ব্যাখ্যা
Two numbers x & y
ATQ,
x2 + y2 = 80
And,
(x – y)2 = 36
Or, x2 + y2 – 2xy = 36
Or, 80 – 2xy = 36
Or, 2xy = 44
Or, xy = 22
∴ The product of two numbers is: 22
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PrepBank · পাতা ৩ / ১৮ · ২০১–৩০০ / ১,৭৩৬
Question: Which of the following digits will replace the M marks in the following equation ?
9M + M8 + M6 = 251
9M+ M8 + M6 = 251
⇒ {(9 × 10) + M} + (10M + 8) + (10M + 6) = 251
⇒ 21M + 104 = 251
⇒ 21M = 251 - 104
⇒ 21M = 147
⇒ M = 7
By hit and trial,
we put x = 5 and y = 1
so that,
(3x + 7y) = (3 x 5 + 7 x 1) = 22, which is divisible by 11.
(4x + 6y) = ( 4 x 5 + 6 x 1) = 26, which is not divisible by 11;
(x + y + 4 ) = (5 + 1 + 4) = 10, which is not divisible by 11;
(9x + 4y) = (9 x 5 + 4 x 1) = 49, which is not divisible by 11;
(4x - 9y) = (4 x 5 - 9 x 1) = 11, which is divisible by 11.
Question: If m and n are even numbers, which of the following is always even?
Solution:
Let m = 2 and n = 4 (both are even numbers)
a) m + n + 1 = 2 + 4 + 1 = 7 ......... Odd
b) mn + 3 = (2 × 4) + 3 = 8 + 3 = 11 ......... Odd
c) 2m + n = (2 × 2) + 4 = 4 + 4 = 8 ......... Even
d) m2 + n + 1 = (2)2 + 4 + 1 = 4 + 4 + 1 = 9 ......... Odd
Answer: c) 2m + n is always even
Question: A mechanic charges a fixed service fee of Tk. 80, along with Tk. 40 for each hour of labor. If a customer’s total bill must not exceed Tk. 320, determine the maximum number of full hours the mechanic can work.
Solution:
Given that,
Fixed service fee = Tk. 80
Charge per hour = Tk. 40
Maximum total bill = Tk. 320
Let h = number of full hours the mechanic works.
Now, total cost equation,
80 + 40h ≤ 320
⇒ 40h ≤ 320 - 80
⇒ 40h ≤ 240
⇒ h ≤ 240/40
⇒ h ≤ 6
So, the mechanic can work a maximum of 6 full hours.
=> 3(2x+9) = 75
=> 2x + 9 = 25
=> x = 8
Answer : 8
Question: If x = 12, which of the following has the least value?
Solution:
ক. x - 2 = 12 - 2 = 10
খ. x/2 = 12/2 = 6
গ. 2/x = 2/12 = 1/6 ≈ 0.17
ঘ. 2 - x = 2 - 12 = - 10
The least value among the calculated options is - 10 or 2 - x
Question: If 7543P is divisible by 9, what is the value of P?
Solution:
একটি সংখ্যা 9 দ্বারা বিভাজ্য হবে যদি সংখ্যাটির অঙ্কগুলোর সমষ্টি 9 দ্বারা বিভাজ্য হয়।
7 + 5 + 4 + 3 = 19 ; এর সাথে P যোগ করলে (19 + P) হবে, যা 9 দ্বারা বিভাজ্য হতে হবে।
19 + P = 27 (যা 9 দ্বারা বিভাজ্য নিকটতম সংখ্যা)
∴ P = 27 - 19 = 8
∴ P = 8
Question: If 5 ≥ x ≥ - 1 and y ≥ - 1, which of the following cannot be a value of x - y?
Solution:
Here, 5 ≥ x ≥ - 1 and y ≥ - 1
Now,
i) If, x = - 1 and y = - 1 then, x - y = - 1 - (- 1) = -1 + 1 = 0
ii) If, x = 2 and y = 1 then, x - y = 2 - 1 = 1
iii) If, x = 5 and y = 0 then, x - y = 5 - 0 = 5
iv) If, x = 5 and y = -1 then, x - y = 5 - (- 1) = 5 + 1 = 6
∴ Any value greater than 6 cannot be a value of x - y.
ধরি, সংখ্যাটি x
প্রশ্নমতে, (2x + 9) × 3 = 75
⇒ 2x + 9 = 25
⇒ 2x = 16
⇒ x = 8
Question: Determine the largest number among four multiples of 5 in a row that equal 170 together.
Solution:
Let the four consecutive multiples of 5 be x, (x + 5), (x + 10) and (x + 15)
According to the question,
x + (x + 5) + (x + 10) + (x + 15) = 170
⇒ 4x + 30 = 170
⇒ 4x = 170 - 30
⇒ 4x = 140
⇒ x = 140/4
⇒ x = 35
Therefore, the largest number is = x + 15 = 35 + 15 = 50
Place value of 6 = 6000
Face value of 6 = 6
Difference = 6000 - 6
= 5994.
Let x be the required number.
Given, x is as much greater than 36 as is less than 86.
ATQ, x - 36 = 86 - x
Or, x + x = 86 + 36
Or, 2x = 122
Or, x = 122/2 = 61
Question: Three numbers are in the ratio 2 : 3 : 5, and the sum of their squares is 608. Find the smallest number.
