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Practice Questions: Simplification

Practice Questions: Simplification

Q1. $(18)^2 + (4)^3 - 150\% \text{ of } 110 + ? = 373$

(a) 120
(b) 150
(c) 110
(d) 130
(e) 140

Ans: (b)
Solution: 

The Fast-Track Solution Break the equation into three main components and solve them mentally:

  1. Squares and Cubes:
    • $(18)^2 = 324$
    • $(4)^3 = 64$
    • Sum: $324 + 64 = \mathbf{388}$
  2. Percentage Calculation:
    • $150\% \text{ of } 110$ is just $110 + 55$ (which is half of 110).
    • Result: $\mathbf{165}$
  3. Combine and Solve for ?:
    • $(388 - 165) + ? = 373$
    • $223 + ? = 373$
    • $? = 373 - 223$
    • $? = 150$

Correct Option: (b). 150

Exam Pro-Tips for Speed
  • Unit Digit Check: If you are in a rush, look at the last digits: $4 + 4 - 0 + ? = 3$. This gives $8 + ? = 3$. The only way to get a last digit of 3 is if $?$ ends in 0 (e.g., $8 + 5 = 13$
  • The "150%" Trick: Always treat $150\%$ as $1.5$ or $1 + \frac{1}{2}$. It is much faster to add half of the number back to itself than to multiply by 1.5.

  • Memorization: For banking exams, ensure you have squares memorized up to 30 and cubes up to 15. It saves the 5-10 seconds that usually make the difference.


Q2. $16 \times 15 \div 20 \times 48 - 22^2 = ?$

(a) 72
(b) 56
(c) 92
(d) 96
(e) 48

Ans: (c)
Solution: 

The Fast-Track Solution

1. Simplify the Multiplication/Division Block:

Don't do $16 \times 15$ first. Write it like this in your head:

$\frac{16 \times 15}{20} \times 48$
  • Divide $15$ and $20$ by $5$: You get $\frac{16 \times 3}{4} \times 48$.
  • Divide $16$ by $4$: You get $4 \times 3 \times 48$.
  • Now you have: $12 \times 48$.

2. Multiply $12 \times 48$ (The "Double & Half" or Distribution Trick):

  • $12 \times 50 = 600$
  • $12 \times 2 = 24$
  • $600 - 24 = \mathbf{576}$

3. Subtract the Square:

  • $(22)^2 = \mathbf{484}$ (This should be memorized for speed).
  • $576 - 484 = \mathbf{92}$

Correct Option: (c). 92

Banker's Shortcut: The Last Digit Method

If you are down to the last 10 seconds:

  1. Term 1: $(6 \times 5) \div 0$ is tricky, so look at $\frac{16 \times 15}{20} \times 48 \rightarrow \frac{...0}{20} \times 8 \rightarrow \frac{240}{20} \times 48 = 12 \times 48$. The last digit is 6 ($2 \times 8 = 16$).

  2. Term 2: $22^2$ ends in 4 ($2 \times 2 = 4$).

  3. Final Digit: $6 - 4 = \mathbf{2}$.

  4. Looking at the options, only 92 and 72 end in 2. Since $12 \times 48$ is roughly $12 \times 50 = 600$, and $600 - 484$ is clearly around $100$, 92 is the only logical choice.


Q3. $?^2 + \{20 + (98 \div 14)\} = 91$
(a) 9
(b) 6
(c) 8
(d) 7
(e) 5

Ans: (c)
Solution: 

The Fast-Track Solution

1. Tackle the Parentheses (Mental Math):

  • $98 \div 14$
  • Trick: If you don't know the table of 14, look at the units digit. $4 \times ? = \text{ends in } 8$. It's either $4 \times 2$ or $4 \times 7$. Since $14 \times 2$ is too small, try $14 \times 7$.
  • $10 \times 7 = 70$, $4 \times 7 = 28 \rightarrow 70 + 28 = 98$.
  • Result: 7

2. Solve the Brackets: $20 + 7 = \mathbf{27}$

3. Isolate the Square:

  • $?^2 + 27 = 91$
  • $?^2 = 91 - 27$
  • Subtraction Trick:$91 - 20 = 71$, then $71 - 7 = \mathbf{64}$.

