Question

Shekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji.he has 7 sweets left . how many did he have to start 3?

Answer 1

Let sweet he had at first be x.
Now, the no. of sweets he has given to renu is x/4
Therefore, x - x/4 + 5 = 3
=>(4x-x)/4 = 3-5
=>3x/4 = - 2
=>3x = - 2×4
=>x = - 8/3
Ans) To have 3 left he must stat with - 8/3.

Answer 2

Let the sweets Shekhar have be x

Sweets given to Renu =
$\frac{1}{4} \times x$

Sweets given to Raji = 5

Sweets left = 7

Total sweets=
$\frac{ x}{4} + 5 + 7$

$x = \frac{x}{4} + 12$

$x = \frac{x + 48}{4}$

$4x = x + 48$

$4x - x = 48$

$3x = 48$

$x = \frac{48}{3}$

$x = 16$
THUS SHEKHAR HAS 16 SWEETS