A seller has two types of items in a store. He allows a discount of 15% on an item whose marked price is Rs. 480 and 25% on an item whose marked price is Rs. 720. If a customer paid total of Rs. 10824 and gets a total discount of Rs. 3096, then how many items of marked price Rs. 720 he has purchased ?
B is more efficient than A and they together can complete a work in 18 days. Had A done 60% of the work and then B, the remaining work, then the work would have been completed in 39 days. B alone will complete 60% of the same work in
OPtion
1) 21 days
2) 24 days
3) 30 days
4) 27 days
5) 36 days
6) 45 days
7) 12 days
8) 18 days
9) 15 days
10) None of these
Solution
Let A and B alone takes 'a' and 'b' days respectively, then 1 day work of A & B = 1/a + 1/b = 1/18 ---(i)
When A does 60% or 3/5 work and B remaining 2/5 work, total days required = 39
⇒ (3/5)/(1/a) + (2/5)/(1/b) = 39 ---(ii)
Substituting 1/a=(1/18 - 1/b) from (i) into Eqn (ii), we get (3/5)/[(1/18 - 1/b)] + (2/5)/(1/b) = 39
54b/[5(b-18)] + 2b/5 = 39
54b + 2b(b-18) = 39*5(b-18)
2b² - 177b + 3510 = 0
2b² - 117b - 60b + 3510 = 0
b(2b - 117) - 30(2b - 117) = 0
(b - 30) (2b - 117) = 0
Considering +ve integer value of the days, b=30
⇒ B alone can complete the work in 30 days.
.'. For 60% of the work B require = (60/100)*30 = 18 days.
A rectangular tank is 20 m long and 21 m deep. If 10,000 liters of water is drawn off the tank, the level of the water in the tank goes down by 1.5 m. How many liters of water can a tank hold ?
1 m³ = 1000 litres
.'. 10,000 litres = 10 m³
Let the breadth of a tank = B meters, then volume of water in 1.5 m height = 20*1.5*B = 10 m³
30B=10 ⇒ B=1/3 m
Volume of a tank = L*B*H = 20*(1/3)*21 = 140 m³
Capacity in Litres = 140*1000 = 1,40,000
Option 2) is correct.
A man travels 840 km in 12 hours, partly by train and partly by car. If he had travelled all the way by train, he would have saved 1/4 of the time he would have taken to travel by car only and would have arrived at his destination 1 hour 30 minutes early. Find the distance he covered while travelling by car.
OPtion
1) 480 km
2) 380 km
3) 520 km
4) 540 km
5) 420 km
6) 360 km
7) 600 km
8) 270 km
9) 300 km
10) None of these
Solution
Total Distance = 840 km.
Total Time taken by Train and Car = 12 Hours
By travelling all the distance by train, he would have reached 1.5 hrs early i.e. He would have taken 12 - 1.5 = 10.5 hrs.
.'. Speed of the train = 840/10.5 = 80 kmph
Time taken by train to travel 840 km = (1 - 1/4)* Time taken by car to travel 840 km = 10.5
⇒ Time taken by car to travel 840 km = 10.5/(3/4) = 14 hrs
.'. Speed of car = 840/14 = 60 kmph
Let the distance covered by car = x km and by train = (840-x) km
Then, x/60 + (840-x)/80 = 12
Solving, we get x=360 km
The word GOODFRIDAY has 10 different letters
When the letters DAY are kept together they can be supposed to form one letter.
Then we have to arrange letters GOODFRI (DAY)
Now 8 letters (7+1) can be arranged in 8!/2! (O being repeated twice)
The letters DAY can be arranged among themselves in 3! ways.
.'. Required number of ways = (8!/2!)*3! = 120960
Let the numbers be x and y, then (x-y) : (x+y) : xy = 1 : 5 : 24
(x-y)/(x+y)=1/5 ⇒ 2x=3y ---(i)
(x+y)/xy=5/24 ⇒ 24x + 24y = 5xy ----(ii)
On solving Eqn. (i) & (ii), we get x=12, y=8
.'. Required product x*y = 12*8 = 96
A packet contains some blue, red and green pens such that the probability of picking a blue pen is 2/9 and probability of picking a green pen is 4/9. If number of red pens is 15 and if all the pens are numbered starting from 1, 2, 3, ... and so on, then what is the probability of getting one pen numbered as multiple of 7 or 11 ?
Given, P(Blue)=2/9, P(Green)=4/9 and number of red pens=15
As we know, P(Blue) + P(Red) + P(Green) = 1
P(Red) = 1 - P(Blue) - P(Green) = 1 - (2/9) - 4/9 = 3/9 = 1/3
As number of red pens = 15 = 1/3 of total pens
.'. Total number of pens=15*3=45 ⇒ Pens are numbered from 1 to 45
Numbers that are multiple of 7 = 7, 14, 21, 28, 35, 42
And numbers that are multiple of 11 = 11, 22, 33, 44
So, favourable outcomes = 6+4 = 10
So, probability of getting pen of multiple 7 or 11 = 10/45 = 2/9
Train A whose length is four-fifth of that of train B crosses it travelling in opposite direction in a time which is 4/7 th of the time taken by train A to cross it when travelling in same direction. Calculate the ratio of the speeds of train A and train B.
Let lengths of trains B = x, then length of train A = (4/5)*x
Distance covered in each case =x + (4/5)*x = (9/5)*x
Let the speed of train A and train B be S1 and S2 respectively.
Relative speed in same direction = (S1-S2)
Relative speed in opposite direction = (S1+S2)
According to given condition, Time taken to cross train B in opposite direction = (4/7)*Time taken to cross train B in same direction
(9/5)*x / (S1 + S2) = (4/7)*[(9/5)*x / (S1 - S2)]
7(S1 - S2) = 4(S1 + S2)
3S1 = 11S2
S1/S2 = 11/3
The person who can solve mathematical problems,can lead life easily ,Maths tellu that every problem as solution...we can find number of solutions for one problem ...Aplly it in the life also we can lead our life happily...
swetha
"IF YOU DON'T KNOW MATHS , IT MEAN YOU ARE FAR AWAY FROM LIFE."