Question

If the given number 925x85 is divisible by 11, I then the smallest value of x is: 2.

Correct
option
is 4

11 से विभाज्यता के लिए, $\dfrac{(5+x+2)\sim (8+5+9)}{11}$ $\dfrac{(7+x)\sim 22}{11} = \dfrac{15-x}{11}$ $x = 4$

11 से विभाज्यता के लिए, $\dfrac{(5+x+2)\sim (8+5+9)}{11}$ $\dfrac{(7+x)\sim 22}{11} = \dfrac{15-x}{11}$ $x = 4$

Question

If the 6-digit numbers x35624 and 1257y4 are- divisible by 11 and 72, respectively, then what is the value of (5x-2y)?

Correct
option
is 14

$\because $x35624 is divisible by 11. . $\dfrac{(7+x)\sim13}{11} = \dfrac{x-6}{11}$ $\because x = 6$ 1257y4 is divisible by 72. Hence it will also be divisible by 9. . y=8 5x - 2y= 30 - 16= 14

$\because $x35624 is divisible by 11. . $\dfrac{(7+x)\sim13}{11} = \dfrac{x-6}{11}$ $\because x = 6$ 1257y4 is divisible by 72. Hence it will also be divisible by 9. . y=8 5x - 2y= 30 - 16= 14

Question

The digit in the unit place of the product is $81\times 82 \times 83\times84 .. ..89 $

Correct
option
is 0

The digit in units place = units digit in the product $1\times 2 \times 3 .... 9 = 0$

The digit in units place = units digit in the product $1\times 2 \times 3 .... 9 = 0$

Question

What is the unit digit $3 \times 38 \times 537 \times 1256$

Correct
option
is 8

Answer: (a) Units digit in $3 \times 38 \times 537 \times 1256$ = Units digit in $ 3\times 8 \times 7\times 6$ = $ 4\times 2 = 8$

Answer: (a) Units digit in $3 \times 38 \times 537 \times 1256$ = Units digit in $ 3\times 8 \times 7\times 6$ = $ 4\times 2 = 8$

Question

The sum of two numbers is 50 and their difference is 30. Find the numbers.

Correct
option
is 10,40

Let the numbers be x and y. Now, as per the given situation, x + y = 50(i) and x - y = 30(ii) We can write, x = 50-y, from eq.(i), Therefore, putting the value of x in eq(ii), we get, 50 - y - y = 30 50 -2y = 30 2y = 50- 30= 20 y = 20/2 = 10 and x = 50 - y = 50-10 =40 Therefore, the two numbers are 40 and 10.

Let the numbers be x and y. Now, as per the given situation, x + y = 50(i) and x - y = 30(ii) We can write, x = 50-y, from eq.(i), Therefore, putting the value of x in eq(ii), we get, 50 - y - y = 30 50 -2y = 30 2y = 50- 30= 20 y = 20/2 = 10 and x = 50 - y = 50-10 =40 Therefore, the two numbers are 40 and 10.

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