Math : RRB, SSC, Banking, and Other Competitive Exams

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Q. A gold ornament weighs 16 g with gold:copper = 3:1. How many grams of pure gold must be added to make the ratio 4:1 ?

Solⁿ:

Let gold to copper = 3 : 1
→ Total parts = 4

So,
Gold = (3/4) × 16 = 12 g
Copper = (1/4) × 16 = 4 g

Let x g of pure gold be added.
Then new ratio = (12 + x) : 4 = 4 : 1

⇒ (12 + x) / 4 = 4 / 1
⇒ 12 + x = 16
⇒ x = 4 g

Ans: Add 4 grams of gold.

Q) Karim and Rahim’s salaries are in the ratio 7 : 5.
If Karim’s salary is ₹400 more than Rahim’s, find Rahim’s salary.

Solⁿ:
Let Karim’s salary = 7x and Rahim’s salary = 5x.
According to the question:
7x – 5x = 400
⇒ 2x = 400
⇒ x = 200

Rahim’s salary = 5 × 200 = ₹1000

Ans: Rahim’s salary = ₹1000

Q: How many numbers between 100 and 300 either start with 2 or end with 2 ?

Solⁿ:

·  Numbers starting with 2: 200–299 → 100 numbers

·  Numbers ending with 2: 102, 112, 122, …, 292 → 20 numbers

·   Numbers counted twice (both start with 2 and end with 2): 202, 212, …, 292 → 10 numbers

Total numbers = 100 + 20 – 10 = 110

Ans: 110

Q: In a school, each student is assigned a unique ID number from 1 to 100.

·  A student plays football if and only if their ID is divisible by 4.

·  A student plays cricket if and only if their ID is divisible by 6.

How many students play both football and cricket ?

Solⁿ:

Find numbers divisible by both 4 and 6

• LCM of 4 and 6 = 12


• So, students playing both → multiples of 12

List multiples of 12 up to 100
12, 24, 36, 48, 60, 72, 84, 96

Count them : Total = 8

Ans: 8 students

Q: If in a proper fraction, both numerator and denominator are increased by the same positive number, what happens ?

Options:

1. Always less


2. Always greater


3. Always equal


4. Cannot say

Ans: Always greater

Example:

• Take a proper fraction: 2 / 5 ​


• Increase numerator and denominator by 1: 2+1/5+1 = 3/6 =0.5


• Original fraction 2 / 5 = 0.4


• New fraction 0.5 > 0.40

Q: A hostel has 12 students. When 13 new students join, the daily expense per student decreases by ₹3. Find the original total daily hostel expense.

Solⁿ :

• Difference per student = ₹3


• Old students = 12


• Total after joining = 12 + 13 = 25


• Number of new students = 13

Original Total Expense = Difference per student × Old students × Total students after joining / Number of new students

Substitute numbers:

Original Total Expense = 3 × 12 × 25 / 13 = 900 / 13 ≈ 69

Ans: ₹69 per day

    Q. 0.2̅5 bar on 5 (that is 0.25555…)

0.25 bar on 5  = 0.2 + 0.05 bar on 5  = 1/5 + 1/18 = 23/90

Short Trick:
“Subtract, then divide by 9 and 0 rule”
→ 0.ab‾ bar on b = (ab−a)/90

Q. 0.375 bar on 75 = ?

Fraction = (Number without decimal)−(Number before repeating part) / 2 nines _Same number of 9’s as repeating digits​​ + 1 zero_ Same number of 0’s as non-repeating digits​​

Number without decimal = 375
Number before repeating part = 3, Here 75 repeating no

So the number = 0.375 bar on 75

Fraction = 375 - 3 / 990

Simplify:

372 / 990 = 62 / 165

Answer: 0.375‾=62 / 165

Short trick memory:
If bar on two digits after one non-repeating digit → 0.abc bar on two digits : 0.abc = (abc - a) / 9900

Q. 0.67 (bar only on 7) = ?

 Identify parts

·         Non-repeating part = 6

·         Repeating part = 7

     Number = 0.67 bar on 7

Fraction = (Number without decimal)−(Number before repeating part) / 2 nines _Same number of 9’s as repeating digits​​ + 1 zero_ Same number of 0’s as non-repeating digits

Apply it

Number without decimal = 67
Number before repeating part = 6

Repeating digits = 1 → one 9
Non-repeating digits = 1 → one 0

Fraction = 67 − 6 / 90 = 61 / 90

Answer = 61 / 90

Short Trick:
If bar on one digit after one non-repeating digit →0.ab bar on b = (ab-a) / 90

Find the value of 4 + 0.1 overline 3 : Click Here

Q. Find the sum of numbers from 1 to 35.

Solⁿ.
We know,
Sum of first n natural numbers = n(n + 1) / 2

Here, n = 35

So,
= 35 × (35 + 1) / 2
= 35 × 36 / 2
= 35 × 18
= 630

Therefore, the sum of numbers from 1 to 35 = 630