Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
P (drawing a white ball) =8/14=4/7.
Weight of 1 apple = 250 g
Weight of 6 apples = 250 * 6= 1.5 kg
Initial weight = 14.5 – 1.5= 13 kg
Number of apples in his carton initially :
= 13000/250
=52 apples
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then,120/x+480/y= 8 -> 1/x+4/y =1/15...(1)
And,200/x+400/y=25/3-> 1/x+2/y =1/24....(2)
solving(1)and(2)we get: x = 60 and y = 80.
Ratio of speeds=60:80=3:4.
Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/ n(S)=9/20.
n(n+1)/2
1+2+3+....+n
therefore average of these numbers=n+1/2
therefore required average
100+1/2=50.5
Let the number of students in rooms A and B be x and y respectively.
Then, x - 10 = y + 10 x - y = 20 .... (i)
and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
The required answer A = 100.
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, 5x + 3/4x + 3=11/9
9(5x + 3) = 11(4x + 3)
45x + 27 = 44x + 33
45x - 44x = 33 - 27
x = 6.
Anand's present age = 4x = 24 years.
From the given data, after servicing speed of the bike = 60 km/h
Distance covered in 6h. = (60 * 6) km = 360km
When it’s not service,the time taken to cover 360 km = (360 / 50) = 7.2h
Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 6 = 5(x - 6)
54 - x = 5x - 30
6x = 84
x = 14.
Son's age after 6 years = (x+ 6) = 20 years..
The tailor has to make 8 piece from 1 metre.
So for 1 piece ,the measurement will be 1/8=0.125 m
from 37.5 m tailor can get 37.5/0.125 = 300 pieces
Let the duration of the flight be x hours.
Then,600/x-600/x + (1/2)= 200
600/x-1200/2x + 1= 200
x(2x + 1) = 3
2x2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x= 1 hr.
Let distance = x km and usual rate = y kmph.
Then,x/y-x/y+3=40/60 -> 2y(y + 3) = 9x ....(i)
And,x/y -2-x/y=40/60 -> y(y - 2) = 3x ....(ii)
On dividing (i) by (ii), we get: x = 40.
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, (6x + 6) + 4/(5x + 6) + 4 =11/10
10(6x + 10) = 11(5x + 10)
5x = 10
x = 2.
Sagar's present age = (5x + 6) = 16 years.
Let the son's present age be x years. Then, man's present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.
Average marks obtained in Physics by all the seven students
=1/7x [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ]
=1/7x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ]
=1/7x [ 520% of 120 ]
=624/7
= 89.14.
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle=(61 -x)km.
So,x/4+(61 -x)/9= 9
9x + 4(61 -x) = 9 x 36
5x = 80
x = 16 km.
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) = n(E)/ n(S)=4/36 =1/9.
Aggregate marks obtained by Sajal
= [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ]
= [ 135 + 78 + 84 + 70 + 54 + 28 ]
= 449.
By solving the given two equations, we get the intersection point (12, - 12).
So, m = 12, n = -12
Hence, m + n = 12 – 12 = 0.
Total Score = Average * Number of matches
Total score of 13 matches = 13 × 42 = 546
Total score of first 5 matches = 5 × 54 = 270
Therefore, total score of last 8 matches = 546 - 270 = 276
Average = 276/8 = 34.5
Let the annual installment be rs. x
therefore (x + x*3*5/100)+(x + x*2*5/100)
+(x + x*1*5/100)+x =6450
=>115x/100+110x/100+105x/100+x=6450
=>115x+110x+105x+100x=6450*100
=>430x=6450*100
x=6450*100/430=rs.1500
Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20),
so there will be 6 significant digits to the right of the decimal point.
Required average=8100 + 9500 + 8700 + 9700 + 8950/5
=44950/5
= 8990.
Let the actual distance travelled be x km.
Then, x/10=x + 20/14
14x = 10x + 200
4x = 200
x= 50 km.
Let the distance travelled by x km.
Then,x/10-x/15= 2
3x - 2x = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr =[60/10]hrs= 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed =[60/5]kmph.= 12 kmph.
Let the number of hens be x and the number of cows be y.
Then, x + y = 48 .... (i)
and 2x + 4y = 140 x + 2y = 70 .... (ii)
Solving (i) and (ii) we get: x = 26, y = 22.
The required answer = 26.
No. of revolutions = Distance/circumference.
Distance = 6.6 × 1000 = 6600 metres.
Circumference = 2 × 22/7 × 21/2 = 66 metres.
No. of revolutions = 6600/66 = 100 revolutions.
Let Abhay's speed be x km/hr.
Then,30/x-30/2x= 3
6x = 30
x = 5 km/hr.
Apply the percentage formula.
The percentage increase in the area will be P + Q + PQ / 100.
So we get the answer as 20 +40 + 20 × 40/100 = 68%.
Father+Mother=2*35=70 years
Father+Mother+Son=27*3=81 years
therefore Son's age=81-70=11 years