2 two digit numbers chosen in such a way that all 4 digits used are distinct, are such that their su

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25 Jun 2015 Two digit number

2 two digit numbers chosen in such a way that all 4 digits used are distinct, are such that their sum is 90. How many such pairs are possible if in both numbers units place is 1 more than tens place.

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10)0

Solution

Let given number are
10x + (x+1) and 10y + (y+1)
then as given
10x + (x + 1) + 10y + (y + 1) = 90
11x + 11y = 88
x + y = 8
So 12 & 78, 23 & 67, 34 & 56.

Option 3)

Correct Option: 3 Best Solution (27)

SURAJ KUMAR   9 years AGO

12 +78,23+ 67, 34+56 these are the the 3 pairs

Like? Yes (3) | No (1)  

viswambher   9 years AGO

let the numbers be(x x+1) and ( y y+1)
and the sum is 90 there fore
x+y+2 = 10
I.e x+y = 8
therefore the pairs are
(12,78) (23,67) , (34,56)

Like? Yes (2) | No (1)  

Chandrima Goswami   9 years AGO

12+78=90
23+67=90
34+56=90
So ans is 3

Like? Yes (1) | No   

RAJAT KHANDELWAL   9 years AGO

the nos are
12+78=90
23+67=90
34+56=90

Like? Yes (2) | No (1)  

Surendhar Ramanujam   9 years AGO

the three possibilities such as:
12+78,
23+67 and
34+56 are the only possibilities for the given conditions that are as follows
1. units place should be 1 more than tens place
2. sum of the numbers should be equal to 90
3. the numbers used should be distinct

Like? Yes (1) | No   

SANTOSH   9 years AGO

34 56
23 67
12 78

Like? Yes (1) | No   

darladiwakar   9 years AGO

a b
c d
-----
b+d sum must equal to 10 ;; so total sum=90 gives
so it seems a+c=8. then check by the given condition there three such kind of pairs.
12+78
23+67
34+56

Like? Yes (1) | No   

vamsi krishna   9 years AGO

let two digit numbers
first number = ab
second number =cd
require answer is = 90
so b+d=10;
s0 we need digits for a,b, positions such that a+b=8
so pairs are (1,7)(2,6)(3,5) bcz digits are distinct
ans=3 pairs are possible

Like? Yes (1) | No   

Deepthi   9 years AGO

12+78=90
23+67=90
34+56=90

Like? Yes (1) | No   

Harihara Panigrahy   9 years AGO

the range of the smaller number is 12 - 45 as 45*2=90 but 45 isn't an option cause 4 digits wont be distinct so range is reduced to 12 to 34 and so only 3 pairs possible 12 78,23 67 and 34 56

Like? Yes (1) | No   

Ramdas Singathiya   9 years AGO

12,78 23,67 56,34

Like? Yes (1) | No   

RAVAL   9 years AGO

as per given condition,
ans. is 3
there pairs are....
1) 1278
2) 2367
3) 3456

Like? Yes (1) | No   

Dhanraj Wanjare   9 years AGO

12 + 78 = 90;
23 + 67 = 90;
34 + 56 = 90; that's why option 3 pairs

Like? Yes (1) | No   

vivek kr. chatterjee   9 years AGO

four possible cases possible
1> 01 and 89
2> 23 and 67
3> 34 and 56

Like? Yes (1) | No   

p.supriya   9 years AGO

12+78=90
23+67=90
34+56=90
these series continue in reverse order so only 3 paires

Like? Yes (1) | No   

shree   9 years AGO

suppose two numbers 10's place are x and y
therefore 10x+ x+1+10y+y+1=90
11x+11y+2=90
x+y=8
therefore 3 pairs are available that are(1, 7), (2, 5)
(3, 5)

Like? Yes (1) | No   

Math Lovers   9 years AGO

Let ten digit place numbers are x and y
then,
numberes are 10x+(x+1) and 10y+(y+1) where x!=y
sum = 11 (x+y)+2 =90
=> x+y = 8
possible value of (x,y) are (1,7), (2,6), (3,5)
hence possible pair of numbers are 3

Like? Yes (1) | No   

Purnima Paliwal   9 years AGO

pairs are formed by :
12
23
34
56
67
78
89
in which 45 and 89 are not fulfilled the condition of the question so 12&78, 23&67,34&56 are the only three pairs.

Like? Yes (1) | No   

ajit kumar sharma   9 years AGO

12+78=90(1st pair)
23+67=90(2nd pair)
34+56=90(3rd pair)

Like? Yes (1) | No   

akshay ishwarkar   9 years AGO

12+78,
23+67,
34+56,

Like? Yes (1) | No   

RUPALI DADHE   9 years AGO

12+78=90;
23+67=90;
34+56=90.

Like? Yes (1) | No   

veeramanikandan   9 years AGO

12+78 = 90
23+67= 90
34+56=90

Like? Yes (1) | No   

Payal meshram   9 years AGO

23+67
12+78
34+56

Like? Yes (1) | No   

selva kumar R R   9 years AGO

12+78=90
23+67=90
34+56=90

Like? Yes (1) | No   

swetha   9 years AGO

01+89=90
12+78=90
23+67=90
34+56=90

Like? Yes (1) | No   

shabana banu shaik   9 years AGO

the numbers which satisfy above conditions are (56,34),(78,12),(23,67)......so three is the answer

Like? Yes (1) | No   

Mohammad Husain   9 years AGO

(12,78),(23,67),(34,56) not (45,45) they r not distinct

Like? Yes (1) | No