Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K.
(a) 9 (b) 8 (c) 7 (d) none of these.
is the ans 8 ??
Q69 -> For the above questions, for how many values of K wil the ratio be highest.
(a) 9 (b) 8 (c) 7 (d) none of these
i think...this is... only 1 value..i.e., 100 ??
Q85 -> Find last two digits of the following number
(201*202*203*204*246*247*248*249)^2
(a) 36 (b) 56 (c) 76 (d) 16
i think its multiplication ..after division by 100..so keep on considering last two digits..
(1*2*3*4*46*47*48*49)^2
(24*46*47*48*49)^2
(24*62*52)^2
(24*24)^2
76^2
76
shivam_01 Saysnumber is 199 and the answer is 8
lostanand Sayscan u pls explain the steps?
no solid method i think..for this..
but still..
take 999..as 3 digit number
999/27 = 37
i need to chk if any has still less so reduce to
899/26 799/25 n so ..
199/19 = 10.47
hi guys..
i have some doubts in logarithms:
1) If log 3= .4771, find log(.81)^2* log(27/10)^2/3 divided by log 9.
pls help me solve this.... somehow i'm not able to arrive at the answer..:huh:
options are:
(a) 2.689
(b)-0.0552
(c)2.2402
(d)None of these
can someone help me with the following probs..
Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K.
(a) 9 (b) 8 (c) 7 (d) none of these.
hi i did this through a little bit of error and trial method
let x, y, z, be the hundreds, tens and units digit.
clearly we want the ratio of 100x+10y+z/x+y+z
now we want this ratio to be least .
for this two objectives must be fulfilled
numerator should be least and denominator should be max
the max effect will be when we when the units digit is highest eg 100/1 and 109/10.hence deno is max
similiary in order to have less numerator to be min let the hundred digit be 1
now in case of tens digit just check a litle by putting 1 and 9 .thus we see 199 is the no ..
hope this solves some of the doubt
can someone help me with the following probs..
Q89 -> The super computer at Ram Mohan Roy Seminary takes an i/p of a no. N & a X where X is a factor of N. In a particular case N is equal to 83p796161q & X is equal to 11 where 0
p
therefore
q-p+3>3
now the max value that could occur is 11
for this q=9 and p=1
thus when divided by q+p=10 the remainder will be the last digit that is 9
and when divided by p= 1 it will be zero there by sum of remainders will be 9(d)
hi guys..
i have some doubts in logarithms:
1) If log 3= .4771, find log(.81)^2* log(27/10)^2/3 divided by log 9.
pls help me solve this.... somehow i'm not able to arrive at the answer..:huh:
options are:
(a) 2.689
(b)-0.0552
(c)2.2402
(d)None of these
Q89 -> The super computer at Ram Mohan Roy Seminary takes an i/p of a no. N & a X where X is a factor of N. In a particular case N is equal to 83p796161q & X is equal to 11 where 0
1) The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.
Lactose 0.16
Maltose 0.32
Glucose 0.74
Sucrose 1.00
Fructose 1.70
Saccharin 675.00
What is the minimum amount of sucrose ( to the nearest gram) that must be added to one-gram of saccharin to make a mixture that will be at least 100 times as sweet as glucose?
a) 7
b) 8
c) 9
d) 100
2) A young girl Roopa leaves home with x flowers, goes to the bank of nearby river. On the bank of river, there are four places of worship, standing in a row. She dips all the x flowers into the river.The number of flowers doubles. Then she enters the first place of worship, offers y flowers to the diety. She dips the remaining flowers into river, and again the number of lowers doubles. She goes to the second place of worship, offers y flowers to the diety. She dips the remaining flowers again into the river, and again the number doubles. She goes to the third place of worship, offers y flowers to the diety. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the fourth place of worship, offers y flowers to the diety. Now she is left with no flowers in hand.
The minimum number of flowers that ould be offered to each deity is:
a) 0
b) 15
c) 16
d) cannot b determined
Hi
Solutions for both the questions are
1. Sweetness of the resultant mixture will be 74 as sweet as sucrose. So after applying allegation rule the sucrose and saccharine should be mixed in the ratio 601:73. Since we are taking 1 gm of saccharine we will have the quantity of sucrose as 8.2 gm nearly.
2. After devoting flowers to the last god she will be left with
16x-15y, number of flowers.
where, x=Initial number of flowers with her
y=Number of flowers she devoted to each God
so, 16x-15y=0
For y to be minimum x=15 and y=16
therefore, ans, is 16 flowers.
hi guys i have this question please help me this
How many numbers smaller than 2.10^8 and are divsible by 3 can be written by means of the digits 0,1 and 2 (exclude single digit and double digit numbers)?
hi guys i have this question please help me this
How many numbers smaller than 2.10^8 and are divsible by 3 can be written by means of the digits 0,1 and 2 (exclude single digit and double digit numbers)?
hey ppl wantd 2 kno if der was a link wher i cud get concepts n sums on inequalities chapter...plz if u cud help me wid dat.. i face probs solvin d inequa ch sums very often...
thnx in advance...
MaskedMenace Sayscan u be clear wid tht....is tht 2*10^8 ??
find the sum of first 50 terms of series 1,2,5,10,17......
find the sum of first 50 terms of series 1,2,5,10,17......
divyaakanksha Saysfind the sum of first 50 terms of series 1,2,5,10,17......