Quant Basics for Zeros in Quant - CAT 2012

MBA CAT 2024
201

The remainder obtained when 12345678987654321 is divided by 1001
a)10
b)309
c)692
d)995
easy approach:divide 1001 by 11 its perfectly divisible and 12345678987654321 leaves a remainder of 1 when divided by 11 so the resulting remainder shud also leave remainder 1 when divided by 11 which is only satisfied in d case of 309

202

Hi.... Last two digits of the following numbers.....
65*29*37*63*71*87(units will be 5,but how to find tens?:|)
a.05 b.95 c.15 d.25
75*35*47*63*71*87*82...
a.50 b.70 c.30 d.90
Do let me know the solution puys.

203
Hi.... Last two digits of the following numbers.....
75*35*47*63*71*87*82...
a.50 b.70 c.30 d.90
Do let me know the solution puys.


Sol:- The concept here is to just find last 2 digits in any multiplication. Then multiply the 2 digit numbers we get till we are left with only one 2 digit number.

(75*35)*(47*63)*(71*87)*82
Now i am finding only last two digits when the numbers in braces are multiplied.
25*61*77*82
(25*61)*(77*82)
Now repeat the same process
25*14 = 50
So 50 are the last 2 digits of given product.
204
Hi.... Last two digits of the following numbers.....
65*29*37*63*71*87(units will be 5,but how to find tens?:|)
a.05 b.95 c.15 d.25
Do let me know the solution puys.


Sol :- Similarly we solve this one now.
(65*29)*(37*63)*(71*87)
85*31*77 = (85*31)*77 = 35 * 77 = 95
205

:thumbsup::cheerio: thank you very much...

206

@HYDROCKER....... u r truly rocking sir...... plz post more concepts...... U have to keep this thread and our CAT hopes alive......:cheers: .......

207
:clap:Agreed....
208

Find the number of terms in the expansion of (a+b+c+d+e)^22.........
:cheers:

209
Find the number of terms in the expansion of (a+b+c+d+e)^22.........
:cheers:


Sol :- In the expansion of (a+b+c+d+e)^22, every term has a combination of the four variables of the form
(a^i)(b^j)(c^k)(d^l)(e^m)
where i, j, k, l and m are non-negative integers and
i+j+k+l+m = 22
So our problem is to find the number of solutions to this equation.
Note that a similar classic problem is the following:
How many ways can 22 pieces of candy be divided among 5 children?

Using formula we get Number of terms = (22+5-1) C (5-1) = 26C4 = 14950 ??
210

a person bought some oranges at the rate of 5 per rupee . he bought the same number of oranges at the rate of 4 per rupee . he mixes both the type of and sells at 9 for 2 rupee in this business he bears a loss of 3 rupee find out hw many oranges he bought in all

211
bankaspirant Says
a person bought some oranges at the rate of 5 per rupee . he bought the same number of oranges at the rate of 4 per rupee . he mixes both the type of and sells at 9 for 2 rupee in this business he bears a loss of 3 rupee find out hw many oranges he bought in all


My take : 1080
212
bankaspirant Says
a person bought some oranges at the rate of 5 per rupee . he bought the same number of oranges at the rate of 4 per rupee . he mixes both the type of and sells at 9 for 2 rupee in this business he bears a loss of 3 rupee find out hw many oranges he bought in all




Hii...

1rs--->5oranges ( type 1)
1orange--->1/5rs

1rs---->4 oranges(type 2)
1orange--->1/4rs

2rs---->9oranges (selling)
1orange---->2/9rs
let there are n oranges of type 1 and n type 2 then
(n/5+n/4)-2/9(2n)=3
n(9/20-4/9)=3
n=540

then 2n=1080
213

yup ur ans is correct

214
Hii...

1rs--->5oranges ( type 1)
1orange--->1/5rs

1rs---->4 oranges(type 2)
1orange--->1/4rs

2rs---->9oranges (selling)
1orange---->2/9rs
let there are n oranges of type 1 and n type 2 then
(n/5+n/4)-2/9(2n)=3
n(9/20-4/9)=3
n=540

then 2n=1080


yup ur ans is correct
n thanx 4 d explanation
215
sol :- similarly we solve this one now.
(65*29)*(37*63)*(71*87)
85*31*77 = (85*31)*77 = 35 * 77 = 95


dnt get plz explain in details.....
216
gemini11 Says
dnt get plz explain in details.....


Just multiply both the numbers which are in braces and get only last 2 digits of the product. Don't solve it for the entire result.
217

Lets Just Solve some Simple Interest - Compound Interest Problems :-

SET - 1

1. Find the Simple Interest on Rs 750 in 4 years at 15% per annum
2. On what sum of money will the Simple Interest be Rs. 2000 in 5 years at 8% Per Annum?
3. A man invested Rs. 20,000 at 10% P.A. Rs. 15,000 at 12% P.A. and some money at 15% P.A. If the total annual interest recieved is Rs. 56,000, find the money invested by him at 15% P.A ?
4. On a sum of money the rate of interest is 5% Per Annum for the first 3 years, 6% Per Annum for the next 4 years, and 8% Per Annum for the next years beyond the first 7 Years. If the interest obtained in 12 Years is Rs. 3,950, Find the Sum?
5. A sum was put at 5% at a certain rate for 5 Years. Had it been put at 3% Per Annum higher rate, it would have fetched Rs. 900 more. Find the Sum?

218

SET - 2 :-

6. On a certain Sum of Money the Simple Interest in 2 years at 10% P.A is Rs. 125, what would be the S.I if the rate of Interest will be 12% P.A in 3 Years?
7. The S.I on Rs 6400 at 12 1/2 % per annum is Rs. 2000. Find the Period ?
8. The S.I on Rs. 6000 in 3 Years and 4 Months is Rs. 3000. Find the rate percent Per Annum ?
9. Manish took a loan of Rs. 4000 at S.I. After 2 Years he cleared the loan by paying Rs. 5600. Find the Rate % P.A?
10. A lent Rs. 25000 to B for 4 years and Rs. 40,000 to C for 3 1/2 years and got Rs. 24,000 S.I from both B and C. Find the rate PCPA

219

1. Rs 450
2. Rs 5000
4. Rs 5000(not sure)
6. Rs 225
7.2.5 years
9.20%

220

My take
1. Rs 450
2. Rs 5000
4. Rs 5000(not sure)
6. Rs 225
7.2.5 years
9.20%;)