solve 33 , 34 and 35
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In question no. 33 we know for some digit B
we are getting B×B = XB (here X can be any digit from 0-9)
So possible Values of B will be
1×1 = 1
5 × 5 = 25
6×6 = 36
B = 0 not possible
So we can easily Find 5 fits perfectly
and Now you can easily Find A which is equals to 4.
In Question No. 34 You can easily Check which option is Correct. The correct option is 1089.
In question No. 35
After simplifying we , get
E = OY
we know Range of E is 0-10
we are given , option , 3 , 4 ,8 ,2
Case -1
E = 3Y
but we know Y > O
so , y can be (4to10)
But , E ranges from 1-10
so , O does not equal 3
Case - 2
E = 4Y
same condition applied here
So , O does not equal to 4
Case - 3
E = 8Y
same condition Applied here
So , O does not equal to 8
Case - 4
E = 2Y
Now Y Can take possible value of 3 , 4 ,5
So O = 2
Hope it didn't helped you!!
we are getting B×B = XB (here X can be any digit from 0-9)
So possible Values of B will be
1×1 = 1
5 × 5 = 25
6×6 = 36
B = 0 not possible
So we can easily Find 5 fits perfectly
and Now you can easily Find A which is equals to 4.
In Question No. 34 You can easily Check which option is Correct. The correct option is 1089.
In question No. 35
After simplifying we , get
E = OY
we know Range of E is 0-10
we are given , option , 3 , 4 ,8 ,2
Case -1
E = 3Y
but we know Y > O
so , y can be (4to10)
But , E ranges from 1-10
so , O does not equal 3
Case - 2
E = 4Y
same condition applied here
So , O does not equal to 4
Case - 3
E = 8Y
same condition Applied here
So , O does not equal to 8
Case - 4
E = 2Y
Now Y Can take possible value of 3 , 4 ,5
So O = 2
Hope it didn't helped you!!
SeemaJain:
tnq
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