CDS Elementary Mathematics 14 Feb 2016 Official Paper (Previous Year Paper)

Calculation:

the roots be p and q respectively

Sum of roots = -b/a

(p + q) = -m/l          -----(1)

Product of roots = c/a

pq = m/l              -----(2)

$\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} + \sqrt {\frac{m}{l}} = [(p + q)/\sqrt {{pq}}] + \sqrt {\frac{m}{l}}$

$\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} + \sqrt {\frac{m}{l}} = [(p + q)/\sqrt {{pq}}] + \sqrt {\frac{m}{l}}$

Using equation (1) & (2)

$\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} + \sqrt {\frac{m}{l}} =\frac{-m/l}{\sqrt(m/l)} + \sqrt {\frac{m}{l}}$

$\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} + \sqrt {\frac{m}{l}} =- \sqrt {\frac{m}{l}} + \sqrt {\frac{m}{l}}$

$\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} + \sqrt {\frac{m}{l}} =0$

∴ $\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} {\rm{;}} + \sqrt {\frac{m}{l}} = 0$

Given:

√(3x2 − 7x − 30) − √(2x2 − 7x − 5) = x − 5 has α and β as its roots

Concept used:

(a − b)² = a² + b² − 2ab

If the quadratic equation ax² + bx + c = 0

Then, a product of roots, (αβ) = c/a

Calculation:

Let, 

3x2 − 7x − 30 = t2   and

2x2 − 7x − 5 = u2     

t² - u² = (3x2 − 7x − 30) - (2x2 − 7x − 5)

t2 - u2 = x² - 25

We know that, (a - b)(a + b) = a² - b²

⇒ (t - u)(t + u) = (x - 5)(x + 5)     ----(1)

Also, according to the question, 

t − u = x − 5              -----(2)

Doing (1) ÷ (4)    

⇒ (t - u)(t + u) / (t - u) = [(x - 5)(x + 5) / (x - 5)]

⇒ t + u = x + 5         -----(3)

Adding equation (2) & (3)

2t = 2x

⇒ t = x

⇒ t2 = x2

3x2 − 7x – 30 = x2  (∵ t² = 3x2 − 7x – 30)

2x2 − 7x – 30 = 0

Hence, a product of the root 

(c/a) = -30/2 = -15

∴ The required value is −15.

Given:

$\frac{p}{{x}}$ + $\frac{q}{{y}}$ = m and $\frac{q}{{x}}$ + $\frac{p}{{y}}$ = n

Calculation:

$\frac{p}{{x}}$ + $\frac{q}{{y}}$ = m 

⇒$\frac{py + qx}{{xy}}$ = m     -----(1)

And, $\frac{q}{{x}}$ + $\frac{p}{{y}}$ = n

⇒$\frac{qy + px}{{xy}}$ = n      ----(2)

Dividing equation (1) by (2) we get,

⇒ $\frac{py + qx}{{xy}}$/$\frac{qy + px}{{xy}}$ = $\frac{m}{{n}}$

⇒ $\frac{py + qx}{{qy + px}}$ = $\frac{m}{{n}}$

⇒ npy + nqx = mqy + mpx

⇒ mpx - nqx = npy - mqy

⇒ x(mp - nq) = y(np - mq)

⇒ x/y = $\rm\frac{n p−m q}{m p−n q}$

∴ The required value is $\rm\frac{n p−m q}{m p−n q}$.

Given:

a2 − by − cz = 0, ax − b2 + cz = 0 and ax+ by − c2 = 0

Calculation:

According to the question

a2 − by − cz = 0

⇒ a2 = by + cz

⇒ a2 + ax = ax + by + cz

⇒ a(a + x) = ax + by + cz

⇒ $\frac{1}{a+x} = \frac{a}{ax+by+cz}$

Similarly,

⇒ $\frac{1}{b+y} = \frac{b}{ax+by+cz}$

⇒ $\frac{1}{c+z} = \frac{c}{ax+by+cz}$

Therefore, the required value

⇒ $\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$

⇒ $\frac{ax}{ax+by+cz}+ \frac{by}{ax+by+cz}+ \frac{cz}{ax+by+cz}$

⇒ $\frac{ax+by+cz}{ax+by+cz}=1$

∴ The required value is 1.

Given:

The given equations = x2 − px + q = 0 and x2 + qx − p = 0

Calculation:

Let the common root be t.

Then, t² − pt + q = 0        ----(1)

⇒ t² + qt − p = 0             -----(2)

Now, subtracting equation (1) from (2), we get

⇒ (q + p)t − (q + p) = 0

⇒ t = 1

Now, substituting the value of t in equation (1), we get

⇒ 1 - p + q = 0

⇒ p − q − 1 = 0

∴ The required value is p − q − 1 = 0.

Given:

x = 21/3 + 2^-1/3

Concept used:

(a + b)³ = a³ + b³ + 3ab(a + b)

(a^m)ⁿ = a^mn

Calculation:

Let a = 2^1/3 and b = 2^-1/3

⇒ x = 2^1/3 + 1/2^1/3       ----(1)

Now, on cubing both sides, we get

x³ = (2^1/3 + 1/2^1/3)³

By using the above concept,

⇒ x³ = $(2^\frac{1}{3})^3+ \frac{1}{(2^\frac{1}{3})^3} + 3\times 2^\frac{1}{3} \times \frac{1}{2^\frac{1}{3}}[2^\frac{1}{3}+\frac{1}{2^\frac{1}{3}}]$

⇒ x3 = 2 + $\frac{1}{2}$ + 3 × 2^1/3 × 1/2^1/3 × (x)

⇒ x³ = 2 + $\frac{1}{2}$ + 3x

⇒ x³ - 3x = 2 + $\frac{1}{2}$

⇒ x³ - 3x = $\frac{5}{2}$

⇒ 2x³ - 6x = 5

⇒ 2x³ - 6x - 5 = 0

∴ The required value is 0.