4. Final Step:

  • $?^2 = 64$
  • $\mathbf{?} = 8$

Correct Option: (c). 8

Exam Pro-Tips for Speed
  • Tables up to 20: Knowing that $14 \times 7 = 98$
  • Unit Digit Shortcut: Look at the equation as $?^2 + (0 + 7) = 1$. This means $?^2$ must end in a 4 ($11 - 7 = 4$). Look at your options:
    • $9^2 = 81$ (ends in 1)
    • $6^2 = 36$ (ends in 6)
    • $8^2 = 64$ (ends in 4)$\rightarrow$Winner!
    • $7^2 = 49$ (ends in 9)

Q4.  $208 \div 52 \times ? \div 64 = 28$
(a) 345
(b) 456
(c) 448
(d) 468
(e) 480

Ans: (c)
Solution:

The Fast-Track Solution

1. Simplify the first division (Mental Math):

  • $208 \div 52$
  • The Trick: Look at the relationship. $50 \times 4 = 200$ and $2 \times 4 = 8$.
  • So, $208 \div 52 = \mathbf{4}$.

2. Rewrite the equation to isolate the unknown: $4 \times \frac{?}{64} = 28$

3. Cancel out the terms:

  • Divide both sides by 4: $\frac{?}{64} = 7$
  • Now, $? = 64 \times 7$

4. Final Multiplication (The Break-up Trick):

  • $60 \times 7 = 420$
  • $4 \times 7 = 28$
  • $420 + 28 = \mathbf{448}$

Correct Option: (c). 448

Speed Hack: Unit Digit & Estimation

  • Unit Digit: From the step $64 \times 7$, calculate the last digit only: $4 \times 7 = 2\mathbf{8}$
  • Look at the options: Only C (448) and D (468) end in 8.
  • Estimation:$60 \times 7$ is $420$. Since we are multiplying $64$, the answer must be slightly higher than $420$. 448 fits perfectly, while $468$ is too far away.

Q5. $68 \times 42 - 14 = ? \times 203$
(a) 21
(b) 20
(c) 28
(d) 14
(e) 24

Ans: (d)
Solution: 

The Fast-Track Solution

Instead of multiplying $68 \times 42$, look for a common number you can pull out to simplify the equation.

1. Spot the Factor of 14:

  • Notice that $42$ is $14 \times 3$.
  • The equation becomes: $68 \times (14 \times 3) - 14 = ? \times 203$
  • Pull out 14 from the left side: $14 \times (68 \times 3 - 1)$

2. Simplify the Brackets:

  • $68 \times 3$: Think $(70 \times 3) - (2 \times 3) = 210 - 6 = 204$.
  • Now subtract the 1: $204 - 1 = \mathbf{203}$.

3. The "Aha!" Moment:

  • Your equation is now: $14 \times 203 = ? \times 203$
  • Cancel 203 from both sides.
  • $? = 14$

Correct Option: (d). 14

Speed Hack: The Unit Digit Check

If you didn't see the factor of 14, you could use the last digit:

  1. Left Side: $(8 \times 2) - 4 \rightarrow 16 - 4 = 12$. The last digit is 2.

  2. Right Side: $? \times 3$ must end in 2.

  3. Check Options:

    • $21 \times 3 = ...3$

    • $20 \times 3 = ...0$

    • $28 \times 3 = ...4$

    • $14 \times 3 = ...12$ (Ends in 2!) $\rightarrow$ Matches.

    • $24 \times 3 = ...12$ (Also ends in 2, but 14 is a more likely factor given the 203).

By spotting the 14, you solved a "hard" multiplication problem in about 10 seconds without actually doing any long multiplication!

The document Practice Questions: Simplification is a part of the Bank Exams Course IBPS PO Prelims & Mains Preparation.
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FAQs on Practice Questions: Simplification

1. What are the key steps involved in simplification for bank exams?
Ans. The key steps involved in simplification for bank exams include identifying the mathematical operations required, applying the order of operations (BODMAS/BIDMAS), simplifying expressions by combining like terms, and performing calculations systematically to arrive at the final answer.
2. How can understanding fractions improve simplification skills?
Ans. Understanding fractions is crucial for simplification as it allows candidates to manipulate ratios, convert mixed numbers to improper fractions, and simplify complex fractional expressions. This knowledge helps in accurately solving problems that involve division and multiplication of fractions.
3. What role does approximation play in simplification during exams?
Ans. Approximation plays a significant role in simplification during exams as it allows candidates to quickly estimate answers without performing detailed calculations. This skill is particularly useful in multiple-choice questions where an approximate answer can help in selecting the correct option efficiently.
4. Why is it important to practice simplification regularly for bank exams?
Ans. Regular practice of simplification is important for bank exams as it enhances speed and accuracy in solving numerical problems. Familiarity with different types of simplification problems helps candidates to develop strategies for tackling complex questions under timed conditions, thus improving overall exam performance.
5. What common mistakes should candidates avoid while simplifying expressions?
Ans. Common mistakes candidates should avoid while simplifying expressions include neglecting the order of operations, miscalculating basic arithmetic operations, failing to simplify fractions fully, and overlooking negative signs. Being mindful of these errors can lead to more accurate and efficient simplification in exams.
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