Question

If X is 80% more than Y, then Y is how much percentage less than X?

Correct
option
is 44.44%

X की तुलना में Y में प्रतिशत कमी $=\left(\dfrac{80}{100 + 80}\right)\times 100$ $ = \dfrac{80}{180}\times 100$ $ = 44.44%$

X की तुलना में Y में प्रतिशत कमी $=\left(\dfrac{80}{100 + 80}\right)\times 100$ $ = \dfrac{80}{180}\times 100$ $ = 44.44%$

Question

A is 20% less than B. B is 10% less than C. C is 30% more than D. If D is 400, then what is the value of A?

Correct
option
is 374.4

दिया गया है - D = 40 $\therefore C = 400\times \dfrac{130}{100} = 520$ $\therefore B = 520\times \dfrac{90}{100} = 468$ $\therefore A = 468\times \dfrac{80}{100} = 374.4$

दिया गया है - D = 40 $\therefore C = 400\times \dfrac{130}{100} = 520$ $\therefore B = 520\times \dfrac{90}{100} = 468$ $\therefore A = 468\times \dfrac{80}{100} = 374.4$

Question

What is the difference between 0.9 and 0.9%?

Correct
option
is 0.891

अभीष्ट अन्तर = 0.9 - $\dfrac{0.9}{100}$ = 0.9-0.009 = 0.900-0.009 - = 0.891

अभीष्ट अन्तर = 0.9 - $\dfrac{0.9}{100}$ = 0.9-0.009 = 0.900-0.009 - = 0.891

Question

Amar, Bhavan, chetan and Dinesh have a total of 150 with them, Amar has one-fourth of the total amount with the other. The amount with Amar (int)is:

Correct
option
is 30

Let Bhawan, Chetan and Dinesh have total Rs.x near immortal = $\dfrac x4$ Total amount = $x + \dfrac x4 = 150$ 5x = 600 x = 120 Amount of immortal = $\dfrac x4 = \dfrac{120}{4}= 30$

Let Bhawan, Chetan and Dinesh have total Rs.x near immortal = $\dfrac x4$ Total amount = $x + \dfrac x4 = 150$ 5x = 600 x = 120 Amount of immortal = $\dfrac x4 = \dfrac{120}{4}= 30$

Question

A is 60% more than B, C is 35% less than D and D is 30% more than A, Which of the following is true?

Correct
option
is 169 B = 125 C

प्रश्नानुसार, $B \times \dfrac{160}{100} = A$ $B \times \dfrac {160}{100}\times \dfrac{130}{100} = D$ $B \times \dfrac {160}{100}\times \dfrac{130}{100} \times \dfrac{65}{100}= C$ 169B = 125C

प्रश्नानुसार, $B \times \dfrac{160}{100} = A$ $B \times \dfrac {160}{100}\times \dfrac{130}{100} = D$ $B \times \dfrac {160}{100}\times \dfrac{130}{100} \times \dfrac{65}{100}= C$ 169B = 125C

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