Percentage Questions Moderate

Solution:

Solution:

Let the repayments be Rs "a – d", Rs "a" and Rs. "a + d" 

a – d + a + a + d = 54000 

3a = 54000 

a = 18000 

The payment at the end of year 2 is Rs. 18,000. 

Borrowed amount = Rs. 45,000 

Amount outstanding at the end of Year 1 = (45000 × 1.1) – (18000 – d) = 31500 + d 

Amount outstanding at the end of Year 2 = ((31500 + d) * 1.1) – 18000 = 34650 + 1.1d – 18000 = 16650 + 1.1d 

Amount outstanding at the end of Year 3 = ((16650 + 1.1d) * 1.1) = 18000 +  d 

18315 + 1.21d = 18000 + d 

0.21d = – 315 

d = –1500 

The payments are Rs. 19500, Rs. 18000 and Rs. 16500 

He paid Rs. 19500 in year 1

Solution:

a = (1 + x) b 

b = c (1 - y) 

c = b/1−y 

a is greater than c or b (1 + x) > b/1−y 

(1 + x) (1 - y) > 1 → x > y (1 + x) 

Converting back to percentages... 

x/100 > y/100 (100/100 + x/100) 

Therefore, y (100 + x) < 100x 

Hence, the answer is y (100 + x) < 100x

Solution:

CP = 80 × 40 

Profit from the n objects = n% × 40 × n. 

Profit from the remaining objects = (100 – n) % × 40 × (80 – n). We need to find the minimum possible value of n% × 40 × n + (100 – n) % ×  40 × (80 – n). 

Or, we need to find the minimum possible value of n2 + (100 – n) (80 – n). Minimum of n2 + n2 – 180n + 8000 

Minimum of n2 – 90n + 4000 

Minimum of n2 – 90n + 2025 – 2025 + 4000 

We add and subtract 2025 to this expression in order to create an expression that can be expressed as a perfect square. 

This approach is termed as the "Completion of Squares" approach. We keep revisiting this in multiple chapters. 

Minimum of n2 – 90n + 2025 + 1975 = (n – 45) 2 + 1975 

This reaches a minimum when n = 45. 

When n = 45, the minimum profit made 

45% × 40 × 45 + 55% × 40 × 35 

18 × 45 + 22 × 35 = 810 + 770 = 1580 

Hence, the answer is Rs. 1580.

Solution:

a = bx 

b = a (1 + x), substituting this in the previous equation, we get x (x + 1) = 1

x2 + x - 1 = 0 or 

x is approximately 0.62 or 62%. The ration 1.618 is also called the golden ratio, and is the conjugate and reciprocal of 0.618. 

Hence, the answer is 62%

Solution:

Solution:

Let the number of girls in class A = x 

Let the number of boys in class A = y 

Total number of students = x + y 

Proportion of girls = x/x+y 

Number of boys in class B = x 

Total number of students in class B = 1.5 (x + y) 

Proportion of girls = 1 – x/1.5 (x+y) 

Percentage of boys in the overall student community = x + y/2.5* (x + y) *  100 = 40%

Solution:

In the final state, the number of girls should be 1.5 * the number of boys. When 50% of the boys are taken as girls, let the number of boys = x Number of girls = 1.5x 

Total number of students = 2.5x 

Original number of boys = 2x (50% of boys = x) 

Original number of girls = 0.5x 

Girls form 20% of the overall class.

Solution:

Let total amount = T 

A would have got T * a. After A takes his share, there would be T (1 - a) left B would have got T (1 - a) * b 

C would have got T (1 - a) (1 - b) * c 

D would have got T (1 - a) (1 - b) (1 - c) 

D gets a % less than what A gets, or, T *a (1 - a) = T (1 - a) (1 - b) (1 - c), or a = (1 - b) (1 - c) ------(1)

B and C get the same amounts, or (1 - a) * b = (1 - a) * (1 - b) * c, or b = (1 - b) c, or, b = c - cb, or 

b = c/c + 1 

b = 2a -------(3) 

Substituting, Equations (2) in and (3) in (1), we get 

c/2 (c+1) = 1−c/c+1, or c = 2 - 2c, or c = 23; b = c/c+1 = 2/5 , a = 1/5 A gets 20% of the loot, B gets 40% of the remaining 80%, or 32% of the loot. C gets 66.6% of 48%, or 32% of the overall loot. And D gets the final  16% of the overall amount (20% lesser than A's share). 