Let the two expressions be P and Q

P + Q = 5x² − x − 4

P − Q = x² + 9x − 10

Adding two equations 2P = 6x² + 8x −14

∴ P = 3x² + 4x − 7

⇒ 3x² + 7x − 3x − 7

⇒ x(3x + 7) − 1(3x + 7)

∴ (x − 1) (3x + 7)

Subtracting two equations

Q = 2x² − 5x + 3

⇒ 2x² − 3x − 2x + 3

⇒ x(2x − 3) − 1(2x − 3)

⇒ (x − 1) (2x − 3)

∴ HCF = (x − 1)

Given:

(s − a) + (s − b) + (s − c) = s

Calculation:

According to the question

(s − a) + (s − b) + (s − c) = s

⇒ 3s - (a + b + c) = s

⇒ 2s = a + b + c       ----(1)

$\rm\frac{(s−a)^2+(s−b)^2+(s−c)^2+s^2}{a^2+b^2+c^2}$

$\rm\frac{s^2+a^2-2as+s^2+b^2-2bs+s^2+c^2-2cs+s^2}{a^2+b^2+c^2}$

$\rm\frac{4s^2 + a^2+b^2+c^2-2s(a+b+c)}{a^2+b^2+c^2}$

$\rm\frac{4s^2 + a^2+b^2+c^2-2s\times2s}{a^2+b^2+c^2}$      [From equation (1)]

$\rm\frac{a^2+b^2+c^2}{a^2+b^2+c^2} =1$

∴ The required value is 1.

Given:

x6 + px5 + qx4 − x2 − x − 3 is divisible by (x4 − 1)

Concept used:

Factor theorem:  It is a theorem that links factors and zeros of the polynomial

Calculation:

According to the question

Here, divisor x⁴ − 1 = (x − 1) (x + 1 (x² + 1)

Now, by factor theorem, f(1) = f(−1) = 0

Case I:

If x = 1

x6 + px5 + qx4 − x2 − x − 3 = 0

⇒ 1 + p + q − 1 − 1 − 3 = 0

⇒ p + q = 4      ----(1)

Case II:

If x = −1

x6 + px5 + qx4 − x2 − x − 3 = 0

⇒ 1 − p + q − 1 + 1 + 3 = 0

⇒ p − q = −2       -----(2)

Now, adding equations (1) and (2)

⇒ (p + q) + (p - q) = 4 - 2

⇒ 2p = 2

⇒ p = 1

Putting the value of p in equation (1), we get q = 3

Now, the value of p2 + q2 = (1² + 3²) = 10

∴ The required value is 10.

Given:

The given polynomial is A(x) = x2 + px + q.

(x − m) and (x − km) are simple factors of A(x), where m is a non-zero integer and k is positive integer, k ≥ 2.

Concept used:

In a quadratic equation ax² + bx + c = 0

Sum of roots = −b/a

Product of roots = c/a

Calculation:

If (x − m) and (x − km) are the factors of the polynomial, then m and km are roots of the polynomial

According to the above concept

Sum of roots = −b/a

⇒ m + km = −p

⇒ m(k + 1) = −p       ---- (1)

Now, a product of roots = c/a

⇒ m × km = q

⇒ km² = q       ---- (2)

Now, divide equation (2) by the square of equation (1)

⇒ $\frac{km^2}{m^2(k+1)^2} = \frac{q^2}{(-p)^2}$

⇒ (k + 1)²/k = p²/q

⇒ (k + 1)²q = kp²

∴ The required value is (k + 1)2q = kp2

Concept :

Remainder Theorem: 

The remainder theorem states that when a polynomial p(x) is

divided by a linear polynomial (x - a), then

The remainder is equal to p(a).

Calculation:

Let P(x) = xⁿ + mⁿ

If P(x) is divided by (x – m).

By using the remainder theorem, the remainder

⇒ R = P(m) = mⁿ + mⁿ

⇒ R = 2mⁿ

∴ R is a non-zero even integer.

4^x2^y = 128

2^2x 2^y = 128 [∵ 4 = 2²]

2^{(2x + y)} = 2⁷ [∵ a^m × aⁿ = a^{(m + n)} and 128 = 2⁷]

Equating the powers

2x + y = 7       ---- (1)

3^3x 3^2y − 9^xy = 0

3^3x 3^2y − 3^2xy = 0 [∵ 9 = 3²]

3^{3x + 2y} = 3^2xy

Equating powers

3x + 2y = 2xy

3x = 2xy − 2y

3x = 2y (x − 1)

∴ y = 3x/[2(x − 1)]

Substituting in equation 1

2x + 3x/[2(x − 1)] = 7

 [(4x² − 4x + 3x)] / [2(x − 1)] = 7

∴ 4x² − x = 14x − 14

4x² − 15x + 14 = 0

4x² − 7x − 8x + 14 = 0

x(4x − 7) − 2(4x − 7) = 0

(x − 2) (4x − 7) = 0

∴ x = 2 or x = 7/4

Substituting in equation 1

y = 3 or y = 7/2

∴ x + y = 2 + 3 = 5 or 7/4 + 7/2 = 21/4

From the given option x + y = 5 is the right choice

ax² − (a² + 1)x + a = ax² − a²x − x + a

⇒ ax(x − a) − 1(x − a)

⇒ (x − a) (ax − 1)

∴ The linear factors are (x − a) and (ax − 1)

Sum of the factors = x − a + ax − 1

⇒ x − 1 + ax − a

⇒ (x − 1) + a (x − 1)

∴ (x − 1) (1 + a)

Given:

x = [√(a + 2b) + √(a − 2b)]/[√(a + 2b) − √(a − 2b)]

