A concave mirror of focal length 20cm produces an erect image that is four times the object, the object distance from the mirror is

The decisive word in this question is "erect". A concave mirror forms an upright (and therefore virtual) image in one situation only: when the object lies between the pole and the principal focus. Any object placed at or beyond the focus gives a real, inverted image. So even before calculating, the object distance must be smaller than the focal length of 20 cm.

The arithmetic confirms it. Magnification is \(m = \dfrac{v}{u}\) in size, and for an erect image from a concave mirror the image is virtual, so the image distance is negative: \(v = -4u\) when the image is four times the object. Substituting into the mirror formula \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\): \[\frac{1}{20} = \frac{1}{u} + \frac{1}{-4u} = \frac{4 - 1}{4u} = \frac{3}{4u}.\] Cross-multiplying gives \(4u = 60\), so \(u = 15\ \text{cm}\), which is indeed less than 20 cm. The image is then 60 cm behind the mirror, virtual, erect and magnified, which is how a shaving or make-up mirror works.

The usual error is to take \(v = +4u\), which gives \(\frac{1}{20} = \frac{5}{4u}\) and \(u = 25\ \text{cm}\), an object distance between the focus and the centre of curvature. That answer describes a real, inverted, magnified image and so contradicts the word "erect" in the question. Exam takeaway: read the image description first, use it to fix the sign of \(v\) before substituting, and remember that for a concave mirror upright means virtual and means the object is inside the focal length.