A got 20% of overall, C got 32% of overall. Or, C got 160% of A A would have got Rs. 200, D Rs. 160. A would have got Rs. 40 more than D

Solution:

a = b (1 + x) => a/b = 1 + x 

a = x (a + b), dividing by a throughout 

1 = x (1 + b/a) → 1 = x (x + 2/x + 1)→ x + 1 = x2 + 2x →x2 + x - 1 = 0 Now, we need to solve this equation. Using the discriminant method, when we solve this, x turns out to be (−1+√5)/2. x has to lie between 0 and 1 and therefore cannot be (−1−√5)/2 

the only solution is (−1+√5)/2 → This is roughly 0.62. Hence, the answer is  62% 

Solution:

A = 5/4 * B or B = 4/5 * A  

C = 5/4 * A 

A = 6/5 * D or D = 5/6 * A and E = 6/5 * A 

In order for the amounts earned by A, B, C, D and E to be integers, the amount earned by A should be a multiple of 4, 5 and 6. Or, the amount earned by A should be a multiple of 60. The only multiple of 60 that is a 2– digit number is 60. 

So, if A earns Rs. 60, B should earn Rs. 48. C should earn Rs. 75, D should earn Rs. 50 and E should earn Rs. 72. 

The total income earned by all 5 of them should be equal to 

Rs. 60 + Rs. 48 + Rs. 75 + Rs. 50 + Rs. 72 => Rs. 305 

The total income earned by all 5 of them should be equal to Rs. 305

Solution:

Alloy 1 - Copper = 2x ; Aluminium = 3x 

Alloy 2 – Copper = 2y ; Aluminium =7y 

If Aluminium and Zinc have to be equal, 3x = 7y. The simplest case occurs at the LCM, when both Aluminium and Zinc are 21kg. 

Alloy 1 → Copper = 14 ; Aluminium = 21 

Alloy 2 → Copper = 6 ; Zinc = 21 

Even a gram of alloy 1 above this level would mean that there is more Aluminium than Zinc in the total alloy. So, this table is a limit on the percentages of different metals. 

The percentage of Copper in the total alloy would be (14 + 6) * 100/62 =  32.25% 

On the other hand, if we want more Aluminium than Zinc, we can use tons and tons of alloy 1, and only a microgram of Alloy 2. The quantity of alloy 2  can be made so small that its presence can be almost neglected, as it will have as good as zero impact on the overall percentage of copper. This is another extreme case, in which Copper will have 40% weight in the alloy  2? (2+3). 

Thus, the percentage of Copper ranges from 32.25% to 40% in this alloy.

Solution:

P = Q (1 + (c/100)) ; Q = R (1 – ((x - 10) / 100)) 

P > R → Q(1 + (c/100)) > Q/(1 – ((x - 10)/100))→ (100 + x) (110 – x) > 100 *  100  

1000 + 10x – x2 > 0 → (x – 5)2 < 1025 →x < 37  

x could range from 10% to 37%

Solution:

Let us assume he buys n goods. 

Total CP = 20n 

Total SP = 2 + 4 + 6 + 8 ….n terms 

Total SP should be at least 40% more than total CP 

2 + 4 + 6 + 8 ….n terms ≥ 1.4 * 20 n 

2 (1 + 2 + 3 + ….n terms) ≥ 28n 

n (n + 1) ≥ 28n 

n2 + n ≥ 28n 

n2 - 27n ≥ 0 

n ≥ 27 → He should sell a minimum of 27 goods