Test: Number bases - 4 - JAMB MCQ

Pick up the unit digit of each number and multiply them;

0² in (1570)² = 0

1² in (1571)² = 1

2² in (1572)² = 4

3² in (1573)² = 9

On adding the unit digits, we will get,

0 + 1 + 4 + 9 = 14 (unit digit = 4)

So, on adding (1570)² + (1571)² + (1572)² + (1573)², the unit digit will be 4.

Suppose the numbers be a, b, and c. So, according to the question,

a + b + c = 2

and, a = ½b ….(i)

c = ¼b ….(ii)

From equation (i),

a/b = ½

and from equation (ii)

b/c = 4/1

Ratio between a, b, c is

a : b : c = 2x : 4x : 1x

(2x + 4x + 1x) = 2

7x = 2

So, x = 2/7

Then a = 2x = 2 * 2/7 = 4/7

b = 4x = 4 * 2/7 = 8/7

c = x = 2/7

So, the second number is 8/7.

Suppose the three numbers be a, b, and c

According to the question,

a = 3b or, a/b = 3/1

c = 2/3a or, a/c = 3/2

Ratio between a, b, c is

a : b : c = 3x : x : 2x

(3x + x + 2x) = 252

6x = 252

So, x = 252/6 = 42

Then, first number a = 3x = 3 * 42 = 126

Second number, b = x = 42

Third number, c = 2x = 42 * 2 = 84

So, the second number is 42.

We can solve it via two ways. Let's see the ways.

Shortcut method:

We can also get the answer using the specified options. We can check the digits of the numbers mentioned in the options. As in the option 'C', there are 39 or 93, and on adding their digits, we will get 12, and on subtracting the digits, we will get 6.

Detailed method:

Suppose the digits of the two-digit number be 'a' and 'b'.

So, according to the question

a + b = 12 ….(i)

a - b = 6 ……(ii)

Add equation (i) and (ii), we get

2a = 18

Or, a = 9

Put a = 9 in eqⁿ (i), we will get

b = 3

So, the two-digit number is 93 (when a > b)

Or 39 (when b > a)

Given ratio is 1 : 2 : 3

So, the numbers will be 1a : 2a : 3a

Then, a³ + 8a³ + 27a³ = 4500

36a³ = 4500

Or, a³ = 4500/36 =125

So, a = 5

The smallest number is 5.

Given: 1505 x 1505

= 1505²

= (1500 + 5)²

Apply formula; (a + b)² = a² + b² + 2ab

= 1500² + 5² + 2 *1500* 5

= 2250000 + 25 + 15000

= 2265025

a = (0.4)² = 0.16

b = 0.04

c = 2/5 = 0.4

So, the relationship is c > a > b.

Sum of two odd number is even number and multiplication of two odd number with 2 always given an even number.
Alternate:
Put , a = 3,   b = 5, option (D) will satisfy

Let the numerator of the fraction be x, so the denominator will be 11 - x

So, the fraction = x/11 - x

On adding 2 in both numerator and denominator, according to the question, the fraction will be -

(x + 2)/(11 - x + 2) = (x)/(11 - x) + 1/24

(x + 2)/(13 - x) - (x)/(11 - x) = 1/24

[11x + 22 - x² -2x - 13x + x²]/(13 - x) (11 - x) = 1/24

After solving the above equation, we will get

=> 528 - 96x = 143 - 24x + x²

x² + 72x - 385 = 0

(x + 77) (x - 5) = 0

So, x =5

So, numerator = 5

and, denominator = 11 - 5 = 6

Difference between both is = 6 - 5 = 1

The required number = Increase in result/(wrong multiplier - correct multiplier)

= 200/45 - 25

= 10

Let the numerator be 'a' so, denominator will be 'a + 3'.

According to the question,

(a + 7)/ ((a + 3) - 2) = 2/1

(a + 7)/(a + 1) = 2/1

2a + 2 = a + 7

=> a = 5

So, the fraction is a/a + 3 = 5/ 5 + 3

= 5/8

And the sum of the numerator and denominator of the fraction is 13.

Let the numbers be a and b.

So, according to the question,

a * 3/5 = 70% of b

3a/5 = 70b/100

a/b = (70 * 5)/(100 * 3) = 7/6

So, the ration between a and b is 7 : 6.

Suppose a = (x - y), b = (y - z), and c = (z - x)

On adding a, b, and c we will get

a + b + c = x -y + y - z + z - x

=> a + b + c =0

So, a³ + b³ + c³ = 3abc [because if a + b + c = 0, then a³ + b³ + c³ = 3abc]

We can say that (x - y)³ + (y - z)³ + (z - x)³ = 3 (x - y) (y - z) (z - x)

Suppose the three consecutive positive numbers be x, x + 1, and x + 2

According to the question,

(x)² + (x + 1)² + (x + 2)² = 365

On expanding, we will get,

x² + x² + 1 + 2x + x² + 4 + 4x = 365

3x² + 6x = 360

Or, x² + 2x - 120 = 0

=> (x - 10) (x + 12) = 0

So, x = 10

First number x = 10

Second number x + 1 = 11

Third number x + 2 = 12

So, sum of the numbers is = 10 + 11 + 12 = 33

Given x = a(b - c), y = b(c - a), z = c(a - b)

x = a(b - c)

or, x/a = (b - c) …..(i)

y = b(c - a)

or, y/b = (c - a) …..(ii)

z = c(a - b)

or, z/c = (a - b) …..(iii)

Add equation (i), (ii), and (iii), we will get

x/a + y/b + z/c = b - c + c - a + a - b

x/a + y/b + z/c = 0

So, (x/a)³ + (y/b)³ + (z/c)³ = 3 (x/a) * (y/b) * (z/c) [because if a + b + c = 0, then a³ + b³ + c³ = 3abc]

= 3xyz/abc