PE 1 - Ratio

Income : A_i/B_i = 3/2    Expense : A_e/B_e = 5/3

Also, A_i - A_e = 1000; B_i - B_e = 1000.

If I is the combined income and E the combined expense, then,

A_i = 3/5 × I, A_e = 5/8 × E, B_i = 2/5 × I, B_e = 3/8 × E

It is given that, 3/5 × I - 5/8 × E = 1000        …(i)

2/5 × I - 3/8 × E = 1000        …(ii)

Solving (i) and (ii) simultaneously, E = Rs. 8,000 and I = 10,000

∴ A_i = 3/5 × 10,000 = Rs. 6,000 and B_i = 10,000 -6,000 = Rs. 4,000.

Hence, option (c).

Let's assume the bag has Rs. 33x (∵ 10 + 15 + 8 = 33).

The value of one-rupee coins ⇒ Rs. 10x i.e. 10x coins

The value of 50ps. coins ⇒ Rs. 15x i.e. 30x coins

The value of 25ps. coins Rs. 8x i.e. 32x coins

⇒ 10x + 30x + 32x = 144

⇒ 72x = 144

⇒ x = 2

∴ Number of 25ps. coins is 32x = 64

Alternately,

Ratio of number of coins = value/(face value)⇒Ratio of number of coins of 1 rupee, 50 paise, 25 paise =

10/100 : 15/50 : 8/25 = 10 : 30 : 32

∴ Number of 50 paise coins = (32(144))/(10 + 30 + 32) = 64

Hence, 64.

Let the strength of the class be x.

Number of girls = 3/5 × x∙

1/3^rd of the girls travel to school by bus. 

∴ 2/3rd of them do not travel to school by bus.

⇒ x × 3/5 × 2/3 = 36

⇒ x = 90

Hence, 90.

Given that  a/b = 2/1 and p/q = 3/2

a = 2b and p = 3/2 q

∴ $\frac{2ap-bq}{3ap+2bq}$ = $\frac{2\times 2b\times {\displaystyle \frac{3}{2}}q-bq}{3\times 2b\times {\displaystyle \frac{3}{2}}q+2bq}$ = $\frac{6bq-bq}{9bq+2bq}=\frac{5bq}{11bq}$ = $\frac{5}{11}$

Hence, option (c).

X/Y = 4/3

⇒ X= 4Y/3

⇒ X² = (16Y²)/9 

Then, $\frac{(3\times {\displaystyle \frac{16}{9}}{Y}^2-2Y^2)}{(3\times \frac{16}{9}{Y}^2+2Y^2)}=\frac{10Y^2}{22Y^2}=\frac{5}{11}$ or 5 : 11.

Hence, option (c).

Given that, the ratio of the ages of Akanksha and Netra is 2 : 1.

Let their ages be 2x and x.

After 8 years Netra's age is equal to the present age of Akanksha.

So, x + 8 = 2x 

⇒ x = 8.

Ages of Netra and Akanksha are 8 and 16 years respectively.

Hence, option (a).

x + 2y – 3z = 0    …(1)

2x – 3y – 5z = 0    …(2)

(1)    × 2 – (2)

7y = z 

Substituting z = 7y in equation (1).

x + 2y – 21y = 0

⇒ x = 19y

∴ x : y : z = 19y : y : 7 = 19 : 1 : 7

Hence, option (d).

Let the required quantity be t

(x + t) : (y + t) = p : q

⇒ (x + t)q = (y + t)p 

⇒ xq + qt = yp + pt

⇒ (p – q)t = xq – yp

∴ t = (xq - yp)/(p - q)

Hence, option (c).

Let us find the share of the remainder money amongst A, B and C. Then, since remainder money = 375 – (4 + 5 + 6) = 360

A’s remainder share = 3/12 × 360 = 90

B’s remainder share = 4/12 × 360 = 120

C’s remainder = 5/12 × 360 = 150

Thus,

A’s share = Rs. 90 + Rs. 4 = Rs. 94

B’s share = Rs. 120 + Rs. 5 = Rs. 125

C’s share = Rs. 150 + Rs. 6 = Rs. 156

Hence, option (c).

$\frac{4a^2+2ab}{2ab+b^2}=\frac{5}{1}$

⇒ $\frac{2a(2a+b)}{b(2a+b)}=\frac{5}{1}$

⇒ 2a/b = 5/1

⇒ a/b = 5/2

Hence, option (c).