বিগত সালের প্রশ্ন ও প্রশ্ন ব্যাংক
বিসিএস, ব্যাংক, প্রাথমিক শিক্ষক ও সকল সরকারি চাকরির বিগত সালের প্রশ্ন ও অধ্যায়ভিত্তিক নির্ভুল সমাধান
Solution:
Let, the two-digit number be 10a + b where a > b.
ATQ,
a + b = 12 ------- (1)
a - b = 6 --------- (2)
On adding equation (1) & (2)
2a =18
∴ a = 9
Putting this value in (1) we get,
9 + b = 12
∴ b = 3
So the number is 10a + b = 9 . 10 + 3 = 93
When, a < b, Then the required number is 39
Solution:
A number is divisible by 11 when the difference between the total number of odd places and the total number of even places is equal to zero or multiple of 11
Therefore, M + M + 3 = 7 + 8 + 6
⇒ 2M = 18
∴ M = 9
Solution:
Let the two positive numbers be x and y; x > y
ATQ,
x + y = 151 ---------- (1)
y = 19 + √x ---------- (2)
Putting the value y = 19 + √x in equ (1)
x + 19 + √x = 151
⇒ x + √x = 151 - 19
⇒ x + √x = 132
⇒ √x = 132 - x
⇒
⇒
⇒ - 265x + 17424 = 0
∴ x = 121, 144
The value of x can be 121 and 144.
The value of x = 144 cannot be possible as it doesn’t satisfies the two condition.
Therefore the value of x is 121.
From equ. (1) we get,
y = 151 - 121
y = 30
∴ The value of greatest number minus the lesser would be x-y, i.e, 121 - 30 = 91
Solution:
We have to find the Least number, therefore we find out the LCM of 8, 12, 16 and 20.
8 = 2 × 2 × 2;
12 = 2 × 2 × 3;
16 = 2 × 2 × 2 × 2;
20 = 2 × 2 × 5;
LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240;
This is the least number which is exactly divisible by 8, 12, 16 and 20
Thus, The required number which leaves the remaining 5 is,
240 + 5 = 245