CDS Maths(2) Questions and answers

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These questions were asked in previous year CDS General Maths exam.You can practice these questions as real time exam in this link.

11). 15 men complete a work in 16 days. If 24 men are employed, then the time required to complete that work will be

A). 7 days

B). 8 days

C). 10 days

D). 12 days

View Answer

Correct Answer: 10 days

According to 'MAN-DAY-WORK' concept

\( \Large \frac{M_{1} \times D_{1}}{W_{1}} = \frac{M_{2} \times D_{2}}{W_{2}} \)

\( \Large \frac{15 \times 16}{1} = \frac{24 \times D_{2}}{1} \)

\( \Large D_{2} = \frac{15 \times 16}{24} = \frac{5 \times 16}{8} \)

\( \Large D_{2} = 10 \)days

12). A train takes 9 seconds to cross a pole. If the Speed of the train is 48 km/hr, the length of the train is

A). 150 m

B). 120 m

C). 90 m

D). 80 m

View Answer

13). Ravi's brother is 3 years elder to him. His father was 28 years of age when his sister was born while his mother was 26 years of age when he was born. If his sister was 4 years of age when his brother was born, the ages of Ravi's father and mother respectively when his brother was born were

A). 32 years and 23 years

B). 32 years and 29 years

C). 35 years and 29 years

D). 35 years and 33 years

View Answer

Correct Answer: 32 years and 23 years

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years

Then sister's age = (x - 28) years

x - 28 = 4

x = 32

Hence when Ravi's brother was born his father was 32 years of age

Now, Ravi's age = (y - 26) years

Age of Ravi's brother = (y - 26 + 3) years

= (y - 23) years

Now, when Ravi's brother was born his age = 0 years

y - 23 = 0

y = 23

Hence when Ravi's brother was born his mother age was 23 years

Hence option (a)is correct

14). In solving a problem, one student makes a mistake in the coefficient of the first degree term and obtains 9 and -1 for the roots. Another student makes a mistake in the constant term of the equation and obtains 8 and 2 for the roots. The correct equation was

A). \( x^{2}\)+\( 10x\)+\( 9\)=\( 0\)

B). \( x^{2}\)-\( 10x\)+\( 16\)=\( 0\)

C). \( x^{2}\)-\( 10x\)+\( 9\)=\( 0\)

D). None of the above

View Answer

Correct Answer: \( x^{2}\)-\( 10x\)+\( 16\)=\( 0\)

The equation with roots -9 and -1 is , \( \Large (x+9)(x+1) = x^{2} + 10x +9 \) The equation with roots 8 and 2 is (x - 8)(x - 2) = \( \Large x^{2} - 10x + 16 \) The correct equation was \( \Large x^{2} - 10x + 16 \)

Hence option (B)is correct.

15). If m and n are the roots of the equation \( \large ax^{2}\)+\(bx \)+\( c\) =0 then the equation whose roots are \( \Large \frac{\left(m ^{2}+ 1\right)}{m} \) and \( \Large \frac{\left(n^{2}+1\right)}{n} \) is

A). \( acx^{2}\)+\( \left( ab+bc\right)\)\( x\)+\( b^{2}\)+\( \left( a-c\right)^{2}\)=\( 0\)

B). \( acx^{2}\)+\( \left( ab-bc\right)\)\( x\)+\( b^{2}\)+\( \left( a-c\right)^{2}\)=\( 0\)

C). \( acx^{2}\)+\( \left( ab-bc\right)\)\( x\)+\( b^{2}\)-\( \left( a-c\right)^{2}\)=\( 0\)

D). \( acx^{2}\)+\( \left( ab+bc\right)\)\( x\)+\( b^{2}\)-\( \left( a-c\right)^{2}\)=\( 0\)

View Answer

Correct Answer: \( acx^{2}\)+\( \left( ab+bc\right)\)\( x\)+\( b^{2}\)+\( \left( a-c\right)^{2}\)=\( 0\)

Let m = n = 1

Equation with root 1 and 1 is \( \Large (x - 1)(x - 1)= x^{2} - 2x + -1 \) ...(i) Comparing equation (i) with \( \Large ax^{2} + bx + c = 0 \)

We get a = 1, b = -2 and c = 1.

