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17 Jul 2026, 11:17
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The correct answer is C. 36
Lets look at the question.
3v+14b=vb
b<15
Lets solve the equation,
vb-3v-14b=0
to make the last part contain the same term, lets add 42 (14*3)
v(b-3)-14b+42=42
v(b-3)-14(b-3)=42
(v-14)(b-3)=42
here v-14 and b-3 should be the factor of 42
Now lets look at the factors of 42- (42,1),(21,2),(14,3),(7,6),(6,7)
1. 42,1
v-14=42, b-3=1
v=56,b=4
total 60- given in the option
2. 21,2
v-14=21,b-3=2
v=35,b=5
total 40, given in the option
3.14,3
v-14=14,b-3=3
v=28, b=6
total 34, given in the option
4. 7,6
v-14=7,b-3=6
v=21,b=9
total= 30
5. 6,7
v-14=6, b-3=7
v=20, b=10
total 30
All the possible options are 30,34, 40, 60. only 36 is left.
Lets look at the question.
3v+14b=vb
b<15
Lets solve the equation,
vb-3v-14b=0
to make the last part contain the same term, lets add 42 (14*3)
v(b-3)-14b+42=42
v(b-3)-14(b-3)=42
(v-14)(b-3)=42
here v-14 and b-3 should be the factor of 42
Now lets look at the factors of 42- (42,1),(21,2),(14,3),(7,6),(6,7)
1. 42,1
v-14=42, b-3=1
v=56,b=4
total 60- given in the option
2. 21,2
v-14=21,b-3=2
v=35,b=5
total 40, given in the option
3.14,3
v-14=14,b-3=3
v=28, b=6
total 34, given in the option
4. 7,6
v-14=7,b-3=6
v=21,b=9
total= 30
5. 6,7
v-14=6, b-3=7
v=20, b=10
total 30
All the possible options are 30,34, 40, 60. only 36 is left.
i did it like this:
3v+14b = v*b
rearrange to vb-3v=14b
v(b-3)=14b
v = 14b/(b-3)
b<15 & b cannot be 0 & b is positive integer.
list out possible values of b-3 = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
are any of these divisors of 14? 7, 2, 1 are divisors of 14
list values for b for these cases: 10, 5 or 4 hence v = 20, 35 or 56 hence v+b = 30, 40 or 60 ruling out options A, D & E
now checking for cases when b-3 is not divisor of 14 hence it is divisor of b to get integer value for v
possible such values of b-3: 11, 10, 9, 8, 7, 6, 5, 4, 3 hence b = 14, 13, 12, 11, 10, 9, 8, 7, 6
how many of these cases are such that b-3 is divisor of b? 3 is divisor of 6, b=6 hence v = 28, v+b = 28+6=34 rule out B
by process of elimination answer correct: C
another way to look at it is:
b and (b-3) are integers. any common factor of two numbers has to be a factor of difference between them. b & (b-3) has a difference of 3. so any common factor they share must be factor of 3. 3 only has 1 and itself as factors therefore b & (b-3) cannot share any common factor other than 1 and 3.
14 is prime factorized into 2*7
so (b-3) must have any one or more combinations of prime factors from: {2,7,3}. since the least common multiple (LCM) each of these share is 42, even if all three are present in (b-3), it would still be factor of 42. and any combination of 2,7,3 that (b-3) may be will be factor of 42. thus (b-3) must be factor of 42 .
v = 14b/(b-3)
we know b<15
factors of 42 are : 1,2,3,6,7,14,21,42
check each case:
b-3 = 1 => b=4 => v = 14(4) = 56 => v+b = 56+4 = 60 E is out
b-3 = 2 => b=5 => v=14(5)/2 = 35 => v+b = 35+5 = 40 D is out
b-3 = 3 => b=6 => v=14(6)/3 = 28 => v+b = 28+6 = 34 B is out
b-3 = 6 => b = 9 => v=14(9)/6 = 21 => v+b = 21+8 = 30 A is out
b-3 = 7 => b = 10 => v=14(10)/7=20 => v+b = 20+10 = 30
b-3 = 14 => b = 17 is not valid. correct answer is C
3v+14b = v*b
rearrange to vb-3v=14b
v(b-3)=14b
v = 14b/(b-3)
b<15 & b cannot be 0 & b is positive integer.
list out possible values of b-3 = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
are any of these divisors of 14? 7, 2, 1 are divisors of 14
list values for b for these cases: 10, 5 or 4 hence v = 20, 35 or 56 hence v+b = 30, 40 or 60 ruling out options A, D & E
now checking for cases when b-3 is not divisor of 14 hence it is divisor of b to get integer value for v
possible such values of b-3: 11, 10, 9, 8, 7, 6, 5, 4, 3 hence b = 14, 13, 12, 11, 10, 9, 8, 7, 6
how many of these cases are such that b-3 is divisor of b? 3 is divisor of 6, b=6 hence v = 28, v+b = 28+6=34 rule out B
by process of elimination answer correct: C
another way to look at it is:
b and (b-3) are integers. any common factor of two numbers has to be a factor of difference between them. b & (b-3) has a difference of 3. so any common factor they share must be factor of 3. 3 only has 1 and itself as factors therefore b & (b-3) cannot share any common factor other than 1 and 3.
14 is prime factorized into 2*7
so (b-3) must have any one or more combinations of prime factors from: {2,7,3}. since the least common multiple (LCM) each of these share is 42, even if all three are present in (b-3), it would still be factor of 42. and any combination of 2,7,3 that (b-3) may be will be factor of 42. thus (b-3) must be factor of 42 .
v = 14b/(b-3)
we know b<15
factors of 42 are : 1,2,3,6,7,14,21,42
check each case:
b-3 = 1 => b=4 => v = 14(4) = 56 => v+b = 56+4 = 60 E is out
b-3 = 2 => b=5 => v=14(5)/2 = 35 => v+b = 35+5 = 40 D is out
b-3 = 3 => b=6 => v=14(6)/3 = 28 => v+b = 28+6 = 34 B is out
b-3 = 6 => b = 9 => v=14(9)/6 = 21 => v+b = 21+8 = 30 A is out
b-3 = 7 => b = 10 => v=14(10)/7=20 => v+b = 20+10 = 30
b-3 = 14 => b = 17 is not valid. correct answer is C
17 Jul 2026, 11:41
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i am going with option c.
as (v-14)(b-3)=42
b<15, b-3 must be a positive divisor of 42, that is less than 12 (1,2,3,6,7)
the only possible totals are 30,34,40 and 60.
hence, all other options are possible, and c is the total number of vehicles that cannot be used.
as (v-14)(b-3)=42
b<15, b-3 must be a positive divisor of 42, that is less than 12 (1,2,3,6,7)
the only possible totals are 30,34,40 and 60.
hence, all other options are possible, and c is the total number of vehicles that cannot be used.
