Test: Arun Sharma Based Level 1: Percentages - CAT MCQ

Assume that he had initially Rs 100.

After spending on food, money left = Rs 75

After spending 50% of Rs 75, he had Rs 37.5

Let us solve this question by assuming values(multiples of 100) and not variables(x).

Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.

Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.

Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.
Hence, the female population in 1980 = 200/1.25 = 160 Also, the female population became 1.2 times itself in 1980 from what it was in 1970.
Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3

Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.

Now, the total population in 1980 = 1.25 times the total population in 1970.
Hence, 1.25 (x + 400/3) = 1.4x + 160
Hence, x = 400/9.
Population change = 300 - 400/9 - 400/3 = 300 - 1600/9 = 1100/9
percentage change = 

Given Data:

Compound Interest Rate (r): 20% or 0.20

The sum of money is more than doubled.

Concept:

The formula for compound interest is A = P(1+r/n)^(nt), where P is the principal amount, r is the annual interest rate, n is the number of times interest is compounded per year, t is the time the money is invested for in years. When compounded annually (n=1), this simplifies to A = P(1+r)^t.

Solution:

We know that the formula for compound interest, when compounded annually, is A = P(1 + r)^t. Here, we want to find the smallest whole number for time t so that A > 2P.

Substituting A as 2P and r as 0.20 into the formula, we get, ⇒ 2 = (1 + 0.20)^t ⇒ 2 = 1.20^t

We can now try different values for t until we reach a number where 1.20^t > 2.

Using t = 1, we get 1.20¹ = 1.20 which is not greater than 2.

Using t = 2, 1.20² = 1.44 which is not greater than 2.

Using t = 3, 1.20³ = 1.728 which is not greater than 2.

Using t = 4, 1.20⁴ = 2.0736 which is greater than 2.

Therefore, the least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is 4 years.

Let salaries of A and B be Rs. x and salary of C be Rs. y, then

8x/100 – 7x/100 = 259

⇒ x/100 = 259

⇒ x = 25900

Salaries of A and B is Rs. 25900.

Donation of A and B = 25900 × (8 + 7)/100 = 25900 × 15/100 = 3885

Donation of C = 3885 – 1185 = 2700

y × (9/100) = 2700

⇒ y = 30,000

Sum of salaries of A, B and C = 25900 + 25900 + 30000 = 81800

Donation of A = 259 × (8/100) = 2072

Total donation of A and C = 2072 + 2700 = 4772

Percentage of total donation of total salaries of A, B and C = (4772/81800) × 100 = 5.8%

Short Trick:

Let salaries of A, B be 100%

⇒ 8% – 7% = 259

⇒ 1% = 259

⇒ 8% = 2072

⇒ 15% = 3885

⇒ 100% = 25900

Donation of C = 3885 – 1185 = 2700

⇒ 9% = 2700

⇒ 100% = 30,000

Total donation of A and C = 2072 + 2700 = 4772

Total salaries of A, B and C = 25900 + 25900 + 30000 = 81800

∴ Required percentage = 4772/81800 × 100 = 5.8%

Weight of solution = (1+4) = 5 units weight of sugar = 1 unit weight of water after heating =1.5 units Required ration =1 : 1.5 = 2 : 3

Turnover in 1998 = Rs 100 cr Turnover in 1999 = Rs 300 cr Turnover in 2000 = Rs 450 cr