ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: The difference between a number and its three-fifths is 150. What is the number?

Solution:
Let the number be x
Three-fifths of x is 3x/5

According to the question:
x - (3x/5) = 150
⇒ (5x - 3x)/5 = 150
⇒ 2x/5 = 150
⇒ 2x = 150 × 5
⇒ x = (150 × 5)/2
∴ x = 375

Question: The HCF and LCM of two numbers are 8 and 48 respectively. If one of the numbers is 16, then the other number is = ?

Solution:
Here,
HCF = 8
LCM = 48

One number = 16

Let the other number be = p
∴ 16p = 48 × 8
⇒ p = 24

Hence, the other number is = 24

Question: Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is :

Solution:
Let the number be x

Then,
3x² - 4x = x + 50
⇒ 3x² - 4x - x - 50 = 0
⇒ 3x² - 5x - 50 = 0
⇒ 3x² - 15x + 10x - 50 = 0
⇒ 3x(x - 5) + 10(x - 5) = 0
⇒ (x - 5)(3x + 10) = 0

∴ x = 5

Hence, the number is 5

Question: The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

Solution:
We know that,

L.C.M × H.C.F. = Product of two numbers

⇒ 7700 × 11 = 275 × other number

⇒ Other number = (7700 × 11) ÷ 275

∴ Other number = 308

Question: The sum of twice a number and three times of 42 is 238. What is the sum of thrice the number and two times of 42?

Solution:
Let the number be 'p'

According to the question,
⇒ 2p + 3 × 42 = 238
⇒ 2p + 126 = 238
⇒ 2p = 112
⇒ p = 56 

∴ Required sum:
= 3p + 2 × 42
= 3 × 56 + 2 × 42
=168 + 84
= 252

Question: The L.C.M. of two numbers is 96. The numbers are in the ratio 2 : 3. Then the sum of the numbers is-

Solution:
Let the numbers be 2x and 3x
Then, their L.C.M. = 6x

So,
6x = 96
∴ x = 16

The numbers are 2x = 32 and 3x = 48

Hence, required sum = (32 + 48) = 80

Question: Find the square of a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

Solution: 
Let the number be x

Then, x + 17 = 60/x
⇒ x² + 17x − 60 = 0
⇒ x² + 20x − 3x − 60 = 0
⇒ (x + 20)(x − 3) = 0
∴ x = - 20, 3 

The positive number = 3
Hence, the square of the positive number = 9

Question: Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

Solution:
Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm 
78 = 2 × 3 × 13
104 = 2 × 2 × 2 × 13 
117 = 3 × 3 × 13
169 = 13 × 13 

HCF of 78, 104, 117 and 169 = 13 

Maximum length of each part = HCF of 78 cm, 104 cm, 117 cm, 169 cm = 13 cm

The maximum number of pieces, 
78/13 = 6
104/13 = 8
117/13 = 9
169/13 = 13

The maximum number of pieces = 6 + 8 + 9 + 13 = 36 

∴ The maximum number of pieces is 36.

Question: A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The unit digit of the number is:

Solution: 
Let the digits be x, y, z 

x + y + z = 10 
y = x + z

y + y = 10 
⇒ y = 5

So, x + z = 5

Now,
(100z + 10y + x) - (100x + 10y + z) = 99
⇒ 99z - 99x = 99
⇒ z - x = 1

x + z + z - x = 5 + 1
⇒ 2z = 6
∴ z = 3

∴ x = 5 - 3 = 2

The number is = 253

Hence, The unit digit of the number is: 3

Question: Find the HCF of 3/4, 5/6 and 6/7 = ?

Solution:
For the HCF of fractions, it has to be taken the HCF of numerators and LCM denominators.

HCF of 3, 5, 6 = 1
LCM of 4, 6, 7 = 84

HCF of numerators/LCM of denominators = 1/84

Hence, the HCF of 3/4, 5/6 and 6/7 = 1/84

Question: The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

Solution:
Let the ten's digit be x and the unit's digit be y.
So, the number = 10x + y

After interchanging the positions of the number's digits, the number will be = 10y + x

Then, (10x + y) - (10y + x) = 36
⇒ 9(x - y) = 36
⇒ x - y = 4

Question: The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15, and 18 is:

Solution: 
L.C.M. of 6, 9, 15 and 18 is 90.
Let the required number be 90k + 4, which is a multiple of 7

The least value of k for which (90k + 4) is divisible by 7 is k = 4

Hence, the equired number = (90 × 4) + 4 = 364

Question: The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. The square of their sum is:
 
Solution:
Let the numbers be a, b and c

Then,
a² + b² + c² = 138
(ab + bc + ca) = 131
 
Now,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (a + b + c)² = 138 + 2 × 131
∴ (a + b + c)² = 400

Question: The sum of L.C.M. and H.C.F. of two numbers is 1260. If their L.C.M. is 900 more than their H.C.F., find the product of two numbers.

Solution:
Let the HCF be x

LCM = HCF + 900
LCM = x + 900 ...............(1)


And,
LCM + HCF = 1260
LCM + x = 1260 .................(2)

From (1) and (2) equation,
(x + 900) + x = 1260
⇒ 2x + 900 = 1260
⇒ 2x = 1260 - 900
⇒ 2x = 360
⇒ x = 360/2
⇒ x = 180

∴ HCF = 180

And, from equation (1),
LCM = HCF + 900
LCM = 180 + 900
∴ LCM = 1080

By formula, the product of the numbers is equal to the product of their HCF and LCM.

Product of numbers = HCF × LCM
= 180 × 1080
∴ Product = 194400

Question: If one-third of one-fourth of a number is 25, then three-tenths of that number is:

Solution:
Let the number be 'p'

Now,
(1/3) × (1/4) × p = 25
⇒ p/12 = 25
⇒ p = 12 × 25
∴ p = 300

∴ Three-tenths of that number will be = (3/10) × p
= (3/10) × 300
= 3 × 30
= 90