Question 1

Surd decomposition and parameter sum Given that √28 − 6√3 is attempted to be written in the form √3·a + b with a, b rational numbers, determine the value of (a + b).

Accepted Answer

Introduction / Context: This question explores expressing a radical combination in a restricted “linear in √3” form. We test whether √28 − 6√3 can be written as √3·a + b where a and b are rational numbers, and then interpret what (a + b) would be. The core idea is to compare like radicals and rational parts consistently. Given Data / Assumptions: Expression: √28 − 6√3. Target form: √3·a + b with a, b ∈ rationals. We treat √7 and √3 as linearly independent over the rationals (they cannot be converted into each other with rational coefficients). Concept / Approach: Break √28 as 2√7. Then √28 − 6√3 = 2√7 − 6√3. If the target is √3·a + b, then any √7 term must vanish or be representable using √3 and rationals, which is not possible since √7 is irrational and independent from √3 over Q. Step-by-Step Solution: Rewrite √28 = 2√7, so the expression is 2√7 − 6√3.Target form is a√3 + b with a, b rational.There is no way to express 2√7 as q√3 + r for rational q, r, because √7 and √3 are distinct quadratic irrationals.Therefore, the proposed representation does not exist with rational a and b. Verification / Alternative check: Assume, for contradiction, 2√7 = a√3 + b. Squaring gives 28 = 3a^2 + b^2 + 2ab√3. The left side has no √3 term; hence 2ab must be 0. If ab = 0, then either a = 0 or b = 0, which still cannot yield 28 from 3a^2 + b^2 without contradicting the absence of √3 and the presence of √7 upon taking square roots. Hence impossible. Why Other Options Are Wrong: Numerical choices like −2, 2, 1, −1 assume a valid (a, b) exists; but no rational pair exists, so those sums are meaningless for this constraint. Common Pitfalls: Confusing √28 with 2√7 and then trying to blend √7 into a combination of √3 and rationals; these radicals cannot be interchanged with rational coefficients. Final Answer: Not possible (no such rational a, b)

Question 2

Cube–reciprocal identity with a = (√2 − 1)^(1/3) Evaluate (a − a^−1)^3 + 3(a − a^−1).

Accepted Answer

Introduction / Context: This problem tests recognition of a compact algebraic identity that relates (a − 1/a)^3 + 3(a − 1/a) to a^3 − 1/a^3. Using the given a = (√2 − 1)^(1/3), we can evaluate the expression exactly without messy decimals. Given Data / Assumptions: a = (√2 − 1)^(1/3). Target: (a − a^−1)^3 + 3(a − a^−1). All algebra performed with exact radicals. Concept / Approach: Identity: (t)^3 + 3t = a^3 − 1/a^3 where t = a − 1/a. Expand (a − 1/a)^3 to verify this, or use the well-known result to avoid expansion. Then compute a^3 and a^−3 directly from the definition of a. Step-by-Step Solution: Let t = a − 1/a. Then t^3 + 3t = a^3 − 1/a^3.Given a = (√2 − 1)^(1/3) ⇒ a^3 = √2 − 1.Compute a^−3 = 1/(√2 − 1) = (√2 + 1)/[(√2 − 1)(√2 + 1)] = √2 + 1.Hence a^3 − a^−3 = (√2 − 1) − (√2 + 1) = −2. Verification / Alternative check: Expand (a − 1/a)^3 = a^3 − 3a + 3/a − 1/a^3. Adding 3(a − 1/a) gives a^3 − 1/a^3 indeed, confirming the identity. Why Other Options Are Wrong: 2, 2√2, and √2 result from sign or reciprocal mistakes; 0 would require a^3 = 1/a^3, which is false here. Common Pitfalls: Forgetting to rationalize 1/(√2 − 1), or misapplying the identity by using plus instead of minus in the reciprocal term. Final Answer: -2

Question 3

Surd ratios and symmetric powers Let x = (√3 + 1)/(√3 − 1) and y = (√3 − 1)/(√3 + 1). Evaluate x^2/y + y^2/x.

Accepted Answer

Introduction / Context: Expressions featuring x and its reciprocal often reduce via symmetry. Here, y is the reciprocal of x, enabling a simplification from mixed fractions to a sum of cubes of x and 1/x. Given Data / Assumptions: x = (√3 + 1)/(√3 − 1), y = (√3 − 1)/(√3 + 1). We need S = x^2/y + y^2/x. Concept / Approach: Notice y = 1/x. Then S = x^2 * x + (1/x^2) * (1/x) = x^3 + 1/x^3. Next compute x in simplest form, then use the identity x^3 + 1/x^3 = (x + 1/x)^3 − 3(x + 1/x). Step-by-Step Solution: Simplify x: (√3 + 1)/(√3 − 1) = ((√3 + 1)^2)/(3 − 1) = (3 + 1 + 2√3)/2 = 2 + √3.Thus 1/x = 2 − √3 (since (2 + √3)(2 − √3) = 1).Then x + 1/x = (2 + √3) + (2 − √3) = 4.Compute x^3 + 1/x^3 = (x + 1/x)^3 − 3(x + 1/x) = 4^3 − 3*4 = 64 − 12 = 52. Verification / Alternative check: Directly cube x and 1/x numerically using √3 ≈ 1.732; sums confirm the exact value 52. Why Other Options Are Wrong: 76 and 64 come from squaring 4 or forgetting the subtraction term; 4 corresponds to x + 1/x, not the cube sum; 28 is a rough, incorrect estimate. Common Pitfalls: Not noticing y = 1/x; failing to rationalize x first, which hides the elegant cancellation. Final Answer: 52

Question 4

Use x + a/x = b to simplify a rational expression If x + a/x = b, evaluate (x^2 + b x + a) / (b x^2 − x^3).