Formula used:

(u + v) (u − v) = u2 − v²

(u + v)2 = u2 + v2 + 2uv

Calculation:

x = [√(a + 2b) + √(a − 2b)]/[√(a + 2b) − √(a − 2b)]

Multiplying by [√(a + 2b) + √(a − 2b)] in the numerator and denominator

⇒ [√(a + 2b) + √(a − 2b)]²/[a + 2b − a + 2b]   [(u + v) (u − v) = u² − v²]

⇒ [a + 2b + a − 2b + 2√(a² − 4b²)/4b]   [(u + v)² = u² + v² + 2uv]

⇒ [2a + 2√(a² − 4b²)/4b]

⇒ [a + √(a² − 4b²)/2b]

So, 2bx = [a + √(a² − 4b²)]

⇒ 2bx − a = √(a² − 4b²)

Squaring on both the sides, we get

⇒ 4b²x² + a² − 4abx = a² − 4b²   [(u + v)² = u² + v² + 2uv]

⇒ 4b²x² − 4abx = − 4b²

Now, dividing both sides by 4b, we get

⇒ bx² − ax = −b

⇒ bx² − ax + b = 0

∴ The required value is 0.

a³ - b³ = 117 and a - b = 3

(a - b)³ = a³ - b³ - 3ab(a - b)

3³ = 117 - 3ab × (3)

27 = 117 - 9ab

9ab = 90

∴ ab = 10

(a + b)² = (a - b)² + 4ab

⇒ (a + b)² = 3² + 4 × 10 = 49

∴ (a + b)² = 49

a + b = 7

Let the roots be α and β

Sum of roots = −b/a

α + β = −b/a

Product of the roots = c/a

αβ = c/a

Sum of square of their reciprocal = 1/α² + 1/β²

⇒ (α² + β²)/(α²β²)

⇒ [(α + β)² − 2αβ]/(α²β²)

Sum of roots = Sum of square of their reciprocal

α + β = [(α + β)² − 2αβ]/(α²β²)

α²β² (α + β) = (α + β)² − 2αβ

(c/a)² × (−b/a) = (−b/a)² − 2c/a

−bc²/a³ = b²/a² − 2c/a

−bc²/a³ = (b² − 2ac)/a²

−bc² = [a(b² − 2ac)]

Rearranging the terms

∴ ab² + bc² = 2a²c

STATEMENT I:

n(n + 1)/2 = 861

n(n + 1) = 1722

n² + n − 1722 = 0

n² + 42n − 41n − 1722 = 0

n(n + 42) − 41(n + 42) = 0

(n − 41) (n + 42) = 0

∴ n = 41 or −42

There are two integer values of n for which S_n = 861

∴ Statement I is true

STATEMENT II:

 S_n = S_{−(n + 1)} and hence for any integer m, we have two values of n for which S_n = m

It is false, because not for all integral values of m, there exists an integral value of n

For example, If m = 5

n(n + 1)/2 = 5

n² + n − 10 = 0

Finding roots using quadratic equation

n = (1 ± √41)/2, not an integer

Calculation:

Statement:1 p + q < p − q

⇒ q < − q

⇒ 2q < 0

⇒ q < 0

Hence, q must be negative

Statement: II  p + q > p − q

⇒ q > −q

⇒ 2q > 0

⇒ q > 0  

Hence, only q must be positive

∴ Only statement I is correct.

Given:

a/b = b/c = c/d

Calculation:

Let, a/b = b/c = c/d = k

a = bk, b = ck

So, a = ck²

c = dk, b = dk² and a = dk³

Statement I:

$\frac{b^3+c^3+d^3}{a^3+b^3+c^3} = \frac{d}{a}$

⇒ $\frac{b^3+c^3+d^3}{a^3+b^3+c^3} = \frac{d^3k^6+d^3k^3+d^3}{d^3k^9+d^3k^6+d^3k^3}$

⇒ $\frac{b^3+c^3+d^3}{a^3+b^3+c^3} = \frac{d^3(k^6+k^3+1)}{d^3k^3(k^6+k^3+1)}$

⇒ $\frac{b^3+c^3+d^3}{a^3+b^3+c^3} = \frac{1}{k^3}$

⇒ $\frac{b^3+c^3+d^3}{a^3+b^3+c^3} = \frac{d}{dk^3}= \frac{d}{a}$       [∵ a = dk³]

So, Statement I is true.

Statement II:

$\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=\frac{a}{d}$

⇒ $\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=\frac{d^2k^6+d^2k^4+d^2k^2}{d^2k^4+d^2k^2+d^2}$

⇒ $\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=\frac{d^2k^2(k^4+k^2+1}{d^2(k^4+k^2+1)}$

⇒ $\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=\frac{d^2k^2(k^4+k^2+1}{d^2(k^4+k^2+1)}$

⇒ $\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=k^2=\frac{dk^2}{d}$

⇒ $\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=\frac{b}{d}$         [∵ b = dk²]

So, Statement II is wrong

∴ The required answer is 1 only.

Alternate Method

Calculation:

a/b = b/c = c/d

Let a = 1, b = 2, c = 4, d = 8

Statement I:

(b3 + c3 + d3)/(a3 + b3 + c3) = d/a 

⇒ (2³ + 4³ + 8³)/(1³ + 2³ + 4³) = 8/1

⇒ (8 + 64 + 512)/(1 + 8 + 64) = 8/1

⇒ (584/73) = 8/1

⇒ 8/1 = 8/1

So, a statement I is true.

Statement II:

(a2 + b2 + c2)/(b2 + c2 + d2) = a/d

⇒ (1² + 2² + 4²)/(2² + 4² + 8²) = 1/8

⇒ (1 + 4 + 16)/(4 + 16 + 64) = 1/8

⇒ (21/84) = 1/8

⇒ 1/4 ≠ 1/8

So, statement II is not right.

