3
139
4
4
4
3
Let the original solution had A and W litres of acid.
When 2 litres of water is added, then A = W + 2.
Similarly, when further 1 litres of acid is added,
A + 15 = 4(W + 2).
Solve the two equations and get A = 5 and W = 3.
Hence, required ratio is 3 : 5.
14
3
2
2
7
1
f(xy) = f(x)f(y) + f(x) + f(y) = (f(x) + 1)(f(y) + 1) – 1
f(160000) = f(400 × 400) = (f(400) + 1)(f(400) + 1) – 1
Now, f(400) = f(20 × 20) = (f(20) + 1)(f(20) + 1) – 1
Now, f(20) = f(4 × 5) = (f(4) + 1)(f(5) + 1) – 1
Now, f(4) = f(2 × 2) = (f(2) + 1)(f(2) + 1) – 1
Since f(p) = 1 where p is a prime number, so
f(4) = (1 + 1)(1 + 1) – 1 = 3
f(20) = (3 + 1)(1 + 1) – 1 = 7
f(400) = (7 + 1)(7 + 1) – 1 = 63
f(160000) = (63 + 1)(63 + 1) – 1 = 4095.
4
1
First quadrant: Both x and y are positive.
This would change the equation to 2x + y = 15 and
x = 20, giving a negative value of y; hence, this is
not the case.
Second quadrant: x is negative, but y is positive.
This would change the equations to y = 15 and
x = 20, giving a positive value of x, which hence
cannot be the case.
Third quadrant: Both x and y are negative.
This would change the equation to y = 15 and
x – 2y = 20; this gives a positive value of y and
hence cannot be the case.
Fourth quadrant: x is positive, but y is negative.
This would change the equations to 2x + y = 15
and x – 2y = 20; this gives the value of x as 10 and
y as –5, which would lie in the fourth quadrant.
The value of (x – y) would be 10 – (–5) = 15.
7
3
2160
Let the cost price of the item be P.
We are given that Bina sells this at 19% loss or at
0.81 × P at 4860.
This gives us the value of P as Rs. 6,000.
If Bina had sold this at 17% profit, the selling price
would have been 1.17 × 6,000 = Rs. 7,020.
So, Shyam bought the product at Rs. 4,860 and
sold it to Hari at Rs. 7,020, thus making a profit of
7,020 – 4,860 = Rs. 2,160.
340
11
3
2
2