gaurav200x said:
how can i be wrong in the 1st one... the last digits have to be 72 to be divisible by 3. 32, 52 or 92 can't be

and for the last one... I too thought about I and III but i dun think I is completely true... bcoz 1 is not included.

_________________________________________________________________

LET ME ANSWER FOR Q. 1) -

The given number should be divisible by both 9 & 4
Thus, as per the rule ,
I] DIVISIBILITY TEST FOR 4 -
A number is divisible by 4 when the number formed by the last two rigjht hand digits is divisible by 4 or if the last two digits are 0's.

II] DIVISIBILTY TEST FOR 9 -
A number is divisible by 9 when the sum of its digits is divisible by 9.

Therefore, 3+0+5+1+4+3+2+A = 18+A

If we take 9 (Option 4), then sum of digits (i.e. 27/9 = 3) is divisible by 9
[as per II] &
last two right hand digits (i.e. 92/4 = 23) is divisible by 4 [as per I].

Hence, answer is 4] 9. .
 

gaurav200x

Gaurav Mittal
Thanks Rahul, i did give it a deeper thought later on, and realised the mistake but since there was no power, i couldn't come online.
 

neeraj_uchila

New member
If a and b are two prime nos it can take the values
78 as it is the addition of two prime nos 61,17i.e 61+17=78
Similarly 7 can be expressed as the addition of 2 and 5
The last no 23 cannot be expressed as the addition of any two prime nos.
So the correct option is 3 i.e 1 and 2 only
 
SQUARES AND CUBES OF NUMBERS

NOTE : In Q. 1 & Q. 2, (a-b){2} means, square of (a-b), likewise, follow other statements also...

1) a and b are negative integers such that b > a. Which of the following statements(s) is/are true?

I. (a - b){2} > 0
II. (a – b){3} > 0
III. (a – b){3} < 0
IV. (a + b){3} > 0
V. (a + b){3) < 0

1] I and II only.
2] II and III only.
3] I, III and V only.
4] I, III and IV only.

2) If x < 0, y > 0 then, which of the following relations is necessarily true?

1] y/x{2} > 0
2] y/x{6} > x{3}/y
3] y{2} + x{2} > x{3} + y{4}
4] xy/x{2}y > 1

3) Which of the following is true?

I. Sum of two squares can not end in 2 and 8.

II. Sum of two cubes can not end in 3 and 9.

III. Sum of a cube and a square can not end in 0.

1] I only.
2] II only.
3] III only.
4] None of these.

4) Which of the following operations will make the number 847 a perfect square?

I. Add 53 to the number.
II. Multiply the number by 7.
III. Divide the number by 11.
IV. Divide the number by 7.
V. Subtract 6 from the number.

1] I, II, III, IV.
2] I, II, IV, V.
3] I, II, III, V.
4] I, II, III, IV, V.
 

neeraj_uchila

New member
The ans to question no 3 is d)none of these as
9+49=58 which ends in 8 and 81+1=82 which ends in 2 so 1 statement in not true
Similarly 1+8=9 which is the sumof the cubes of 1 and 2 ends in 9 ans 216+27 which is the sum of cubes of 6 and 3 ends in 9
also the sum of 100+1000 ends in 0 so the third statement is false
 

neeraj_uchila

New member
The ans to the 4 problem is option 2)I, II, IV, V.
847+53=900 which is the square of 30.
847*7=5929 which is the sqaure of 77.
847/11=77 which is not a perfect square.
847/7=121 which is the square of 11.
847-6=841 which is the square of 29
PS:plz give the ans for the first 2 questions
 
neeraj_uchila said:
The ans to question no 3 is d)none of these as
9+49=58 which ends in 8 and 81+1=82 which ends in 2 so 1 statement in not true
Similarly 1+8=9 which is the sumof the cubes of 1 and 2 ends in 9 ans 216+27 which is the sum of cubes of 6 and 3 ends in 9
also the sum of 100+1000 ends in 0 so the third statement is false

good, neeraj , answer for Q. 3 is correct...
good going...