∴ 1 only.

Concept used:

(a2 − b2) = (a + b) (a − b)

Divisibility rule of 11: If the difference of the sum of alternative

digits of a number are divisible by 11, then that number is divisible by 11 completely.

Calculation:

Given that,

7¹⁰ − 5¹⁰

⇒ (7⁵ + 5⁵) (7⁵ − 5⁵)    

⇒ 7⁵ = 343 × 49 = 16807

⇒ 5⁵ = 625 × 5 = 3125

⇒ 710 − 510 = 19932(7⁵ − 5⁵)

Now, for the number 19932,

Sum of odd place digits = 2 + 9 + 1 = 12

Sum of even place digits = 9 + 3 = 12

So, the difference = 12 − 12 = 0

∴ The number is divisible by 11.

Given:

Let a two-digit number be k times the sum of its digits.

Calculation:

Let the two-digit number be ab, i.e. 10a + b.

Now, according to the question

⇒ 10a + b = k(a + b)    -------(1)

Now, the number is formed by interchanging the digits

10b + a = m(b + a)     ------(2)

Adding both equations, we get

⇒ 10a + b + 10b + a = K(a + b) + m(b + a)

⇒ 11a + 11b = (k + m) (a + b)

⇒ 11(a + b) = (k + m) (a + b)

⇒ 11 = (k + m)

⇒ m = 11 − k.

∴ The required value of m is 11 − k.

Relative displacement of the train = Length of the train

Relative speed = Relative distance/Time

⇒ 225/9 = 25 m/s = 25 × 18/5 = 90 km/hr

Relative speed = Speed of train + Speed of man

90 = Speed of train + 5

∴ Speed of train = 90 − 5 = 85 km/hr

sin θ = cos (90 − θ)

∴ sin 35° = cos (90 − 35) = cos 55°

∴ sin 35°/cos 55° = cos 55°/sin 35° = 1

(sin 35°/cos 55°)² − (cos 55°/sin 35°)² + 2 sin 30° = 1² − 1² + 2 × ½ = 1

Let the original average speed be x

∴ Reduced speed = x − 1

Difference in time = Time taken at average speed (x − 1) − Time taken at average speed x

20/60 = 14/(x − 1) − 14/x [∵ Time = Distance/Speed]

1/3 = 14 × [1/(x − 1) − 1/x]

1/3 = 14/[x(x − 1)]

x(x − 1) = 42

x² − x − 42 = 0

x² − 7x + 6x − 42 = 0

x(x − 7) + 6(x − 7) = 0

(x + 6) (x − 7) = 0

x cannot be negative

∴ x = 7 km/hr

Given:

A sum of money at a certain rate of simple interest per year becomes doubles in 5 years, and at a different rate of simple interest per year becomes three times in 12 years.

Formula Used:

Interest = P × r × t/100

Where p → Principal

r → Rate of interest

t → Time

Calculation:

Interest/Principal = (Rate/100) × Time

In first case, Amount gets double

Interest = Principal

⇒ 1 = (Rate/100) × 5

⇒ Rate = 20%

In second case, amount gets tripled

Interest = 2 × Principal

⇒ 2 = (Rate/100) × 12

⇒ Rate = 50/3%

Difference = 20 - 50/3 = (10/3)%

∴ The difference in the two rates of simple interest per year is (10/3)%.

STATEMENT II:

In ΔADB

AD² + BD² - 2 × AD × BD × cosθ = AB²                         [∵ Cosine formula]

⇒ AD² + BC²/4 - AD × BC × cosθ = AB²       ---- 1               [∵ BD = BC/2]

In ΔADC

AD² + CD² + 2 × AD × CD × cosθ = AC²             [∵ Cosine formula, cos(180 - θ = -cosθ]

⇒ AD² + BC²/4 + AD × BC × cosθ = AC²        ---- 2              [∵ CD = BC/2]

On adding eq 1 and eq 2.

⇒ 2AD² + BC²/2 = AB² + AC²

⇒ AD² = (2AB² + 2AC² - BC²) /4

Using similar algorithm

⇒ BE² = (2AB² + 2BC² - AC²) /4

⇒ CF² = (2BC² + 2AC² - AB²) /4

⇒ AD² + BE² + CF² = 3(AB² + BC² + CA²) /4

⇒ (AB² + BC² + CA²) / (AD² + BE² + CF²) = 4/3 for any triangle

⇒ Statement II is true

Statement I:

AD² = (2AB² + 2AC² - BC²) /4

Since there is a negative term so no general conclusion can be made. In obtuse angled triangle, statement I will not be true

Let ABC be a triangle such that AB = BC = 5 cm and ∠ABC = 150°

Using cosine formula

⇒ AC = √(AB² + BC² - 2AB × BC cos 150°)

⇒ √(5² + 5² - 2 × 5 × 5 cos 150°) ≈ 9.69

⇒ BE² = (2AB² + 2BC² - AC²) /4

Substituting AB = BC = 5 cm and AC = 9.69 cm

⇒ BE ≈ 1.24 cm

Now A’B’C’ be an equilateral triangle of sides 4 cm

Median of equilateral triangle = √3/2 × side = 2√3 cm

Although all sides of A’B’C’ are less than ABC, its median is greater than the latter’s median

⇒ Statement I is false

∴ Statement I is false but Statement II is true.

