Quant by Arun Sharma

MBA CAT 2024
2118
guys i m gonna prepare for CAT 2011........ do i need 2 go thru books of 8th, 9th and 10th or can i start with Arun Sharma...............

thx

CAToholic 2011

No one gives it to u , u have to take it.....!!!

You can start with Arun Sharma. It has explained the basics pretty well. If you still think you aren't able to follow the concepts well enough, go through the school books. Don't think it would be required though.
2119
haha.. you took 9+x/90+x and I took 10+x/100+x
x's value will keep changing if you took different values..
so there's one missing information in this question.
the answer can indeed be 2.5 if initial vol was 90, if initial vol was 100 ans is 2.85 and so on..
You can't just say how much volume of water to change its concentration from 10 to 12.5% you have to say "in how much volume" ie total volume initially to solve uniquely.
cos for that matter i can take 13/130 initially that will give me a different answer.
check here.


i guess we must assume the initial solution as 100..:lookround: so 2.85 wud be the answer
2120

Please help me with the following problem:

The sum of the series : 1/2 + 1/6 + 1/12 + 1/20 .... + 1/156 + 1/182 is :
a) 12/13 (b) 13/14 (c) 14/13 (d) none of these

Thanks in advance

2121
Please help me with the following problem:

The sum of the series : 1/2 + 1/6 + 1/12 + 1/20 .... + 1/156 + 1/182 is :
a) 12/13 (b) 13/14 (c) 14/13 (d) none of these

Thanks in advance

The answer should be 13/14 (option B).

The series can be looked at in the following way:

1/(1x2) + 1/(2x3) + 1/(3x4) + ....... + 1/(12x13) + 1/(13x14)

Now, reforming it,

(1-1/2) + (1/2-1/3) + (1/3-1/4) + ........ + (1/12-1/13) + (1/13-1/14)

You can clearly see all terms cancel out except the first and the last term, i.e. 1 - 1/14, which is equal to 13/14.
2122

Puys............!!! any 1 starting ARUN SHARMA for CAT 2011.........do reply........!!

2123

Q the dimensions of a room r 16m x 12m x 9m. it has 2 doors of size 4m x 2m n one almirah of size 6m x 4m. find d cost of covering the walls by wallpaper which is 40cm wide at Rs1.25 per m

a) Rs 580 b) Rs 630 C) Rs 1450 D) 1050
ans is c bt how???

2124
Q the dimensions of a room r 16m x 12m x 9m. it has 2 doors of size 4m x 2m n one almirah of size 6m x 4m. find d cost of covering the walls by wallpaper which is 40cm wide at Rs1.25 per m

a) Rs 580 b) Rs 630 C) Rs 1450 D) 1050
ans is c bt how???

We need to cover the four walls of the room, minus the space covered by the doors and the almirah.

Wall 1: 16x9 = 144 sq m
Wall 2: 12x9 = 108 sq m
Wall 3: 16x9 = 144 sq m
Wall 4: 12x9 = 108 sq m

Door 1: 4x2 = 8 sq m
Door 2: 4x2 = 8 sq m
Almirah: 6x4 = 24 sq m

Total: 504 - 40 = 464 sq m

Now, the wallpaper is 40 cm = 0.4 m wide

So, total wallpaper required = 464/0.4 = 1160 m

Thus, cost of wallpaper = 1160x1.25 = Rs. 1450
2125

thnx for the sol bt i hv a doubt..area of four walls is 2(l+b)h...after using this formula , if i subtract doors n almirah, y m i nt gting the ans den??

2126

thnx for the sol bt i hv a doubt if i use 2(l+b)h formula n den subtract will dat b wrong bec i ws using this formula n i ws nt gting d ans..

