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Equations Reducible to a Pair of Linear Equations in Two Variables

Question 18 2...

Question

Question 18

$2 \frac{2}{3} \div 5$ is equal to

(a) $\frac{8}{15}$
(b) $\frac{40}{3}$
(c) $\frac{40}{5}$
(d) $\frac{8}{3}$

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Solution

Given, $2 \frac{2}{3} \div 5 = \frac{(2 \times 3) + 2}{3} \div 5 = \frac{6 + 2}{3} \div 5 = \frac{8}{3} \times \frac{1}{5} [∵ reciprocal of 5 = \frac{1}{5}] = \frac{8}{15}$
Hence, $2 \frac{2}{3} \div 5$ is equal to $\frac{8}{15}$

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Question 2

$3 \frac{3}{4} \div \frac{3}{4}$ is equal to

(a) 3
(b) 4
(c) 5
(d) $\frac{45}{16}$

Question 1

$\frac{2}{5} \times 5 \frac{1}{5}$ is equal to
(a) $\frac{26}{25}$
(b) $\frac{52}{25}$
(c) $\frac{2}{5}$
(d) 6

Solve:

(a) (b)

(c) (d)

(e) (f)

(g) (h)

(i)

$^{5} C_{1} +^{5} C_{2} +^{5} C_{3} +^{5} C_{4} +^{5} C_{5}$ is equal to

If $^{15} C_{3 r} =^{15} C_{r + 3},$ then r is equal to

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Equations Reducible to a Pair of Linear Equations in Two Variables

Standard X Mathematics

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