CRE 1 - Percentage

12.5% = 1/8;

Therefore (1/8) × 2000 = 250.

Hence, option (b).

x = 90% of y

⇒ x = 0.9y 

⇒ x/y = 0.9 or 90% and y/x = 1/0.9 = 1.11 = 111.1%.

∴ y is 111.1% of x.

Hence, option (d).

$\frac{27}{36}$ × 100 = $\frac{300}{4}$ = 75%

Hence, option (d).

$\frac{26}{52}$ × 100 = 50%

Hence, option (e).

$66\frac{2}{3}$% = $\frac{2}{3}$

$66\frac{2}{3}$% of 672 = $\frac{2}{3}$× 672 = 448

Hence, option (c).

37.5% = $\frac{3}{8}$ and 87.5% = $\frac{7}{8}$

Let the number be x. 

Then, from the problem $\frac{3}{8}$× x = 450.    

∴ x = 450 × $\frac{8}{3}$

Hence, 87.5% of x = $\frac{7}{8}$× x = $\frac{7}{8}$× 450 × $\frac{8}{3}$ = 1050.

Hence, option (c).

$\frac{48}{80}\times 100$ = 60%

Hence, option (a).

80% of 80% of 200 = $\frac{80}{100}\times \frac{80}{100}\times 200$ = $\frac{4}{5}\times \frac{4}{5}\times 200$ = 128

Hence, option (b).

2 hours = 120 mins,

∴ $\frac{60}{120}$ × 100 = 50%

Hence, option (c).

Passing marks = 40% of 400 = 160

He obtained 80, i.e. he failed by (160 - 80=) 80 marks.

Percentage by which he failed =  $\frac{80}{400}$ × 100 = 20%

Hence, option (d).

In 50 L, 40% is water i.e. $\frac{40}{100}$ × 50 = 20 L

So, quantity of milk = 50 – 20 = 30

Hence, option (b).

Required Percentage = $\frac{\mathrm{A}}{\mathrm{3A/15}}$ × 100 = 15/3 × 100 = 500%.

Hence, option (c).

30% of 50% of 100 = $\frac{30}{100}$ × $\frac{50}{100}$ × 100 = 15.

Hence, option (e).

Going by options

A. 33% = 33/100 = 0.33

B. 1/3 = 0.3333

C. 0.34

D. 7/20 = 0.35

E. 11/34 = 0.32

So out of this 7/20 is greatest.

Hence, option (d).

20% of 80 = x% of 32. 

⇒ 16 = x/100 × 32

⇒ x = 50

Hence, option (a).

M = 60% of N

M = 3N/5 (60% = 3/5)

∴ N = 5M/3 = 166.66% of M

Hence, option (b).

A's salary as a % of B's salary = 300/500 × 100 = 60%.

Hence, 60.

% by which A's salary is less than B's salary = (500 - 300)/500 × 100 = 40%.

Alternately,
In the previous question we calculated that A's salary is 60% of B's salary.

Hence, A's salary is (100 - 60 =) 40% less than B's salary.

Hence, 40

Let the number of boys be b.

Hence, the number of girls is b/7 (14$\frac{2}{7}$% = $\frac{1}{7}$)

Then, from the problem,

$\frac{\mathrm{b}}{7}$ + b = 560

⇒ 8b/7 = 560

⇒ b = 490

⇒ No. of girls = 560 – 490 = 70

Hence, option (b).