:tea:
 
neeraj_uchila said:
The ans to the 4 problem is option 2)I, II, IV, V.
847+53=900 which is the square of 30.
847*7=5929 which is the sqaure of 77.
847/11=77 which is not a perfect square.
847/7=121 which is the square of 11.
847-6=841 which is the square of 29
PS:plz give the ans for the first 2 questions

Thanx, neeraj. Good answer.
i'm not sure abt this question...
let gaurav give answers...for 1 & 2...

these r so simple yaar... try again !
 

neeraj_uchila

New member
sorry man i did not read the note for the 1st 2 questions.

The ans for the 1 st question is 3] I, III and V only.

For the 2 nd question I am getting the 1) and2)statement as always true
 
neeraj_uchila said:
sorry man i did not read the note for the 1st 2 questions.

The ans for the 1 st question is 3] I, III and V only.

For the 2 nd question I am getting the 1) and2)statement as always true
u r correct for Q. 1, neeraj

try again Q. 2,plz...cheer up!
 

mana

New member
PRIME AND COMPOSITE NUMBERS

5) If a and b are two prime numbers, then, a + b can take value :-
1. 78
2. 7
3. 23

Options :
a} 1 Only
b} 2 Only
3} 1 & 2 only
4} 2 & 3 Only


ans- C

'coz 78 is even no.s thus it shud b sum of 2 odd prime no.s i.e 37 n 41
7 is odd no.s thus shud b sum of 1 even n 1 odd prime no.s i.e 2 n 5
23 is odd no.s thus must be sum of i even n 1 odd prime no.s which is nt possible
 

gaurav200x

Gaurav Mittal
rahul_parab2006 said:
u r correct for Q. 1, neeraj

try again Q. 2,plz...cheer up!
good going neeraj..... u have solved the problems successfully.

Regd the 2nd question, when u see the 1st option, u can immediately say that, that is the correct choice. U don't even have to read the subsequent choices.

So the answer would be 1
 

gaurav200x

Gaurav Mittal
Rahul, please post the next set of questions in numbers.

p.s. If u're going to use this thread for numbers alone, then i suggest that u ask the mods to rename it to something like "Problems on Number Theory" or something, for the record.
 
Last edited:

neeraj_uchila

New member
The answer to the question 2 is 1) cant say ...I do not have a specific formula to calc tha ans i substituted the options....If you have one please post it
 

gaurav200x

Gaurav Mittal
neeraj_uchila said:
The answer to the question 2 is 1) cant say ...I do not have a specific formula to calc tha ans i substituted the options....If you have one please post it
neeraj_uchila said:
The correct ans to question no 3 is 36/125...again i am noe sure abt the correct formula....I think its (n-1)(2n-1)

i have shifted the questions to the other thread called Quantifying Quant so that this thread stays exclusively for numbers.

p.s. the answer is not can't say. think harder

Edit:
Please post ur answers in the other thread.... and give explanation too.
 
Choose the correct alternative.

NOTE : FOR Q. NO. 2 TO 5, "2{2}" MEANS "SQUARE OF 2"

1) Which of the following is/are true about the fractions?

I. The HCF of a set of proper fractions has to be a proper fraction.
II. The LCM of a set of proper fractions has to be a proper fraction.
III. The HCF of a set of improper fractions can be integer.

1] I and II only.
2] I and III only.
3] II and III only.
4] I, II and III.

2) (-3){m+1} = (route 3){n}. If m and n are positive integers, which of the following is true?

1] m and n are odd.
2] m is even and n is odd.
3] m is odd and n is even.
4] m and n are even.

3) 4{m} + 3{n} = 43, where m and n are natural numbers. Find the value of
(m + n){2} – 4mn.

1] 1.
2] 24.
3] 37.
4] 25.

4) Which of the following is false?

1] (24691){m} will always end in 1 where m is a natural number.
2] (24691){m} will always end in 1 where m is a whole number.
3] (63456){m} will always end in 6 where m is a natural number.
4] (63456){m} will always end in 6 where m is a whole number.

5) What is the value of
(a{1/2} + b{1/2}) (a{1/2} - b{1/2}) +
(a{1/3} + b{1/3}) (a{2/3} – a{1/3}b{1/3} + b{2/3}) ?