Calculation:

Let their incomes be 9x and 7x whereas their expenses be 4y and 3y respectively

Saving = Income – Expenses

⇒ 2000 = 9x – 4y         ----(1)

⇒ 2000 = 7x – 3y         ----(2)

Comparing both the equations

⇒ 9x – 4y = 7x – 3y

⇒ 2x = y

Substituting value of y in first equation

⇒ 2000 = 9x – 4 × 2x

⇒ x = 2000

∴ Difference in their annual income = 9x – 7x = 2x = 4000

Given:

S = {1, 2, 3, 4, 5, - - - - - - - - , 14}

Divisor = 15

Remainder = 1

Concept:

Dividend = (Divisor × Quotient) + Remainder

Calculation:

Maximum product of the numbers = 13 × 14 = 182

Number of terms such which leaves a remainder 1,

when divided by 15 from 2 to 182

= (15 × x) + 1

⇒ (15 × 1) + 1, - - - - - - - - , (15 × 12) + 1

⇒ 16, 31, 46, 61, 76, 91, 106, 121, 136, 151, 166, 181,

Now,

16 = (2 × 8) or (8 × 2) can be expressed as product of two natural numbers less than 15

31 cannot be expressed as product of two natural numbers less than 15

46 cannot be expressed as product of two natural numbers less than 15

61 cannot be expressed as product of two natural numbers less than 15

76 cannot be expressed as product of two natural numbers less than 15

91 = (7 × 13) or (13 × 7) can be expressed as product of two natural numbers less than 15

106 cannot be expressed as product of two natural numbers less than 15

121 cannot be expressed as product of two natural numbers less than 15

136 cannot be expressed as product of two natural numbers less than 15

151 cannot be expressed as product of two natural numbers less than 15

166 cannot be expressed as product of two natural numbers less than 15

181 cannot be expressed as product of two natural numbers less than 15

Hence, sets = (2, 8), (8, 2), (7, 13), (13, 7)

∴ Number of sets = 4

Given:

Clock strike once at 1 o'clock, twice at 2 o'clock, thrice at 3 o'clock, and so on

Time taken to strike 5 times at 5 o'clock = 8 seconds 

Calculation:

The clock strikes for 8 seconds at 5 o'clock.

Let the interval between two consecutive bells be t.

Time taken for the bell to ring 5 times = 4t [∵ there are only 4 intervals here]

⇒ 8 = 4t [∵ the bell rings for 8 seconds at 5 o'clock]

⇒ t = 2 seconds

∴ Time taken for the bell to ring 10 times at 10 o'clock = 9t = 18 seconds

Hence, the correct answer is option (3).

Let the centre of the circle be O

PC = PD = CD/2 [∵ Perpendicular drawn from centre bisects the chord]

⇒ PC = 1 unit

In ΔOPC

OC² = OP² + PC² [∵ Pythagoras theorem]

⇒ 2² = OP² + 1²

⇒ OP = √3

∴ AP = OA + OP = 2 + √3 units [∵ OA = radius]

Given:

x2 + px + q = 0

Concept used:

ax² + bx + c = 0

Sum of the roots = −b/a

Product of the roots = c/a

Calculation:

x2 + px + q = 0

Let the roots of the quadratic equation be t and t²

Sum of roots = (t + t²) = −p/1

⇒ t + t² = − p

Product of roots = (t × t²) = q/1

⇒ t³ = q

By observing option, we have to calculate,

p³ + q² = −(t + t²)³ + t⁶

⇒ −[t³ + t⁶ + 3t³(t + t²)] + t⁶ [∵ (a + b)³ = a³ + b³ + 3ab(a + b)]

⇒ −[t³ + 3t³(t + t²)]

⇒ −[q − 3qp] [∵ t³ = q, t + t² = −p]

⇒ p³ + q² = −q + 3pq

∴ p³ + q + q² = 3pq

Given:

N = 90 × 42 × 324 × 55

Concept used:

Break the number is multiple of 3

Calculation:

⇒ (9 × 10) × (3 × 14) × (81 × 4) × 55

⇒ (9 × 3 × 81) × (10 × 14 × 4 × 55)

⇒ (3² × 3 × 3⁴) × (10 × 14 × 4 × 55)

⇒ 3⁷ × (10 × 14 × 4 × 55)

∴ Maximum value of m for which 3^m divides N = 7

Important Points Dividend = (Divisor × Quotient) + Remainder

Given:

Speed range 25 - 50 km/hr, fuel consume 20 mL per km

Speed range other than 25 - 50 km/hr, fuel consume 40 mL per km

Distance travelled at a speed of 40 km/hr = 10 km

Distance travelled at a speed of 60 km/hr = 30 km

Distance travelled at a speed of 30 km/hr = 10 km

Total distance travelled = 50 km

Concept used:

Speed = distance/time

Fuel consumed = fuel consumption rate per km × distance travelled 

Calculation:

Distance travelled above the speed range of 25 – 50 km/hr = 30 km

⇒ Petrol consumed = 30 × 40 = 1200 mL = 1.2 L

Distance travelled other than speed range of 25 - 50 km/hr 

= (10 + 10) = 20 km

⇒ Petrol consumed = 20 × 20 = 400 mL = 0.4 L

∴ Total petrol consumed = 1.2 + 0.4 = 1.6 L

Concept:

Calculation:

Statement 2:  Every real number is a rational number

This is not true because a real number is a collection of all rational &

irrational numbers. 

∴ The correct answer is 1,3 and 4

If (x + 1) is a factor, then −1 is the root of the equation

⇒ (−1)³ + k (−1)² – 1(−1) + 2 = 0

⇒ −1 + k + 1 + 2 = 0

⇒ k + 2 = 0

∴ k = − 2

Statement I:

For cosx ≤ 1 [∵ -1 ≤ cosx ≤ 1]

⇒ 2^{m + 1} ≤ 1(2⁰)

⇒ m ≤ -1

⇒ There exists no positive real number for which statement I is true

Statement II:

mn ≥ m + n

If we take m = 1

⇒ n ≥ n + 1, which is not possible for any natural number

⇒ Statement II is wrong

∴ Neither 1 nor 2 is true.

