Question

2/3 of a beads in a box are red, 1/4 are yellow and the rest are blue . there are 35 more red beads than blue beads . how many beads are there altogether?

Answer 1

Answer :-

The total beads altogether are 60 beads.

Step-by-step explanation:

To Find :-

• Total beads.

★ Solution

Given that,

• 2/3 beads in a box are red.
• 1/4 beads in box are yellow.

∴ Fraction of blue beads in box :-

⇒ 1 - ( 2/3 + 1/4 )

⇒ 1 - ( 11/12 )

⇒ (11 - 12)/12

⇒ 1/12 beads are blue

Also given,

There are 35 more red beads than blue.

Let us assume the total no. of beads as (x).

∴ (2/3)x - (1/12)x = 35

By simplifying,

⇒ (2/3)x - (1/12)x = 35

⇒ 2x/3 - x/12 = 35

⇒ 7x/12 = 35

⇒ 7x = 35 × 12

⇒ 7x = 420

⇒ x = 420/7

⇒ x = 60

Hence, Number of beads are 60.

Answer 2

☆Given that

• 2/3 of a beads in a box are red, 1/4 are yellow and the rest are blue there are 35 more red beads than blue beads

☆To Find

• How many beads are there altogether?

☆Solution

given data - 2/3 of a beads in a box are red, 1/4 are yellow and the rest are blue are 35 more red beads then blue beads

Fraction that are blue

$\longmapsto \tt \:beads _{(blue)} = 1 - \bigg \lgroup \frac{2}{3} + \frac{1}{4} \bigg \rgroup$

$\sf \: now \: we \: take \: LCM \: of \: 3 \: and \: 4 = 12\\ \\ 》\bf \frac{2 \times 4}{3 \times 4} = \frac{8}{12} \\ \\》 \bf \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

$\longmapsto \tt \: beads _{(blue)} = 1 - \bigg \lgroup \frac{8}{12} + \frac{3}{12} \bigg \rgroup \\ \\ \\ \longmapsto \tt \: beads _{(blue)} =1 - \frac{8 + 3}{12} \\ \\ \\ \longmapsto \tt \: beads _{(blue)} =1 - \frac{11}{12} \\ \\ \\ \longmapsto \tt beads _{(blue)} =\frac{11 - 12}{12} \\ \\ \\ \red{ \boxed{\longmapsto \tt \blue{beads _{(blue)} = \frac{1}{12} }}}$

As a given data there are 35 more red beads then blue . let the assume the total number of beads as b

$\sf \: therefore \: \bigg ( \frac{2}{3} \bigg )b - \bigg( \frac{1}{12} \bigg)b= 35$

now solving

$\qquad\longmapsto \tt \: \bigg( \frac{2}{3} \bigg)b - \bigg( \frac{1}{12} \bigg)b = 35 \\ \\ \\ \qquad\longmapsto \tt \: \frac{2 \times b}{3} - \frac{1 \times b}{12} = 35 \\ \\ \\ \qquad\longmapsto \tt \: \frac{2b}{3} - \frac{b}{12} = 35 \\ \\ \\ \qquad\longmapsto \tt \: \frac{7b}{12} = 35 \\ \\ \\ \qquad\longmapsto \tt \: 7b = 35 \times 12 \\ \\ \\ \qquad\longmapsto \tt \: 7b = 420 \\ \\ \\ \qquad\longmapsto \tt \: b = \cancel\frac{420}{7} \\ \\ \\ \qquad\longmapsto \tt \: \blue{b = 60}$

$\large \: \rm \: hence \: numbers \: are \: beads \: altogether \: be \: 60 \:$