SRCC GBO 2010 Question Paper

We have $$\frac{a^2-b^2-c^2-2bc}{a^2+b^2-c^2-2ab}=\frac{a^2-\left(b+c\right)^2}{\left(a-b\right)^2-c^2}=\frac{\left(a+b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)\left(a-b-c\right)}$$

Substituting the values of a,b and c we get 

$$\frac{\left(a+b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)\left(a-b-c\right)}=\frac{0.7\times\ \left(-0.2\right)}{\left(0.35\right)\left(-0.2\right)}=2$$

The addition of 4a3 and 984 gives 13b7.

We shall refer to the divisibility rule of 11, which states that |sum of alternate digits from unit digit-sum of rest of the digits| shall be 0 or a multiple of 11. 

From the similar logic, 13b7 has 7+3 as the sum of alternate digits starting from unit digit, and 1+b as the remaining ones. 

It is clear that the difference cannot touch any higher value (starting from 11) as none of the two values (3+7 or 1+b) can be greater than 10, hence it has to be 0. 

(3+7)- (1+b) = 0

b=9

Now, we shall go back to the original equation: 4a3 + 984 = 1397

from this, 4a3= 413

Hence, a = 1

Substituting the values of a and b in |2a-b|, we get the final answer as 7. 

Let the four numbers be a,b,c,d. Hence a+b+c=180  a+b+d=197    a+c+d=208 and b+c+d=222

Adding all the four equations 3(a+b+c+d)= 807 or a+b+c+d= 269. To find the smallest number subtract the maximum sum i.e. 222

So b+c+d=222 and a= 269-222=47

Hence 47 is the smallest of the four numbers

Take $$1.331^{-1}$$ common from the denominator. As a result of which the numerator and denominator effectively cancel out. Hence the fraction effectivel becomes

$$\left(\frac{1}{1.331^{-1}}\right)^{\frac{1}{3}}$$

or $$\left(1.331^1\right)^{\frac{1}{3}}=1.331^{\frac{1}{3}}=1.1$$ which is the correct answer

Let the total number of objects in the box be "p". 0.2p objects are beads and 0.8p are coins. Out of 0.8p, 0.4(0.8p) or 0.32p coins are silver and 0.48p are gold coins.

Hence 48% of the total objects are gold coins.

Tip: It is always convenient to use fractions in place of percentages. For example, 37.5% can be written as $$\frac{3}{8}$$

As per the question, the initial expenditure was 80% of her income. 

Let us assume the income as x, and her initial expenditure as $$\frac{8}{10}$$ of x. This implies that the initial saving was $$\frac{2}{10}$$x.

Now, her expenditure is increased by 37.5%, which means it becomes $$\frac{11}{10}$$x or 1.1x.

For her income, it is given that the same increased by 25%. This means that the income is now $$\frac{5}{4}$$x or 1.25x.

Subtracting her new expenditure from new income, we get the savings as 0.15x, which is 12% of 1.25x.

Let the mixture initially has "a" litres acid and "b" liters water. Now you add 1-litre water, hence the total water is "b+1" litres.

According to question    a=0.2(a+b+1) or 0.8a=0.2b +0.2  or 4a=b+1

Now one litre of acid is added to the mixture. Hence acid content is "a+1"  and water content is "b+1"

Now a+1= 1/3(a+b+2) or 3a+3 = a+b+2 or 2a+1=b

Substituting above 4a= 2a+1+1 or 2a=2 or a=1 and b=3 

Hence percentage of water in the original mixture is 3/(3+1) =3/4 or 75%

Let the cost price of each pen be "p". Hence total cost price of the pens is 749p. He sold 700 pens for 749p. Selling price per pen is $$\frac{749}{700}p\ or\ 1.07p$$

Now selling price of the remaining 49 pens is 1.07p(49) or 52.43p.

Hence total selling price is 749p+52.43p = 801.43p

Total profit made isc 801.43p-749p =52.43p

Profit percentage $$\frac{52.43p}{749p}\times\ 100=7\%$$

Hence net profit percentage is 7%

Let the Cost Price of each dress be C and let the marked price of each dress be M.

She will be selling the dress at 1/4th off i.e at 3/4(M).

According to the question C= (2/3)(3/4)M= 1/2(M)

Hence C:M=1/2

Let us assume that the first candle burns at a rate of X units per hour, and the second at Y units per hour. 

Hence, the length of the shorter candle is 11X and the longer one is 7Y. 

After 3 hours, the remaining lengths would be 8X and 4Y, which is to assumed as equal in value. 

Hence, 8X = 4Y, which means 2X = Y

We shall substitute this in the initial expressions of length that gives us the length of shorter candle as 11X, and the longer as 14X.

Hence, the ratio of lengths is 11:14.