Let the age of mother be M and that of her daughter be D Therefore, [M+D]/2 = 42 [M/D] =2/1 and Solving the above equations we get D = 28 yrs.
54 km/hr = 54 x (5/18) = 15m/s
72 km/hr = 72 x (5/18) = 20 m/s
Total distance = 150 + 200 = 350 m
Relative speed = 15 + 20 = 35 m/s
Total time = Total distance/Relative speed
= 350/35 = 10s
S.I on Rs.1800 = T.D on Rs.1872.
P.W on Rs.1872 is Rs.1800.
Rs.72 is S.I on Rs. 1800 at 12%.
Time = (100x72 / 12x1800)
= 1/3 year = 4 months.
Amount = Rs [7500x (1+4/100)²]
= Rs.(7500 x 26/25x26/25Rs)
= Rs.8112.
C.I = Rs (8112 - 7500)
= Rs.612.
Using the Pythagoras theorem, the radius of the cone = √ (172 - 152) = 8 cm.
Volume of cone = (1/3) x π x r2 x h= (1/3) x π x 82 x 15
Radius of hemisphere=radius of cone
Volume of hemisphere = (2/3) x π x r3 = (2/3) x π x 83
Total volume of figure = Volume of cone + Volume of hemisphere = 4233.58 cu. cm.
A deck of card has 52 cards of which 26 are black and the other 26 are red.
The number of face cards is 12, but only 6 of them are red.
So, required probability is 6/52 = 3/26
Let SP of each article be Re. 1
So, SP of 100 articles =Rs. 100
Implies profit = Rs. 50
So, CP = 100 – 50 = Rs. 50
Therefore, profit percentage of the shopkeeper is 50/50 × 100 = 100%
Speed Downstream = (15 + 3) km/hr
= 18 km/hr.
Distance travelled = (18 x 12/60) hrs
= 3.6km.
T3 = a + 3d
T6 = a + 6d
T3 + T6 = 2a + 9d = 27
Sum of first 10 terms of an AP is:
S= (10/2) (2a+9d)
= 5 x 27
= 135
Let x be the marked price,
So x - 32 = 320 X 1.15
x = 400.
So required value is
400 = 320 (1 + profit/100),
So profit is 25%
Let the number be x and (28 - x) = Then,
x (28 - x) = 192
‹=›x2 - 28x + 192 = 0.
‹=›(x - 16) (x - 12) = 0
‹=›x = 16 or x = 12.
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of An in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4) +9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.
As A completes half of the work in 10 days, he/she will complete the work in 20 days.
As B is half as efficient as A and C is half as efficient as B,
They will complete the work in 1/20 + 1/40 + 1/80 = 7/8@So 7/80 of the work in a single day.
So, they can complete the entire work in 80/7 days
Therefore, they can complete the remaining 50% of the work in (1/2) x 80/7 = 40/7 days.
We know, Average speed = Total distance travelled / Total time taken Average = [45+50+25]/ [3+2+ {25/15}] = 18 kmph
C.P = Rs. (100 / 75×34.80)
= Rs.46.40.
Let the three numbers be x, x+2, x+4
Then 3x = 2(x+4) + 3
‹=›x = 11
Third integer = x + 4 = 15.
Let log216 = n.
Then, 2n = 16 = 24
‹=› n=4.
Total weight of (36+44) Students = (36x40+44x35) Kg = 2980 kg.
Average weight of the whole class = (2980 / 80) = 37.25.
Aman : Rakhi : Sagar =(70,000 x 36):(1,05,000 x 30):(1,40,000 x 24)
=12: 15: 16.
We will group the letters that need to come together (A & E) and consider them as a single letter. So, here the letters are O, R, C, L, and AE.
Number of ways A & E can be arranged is 2!
So, the total number of ways in which the words can be formed so that all vowels are together is 5! x 2! = 240 ways.
T.D = [B.G x 100 / R x T]
= Rs. (6 x 100 / 12 x 1)
= Rs.50.
A/b=9/5; i.e. 5a = 9B.......... (i)
A-5/B-5 = 2/1; i.e. A-5 = 2B .........10
∴A - 2B = -5................... (ii)
From equation (i) and (ii), A = 45, B = 25
∴ Average =45+25/2 = 70/2 = 35.
Let breadth = x metres
Then, length = (x + 20) metres.
Perimeter = (5300 / 26.50) m
= 200m
Let the average after 17th inning = x. Then, average after 16th inning = (x - 3)
Average =16 (x-3) +87
= 17x or x= (87-48)
= 39.
Time from 5 a.m on a day to 10 p.m.on 4th day = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
Therefore 356 / 15 hrs of this clock = 24 hours of correct clock.
89 hrs of this clock = (24 x 15/356 x 89) hrs
= 90 hrs
So, the correct time is 11 p.m.
S.P = 85% of Rs.1400
= Rs. (85/100×1400)
Rs.1190.
Suppose he covers x km at 100 kmph
So he covers 170-x at 50 kmph
So {X/100}+{170-X}/50=2
Solving this equation, we get x = 140.
So he covers 30km at 50 kmph.
Clearly, we have l=9 and l+2b=37
Area = (l x b)
= (9 x 14) sq.ft = 126 sq.ft.
Let x be the new boys as well as girls, Therefore
[20+X]/ [32+X] =3/4
Solving this we get x = 16
So total will be 36 + 48 = 84.
Ratio of their shares = (35000×8): (42000×10)
= 2: 3.
Suganya's share = Rs. (31570 ×2/5)
= Rs.12628.
Principal = Rs.16, 000;
Time=9 months = 3 quarters;
Amount = Rs. [16000x (1+5/100)³]
= [16000x21/20x21/20x21/20]
= Rs.18522.
C.I = Rs. (18522 - 16000)
= Rs.2522.
Suppose the vessel initially contains 8 litres of liquid. Let x litters of this liquid be replaced with water.
Quantity of water in new mixture = (3 - 3x/8 + x) litres.
Quantity of syrup in new mixture = (5 - 5x/8) litres.
(3 - 3x/8 + x) = (5 - 5x/8) = 5x + 24 = 40 - 5x
=› 10x = 16 =› x = 8/5
So, part of the mixture replaced = (8/5 x 1/8) = 1/5.
Angle traced by hour hand in
5 hrs 10 min. = (360/12 x 6) °
= 180°.
Required decimal = 1/ 60 x 60
= 1/ 3600
= .00027.
Log 520 =20 log 5
=20 × [log (10/2)]
=20 (log 10 - log 2)
=20 (1 - 0.3010)
=20×0.6990
=13.9800.
Characteristics = 13.
Work done by A and B in 12 days is = 12 * 5/120 = 1/2
Therefore Remaining work = 1- 1/2 = 1/2work
B does 1/40 work in one day
Therefore B does 1/2 work in 40*1/2 =20days
Using the formula to calculate % decrease as [R/(100+R)]x100 where R = percentage increase in price, we get
Required % decrease in consumption = 25/125 MULTIPLIED 100 = 20%.
Speed Downstream = (13 + 4) km/hr
= 17 km/hr.
Time taken to travel 68 km downstream = (68 / 17) hrs
= 4 hrs.