SSC CGL Tier-2 2010 Maths

Instructions

For the following questions answer them individually

Question 1

A General, while arranging his men, who were 6000 in number, inthe form of a square,found that there were 71 men left over. How many were arranged in each row?

Video Solution

Question 2

A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4 respectively. The least such number is

Video Solution

Question 3

A number, when divided by 296, gives 75 as the remainder. Ifthesamenumberis divided by 37 then the remainder will be

Video Solution

Question 4

The square root of $$\frac{(0.75)^{3}}{1-0.75}$$+$$[0.75+(0.75)^{2}+1]$$ is

Video Solution

Question 5

The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be

Video Solution

Question 6

If $$a^{2}+b^{2}$$+$$\frac{1}{a^{2}}$$+$$\frac{1}{b^{2}}$$=4 then the value of $$a^{2}+b^{2}$$ will be

Video Solution

Question 7

If $$(x+\frac{1}{x})^{2}$$=3 then the value $$x^{3}$$+$$\frac{1}{x^{3}}$$ is

Video Solution

Question 8

$$\frac{0.1\times0.1\times0.1+0.02\times0.02\times0.02}{0.2\times0.2\times0.2+0.04\times0.04\times0.04}$$ is equal to

Video Solution

Question 9

If x+$$\frac{1}{x}$$=2 then the value of $$x^{100}$$+$$\frac{1}{x^{100}}$$ is

Video Solution

Question 10

If $$x^{3} + 3 x^{2}+ 3x$$ = 7, then x is equal to

Video Solution

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SSC CGL Tier-2 2010 Maths

SSC CGL Tier-2 2010 Maths

A General, while arranging his men, who were 6000 in number, inthe form of a square,found that there were 71 men left over. How many were arranged in each row?

A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4 respectively. The least such number is

A number, when divided by 296, gives 75 as the remainder. Ifthesamenumberis divided by 37 then the remainder will be

The square root of $$\frac{(0.75)^{3}}{1-0.75}$$+$$[0.75+(0.75)^{2}+1]$$ is

The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be

If $$a^{2}+b^{2}$$+$$\frac{1}{a^{2}}$$+$$\frac{1}{b^{2}}$$=4 then the value of $$a^{2}+b^{2}$$ will be

If $$(x+\frac{1}{x})^{2}$$=3 then the value $$x^{3}$$+$$\frac{1}{x^{3}}$$ is

$$\frac{0.1\times0.1\times0.1+0.02\times0.02\times0.02}{0.2\times0.2\times0.2+0.04\times0.04\times0.04}$$ is equal to

If x+$$\frac{1}{x}$$=2 then the value of $$x^{100}$$+$$\frac{1}{x^{100}}$$ is

If $$x^{3} + 3 x^{2}+ 3x$$ = 7, then x is equal to

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