Solution:
Let the numbers be 2x, 3x, 5x
ATQ,
(2x)2 + (3x)2 + (5x)2 = 608
⇒ 4x2 + 9x2 + 25x2 = 608
⇒ 38x2 = 608
⇒ x2 = 608/38
⇒ x2 = 16 = 42
∴ x = 4
∴ Smallest number = 2x = 2 × 4 = 8
There are two possible values of k 7 and 14.
So, their summation is 21 and 4 will be the remainder when 21 is divided by 17.
100 years contain 5 odd days.
∴ Last day of 1st century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3rd century is Monday.
400 years contain 0 odd day.
∴ Last day of 4th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.
Question: If Kamal gves 20 marbles to Shuvo, then, both of them will have equal numbers of marbles in their possessions. If Shovo gives 40 marbles to Kamal then Kamal will have twice the number of marbles that Shuvo will retain. What is the number of marbles that Kamal has?
Solution:
কামালের মার্বেল আছে = x টি
শুভর মার্বেল আছে = y টি
১ম শর্তমতে
x - 20 = y + 20
x - y = 20 + 20
x - y = 40 ......................(1)
২য় শর্তমতে
x + 40 =2(y - 40)
x + 40 = 2y - 80
2y - x = 40 + 80
2y - x = 120 ......................(2)
(1) × 2 + (2) ⇒
2x - 2y + 2y - x = 80 + 120
x = 200
কামালের মার্বেল আছে = 200 টি
Here, a + b + c = 12 & a + b = 4
So, c = 8
As, a + c = 7
∴ a = 7 - 8 = -1
Question: What is the smallest number of soldiers that can be arranged in groups of 12, 15, 18, and 20, and also arranged to form a perfect square?
Solution:
LCM of 12, 15, 18, and 20 is-
12 = 2 × 2 × 3
15 = 3 × 5
18 = 2 × 3 × 3
20 = 2 × 2 × 5
∴ LCM = 2 × 2 × 3 × 5 × 3
Since the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, we have to multiply it by 5,
∴ The required number of soldiers = 2 × 2 × 3 × 3 × 5 × 5
= 900
Let x be the larger number and y be the smaller number
Therefore x - y = 20%of x
Now put the value of the smaller number in the equation
x - 20 = (20/100) × x
⇒ x - 20 = (1/5) × x
Here 5 is in the denominator.
As we bring 5 on the left side of the equation it will be multiplied by x - 20
Now the equation will be 5 × (x - 20) = x
⇒ 5x - 100 = x
Bring 100 on the right side of = and x on the left side of =
So it will become 5x - x=100
⇒ 4x = 100
⇒ x = 100/4
⇒ x = 25
Therefore the larger number is 25.
let, the smaller number be x and bigger number be y
So, y + x = 55 ..... (i) and y - x = 9 ...... (ii)
by solving equation (i) and (ii) we get y = 32 and x = 23
Let the number be x
According to the question,
91 - (30/100) = x
⇒ 9100 - 30x = 100x
⇒ 9100 = 130x
⇒ x = 9100/130
Hence, x = 70.
Question: a, b, c, d and e are five consecutive integers in increasing order of size. Which of the following is always even?
Solution:
ধরি
a = 1, b = 2, c = 3, d = 4, and e = 5,
অপশন (ক) ac + e = 1 × 3 + 5 = 3 + 5 = 8
অপশন (খ) ac + d = 1 × 3 + 4 = 3 + 4 = 7
অপশন (গ) a + b + c = 1 + 2 + 3 = 6
অপশন (ঘ) ab + c = 1 × 2 + 3 = 2 + 3 = 5
............................
...........................................
আবার
ধরি
a = 2, b = 3, c = 4, d = 5, and e = 6,
অপশন (ক) ac + e = 2 × 4 + 6 = 8 + 6 = 14
অপশন (খ) ac + d = 2 × 4 + 5 = 8 + 5 = 13
অপশন (গ) a + b + c = 2 + 3 + 4 = 9
অপশন (ঘ) ab + c = 2 × 3 + 4 = 6 + 4 = 10
উভয় ক্ষেত্রে অপশন (ক) জোড় সংখ্যা। তাই সঠিক উত্তর হিসেবে অপশন (ক) নেওয়া হয়েছে।
The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
Question: A positive number, when decreased by 4, is equal to 21 times the reciprocal of the number. The number is:
Solution:
মনেকরি
সংখ্যাটি = x
প্রশ্নমতে
x - 4 = 21/x
বা, x(x - 4) = 21
বা, x2 - 4x - 21 = 0
বা, x2 - 7x + 3x - 21 = 0
বা, x(x - 7) + 3(x - 7) = 0
(x - 7)(x + 3) = 0
হয়
x - 7 = 0
x = 7
অথবা
x + 3 = 0
x = - 3
সংখ্যাটি = 7
Question: If √(7m) = 343, then the value of m is -
Solution:
Given that, √(7m) = 343
⇒ √(7m) = 73 (since, 343 = 73)
⇒ {√(7m)}2 = (73)2
⇒ 7m = 76
⇒ m = 75
⇒ m = 16,807
∴ The value of m is 16,807.