1] 1.
2] a.
3] b.
4] 2a.
 

gaurav200x

Gaurav Mittal
rahul_parab2006 said:
Choose the correct alternative.

NOTE : FOR Q. NO. 2 TO 5, "2{2}" MEANS "SQUARE OF 2"

1) Which of the following is/are true about the fractions?

I. The HCF of a set of proper fractions has to be a proper fraction.
II. The LCM of a set of proper fractions has to be a proper fraction.
III. The HCF of a set of improper fractions can be integer.

1] I and II only.
2] I and III only.
3] II and III only.
4] I, II and III.

2) (-3){m+1} = (route 3){n}. If m and n are positive integers, which of the following is true?

1] m and n are odd.
2] m is even and n is odd.
3] m is odd and n is even.
4] m and n are even.

3) 4{m} + 3{n} = 43, where m and n are natural numbers. Find the value of
(m + n){2} – 4mn.

1] 1.
2] 24.
3] 37.
4] 25.

4) Which of the following is false?

1] (24691){m} will always end in 1 where m is a natural number.
2] (24691){m} will always end in 1 where m is a whole number.
3] (63456){m} will always end in 6 where m is a natural number.
4] (63456){m} will always end in 6 where m is a whole number.

5) What is the value of
(a{1/2} + b{1/2}) (a{1/2} - b{1/2}) +
(a{1/3} + b{1/3}) (a{2/3} – a{1/3}b{1/3} + b{2/3}) ?

1] 1.
2] a.
3] b.
4] 2a.

Solutions:
First a little theory for the 1st problem.

Proper fractions are those where numerator < denominator
Improper fractions are those where denominator < numerator

1)
I am not sure about the solution, but i think the correct answer might be (4)

2) Equate the powers. m cant be even or else m+1 will become odd and so the eqn will not hold.

So m+1=n/2 hence n is even.

Ans: m is odd, n is even

3) The only values of m and n satisfying the criteria would be m=2, n=3
So input the value and u get the answer as 1 i.e. choice 1

4) the answer would be choiuce 4. Whole numbers are 0,1,2,3.... and natural numbers 1,2,3,......

analysing each choice u would find that they are correct... however, the 4th choice wont be... as 63456 ^ 0 = 1 which doesnt end in 6


5) the product of the first 2 terms yield a-b [ (x-y)(x+y) = x^2-y^2 ]
the second term would be a+b [take root a as x and root b as y]

So answer = 2a
 
gaurav200x said:
Solutions:
First a little theory for the 1st problem.

Proper fractions are those where numerator < denominator
Improper fractions are those where denominator < numerator

1)
I am not sure about the solution, but i think the correct answer might be (4)

2) Equate the powers. m cant be even or else m+1 will become odd and so the eqn will not hold.

So m+1=n/2 hence n is even.

Ans: m is odd, n is even

3) The only values of m and n satisfying the criteria would be m=2, n=3
So input the value and u get the answer as 1 i.e. choice 1

4) the answer would be choiuce 4. Whole numbers are 0,1,2,3.... and natural numbers 1,2,3,......

analysing each choice u would find that they are correct... however, the 4th choice wont be... as 63456 ^ 0 = 1 which doesnt end in 6


5) the product of the first 2 terms yield a-b [ (x-y)(x+y) = x^2-y^2 ]
the second term would be a+b [take root a as x and root b as y]

So answer = 2a

U r wrong dude in 1st question...

I. In proper fractions numerator is smaller than the denominator.
Also, HCF of fractions = HCF of numerators/LCM of denominators

Since, we have to find HCF of a set of a number say A and LCM of a set of numbers say B where atleast one element of B is greater than all the elements of A, LCM of B is greater than HCF of A. Thus, the numerator is smaller yhan yhe denominator and the required HFC is aproper fraction. thus, I is true.

II Consider three proper fractions ,1/2. 3/5and 5/7
LCM = LCM of 1, 3 and 5/hcf of 2, 5 and 7 = 15/1 = 15.
Thus, II is not always true.

III. Consider a set of improper fractions = 4/1, 6/2, 10/2
HCF = HCF of 4, 6 and 10/LCM of 1, 2 and 2 = 2/2 = 1
Thus, III is true.

Hence, 2]
 
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