Q3 . If (a/(b+c)) = (c/(b+a)) = (b/(a+c)), then each fraction is equal to
(a+b+c)^2
1/2
1/4
0
None of these
Q28. A cat take 5 leaps for every 4 leaps of a dog, but 3 leaps of dog are equal to 4 leaps of cat. What is the ratio of the speed of the cat to that of the dog?
11:15
15:11
16:15
15:16
11:16
Q46 . A varies jointly as B and C, and A = 6 when B = 3 and C = 2. Find A when B = 5, C = 7
17.5
35
70
105
Guys please give the explanation of the answer don't just reply back the answer, thanks in advance
Q3) I assumed a=b=c=1 and got all the fractions evaluating to 1/2. I think unequal values of a,b,c will not satisfy the condition given in Qn. So assuming a=b=c = any one number of ur choice wil always give 1/2 as answer. ****************** Q.28) Since cat and dog belong to different species let speed of cat be denoted in CL units and speed of dog be denoted in DL units to avoid any confusion.
Bring dog in cat's world so that DLs have to be converted into CLs GIven that every 3 DLs = 4 CLs So, 1 DL = 4/3 CLs
Given, speed of Cat = 5 CLs and speed of Dog = 4 DLs
Ratio of cat's speed to dog's speed = 5 CL / 4 DL = 5 CL / (4*4/3) CL = 15/16 ....Answer *******************
Q.46) A varies jointly as B and C means that A is (proportional to) B*C ...........means ......... If B and C will rise then A will rise , if B and C will fall then A will fall.(Rule of proportionality) To remove proportionality we put a constant.( Its again a rule) A = K (BC)
Now given in Qn that A = 6 when B = 3 and C = 2. This gives K = 1
So when B = 5 and C = 7 , we already have K = 1 So, A = 1*7*5 = 35.....................should be the answer. ***************************************
Q 2. The duration of a railway journey varies as the distance and inversely as the velocity, the velocity varies directly as the square root of the quantity of coal used per km, and inversely as the no carriages in the train. In a journey of 50 km in half an hour with 18 carriages, 100 kg of coal is required. How much coal will be consumed in a journey of 42 km in 28 min with 16 carriages
53.76
149.33
47.51
48.43
None of these
Q10. Brass is an alloy of copper and zinc. Bronze is an alloy containing 80% of copper, 4% of zinc and 16% of tin. A fused mass of brass and bronze is found to contain 74% of copper, 16% of zinc and 10% of tin. The ratio of copper to zinc in brass is
64 and 36
33 and 67
50 and 75
35 and 65
None of these
Q37. Mr AM, the maganimous cashier of XYZ ltd, while distributing salary, adds whatever money is needed to make the sum a multiple of 50. He adds rs 10 and rs 40 to A's and B's Salary resp. and then he realizes that the salaries of A, B and C are now in the ratio of 4:5:7 . The salary of c could be
Q2) V = Velocity , Q = quantity of coal , C = no. of carriage, K = Constant. Now,
The velocity varies directly as the square root of the quantity of coal used per km, and inversely as the no carriages in the train.
V = K*Q^1/2 / C 50/0.5 = K 100^1/2 / 18 Solving the above u will get the value of K = 180.
42*60/28 = 180*Q^1/2 / 16
Q = 64 answer
Q 10) the Tin has come in the alloy has come only from the bronze i.e 10% from the 16% in bronze. LCM of 10 and 16 = 80 so let the weight of the alloy be 800 grams. (easy for calculation)
copper is = 74/100*800 = 592 grams Zinc = 16*8 = 128 grams Tin = 10* 8 = 80 grams.
From the bronze we will find the contribution of each metal - Tin = 16 * 5 = 80 grams Copper = 80* 5 = 400 grams Zinc = 4*5 = 20 grams.
The residue has come from Brass: - Copper = 592 - 400 = 192 grams Zinc = 128 - 20 = 108 Grams
Copper : zinc = 192:108 = 64 : 36 = 16: 9 Ans
Q37 ) do the questn using options.
A:B: C = 4x: 5x: 7x
taking the option as C's salary as 2100 = 7x then X = 300.
B's salary will be 1500 - 40 = 1460 A's salary will be 1200 - 10 = 1190
the other options are not completely divisible by 7 and so will not meet the other conditions.
Hi I just started Preparing for CAT and i am looking for some shortcut techniques. If anybody has that then please share that with me.. In Addition to that i am facing a problem in Q.19 of number system(Practice Exercise).
N=202 x 20002 x 200000002 x 200000...2(15 zeros) x 20000....2(31 zeros).The sum of digits in this multiplication will be a 112 b 160 c 144 d cannot be determined
7 white balls & 3 black balls are randomly placed in a row. Find the probabilitythat no two black balls are placed adjacent to each other??
For this question after counting the total no. of possible arrangements i.e 120. What I did was clubbed two balls together and counted the no. of arrangements which I thought should have given the right answer but alas! NO.
I understand the other way of arranging the white balls in a row and selecting 3 out of 8 empty spaces i.e 8C3/120 which is the required probability but what was the problem in my approach.
@crazy4pagal said:7 white balls & 3 black balls are randomly placed in a row. Find the probabilitythat no two black balls are placed adjacent to each other??For this question after counting the total no. of possible arrangements i.e 120. What I did was clubbed two balls together and counted the no. of arrangements which I thought should have given the right answer but alas! NO.I understand the other way of arranging the white balls in a row and selecting 3 out of 8 empty spaces i.e 8C3/120 which is the required probability but what was the problem in my approach.i did 10!/3!7! - 9!/2!7! .Any1 ??
I guess you missed the arrangement of 3 white ball together that is why you are not getting right answer. So in my opinion as per your approach.... answer should be like, Total - (3 white together + 2 white together)
I dont think so as the arrangement would contain the possibility where the 2 clubbed balls will lie with the 3rd black ball which is alone. Thanks anyways.
@crazy4pagal said:7 white balls & 3 black balls are randomly placed in a row. Find the probabilitythat no two black balls are placed adjacent to each other??For this question after counting the total no. of possible arrangements i.e 120. What I did was clubbed two balls together and counted the no. of arrangements which I thought should have given the right answer but alas! NO.I understand the other way of arranging the white balls in a row and selecting 3 out of 8 empty spaces i.e 8C3/120 which is the required probability but what was the problem in my approach.i did 10!/3!7! - 9!/2!7! .Any1 ??
Firstly when you know the simpler approach why go for complicated stuff .
Coming to your solution ... this the right way to solve it
Total no of arrangements of 2 types of objects 1)white ball (no: 7) 2)Black ball(no: 3) = 10!/(7!3!) = 120 ..... (1)
When we take 2 black balls as 1 object, we get 3 types of objects basically: 1) White ball 2) one black ball 3) double black ball The number of arrangements possible for this is 9!/7! = 72 ......... (2) (since only 7 similar objects i.e white balls are there ...others are dissimilar)
Now in the above case there will be few arrangements where double black(object type 3) ball gets placed along side single black ball(object type 1) ....
say double ball is object 'A' say single black ball is object 'B'
possible arrangements where A and B are alongside are AB and BA but since they are nothing but 3 black balls we cannot distinguish them therefor must not be calculated separately....so we need to find the no of arrangements where 3 black balls are placed side by side = 8!/7! = 8 ...... (3)
So no of arrangements where no 2 black balls are placed side by side(N) = total no of arrangements - no of arrangements where 2 black balls are placed side by side + arrangements where 3 black balls are placed side by side
From (1) (2) and (3) ....we get N = 120 - 72 +8 = 56