Now \( \Large \frac{m^{2} + 1}{m} = 2 \) and \( \Large \frac{n^{2} + 1}{n} = 2 \)

Now from option (a)

\( \Large acx^{2} +(ab + bc)x + b^{2} +(a - c)^{2} \)

\( \Large = x^{2} + (-2-2)x + 4 + (0) \)

\( \Large = x^{2} - 4x + 4 \)

\( \Large = (x - 2)(x - 2) \)

x = 2,3

Hence equation is satisfied

Option a is correct

16). The value of \( x^{2} - 4x + 11 \) can never be less than

A). 7

B). 8

C). 11

D). 22

View Answer

Correct Answer: 7

The value of \( \Large x^{2} - 4x + 11 \) can never be less than

i.e. we have to find minimum value.

Differentiate the given equation and equate it to zero

\( \Large f(x) = x^{2} + 4x + 11 \)

\( \Large f'(x) = 2x - 4 = 0 \)

x = 2

and f''(x) = 2 > 0 hence at x = 2 function will have minimum value

\( \Large f(x) = x^{2} - 4x + 1 \)

=> \( \Large f(2) = (2)^{2} - 4(2) + 11 \)

= 4 - 8 + 11 = 7

17). If \( \Large \left( x^{2} + \frac{1}{x^{2}} \right) = \frac{17}{4}\), then what is \( \Large \left( x^{3} - \frac{1}{x^{3}} \right) \) equal to?

A). \( \Large \frac{75}{11} \)

B). \( \Large \frac{63}{8} \)

C). \( \Large \frac{95}{8} \)

D). None of these

View Answer

Correct Answer: \( \Large \frac{63}{8} \)

Here \( \Large x^{2} + \frac{1}{x^{2}} = \frac{17}{4} \)

Now \( \Large \left(x - \frac{1}{x}\right)^{2} = x^{2} + \frac{1}{x^{2}} - 2 \)

\( \Large \left(x - \frac{1}{x}\right)^{2} = \frac{17}{2} - 2 \)

\( \Large \left(x - \frac{1}{x}\right)^{2} = \frac{17 - 8}{4} \)

\( \Large \left(x - \frac{1}{x}\right)^{2} = \frac{9}{4} \)

\( \Large \left(x - \frac{1}{x}\right) = \frac{3}{2} \)

Now

\( \Large \left(x - \frac{1}{x}\right)^{2} = \left(x - \frac{1}{x}\right) \left(x^{2} + 1 + \frac{1}{x^{2}} \right) \)

\( \Large = \left(\frac{3}{2}\right) \left(\frac{17}{4} + 1\right) = \left(\frac{3}{2}\right) \left(\frac{21}{4}\right) \)

\( \Large =\frac{63}{8}\)

18). \( 3x^{4}\)-\( 2x^{3}\)+\( 3x^{2}\)-\( 2x\)+\( 3\) divided by \( \left( 3x+2\right)\), then the remainder is

A). \( 0\)

B). \( \Large \frac{185}{27} \)

C). \( \Large \frac{181}{25} \)

D). \( \Large \frac{3}{4} \)

View Answer

Correct Answer: \( \Large \frac{185}{27} \)

Here

\( \Large f(x) = 3x^{4} - 2x^{3} + 3x^{2} -2x + 3 \)

For 3x + 2 = 0

= > x = -2/3

Now remainder

\( \Large =f \left(- \frac{2}{3} \right) = 3\left(- \frac{2}{3} \right)^{4} -2\left(- \frac{2}{3} \right)^{3} +3\left(- \frac{2}{3} \right)^{2} - 2\left(- \frac{2}{3} \right) + 3 \)

\( \Large = 3 \times \frac{16}{81} + \frac{16}{27} +\frac{12}{9} + \frac{4}{3} + 3 \)

\( \Large = \frac{16}{27} + \frac{16}{27} + \frac{12}{9} + \frac{4}{3} + \frac{3}{1}\)

\( \Large \frac{16 + 16 + 36 + 36 + 81}{27} = \frac{185}{27} \)

19). What should be added to the expression\(x \)\( \left( x+a\right)\)\( \left( x+2a\right)\)\( \left( x+3a\right)\) so that the sum may ba a perfect square?

A). \( 9a^{2}\)

B). \( 4a^{2}\)

C). \( a^{4}\)

D). None of these

View Answer

20). If the roots of the equation \(Ax^{2} \)+\( Bx\)+\( C\) =0 are -1 and 1 then which one of the following is correct?

A). A andC are both zero

B). A and B are both positive

C). A and C are both negative

D). A and C are of opposite sign

View Answer

Correct Answer: A and C are of opposite sign

Roots of equation \( \Large Ax^{2} + Bx + C = 0 \) are - 1 and 1

A - B + C= 0 ...(i)

and A + B + C = 0 ...(ii)

From equations (i) and (ii) A + C = 0

A = - C

Hence 'A' and 'C' are of opposite sign.

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