Accepted Answer

Introduction / Context: This problem relies on manipulating the relation x + a/x = b. Converting it to a quadratic in x lets us replace x^2 terms and reduce a complicated rational expression to a simple ratio. Given Data / Assumptions: x + a/x = b, with x ≠ 0 and a ≠ 0. Expression: E = (x^2 + b x + a) / (b x^2 − x^3). Concept / Approach: From x + a/x = b, multiply by x to get x^2 + a = b x ⇒ x^2 − b x + a = 0. Use this to replace x^2 and also to simplify b − x in terms of a/x, which collapses the denominator neatly. Step-by-Step Solution: From x^2 − b x + a = 0, we have x^2 = b x − a.Numerator: x^2 + b x + a = (b x − a) + b x + a = 2 b x.Also from x + a/x = b ⇒ b − x = a/x.Denominator: b x^2 − x^3 = x^2 (b − x) = x^2 * (a/x) = a x.Therefore, E = (2 b x) / (a x) = 2b/a. Verification / Alternative check: Pick sample values a = 2, x satisfying x + 2/x = b; compute both sides numerically to confirm E = 2b/a regardless of the particular root x. Why Other Options Are Wrong: a + b, b/a, and ab ignore the key transformations; a/(2b) flips the correct ratio. Common Pitfalls: Losing track of signs when substituting x^2 = b x − a; not noticing that b − x = a/x. Final Answer: 2b/a

Question 5

D and E are the points on the sides AB and AC respectively of a ΔABC and AD = 8 cm, DB = 12 cm, AE= 6 cm and EC = 9 cm, then BC is equal to:

Accepted Answer

5/2 DE

Question 6

A tailor has 37.5 metres of cloth and he has to make 8 pieces out of a metre of cloth. How many pieces can he make out of half of the cloth he has ?

Accepted Answer

Half of the cloth = 37.5/2 From 1 meter he will make 8 pieces => in 37.5/2 ---- ? 37.5/2 x 8 = 150.

Question 7

D and E are points on side AB and AC of ΔABC. DE is parallel to BC. If AD:DB = 2:5 and area of ΔADE is 8 cm sq, what is the area (in sq cm) of quadrilateral BDEC?

Accepted Answer

90

Question 8

What is the slope of the line perpendicular to the line passing through the points (8,2) and (3,1)?

Accepted Answer

5

Question 9

In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 60 questions and secures 130 marks, the no of questions he attempts correctly is :

Accepted Answer

Let the number of correct answers be X. Number of incorrect answers = (60 – X). 4x – (60 – x) = 130 => 5x = 190 => x = 38

Question 10

If 2 tables and 3 chairs cost Rs, 3500 and 3 tables and 2 chairs cost Rs. 4000, then how much does a table cost ?

Accepted Answer

Let the cost of a table and that of a chair be Rs. x and Rs, y respectively. Then, 2x + 3y = 3500 ...(i) and 3x + 2y = 4000 .....(ii) solving (i) and (ii) we get x = 1000, y = 500

Question 11

A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed :

Accepted Answer

Let the no of shots be x. Then, Shots fired by A =(5/8)x Shots fired by B = (3/8)x Killing shots by A = 1/3 of (5/8)x = (5/24)x Shots missed by B = 1/2 of (3/8)x = (3/16)x (3/16)x = 27 ⇒ x =144 Birds killed by A = (5/24)x = (5/24)*144 = 30

Question 12

If a * b = 2a - 3b + ab, then 3 * 5 + 5 * 3 is equal to :

Accepted Answer

3 * 5 + 5 * 3 = (2 * 3 - 3 * 5 + 3 * 5) + ( 2 * 5 - 3 * 3 + 5 * 3) = (6 + 10 - 9 + 15) = 22.

Question 13

If a + b + c = 13, a2+b2+c2=69 then find the ab + bc + ca :

Accepted Answer

a + b + c 2 = a 2 + b 2 + c 2 + 2 a b + b c + c a 2 a b + b c + c a = a + b + c 2 - a 2 + b 2 + c 2 = 169 - 69 = 100 = a b + b c + c a = 50

Question 14

The difference of 18 and its reciprocal is equal to :

Accepted Answer

Required difference = 19/16 - 16/19 = 105/304..

Question 15

To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of bucket is reduced to two-fifth of its present ?

Accepted Answer

62.5

Question 16

A tank is 2/5 full. If 16 litres of water is added to the tank, it becomes 6/7 full. The capacity of the tank is:

Accepted Answer

Let the capacity of the tank be x litres. Then, (6x/7 - 2x/5) = 16 <=> 30x - 14x = 16*35 <=> x = 35

Question 17

Simplify the following equation. 1943 + 12 + 13 + 11 + 14

Accepted Answer

After solving we get, 62/931. None of the options satisfy.

Question 18

In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?

Accepted Answer

Suppose the man works overtime for x hours. Now, working hours in 4 weeks = (5 x 8 x 4) = 160. Therefore, 160 * 2.40 + x * 3.20 = 432 => 3.20x = 432 - 384 = 48 => x = 15. Hence, total hours of work = (160 + 15) = 175.

Question 19

How many 18 ' s are there in 3712 ?

Accepted Answer

37 1 2 = 75 2 Required number = 75 2 1 8 = 75 x 4 = 300

Question 20

The total of 324 of 20 paise and 25 paise make a sum of Rs. 71. The no of 20 paise coins is :

Accepted Answer

Let the number of 20 paise coins be x. Then the no of 25 paise coins = (324 - x). 0.20 * (x) + 0.25 (324 - x) = 71 => x = 200..

Simplification Questions