Given:

1 + 1/x – 1/x² ≥ 0, x ≠ 0

Formula used:

Quadratic equation: px² + qx + r = 0

Roots of the quadratic equation = $x = {- q\space \pm \space \sqrt{q^2\space -\space4pr} \over 2p}$ 

Calculation:

1 + 1/x – 1/x2 ≥ 0

⇒ (x² + x – 1)/x² ≥ 0

⇒ x² + x – 1 ≥ 0

Roots of the equation = $x = {- 1\space \pm \space \sqrt{1^2\space -\space(4\times 1\times -1)} \over 2\times 1}$

⇒ $x = {- 1\space \pm \sqrt{5} \over 2}$

Let the roots of this equation be a and b, such a > b

= (x-a)(x-b) ≥ 0

Either both the terms can be positive, or both can be negative

For both to be positive, x ≥ a

For both to be negative, x≤ b

For the given equation the roots are

∴ x ≤ (-1 - √5) /2 or x ≥ (-1 + √5) /2

For P:

x + y = 5 – 1 = 4 = 4

x – y = 5 + 1 = 6 > 2

⇒ P lies in the given region

For Q:

x + y = 3 – 2 = 1 < 4

x – y = 3 + 2 = 5 > 2

⇒ Q lies in the given region

For R:

x + y = 1 + 1 = 2 < 4

x – y = 1 – 1 = 0 < 2

⇒ R does not lie in the given region

∴ Only P and Q lie in the given region.

Let the original speed be x km/h

⇒ Increased speed = x + 10 km/h

Time difference = Usual time taken – Time taken when speed is increased

1 = 120/x – 120/ (x + 10) [∵ Time = Distance/Speed]

⇒ 1 = 120(x + 10 – x) / [x(x + 10)]

⇒ x(x + 10) = 1200

⇒ x² +10x – 1200 = 0

⇒ x² + 40x – 30x – 1200 = 0

⇒ x(x + 40) – 30(x + 40) = 0

⇒ (x + 40)(x – 30) = 0

Ignoring negative sign

∴ Usual speed, x = 30 km/hour

Shortcut Trick

For a system of linear equations to have no solution, the lines must be parallel.

The condition for parallel lines is: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Comparing ratios: 3/6 = (−k)/(−10)

⇒ 1/2 = k/10

⇒ k = 10/2 = 5

∴ The correct answer is 5.

Alternate Method

Given:

Equation 1: 3x − ky − 20 = 0

Equation 2: 6x − 10y + 40 = 0

Formula used:

For the system a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition for no solution is:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Calculations:

From the equations, we have:

a₁ = 3, b₁ = −k, c₁ = −20

a₂ = 6, b₂ = −10, c₂ = 40

Applying the no solution condition:

⇒ 3/6 = (−k)/(−10)

⇒ 1/2 = k/10

⇒ k = 10 × (1/2)

⇒ k = 5

Check the second part of the condition:

⇒ a₁/a₂ = 1/2 and c₁/c₂ = −20/40 = −1/2

⇒ 1/2 ≠ −1/2 (Condition satisfied)

∴ The correct answer is 5.

Additional Information

Unique Solution

The system has a unique solution if the lines intersect at one point, which occurs when a₁/a₂ ≠ b₁/b₂.

Infinite Solutions

The system has infinitely many solutions if the lines are coincident, which occurs when a₁/a₂ = b₁/b₂ = c₁/c₂.

Consistency of Equations

A system is consistent if it has at least one solution (unique or infinite) and inconsistent if it has no solution.

Statement II:

Let the HCF of b and c be t

⇒ b = tp and c = tq where p and q are co-prime numbers

⇒ ab = atp and ac = atq

⇒ HCF(ab, ac) = at                  [∵ p and q are co-prime]

⇒ HCF(ab, ac) = a × HCF (b, c)

⇒ Statement II is correct

LCM (ab, ac) = Product of numbers (ab, ac) /HCF (ab, ac)

⇒ [(atp) × (atq)] / (at)

⇒ atpq

⇒ a [(tp) × (tq)] /t

⇒ a × Product of (b, c) /HCF (b, c)

⇒ a × LCM(b, c)

So, Statement I is true

Statement III:

When a = b

⇒ HCF (a, a) = a

⇒ LCM (a, a) = a

⇒ LCM (a, a) = HCF (a, a)

Statement III fails when a = b

Statement IV:

Let the HCF (a, b) = r

⇒ a = rp and b = rq, where r and q are coprime

LCM = Product of the numbers/HCF

⇒ rp × rq/r = rpq

⇒ LCM = HCF × pq

⇒ HCF divides LCM

⇒ Statement IV is correct

∴ Statements 1, 2 and 4 are true.

Given:

Sums of ages of two of them at a time are 4 years, 6 years and 8 years.

Calculation:

Let their ages be a years, b years, and c years

⇒ a + b = 4    ------ (I)

⇒ b + c = 6    ------ (II)

⇒ c + a = 8    ------ (III)

By adding equation (I) , (II), and (III), we get

⇒ 2(a + b + c) = 18

⇒ a + b + c = 9    ------ (IV)

Equation (IV) - equation (I)

So, c = 9 - 4 = 5

Putting value of c in equation (II), 

b + 5 = 6

So, b = 1

Putting value of c in equation (III), 

5 + a = 8

So, a = 3

∴ Difference in the ages of eldest and youngest = 5 – 1 = 4 years

Let OA be position of the hour hand and OB be the position of the minute hand initially

If they get reversed hour hand has to transverse an angle of θ and minute hand had to traverse an angle of 360 – θ

Let the time taken be ‘t’ minutes

Angle swept by hour hand in 1 minute = 360/ (12 × 60) = 0.5° [∵ It sweeps 360° in 12 hours]

