Let the numbers be 6x and 6y where x and y are prime to each other.
Then, LCM = 6xy
(6x + 6y) / 6xy = 12/35
35(x+y) = 12xy
⇒ x=5, y=7
Smallest number=6x=6*5=30
Correct Option 4)
A, B and C together can complete certain work in 24 days. A and C together work twice as much as B while A and B together works four times the as much as C, then A alone can do the same work in
OPtion
1) 72 days
2) 120 days
3) 48 days
4) 30 days
5) 36 days
6) 180/7 days
7) 360/7 days
8) 90/7 days
9) 270/7 days
10) None of these
Solution
Let A, B and C can do the work in a, b and c days respectively.
Then, 1 day's work of A, B & C = 1/a + 1/b + 1/c = 1/24 ----(i)
1/a + 1/c = 2/b ----(ii)
1/a + 1/b = 4/c ----(iii)
Now subtracting (ii) from (i), we get 1/b = 1/72
Similarly subtracting (iii) from (i), we get 1/c = 1/120
Substituting values of 1/b and 1/c in (i), 1/a = 7/360
.'. A alone can do the work in 360/7 days.
Correct Option 7)
A bus left point P for point Q. 70 minutes later train left P for Q and arrived at Q at the same time as the bus. If the bus and the train left simultaneously from the opposite ends P and Q towards each other, they would have met in 84 minutes after the start. How much time did it take the bus to travel from P to Q?
Let distance from P and Q = d ; Speed of Bus = x ; Speed of Train = y
Let, time taken by bus to travel distance d = t min. and by train = (t-70) min.
Speed = Distance/Time
x = d/t
y = d/(t-70)
When travelling towards each other time taken=84 min.
⇒ Distance covered by bus in 84 min + Distance covered by train in 84 min = Total distance
(d/t)*84 + [d/(t-70)]*84 = d
84/t + 84/(t-70) = 1
t² - 238t + 5880 = 0
(t-28) (t-210) = 0
t=28, 210
The time 't' can not be less than 70 min as train starts 70 minutes after the bus starts and they meet after some time, .'. t = 210 min = 3 hrs 30 min.
Correct Option 7)
If 1/x - 1/y = -29 and the value of (x + 12xy - y) / (x - 6xy - y) can be expressed as m/n, where m and n are co prime positive integers. The value of m+n=?
Series is formed by multiplication by 3 and subtraction of 2, 4, 8, ....
7 x 3 - 2 = 19
19 x 3 - 4 = 53
53 x 3 - 8 = 151
151 x 3 - 16 = 437
437 x 3 - 32 = 1279
In a End of Season Sale, to clear stock a shopkeeper offers 50 items in price of 20 items but still there is no loss or gain in this transaction. What is his usual gain percentage when there is no sale ?
Given, shopkeeper sells 50 items in price of 20 items but there is no loss or gain.
⇒ CP of 50 items = SP of 20 items
Let CP of 1 item=1, so CP of 50 items=50=SP of 20 items
.'. SP of 1 item=50/20=2.5
% profit = [(2.5-1)/1]*100 = 150%
Correct Option 4)
Given, one number=45 and LCM=60HCF. Let other number=x
We know, product of LCM & HCF = Product of two numbers
LCM*HCF = 45x
60HCF*HCF = 56x ----(i)
Also, LCM + HCF = 549
60HCF + HCF = 549
61HCF = 549
⇒ HCF=9
Now substituting HCF=9 in Eqn (i), we get 60*9*9 = 45x
⇒ x=108
If P is a three digit number where first digit is three times the last digit and no digit is repeated. How many P are possible which are divisible by 6 ?
Given, 1st digit is 3 times the 3rd digit. .'. The 3-digit number can be of the form 3 _ 1 ; 6 _ 2 ; 9 _ 3
Now, for a number to be divisible by 6, it should be an even number divisible by 2 and 3 both.
.'. Only possible numbers will be even numbers of the form 6 _ 2
So, the numbers are 612, 642, 672, Total 3
Correct Option 2)
An alloy contains magnesium, copper and zinc in the ratio 3 : 4 : 2 and another alloy contains copper, zinc and nickel in the ratio 4 : 5 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of copper per kg in new alloy will be:
In 1st alloy: Ratio of magnesium, copper, Zinc = 3 : 4 : 2
In 2nd alloy: Ratio of copper, zinc, nickel = 4 : 5 : 3
Now, let the weight of first alloy be 36 kg (taken as Magnesium=12 kg, Copper=16 kg, Zinc=8 kg) and
Weight of second alloy = 36 kg (taken as Copper=12 kg, Zinc=15 kg, Nickel=9 kg)
When two alloys are mixed in equal weights, total weight of third alloy=36+36=72 kg
Weights in 3rd alloy: Magnesium=12 kg, Copper=(16+12) kg, Zinc=(8+15) kg, Nickel=9 kg.
Weight of Copper in 3rd alloy = 28 kg per 72 kg
.'. Weight per kg = 28/72 kg = 388.88 gm..
A leak in the bottom of a tank can empty it in 8 hours. An inlet pipe fills the tank at the rate of 6 liters per minute. When the tank is full, the inlet is opened but leak emptied the tank in 9 hours. The capacity of the tank (in liters) is
Let the capacity of tank = x liters
Tank emptied per hour by leakage = x/8
Tank filled per hour by inlet pipe = 6*60 = 360 litres
Effective filling per hour with inlet pipe & leakage = 360 - x/8
From the given data, in 9 hrs, full tank get emptied with inlet and leakage
.'. x + 9*(360 - x/8) = 0
(9/8)*x - x = 3240
x = 25920
Correct Option 5)