CRACK THE CAT - DOUBT OVER MATHS (PROBLEM SOLVING)

Originally Posted by gaurav200x

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.

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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. .

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.

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.

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

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
PSlz give the ans for the first 2 questions

Originally Posted by neeraj_uchila

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...

Originally Posted by neeraj_uchila

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
PSlz 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 !

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

Originally Posted by neeraj_uchila

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!