Angle swept by minute hand in 1 minute = 360/60 = 6°

⇒ θ = 0.5t

⇒ 360 – θ = 6t

⇒ 360 – 0.5t = 6t

⇒ 6.5t = 360

∴ t = 360/6.5 ≈ 55.38 minutes

Given:

Total number of people = 100

Hindi = 70, English = 60, French = 30

Hindi + English = 30

Hindi + Franch = 20

English + French = x

Calculation:

Let the number of people who speak all three languages be y

⇒ 100 = (20 - y) + (30 - y) + y + (20 - y) + (30 - x) + x + (10 - x)

⇒ 100 = 110 - x - 2y

⇒ Least possible value of y = 0

⇒ x = 10 

⇒ x ≥ 1

y cannot exceed 20, because for knowing all three languages, they should know Hindi and French too, only 20 people know Hindi and French

If y = 20

then x ≤ 30

∴ 10 ≤ × ≤ 30 or 9 < × ≤ 30

Shortcut Trick 

Concept used:

Profit % = [(SP – CP) /CP] × 100

Calculation:

Let the actual cost price of 1 m cloth be Rs. a

False cost price of 1.1 m  cloth = Rs. a

So, the false cost price of 1 m cloth = Rs. a/1.1

Actual selling price of 1 m cloth = Rs. a

The false selling price  of 0.9 m cloth = Rs. a/0.9

Profit % = [(SP – CP) /CP] × 100

$⇒ \frac{ [\frac{a}{0.9} -\frac{a}{1.1}]}{\frac{a}{1.1}}\times100$

⇒  200/9 % = $22\frac{2}{9};\%$

∴ Profit % = $22\frac{2}{9};\%$

Concept used:

Efficiency = volume of the tank/time taken

Calculation:

LCM of 50, 60 and 75 = 300

Let the volume of the tank be 300 litres 

Efficiency of pipe A = 300/60 = 5 litres/minute

Efficiency of pipe B = 300/75 = 4 litres/minute

Efficiency of pipes A, B and C together

⇒ 300/50 = 6 litres/minute

So, efficiency of pipe C = (5 + 4 - 6) = 3 litres/minute

∴ Time is taken by pipe C to empty the tank = 300/3 = 100 minutes

Given:

Time taken by B – time taken by A to cover 1000 m = 10 seconds

In the same time A covers 1000 m, B covers (1000 – 100) = 900 meters

B’s speed gets halved after covering 450 m

Concept used:

Speed = distance/time

Calculation:

Let the speed of A and B be ‘a’ and ‘b’ (m/s) respectively

Time taken by B – Time taken by A to cover 1000 m = 10 seconds

⇒ 1000/b – 1000/a = 10    ------ (I) [∵ Time = Distance/Speed]

⇒ In the same time A covers 1000 m, B covers (1000 – 100) = 900 meters

Speed of A/Speed of B = Distance covered by A/Distance covered by B

⇒ a/b = 1000/900

⇒ a = 10b/9    ------ (II)

From equation (I) & (II), 

⇒ 1000/b – 1000/(10b/9) = 10

⇒ 1000/b – 900/b = 10

⇒ 100/b = 10

⇒ b = 10 m/s

⇒ a = 10b/9 = 100/9 m/s

Now B’s speed gets halved after covering 450 m

⇒ Time taken by B = [(450/b) + 550/(b/2)]

⇒ (450/10) + (550/5) = 155 seconds

⇒ Time taken by B (if no injury occurred) = 1000/10 = 100 seconds

⇒ Time taken by A = Time taken by B (without injury) - 10 = 90 seconds

∴ A beat B (during injury) by 155 – 90 = 65 seconds

Given:

Salary increment = 10%

Shortcut TrickConvert percentage into fraction

Concept used:

$Incresed\space percentage = \frac{(Final - Initial)}{Final}\times 100$

Calculation:

Let his initial salary be Rs. S

Salary after the increment

⇒  [(100 + 10)/100] × S = 1.1S

The amount he didn’t receive

⇒ 1.1S – S = 0.1S

Percentage of salary didn't receive

⇒ [(0.1S)/(1.1S)] × 100 = (100/11)%

∴ The percentage of salary didn't receive is 100/11%.

Given:

Distance travelled on circular path = length of the arc = 100 m

Angle formed = 36^o

Concept used:

Length of the arc = (θ/360) × 2πr

Calculation:

Let radius of the circle = r m

100 = (36/360) × 2πr

⇒ 100 = πr/5 

So, r = 500/π 

∴ Radius of the circle = 500/π m

Given:

Width of the road = PQ

PR = RS = x meter and TS = y meter

Q, R and T on same line

Concept:

Congurence of the triangles: Two triangles are said to be congurent if pairs of their corresponding sides and their corrosponding angles are equal.

Calculation:

In ΔTSR & ΔPRQ;

∠RPQ = ∠RST = 90^∘

∠QRP = ∠SRT [Opposite angle]

PR = RS = x meter

Hence,

ΔTSR ≅ ΔPRQ [By AAS congurency]

⇒ PQ/TS = PR/RS

⇒ PQ/y = x/x

So, PQ = TS = y

∴ Distance between P and Q = y

In ΔAPC and ΔDPB

∠APC = ∠DPB [∵ vertically opposite angle]

∠PAC = ∠BDP [∵ Angle subtended by chord BC at every point on circle is same]

⇒ APC ∼ DPB

⇒ AP/DP = CP/BP

⇒ AP × BP = CP × DP            ---- 1

In triangle A’P’B’ and C’Q’D’ to be similar

A’P’/C’Q’ = B’P’/Q’D’

AP/CP = BP/DP [∵ Given A’P’ = AP, B’P’ = BP, C’Q’ = CP, D’Q’ = DP]

But AP/CP = DP/BP [Proved]

⇒ A’P’B’ and C’Q’D’ are not similar

If they are not similar, they are not congruent too

Area of triangle A’P’B’ = ½ × AP × BP

Area of triangle C’Q’D’ = ½ × CP × DP

From equation 1, both the areas are equal

Product of the sides are same, but that need not make perimeter same.

