Quant Booster by Hemant Malhotra  Set 13

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
A GPS device shows that car A travelling at a speed of 60 kph is located at a point whose coordinates are(100,90). Behind car A is car B, travelling in the same direction at a speed of 80 kph, that is located at a point whose coordinates are (20,30). After how many hours will the two cars be at the same point?
The distance between the initial positions of the cars is 100 km as (9030, 10020, 100) = 20(3, 4, 5) is a Pythagorean triple. The second car is traveling 20 kmph towards and relative to the first car so the two cars will be at the same point 100/20 = 5 hours after.
A drives to the station each day to pick her husband B who usually arrives on a train station at 6 o'clock. Last Monday, B finished work earlier, caught an earlier train and arrived at the station at 5 o'clock. He started to walk home and eventually met A who drove him the rest of the way, getting home 20 minutes earlier than usual. On Tuesday, he again finished early and found himself at the station at 5:30. Again he began to walk home, again he met A on the way, and she drove him home the rest of the way. Assume constant speed throughout with no wasted time for waiting, backing of the car, etc. How earlier than the usual time were they home on Tuesday?
No need to form any equations. 1 hour of walking saves 20 minutes of driving so half an hour of walking saves 10 minutes of driving.
Chian is between her home and the stadium. To get to the stadium, she can walk directly to the stadium or else she can walk home and then ride a bicycle to the stadium. She rides 7 times as fast as he walks and both choices require the same amount of time. What is the ratio of Chian's distance from her home to her distance to the stadium?
Let x and 1 be the distances between her current position and her home and the stadium, respectively.
So x/1+1/7 = (1x)/1,
x = 3/7.Kejrival fixes the price of his goods 25% above the cost price & allows 12.5% discount on it. His profit percentage is?
Method1 c.p.100
m.p.125
sp..125 * 7/8==109.375
so profit==9.37%Method2 let CP=1
MP=1+1/4=5/4
and 12.5%=1/8 discount
so 5/4 * 7/8=35/32
so profit=35/321=3/32 * 100=9.37%A car leaves P at 8 AM and travels to Q at a constant speed. A bus leaves Q at 8:45 AM and travels to P at a speed three times that of the car. If they meet at 10:00 AM, find the ratio of the distance traveled by the bus to the distance traveled by the car when they meet.
S = speed of car
D1 = distance of car
D2 = distance bus
D1=2s
D2=(23/4)(3s)
D2=15/4s
D1:D2
2s:15/4s
Ratio is 8:15A, B and C start simultaneously from X to Y. A reaches Y, turns back and meet B at a distance of 11 km from Y. B reached Y, turns back and meet C at a distance of 9 km from Y. If the ratio of the speeds of A and C is 3:2, what is the distance between X and Y
Let a, b, c are the speeds of A, B, C respectively;
let x be the distance between X and Y. Note that speed is proportional to distance.
We have
a/b = (x+11)/(x11) and b/c = (x+9)/(x9) and a/c = 3/2,
(x+11)/(x11) * (x+9)/(x9) = (a/b)(b/c) = a/c = 3/2,
2(x^2+20x+99) = 3(x^220x+99),
x^2100x+99 = 0,
(x1)(x99) = 0.
bcz x > 11, so x = 99 km.A beaker contained V litres of a mixture of milk and water, with milk and water in the ratio of 3 : 2. The total volume of the mixture was increased by 60% by adding water. Next, 38.4 litres of the solution in the beaker was replaced by water. If the final ratio of milk and water in the beaker is 3:7, then find the value of V (in litres)
a) 80
b) 96
c) 120
d) 192Milk=3v/5
and water=2v/5 +(v * 60/100)=2v/5+3v/5=v
now 38.4 liter removed
so MF=(8v/5 38.4)/(8v/5)
so milk=3v/5 * (8v/5 38.4)/(8v/5)'
now equate to ratio and answer = 192 x 5/8 = 120In how many ways can 6 letters A, B, C, D, E and F be arranged in a row such that D is always somewhere between A and B?
a 324
b 240
c 60
d 48ways to arrange D,A,B =3! but we need A,D,B or B,A ,D so out of 3! , 2 favorable cases , so out of 6! we have 2/3! * 6! = 6 * 5 * 4 * 2 = 240
Raju has forgotten his sixdigit id number, he remembers the following: the first two digits are either 1,5 or 2,6,the number is even and 6 appears twice. If Raju uses a trial and error process to find his ID number at the most, how many trial does he need to succeed?
when 1,5 at first two place
then last four places a b c d
when d=6 then out of 3 places one place will be filled by 6 so 3c1 and rest two have 9 choices so
3c1 * 9 * 9 = 243
now when d=0/2/4/8 so 4 choices so out of 3 places 2 places will be filled by 6 and one place have 9 choices so 3c2 * 9 * 4 = 108
now when 2,6 at first two places then d=6 then other 3 places have 9 * 9 * 9 = 729 choices
and when d=0/2/4/8 then one place should have 6 and rest 2 places have 9 choices so 4 * 3c1 * 9 * 9 = 972
so sum=2052If average of 20 double digit numbers is 18 , but becomes 21.8 when we interchange digits of a no , then how many original no's are possible ?
(10a1+b1+10a2+b2+....10a20+b20)=360
10(a1+a2+a3+...a20)+(b1+b2+...b20)=360
now
10*(b1+b2+b3+...b20)+(a1+a2+a3...a20)= 436.0
so 9(b1+b2+...b20)9(a1+a2+....a20)=76
so this is factor of 9 so not possible case