfore, the number of ordered pairs to the equation is 16

Ans: D)

The number of ordered pairs of positive integers (a, b) satisfying the

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GMAT Problem Solving (PS) Questions

The number of ordered pairs of positive integers (a, b) satisfying the

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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