That requires AP = CP and BP = DP

∴ Statement 1, 2 and 4 are wrong

Given:

In ΔABC, AB = 1 unit

BC ∥ DE

area of triangle ABC = 2 × area of triangle ADE

Concept used:

Corresponding angles are the same

Calculation:

In triangle ADE and ABC,

∠A is common

Since, DE ∥ BC, so corresponding angles are the same

∠ADE = ∠ABC 

⇒ ADE ∼ ABC [By A-A similarity]

⇒ Area (ADE) /Area (ABC) = (AD/AB)²

⇒ 1/2 = (AD/1)²

⇒ AD = 1/√2 unit

∴ Length of the AD = 1/√2 unit  

Important Points

When two parallel lines are intersected by a transversal, then corresponding angles are always equal.

Let the sides of the rectangle be l and b respectively

Hence, diagonals of the rhombus = l and b

Now, Area of rhombus = ½ × Product of diagonals

⇒ 2 = ½ × l × b

⇒ lb = 4 square units = Area of rectangle 

∴ Correct option is 4 square units

In a polygon, Exterior angle = 180° – Interior angle

⇒ Exterior angle = 180 – 140 = 40°

Sum of exterior angles = Number of sides × Value of each exterior angle

⇒ 360 = n × 40 [ ∵ Sum of exterior angles of any polygon is 360°]

∴ Number of sides, n = 360/40 = 9

Given:

Let the triangles ABC and DEF be such that

∠ABC = ∠DEF, ∠ACB = ∠DFE and ∠BAC = ∠EDF.

L & M are midpoints of BC & EF respectively.

Concept used:

AA similarity: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.

SAS similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

Congruence of triangles: Two triangles are said to be congruent if all three corresponding sides are equal and all the three corresponding angles are equal in measure.

Statement I:

In ΔABC and ΔDEF

⇒ ∠ABC = ∠DEF and ∠ACB = ∠DFE   [Given]

⇒ ABC ∼ DEF   [AA property]

⇒ AB/DE = BC/EF  

⇒ AB/DE = (BC/2)/(EF/2)

⇒ AB/DE = BL/EM  

Now, In ΔABL and ΔDEM

⇒ AB/DE = BL/EM       [Proved]

⇒ ∠ABC = ∠DEF         [Given]

⇒ ABL ∼ DEM             [SAS property]

So, Statement I is true.

Statement II:

For any two triangles to be congruent, their corresponding sides have to be equal.

Here, one side is not equal.

Then, ΔALC is not congruent to ΔDMF.

So, Statement II is false.

∴ Statement I is true but Statement II is false.

Concept:

Circumference of circle = π × d  where d is the diameter.

Calculation:

Distance traversed = Number of rotations × Perimeter of the wheel

⇒ 4000 = Number of rotations × πd [ ∵ 1 km = 1000m]

⇒ 4000 = Number of rotations × 22/7 × 7/11

⇒ 4000 = 2 × Number of rotations

∴ Number of rotations = 4000/2 = 2000

Given:

The base of an isosceles triangle = 300 units

Each side of equal units = 170 units

Concept used:

 

Area of isosceles triangle = $\frac{a}{4}\sqrt{(4b^2-a^2)}$

Calculation:

According to the above concept

⇒ Area = $\frac{300}{{4}}\sqrt{(4\times (170)^2)-(300)^2}$

⇒ Area = 75√(4 × 28900 - 90000)

⇒ Area = 75√(115600 - 90000)

⇒ Area = 75√25600

⇒ Area = (75 × 160) units

⇒ Area = 12000 square units.

∴ The area of the triangle is 12000 square units.

Let the radius be r

Area of empty space = Area of the square – 4 × Area of quadrant

⇒ Side of square = 2 × radius = 2r

⇒ 150/847 = Side² – 4 × ¼ × π r²

⇒ 150/847 = (2r)² – 22/7 r²

⇒ 150/847 = r² (4 – 22/7)

⇒ 150/847 = 6r²/7

⇒ r² = 25/121

∴ radius, r = 5/11 cm

Given:

The ratio of side AB to side DE is (√2 + 1) ∶ √3

Concept:

Theorem: If two triangles are similar, then the ratio of the area of

both triangles are proportional to the square of the ratio of their

corresponding sides.

Calculation:

According to the question, ΔABC ∼ ΔDEF

By using the above concept

Area of ΔABC/Area of ΔDEF = (AB/DE)²

⇒ Area of ΔABC/Area of ΔDEF = $\frac{ (√2 + 1)^2}{{(√3)^2}}$

⇒ Area of ΔABC/Area of ΔDEF = $\frac{3 + 2√2}{{3}}$

Now, multiply the numerator and denominator by 3 - 2√2.

⇒ Area of ΔABC/Area of ΔDEF = ($\frac{3 + 2√2}{{3}}$ × $\frac{3 - 2√2}{{3-2\sqrt 2}}$)

Since, (a + b)(a - b) = a² - b²

⇒ Area of ΔABC/Area of ΔDEF = $\frac{1}{{9-6\sqrt2}}$

∴ The required ratio is 1 ∶ (9 – 6√2).

In triangle COP

∠CPO = 90° [ ∵ Tangent is perpendicular to the radius]

⇒ CO² = CP² + OP²

⇒ 3² + 4² = 25

⇒ CO = 5 unit

∴ sin ∠COP = CP/CO = 3/5 [ ∵ sin θ = Altitude/Hypotenuse]