2127
Pl help me out.

Ten tickets are numbered 1,2,3....10.Six tickets r selected at random one at a time with replacement.The probability that the large number appearing on the selected tickets is 7 is:

A] (7^6-1)/10^6 B] (7^6 - 6^6)/10^6 C] 6^6/10^6 D] (7^6 + 6^6)/10^6 E] none of these.

pl explain with solution.
thanx

The answer is B) (7^6-6^6)/10^6.
The 6 tickets with 7 as highest can be selected in 7^6 ways. Now these also contain the case in which we have all the tickets lower than 7 i.e. 6^6. So we subtract it from 7^7.
2128
mukherjipooja Says
thnx for the sol bt i hv a doubt..area of four walls is 2(l+b)h...after using this formula , if i subtract doors n almirah, y m i nt gting the ans den??

Use your formula with the figures, you would get 504 sq m as the area of four walls. I guess you must have made some calculation error.
2129

In a right angled tirangle, find the hypotenuse if the base and perpendicular are respectively 36015cm and 48020cm.

a) 69125cm

b) 60025cm

c) 391025cm

d) 60125cm

e) None of these

I think we all know the obvious pythagoras method but that will be very tedious. Is there any other method?

2130
In a right angled tirangle, find the hypotenuse if the base and perpendicular are respectively 36015cm and 48020cm.

a) 69125cm

b) 60025cm

c) 391025cm

d) 60125cm

e) None of these

I think we all know the obvious pythagoras method but that will be very tedious. Is there any other method?

You don't need to be intimidated by the huge numbers, make it small as per your convenience.

Keep the base as 36 and perpendicular as 48. Thus, the sum of their squares:

1296 + 2304 = 3600

So the hypotenuse would be slightly more than 60 (square root of 3600). Most times, you would find the answer this way, but since there are two close options, we need to check further. Check the last couple of digits we left, 15 and 20 respectively. The sum of their squares is:

225 + 400 = 625, whose square root is 25.

Thus, the final answer is 60025 😃 Not very tough if you know how to calculate squares quickly 😁
2131

q1 a cube whose edge is 20cm long has circle oneach of its faces painted black.wat is total area of the unpainted surface of d cube if d circles r of the largest possible area??? ans- 514.28

q2 find d length of d longest pole dat cn b placed in an indoor stadium 24m long , 18m wide n 16m high??ans-34m

help me wid d sol...
thnx

2132
q1 a cube whose edge is 20cm long has circle oneach of its faces painted black.wat is total area of the unpainted surface of d cube if d circles r of the largest possible area??? ans- 514.28

q2 find d length of d longest pole dat cn b placed in an indoor stadium 24m long , 18m wide n 16m high??ans-34m

help me wid d sol...
thnx



Area of cube = 6 * 20 * 20 =2400

Area of circle = (22/7)*10 * 10 radius of circle will be 10


so required area will be 2400-(area of circle)= 514.28
2133

Hi,

Could anyone please tell me the answer to geometry question no. 24 from exercise geometry LOD1 ? It's got a diagram and I can post that here.

2134

The expression 333^555 + 555^333 is divisible by

a. 2
b. 3
c. 37
d. 11
e. All of these

2135

Puys can you tell how to solve this problem:

The sum of the series as represented as:
(1/1*5)+(1/5*9)+(1/9*13)...+(1/221*225)

I am not able to solve any of the sums from 31-36 of Progressions exercise:( If some one can post the solution for all of these 5 sums i will be glad:)
Are there any special formulas for these as they are nt in AP or GP.

2136
The expression 333^555 + 555^333 is divisible by

a. 2
b. 3
c. 37
d. 11
e. All of these

It is divisible by 2.
333^555 gives 3 as last digit and 555^333 gives 5 as last digit. So the last digit is even i.e. 8.
It is not divisible by 11 and hence answer cannot be All of the above.
2137
Puys can you tell how to solve this problem:

The sum of the series as represented as:
(1/1*5)+(1/5*9)+(1/9*13)...+(1/221*225)

I am not able to solve any of the sums from 31-36 of Progressions exercise:( If some one can post the solution for all of these 5 sums i will be glad:)
Are there any special formulas for these as they are nt in AP or GP.

The series can be written as
1/4(1-1/5+1/5-1/9+....+1/221-1/225).
So the answer is 224/4*225 = 56/225