25.675% of 1321 + 64.328% of 4001=

2912

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3016

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3126

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3254

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3348

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Solution

The correct option is A 2912
$25.6751 \approx 25 % 1321 \approx 1320 64.328 \approx 64 % 4001 \approx 4000 r e w r i t i n g t h e e q u a t i o n 25 % o f 1320 + 64 % o f 4000 = \frac{1}{4} \times 1320 + \frac{64}{100} \times 4000 = 330 + 2560 = 2890 S o 2890 \approx 2912 O p t i o n (a) i s a c o r r e c t a n s w e r$

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Similar questions

\frac{(64.328)^{2} - (35.672)^{2}}{28.656} =

x_{1}, x_{2}, x_{3} . . . . . . . . . x_{4001}

\frac{1}{x_{1} x_{2}} + \frac{1}{x_{2} x_{3}} + . . . . . . . . + \frac{1}{x_{4000} x_{4001}} = 10

x_{2} + x_{4000} = 50

| x_{1} - x_{4001} | =

A square pyramid covers an area of $400 m^{2}$ on the ground and height of the pyramid is 10 m. Then find its total surface area of the pyramid

lim x \to \infty [(1 - \frac{1}{2^{2}}) (1 - \frac{1}{3^{2}}) (1 - \frac{1}{4^{2}}) . . . (1 - \frac{1}{n^{2}})] =

x_{1}, x_{2}, x_{3}, . . . . x_{4001}

\frac{1}{x_{1} x_{2}} + \frac{1}{x_{2} x_{3}} + . . . + \frac{1}{x_{4000} x_{4001}} = 10

x_{2} + x_{4000} = 50

| x_{1} - x_{4001} | =

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