CLAT Exam > CLAT Notes > Quantitative Techniques for CLAT > Practice Questions: Simplification - 2
Practice Questions: Simplification - 2 | Quantitative Techniques for CLAT PDF Download
Q1: 5÷√5=?
(a) √5
(b) 5
(c) 25
(d) 0.05
Ans: (b)
Applying BODMAS rule: 5 ÷ √5 = 5/√5 = √5
Q2: 
(a) 2(1/7)
(b) 1/7
(c) 6/7
(d) None of these
Ans: c
= (1/8) + (3/7) - (1/8) + (3/7) = (6/7)
Q3: Solve 5(1/2) + 6(1/2) - 8(1/4)
(a) 2(3/4)
(b) 3(3/4)
(c) 2(1/4)
(d) None of these
Ans: (b)
5(1/2) + 6(1/2) - 8(1/4) = (5+6-8) + ((1/2) + (1/2) - (1/4)) = 3(3/4)
Q4: 22 – [9 – {6 – (10 
]
(a) 12
(b) 14
(c) 10
(d) 8
Ans: c
= 22 - [9 - {6 -) - (10 – 1)}] = 22 - [ 9 - { 6 - 9}]
= 22 - [9 - {-3}] = 22 - [9 + 3] = 22 - [12] = 22 - 12 = 10
Q5: 103 - 33 - 23/102 =?
(a) 9.5
(b) 9.62
(c) 9.61
(d) 9.6
Ans: a
103 - 33 - 23/102
= 1000-27-8/100
=965/100 = 9.65
Q6: (14)2 – (.06)2
(a) 194.36
(b) 193.36
(c) 195.36
(d) None of these
Ans: d
(14)2 – (.06)2
= 196 - .0036 = 195 .9964
Q7: 37.5% of 800 – 6.25% of 1600 =?
(a) 200
(b) 250
(c) 275
(d) 236
Ans: (a)
37.5% of 800 - 6.25% of 1600
3/8 × 800 - 1/16 × 1600 = 300 – 100 = 200
Q8: (√289/34) × (68/√1156) × (√961/62) = ?
(a) 1/2
(b) 3/4
(c) 5/6
(d) 7/8
Ans: a
(√289/34) × (68/√1156) × (√961/62)
= (17/34) × (68/34) × (31/62) = 1/2
Q9: 63 + 83 -73 / 62 + 83 - 72 = ?
(a) 287/385
(b) 223/167
(c) 385/51
(d) 231/378
Ans: c
63 + 83 -73 / 62 + 83 - 72
= (216+512-343)/(36+64-49)
= (728-343)/(100-49)
= 385/51
Q10: (19×9+31×10+32×11)/(233.25×2-50) = ?
(a) 4
(b) 8
(c) 6
(d) 2
Ans: (d)
(19×9+31×10+32×11)/(233.25×2-50)
= 171+310+350/466.5-50
= 833/416.5
= 2
The document Practice Questions: Simplification - 2 | Quantitative Techniques for CLAT is a part of the CLAT Course Quantitative Techniques for CLAT.
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FAQs on Practice Questions: Simplification - 2 - Quantitative Techniques for CLAT
| 1. What is simplification in mathematics? |  |
Ans. Simplification in mathematics refers to the process of performing operations, such as addition, subtraction, multiplication, and division, to simplify an expression or equation to its simplest form.
| 2. How is simplification useful in solving mathematical problems? | |
Ans. Simplification is useful in solving mathematical problems as it helps in reducing complex expressions or equations into simpler forms, making it easier to understand and solve the problem efficiently.
| 3. Can simplification be applied to both numerical and algebraic expressions? | |
Ans. Yes, simplification can be applied to both numerical and algebraic expressions. It involves simplifying numerical expressions by evaluating the given values and simplifying algebraic expressions by combining like terms and applying algebraic rules.
| 4. Are there any specific steps or rules to follow for simplification? | |
Ans. Yes, there are specific steps and rules to follow for simplification. These include performing operations in the correct order (BIDMAS), combining like terms, removing parentheses and brackets, and simplifying fractions or radicals.
| 5. What are some common mistakes to avoid in simplification? | |
Ans. Some common mistakes to avoid in simplification include forgetting to apply the correct order of operations, making errors while combining like terms, incorrectly simplifying fractions or radicals, and overlooking negative signs or exponents. It is important to double-check the steps and calculations to minimize errors.
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