SRCC GBO 2009 Question Paper

Total students = 45, Number of boys = 30 and number of girls = 15

=> Ratio of boys : girls = 2:1

Average weight of girls is 45 kg and that of boys is 52 kg

$$\therefore$$ Average weight of class = $$\frac{52(2)+45(1)}{2+1}$$

= $$\frac{149}{3}=49.67\approx50$$ kg

=> Ans - (C)

Average of first $$n$$ even numbers = $$(n+1)$$ and average of first $$n$$ odd numbers = $$n$$

=> Average of first 10 even numbers = $$(10+1)=11$$

=> Ans - (C)

Terms : $$6, 7, x - 2, x, 17$$ and $$20$$

Since, there are 6 terms, median is the average of 3rd and 4th term

=> Median = $$\frac{x-2+x}{2}=16$$

=> $$x-1=16$$

=> $$x=17$$

=> Ans - (D)

Let the unit's digit is $$y$$ and ten's digit is $$x$$, => Number = $$10x+y$$

=> $$10x+y=7(x+y)$$

=> $$10x+y=7x+7y$$

=> $$3x=6y$$

=> $$x=2y$$ ----------------(i)

Also, $$10x+y-9=10y+x$$

=> $$9(x-y)=9$$

=> $$x-y=1$$

Substituting value from equation (i), => $$2y-y=y=1$$

=> $$x=2$$

$$\therefore$$ Original number = 21

=> Ans - (A)

Let original sum = Rs. $$6x$$

Sum invested at 3% = Rs. $$2x$$ and sum invested at 6% = Rs. $$x$$ and sum invested at 8% = Rs. $$3x$$

Simple interest = $$\frac{P\times r\times t}{100}$$

According to ques,

=> $$\frac{2x\times3\times1}{100}+\frac{x\times6\times1}{100}+\frac{3x\times8\times1}{100}=600$$

=> $$6x+6x+24x=60,000$$

=> $$x=\frac{60000}{36}=\frac{10,000}{6}$$

$$\therefore$$ Original sum = $$6\times\frac{10000}{6}=Rs.$$ $$10,000$$

=> Ans - (D)

Let principal sum = Rs. $$x$$

Rate of interest = 4% for 2 years

Compound interest = $$P[(1+\frac{r}{100})^t-1]$$

=> $$P[(1+\frac{4}{100})^2-1]=102$$

=> $$P[(\frac{26}{25})^2-1]=102$$

=> $$P(\frac{676-625}{625})=102$$

=> $$P=625\times2=Rs.$$ $$1250$$

$$\therefore$$ Simple interest = $$\frac{P\times r\times t}{100}$$

= $$\frac{1250\times4\times2}{100}=Rs.$$ $$100$$

=> Ans - (B)

Expression : $$X^{2}-Y^{2}-Z^{2}+2YZ+X+Y-Z$$

= $$x^2-(y^2+z^2-2yz)+(x+y-z)$$

= $$x^2-(y-z)^2+(x+y-z)$$

= $$[x^2-(y-z)^2]+(x+y-z)$$

Using, $$a^2-b^2=(a-b)(a+b)$$

= $$(x-y+z)(x+y-z)+(x+y-z)$$

= $$(x+y-z)(x-y+z+1)$$

=> Ans - (A)

Given : $$a + b + c = 0$$

=> $$(a+b)=-c$$

Cubing both sides, => $$(a+b)^3=(-c)^3$$

=> $$a^3+b^3+3ab(a+b)=-c^3$$

=> $$a^3+b^3+3ab(-c)=-c^3$$

=> $$a^3+b^3+c^3=3abc$$ --------------(i)

Expression : $$(a+b)^{3}+(b+c)^{3}+(c+a)^{3}$$

= $$[a^3+b^3+3ab(a+b)]+[b^3+c^3+3bc(b+c)]+[c^3+a^3+3ca(c+a)]$$

= $$[a^3+b^3+3ab(-c)]+[b^3+c^3+3bc(-a)]+[c^3+a^3+3ca(-b)]$$

= $$2(a^3+b^3+c^3)-9abc$$

Substituting value from equation (i), we get :

= $$2(3abc)-9abc=-3abc$$

$$\therefore$$ $$abc$$ is a factor of the given expression.

=> Ans - (A)

Let length and breadth of rectangle initially be 10 cm each

Area = $$A=lb=10\times10=100$$ $$cm^2$$

New length after 10% decrease = 9 cm and similarly new breadth = 11 cm

=> New area = $$A'=9\times11=99$$ $$cm^2$$

$$\therefore$$ Decrease in area = $$\frac{100-99}{100}\times100=1\%$$

=> Ans - (B)

Let son's present age = $$x$$ years and father's present age = $$y$$ years

=> $$x+y=99$$

Let $$n$$ years ago, father was same as his son's age, where $$n=y-x$$

=> $$4(x-n)=(y-n)$$

=> $$4x-4(y-x)=y-y+x$$

=> $$8x-4y=x$$

=> $$7x=4y$$

=> $$\frac{x}{y}=\frac{4}{7}$$

=> Ans - (D)