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 = (10x + y) where x < y.
Number obtained on reversing the digits = (10y + x)
ATQ,
10y + x = 4 (10x + y) - 24
⇒ 40x + 4y - 10y - x = 24
⇒ 39x - 6y = 24
⇒ 13x - 2y = 8 ....(i)
Again, y - x = 7
y = x + 7 ....(ii)
⇒ 13x - 2 (x + 7) = 8
⇒ 13x - 2x - 14 = 8
⇒ 11x = 14 + 8 = 22
⇒ x = 22/11
∴ x = 2
From equation (ii),
y - 2 = 7
⇒ y = 2 + 7
∴ y = 9
So, Number = 10x + y = (10 × 2) + 9 = 29
Solution:
Suppose, a = 1 and b = 2
a + b = 1 + 2 = 3
ab + 1 = 1 × 2 + 1 = 3
=10
Solution:
Let the original number be x
ATQ,
{5(x + 7)/9} - 3 = 12
⇒ {5(x + 7) - 27}/9 = 12
⇒ 5(x + 7) - 27 = 108
⇒ 5x + 35 - 27 = 108
⇒ 5x + 8 = 108
⇒ 5x = 100
∴ x = 20
Solution:
Maximum daily wages of the officers = H.C.F of Tk. 4956 and Tk. 3894
H.C.F of 4956 & 3894 = 354
So, he was appointed for 4956/354 = 14 days
But he was present = 3894/354 = 11 days.
So, he was absent for (14 -11) days = 3 days
Solution:
Let the number be x
ATQ,
x - 12 = 20% of x
⇒ x - 12 = x/5
⇒ x - x/5 = 12
⇒ 4x/5 = 12
⇒ x = (12 × 5)/4
∴ x = 15
Solution:
Let the number be = a
The reciprocal of the number is = 1/a
According to the question,
a + (1/a) = 2
⇒ + 1 = 2a
⇒ - 2a + 1 = 0
⇒ = 0
∴ a = 1
Hence, the number = 1
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